Lectures on Topics in Mean Periodic Functions And The Two

Transcription

Lectures on
Topics in Mean Periodic Functions
And The Two-Radius Theorem
By
J. Delsarte
Tata Institute of Fundamental Research,
Bombay
1961
Introduction
Three different subjects are treated in these lectures.
1
1. In the first part, an exposition of certain recent work of J.L. Lions
on the transmutations of singular differential operators of the second
order in the real case, is given. (J.L. Lions- Bulletin soc. Math. de
France, 84(1956)pp. 9 − 95)
2. The second part contains the first exposition of several new results on
the theory of mean periodic functions F, of two real variables, that
are solutions of two convolution equations: T 1 ∗ F = T 2 ∗ F = 0,
in the case of countable and simple spectrum. These functions can
be, at least formally, expanded in a series of mean-periodic exponentials, corresponding to different points of the spectrum. Having
determined the coefficients of this development, we prove its uniqueness and convergence when T 1 and T 2 are sufficiently simple. The
result is obtained by using an interpolation formula, in C 2 , which is
analogous to the Mittag-Leffler expansion, in C 1 .
The exposition and the proofs given here can probably later, be simplified, improved, and perhaps generalized. They should therefore be
considered as a preliminary account only.
3. Finally, in the third part, I state and prove the two-radius theorem,
which is the converse of Gauss’s classical theorem on the spherical 2
mean for harmonic functions. The proof is the same as that recently
published, (Comm. Math. Helvetici, 1959) in collaboration with J.L.
Lions; it uses the theory of transmutations of singular differential
iii
iv
Introduction
operators of the second order, and the fundamental theorem of meanperiodic functions in R1 .
J. Delsarte
Contents
Introduction
iii
I Transmutation of Differential Operators
1
1 Riemann’s Method
1
Riemann’s Method for the Cauchy problem . . . . . . .
2
Proof for the Riemann’s method . . . . . . . . . . . . .
3
3
7
2 Transmutation of Differential Operators
11
3 Transmutation in the Irregular Case
17
1
1
The operator B p for Rep > − . . . . . . . . . . . . . . 18
2
2
Continuation of the operator B p . . . . . . . . . . . . . 23
3
Continuation of B p . . . . . . . . . . . . . . . . . . . . 34
4 Transmutation in the Irregular Case
41
1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3
Continuation of T p . . . . . . . . . . . . . . . . . . . . 47
II Topics In Mean-Periodic Functions
51
1 Expansion of a Mean-periodic Function in Series
53
1
Determination of the coefficients in the formal series . . 55
v
Contents
vi
2
Examples . . . . . . . . . . . . . . . . . . . . . . . . . 57
2
Mean Periodic Function in R2
3
The Heuristic Method
1
. . . . . . . . . . . . . . . . . . . . . . . . . .
2
. . . . . . . . . . . . . . . . . . . . . . . . . .
3
The general formula in R2 by the heuristic process
4
. . . . . . . . . . . . . . . . . . . . . . . . . .
5
. . . . . . . . . . . . . . . . . . . . . . . . . .
6
. . . . . . . . . . . . . . . . . . . . . . . . . .
7
The formula (F ) for a polynomial . . . . . . . .
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. . . . . . . . . . . . . . . . . . . . . . . . . .
9
The fundamental theorem of Mean... . . . . . . .
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III The Two-Radius Theorem
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1
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2
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4
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Study of certain Cauchy-Problems .
The generalized two-radius theorem
Discussion of the hypothesis H . . .
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Part I
Transmutation of Differential
Operators
1
Chapter 1
Riemann’s Method
1 Riemann’s Method for the Cauchy problem
Definition. A function is said to be (C, r) on a subset of A of Rn if all its 3
partial derivatives upto the order r exists and are continuous in A.
Let D be a region (open connected set) in R2 , the (x, y) plane. Let
a, b, c be three functions which are (C, 1) in D and u a function (C, 2) in
D and let L denote the differential operator
Lu =
∂2 u
∂u
∂u
+ a + b + cu.
∂x∂y
∂x
∂y
Let v be a function which is (C, 2) in D and L∗ be a differential operator
of the same type as L:
L∗ v =
∂
∂
∂2 v
− (av) − (bv) + cv
∂x∂y ∂x
∂y
then
vL(u) − uL∗ (v) =
∂M ∂N
+
∂x
∂y
(1)
where M and N are (C, 1) in D and are certain combinations of u, v and 4
their partial derivatives.
vL(u) = v
∂2 v
∂u
∂u
+ av + bv + cuv
∂x∂y
∂x
∂y
3
1. Riemann’s Method
4
!
∂
∂ ∂u
∂
v
= cuv + (auv) + (buv) +
∂x
∂y
∂x ∂y
∂
∂
∂v ∂u
− u (av) − u (bv) −
∂x
∂y
∂x ∂y
#
" 2
∂
∂
∂ v
− (av) − (bv) + cv
=u
∂x∂y ∂x
∂y
#
"
#
"
∂u
∂ ∂v
∂
auv + v
−
u − buv
+
∂x
∂y
∂y ∂x
∂M ∂N
+
where
∂x
∂y
∂v
∂u
M = auv + v , N = buv − u .
∂y
∂x
vL(u) − uL∗ (v) =
The right hand member of (1) does not change if we replace M by
∂v
1 ∂u
∂v
1 ∂u
auv + (v − u ) and N by buv + (v − u ). We prefer to have
2 ∂y
∂y
2 ∂x
∂x
1 ∂
(uv) − uP(v)
2 ∂y
1 ∂
(uv) − uQ(v)
N=
2 ∂x
M=
where
P(v) =
5
(i)
(ii)
∂v
∂v
− av , Q(v) =
− bv.
∂y
∂x
Let C be a closed curve lying entirely in the region D.
By Green’s formula,
!
Z
Z Z
∂M ∂N
(λM + µN)ds =
+
dxdy
∂y
C
A ∂x
where λ, µ denote the direction cosines of the interior normal to C and
A denotes the region enclosed by C.
In view of equation (1),
Z Z
Z
(vL(u) − uL∗ (v))dxdy
(2)
(λM + µN)ds =
C
A
1. Riemann’s Method for the Cauchy problem
5
We shall consider this equation in the case when C consists of two
straight lines AX, AY parallel to the axes of coordinates and a curve Γ,
monotonic in the sense of AX and AY, joining X and Y. Suppose that
R
RA
RX
L(u) = 0 and L∗ (v) = 0 then C (λM+µN)ds = 0 i.e. A Ndx− Y Mdy =
RX
(λM + µN)ds substituting for M and N from (i) and (ii) respectively,
Y
ZX
and
ZX
1
Ndx = [(uv)X − (uv)A ] − uQ(v)dx
2
A
A
Z A
Z A
1
uP(v)dy.
Mdy = [(uv)Y − (uv)A ] +
−
2
Y
Y
If the functions u, v satisfy
Lu = 0, L∗ v = 0, P(v) = 0 on AY and Q(v) = 0 on AX
6
(3)
then we obtain the Riemann’s Resolution formula:
Z X
1
(λM + µN)ds.
(uv)A = [(uv)X + (uv)Y ] +
2
Y
Let A be (x0 , y0 ). Then v = g(x, y; x0 , y0 ) satisfying the conditions (3) is
the Riemann’s function for the equation L(u) = 0.
In the situation
L(u) = f (x, y), L∗ v = 0, P(v) = 0 on AY, Q(v) = 0 on AX,
6
exactly similar computation gives the formula:
Z X
Z Z
1
(uv)A = [(uv)X + (uv)Y ] +
(λM + µN)ds +
v f (x, y)dxdy.
2
Y
A
An important property of the Riemann’s function.
Let now Γ consider of two straight lines XB, Y B parallel to the axes
of coordinates. Then
Z X
Z X
Z B
(λM + µN)ds =
Mdy −
Ndx
Y
We can write
7
where P∗ (u) =
Z
X
B
B
Y
1 ∂
(uv) + vP∗ (u)
2 ∂y
1 ∂
N=−
(uv) + vQ∗ (u)
2 ∂x
M=−
∂u
∂u
+ au , Q∗ (u) =
+ bu.
∂y
∂x
Z X
1
Mdy = − [(uv)X − (uv)B ] +
vP∗ (u)dx
2
B
Z B
Z B
1
Ndx = [(uv)B − (uv)Y ] −
−
vQ∗ (u)dx
2
Y
Y
Thus in this case Riemann’s resolution formula becomes
1
1
[(uv)X + (uv)Y ] − [(uv)X − (uv)B ]
2
2
Z X
Z B
1
∗
+
vP (u)dx + [(uv)B − (uv)Y ] −
vQ∗ (u)dx
2
B
Y
Z X
Z B
vP∗ (u)dx
vQ∗ (u)dx +
i.e., (uv)A = (uv)B −
(uv)A =
Y
B
If u = h(x, y; x1 , y1 ), (x1 , y1 ) being the point B, is such that
Lu = 0, P∗ (u) = 0 on BX, Q∗(u) = 0 on BY,
2. Proof for the Riemann’s method
(uv)A = (uv)B .
we get
8
7
Choosing constant multipliers for u = h(x, y; x1 , y1 ) and v = g
(x, y; x0 , y0 ) in such a way that u = 1 at B and v = 1 at A, we have
h(x0 , y0 ; x1 , y1 ) = g(x1 , y1 ; x0 , y0 )
(4)
This shows that the Riemann’s function g, considered as a function
of (x0 , y0 ) satisfies the differential equation Lu = 0.
2 Proof for the Riemann’s method
We have obtained (4) under the hypothesis that there exists functions
u = h(x, y; x1 , y1 ) and v = g(x, y; x0 , y0 ) which are (C, 2) in D and which
satisfy
∂u
∂u
∂2 u
+ a + b + cu = 0
∂x∂y
∂x
∂y
∂u
P∗ (u) =
+ au = 0 on BX
∂y
∂u
+ bu = 0 on BY and
Q∗ (u) =
∂x
∂v
∂v
∂2 v
+ a∗
+ b∗ + c∗ v = 0 where
L∗ v =
∂x∂y
∂x
∂y
∂a ∂b
a∗ = −a, b∗ = −b, c∗ = − −
+c
∂x ∂y
Lu =
∂v
+ a∗ (v) = 0 on AY
∂y
∂v
+ b∗ (v) = 0 on AX.
Q(v) =
∂x
P(v) =
By change of notation and translation of the origin, the problem for 9
the existence of the Riemann’s function v = g(x, y) in D for the point
8
(x0 , y0 ) ∈ D for the differential equation Lu = 0 reduces to the solution
of the problem 1:
∂2 u
∂u
∂u
+ a + b + cu = 0
∂x∂y
∂x
∂y
with the conditions
u(0, y) = α(y)
u(x, 0) = β(x) I
α(0) = β(0) = 1
where a, b, c are (C, 1) in D and α, β are (C, 1) functions of one real
variable. The solution of the more general problem 2:
"
#
∂2 u
∂u
∂u
= λ a + b + cu
∂x∂y
∂x
∂y
with the conditions I will give for λ = −1 the solution of the problem (1).
We shall now prove the existence of the unique solution for the problem
2 by using Piccard’s method of successive approximations. Consider
the series
u0 (x, y) + λu1 (x, y) + · · · + λn un (x, y) + · · ·
(5)
10
where ui (x, y) are defined by the following recurrence formula:
∂2 u0
= 0 u0 (0, y) = α(y), u0 (x, 0) = β(x)
∂x∂y
∂un−1
∂un−1
∂2 un
=a
+b
+ cun−1 , un (x, 0) = un (0, y) = 0 for n ≥ 1.
∂x∂y
∂x
∂y
R xRy
It suffices to take u0 (x, y) = α(y) + β(x) − 1 and un (x, y) = 0 0
φn−1 (ξ, η)dξdη where
φn−1 = a
∂un−1
∂un−1
+b
+ cun−1
∂x
∂y
We shall now prove the convergence of the series (5) by the process
of majorisation which is classical. Suppose that a, b, c are (C, 0) in D
2. Proof for the Riemann’s method
9
and α, β are (c, 1) of one real variable. Let K be a compact subset D
containing the rectangle with sides parallel to the axes and (0, 0) and
(x, y) as opposite corners. Then there exists an M, A such that |α(ξ) +
β(η) − 1| ≤ M,
∂
(α(ξ) + β(η) − 1)| ≤ M,
∂ξ
∂
| (α(ξ) + β(η) − 1)| ≤ M,
∂η
|
and
for (ξ, η) in K, and |a|, |b|, |c|, ≤ A in K. Then |φ0 (x, y)| ≤ 3 A M. By the 11
recurrence formula for n = 1,
Z xZ y
φ0 (ξ, η)dξdη| ≤ 3AM|x||y|
|u1 (x, y)| ≤ |
0
0
∂u1 ∂u1 ∂u1 ∂u1 ≤ 3AM|y| and, ≤ 3AM|x|. Hence |u1 (x, y)|, , are
∂x ∂y ∂x ∂y each ≤ 3AM(1 + |x|)(1 + |y|).
Computing φ1 (x, y), we have immediately,
!2
!2
1 + |x|
1 + |y|
|u2 (x, y)| ≤ 9A2 M
2!
2!
!2
∂u2 (x, y) ≤ 9A2 M(1 + |x|) 1 + |y|
∂x 2
!2
∂u2 (x, y) ≤ 9A2 M 1 + |x| (1 + |y|)
∂y 2
In general
∂un ∂un M[3A(1 + |x|)(1 + |y|)]n
, , ≤
|un (x, y)|, ∂x
∂y
n!
Comparing with the exponential series, this majorization proves that
the series (3) as also the series obtained from (3) by differentiating each
term once and twice are all convergent uniformly on each compact subset of D and absolutely in D, so that (3) converges to a function (C, 2) in
D. This function u(x, y) is evidently the solution of problem 2 and the
proof for the existence of Riemann’s functions is complete.
10
Remark. From the recurrence formula it is clear that the function u(x, y) 12
satisfies
#
Z xZ y"
∂u
∂u
u(x, y) = α(y) + β(x) − 1 + λ
a + b + cu dξ dη
∂x
∂y
0
0
x=ξ
y=η
This integral equation is equivalent to the differential equation of problem 2 with conditions I. When a, b, c are (C, 0) and α, β are (C, 1), the
solution u(x, y) of the integral equation may not be (C, 2). But then it is
a solution of problem 2 in the sense of distributions.
Chapter 2
Transmutation of Differential
Operators
Let L1 and L2 denote two differential operators on the real line R and 13
E m (x ≥ a) denote the space of functions m times continuously differentiable in [a, ∞) furnished with the usual topology of uniform convergence of functions together with their derivatives upto the order m on
each compact subset of [a, ∞). Let E be a subspace of the topological
vector space E m (x ≥ a).
Definition. A transmutation of the differential operator L1 into the differential operator L2 in E is a topological isomorphism X of the topological vector space E onto itself (i.e. a linear, continuous, one-to-one,
onto map), such that
XL1 = L2 X
X is said to transmute the operator L1 into the operator L2 in E and
L2 = XL1 X −1 on E.
Transmutation in the regular case. Let L1 = D2x − q(x), L2 = D2x
where q satisfies certain conditions of regularity. The construction of
the transmutation operator in this case depends on the consideration of
certain partial differential equation.
Problem 1. To determine Φ(x, y) in x ≥ a, y ≥ a satisfying Φxx − Φyy −
11
12
2. Transmutation of Differential Operators
q(x)Φ = 0 with the boundary conditions
Φ(x, a) = 0 = Φ(a, y); Φy (x, a) = f (x)
14
This mixed problem is equivalent to the Cauchy problem if we set
u(x, y) = Φ(x, y) for x ≥ a
= −Φ(2a − x, y) for x ≤ a
We set without proof the following proposition.
Proposition 1. (a) If q ∈ E ◦ (R), f ∈ E ◦ (x ≥ a) with f (a) = 0, then
problem 1 possesses a unique solution which is (C, 1) in x ≥ a, y ≥ a
and satisfies the differential equation in the sense of distributions.
(b) If q ∈ E 1 (R) and if f ∈ E 2 (x ≥ a) with f (a) = 0, the solution of
problem 1 is (C, 2) in x ≥ a, y ≥ a. In the region y ≥ x(or x ≥ y),
the solution is (C, 3).
(c) If q ∈ E 2 (R) with q′ (a) = 0 and f ∈ E 3 (x ≥ a) with f (a) = f ′′ (a) =
0, then the solution u of the problem is (C, 4) in y ≥ a. [Refer to E.
Picard, ‘Lecons sur quelques types simples d’ equation aux derivces
partielles’, Paris, Gauthier-VIllars, 1927.]
With the help of this proposition we prove
15
Proposition 2. If q ∈ E 2 (R) with q′ (a) = 0 and f ∈ E 4 (x ≥ a) with
f (a) = f ′′ (a) = 0 then D2 A f = AL f where L = D2 = q and A is defined
by
∂
A f (y) =
[Φ(a, y)],
∂x
Φ(x, y) being the solution of problem 1.
2
Let ψ(x, y) = L x [Φ(x, y)] = ∂∂xΦ2 − q(x)Φ and g(x) = L x f (x). As Φ
is (C, 4) by Proposition 1 (c), and f ∈ E 4 (x ≥ a), Ψ(x, y) is (C, 2) and
g ∈ E 2 (x ≥ a) with g(a) = 0. Replacing f by g in proposition 1 (b),
problem 1 possesses a unique solution. We verify below that this unique
solution is Ψ(x, y).
L x Ψ − D2y Ψ = L x L x Φ − D2y L x Φ = L x [L x Φ − D2y Φ] = 0;
13
Ψ(x, a) = L x [Φ(x, a)] = L x [0] = 0;
∂
Ψ(x, y) = Dy L x Φ(x, y) = L x Dy Φ(x, y) so that
∂y
Ψy (x, a) = L x Dy Φ(x, a) = L x f (x) = g(x);
Ψ(a, y) = D2y [Φ(a, y)] = 0.
Now by definition of A, AL[ f (y)] = A.g =
∂
Ψ(a, y)
∂x
Ψ x (x, y) = D x L x Φ = D2y [Φx (x, y)] gives
Ψ x (a, y) = D2y [Φx (x, y)] = D2y A f (y). Hence we have proved that
AL = D2y A.
Computation of the solution u(x, y), y ≥ a of problem 1 by using 16
Riemann’s function.
Let K(x, y; x0 , y0 ) be the Riemann’s function defined in the shaded
∂2 K ∂2 K
− 2 − q∗ (x)K = 0
part of the satisfying the conditions
∂x2
∂y
with q∗ (x) = q(x) if x ≥ a
= q(2a − x) if x < a
14
1
and K on Mm1 = K on Mm2 = − .
2
In this case Riemann’s method gives
Z x0 +y0 −a
u(x0 , y0 ) =
f ∗ (x)K(x, a; x0 , y0 )dx
a−y0 +x0
where
∗
f (x) = f (x)
if x ≥ a
= − f (2a − x)
if x ≤ a
Hence
∂
u(a, y0 )
∂x0
Z
= f (y0 ) − 2
A[ f (y0 )] =
a
17
y0
f (x)
∂
K(x, a; a, y0 )dx.
∂x0
Problem 2. To determine the function Φ(x, y) in x ≥ a, y ≥ a satisfying
the conditions Φxx − Φyy − q(x)Φ = 0; Φ(a, y) = 0; Φx (a, y) = g where
g ∈ E 2 (y ≥ a) with g(a) = 0, and Φ(x, a) = 0. Problem 2 , is the
same as Problem 1 if in the boundary conditions the lines x = a, y = a
are interchanged. If Φ is the solution of Problem 2, we define a g(x) =
∂
Φ(x, a).
∂y
Proposition 3. If q ∈ E 1 (R) and f ∈ E 2 (x ≥ a) with f (a) = 0, then
aA f = Aa f = f .
∂
Φ(a, y).
Let Φ be the solution of Problem 1 and let g(y) = A f (y) =
∂x
By Proposition 1 (b), g is (c, 2) in y ≥ a and g(a) = 0. Hence Φ is the
∂
solution of Problem 2 and a · g(x) = Φ(x, a) = f (x). This shows that
∂y
aA f = f . Similarly Aa f = f .
Proposition 3 together with Proposition 2 shows that if q ∈ E 1 (R)
with q′ (a) =n 0 the map A, which is obviously linear
is one-to-one of the
o
4
′′
space E = f / f ∈ E (x ≥ a), f (a) = f (a) = 0 onto itself and verifies
AL f = D2 A f . Further in view of the formula for A f on page 13 in
items of Riemann’s functions, A is continuous on E with the topology
induced by E 4 (x ≥ a). As E is a closed subspace of the Frechet space
15
E 4 (x ≥ a), A is a topological isomorphism. Thus we have proved the
existence of the transmutation operator A in E transmuting D2 − q(x)
into D2 when q is sufficiently regular.
We now consider the problem of transmuting more general differen- 18
tial operators Li = D2 + ri (x)D + si (x) (i = 1, 2) into each other when ri
and si are regular (e. g. ri , s1 ∈ E (R)).
Proposition 4. There exists an isomorphism AL1 L2 of E which satisfies
A L1 L2 L 1 = L 2 A L1 L2 .
The proposition will be proved if we prove the existence of transmutation X of the operator L1 into the operator D2 − q1 . For then the same
method will give a transmutation of L2 into L∗2 = D2 − q2 and each of the
operators D2 − qi (i = 1, 2) can be transmutated into the operator D2 so
that finally we obtain the required transmutations AL1 L2 by composing
several transmutations.
Rx
Let R1 (x) = a r1 (ξ)dξ then the verification of the following equations is straight forward:
1
1
L1 e− 2 R1 (x) f (x) = e− 2 R1 (x) L1 [ f ].
1
Hence we have X f (x) = e 2
− R1 (x)
f (x).L∗1 = D2 − q1 where
1
1
q1 (x) = r12 (x) + r1 (x) − s1 (x) ∈ E .
4
2
Application of transmutation to the Mixed Problems of differential equations.
If Λ is an elliptic operator inRn ( independent of the variable t which 19
corresponds to time) we consider the problem of finding a function
u(x, t) (x ∈ Rn , t time) which satisfies the differential equation
!
∂
∂2
+ r(t) + s(t) u(x, t) = 0
Λ x (x, t) +
∂t
∂t2
16
with the Cauchy data in a bounded domain Ω ⊂ Rn for t = 0 and also
on the hemicylinder Ω∗ × [t ≥ 0] where Ω∗ denotes the frontier of Ω. In
this problem the variable x which corresponds to space and the variable
t which corresponds to time are strictly separated. Suppose that At is
∂2
∂
a transmutation in the variable t which transmutes 2 + r(t) + s(t)
∂t
∂t
∂2
into 2 and let V(x, t) = At u(x, t) when u(x, t) is the solution of the
∂t
differential equation. Then applying At to the left hand member of the
equation we have
∂
∂2
+ r(t) + s(t))u(x, t) = 0
∂t
∂t2
∂2 v(x, t)
Λ x v(x, t) +
=0
∂t2
At Λ x u(x, t) + At (
i.e.,
and v(x, t) satisfies Cauchy’s data i.e., by means of the transmutation,
consideration of the gives equation is reduced to the consideration of
the wave equation.
Chapter 3
Transmutation in the
Irregular Case
Introduction. Our aim in this chapter is to obtain a transmutation op- 20
erator for a differential operator with regular coefficients. In order to
reduce the mixed problem relative to the operator
Λ+
∂2 2p + 1 ∂
, p real or complex,
+
t ∂t
∂t2
where Λ = −∆(∆ being the Laplacian in the variables x1 , . . . , xn ) to the
mixed problem relative to the operator
Λ+
∂2
∂t2
we shall construct an operator will transmute the operator
L p = D2 +
2p + 1
D
x
into the operator D2 . The difficulty in this case arises due to the pres1
ence of the coefficient which has a singularity at x = 0. If we seek the
x
solution of the problem in x ≥ a, where a > 0, the method of the preceding chapter is perfectly valid without any change. But the important
case is precisely the one in which a = 0.
17
3. Transmutation in the Irregular Case
18
We shall determine isomorphisms B p , B p (of certain space which 21
will be precisely specifies in the sequel ) which satisfy
D2 B p = B p L p ; B p D2 = L p B p .
1
1
The operator B p for −1 < Rep < − and the operator B p for Rep > −
2
2
are classical. B p is the Poisson’s operator and B p is the derivative of the
Sonine operator.
1 The operator B p for Rep > −
1
2
It can be forseen that the operator B p is defined in terms of the solution
for y > 0 of the partial differential equation
Φxx − Φyy +
2p + 1
Φx = 0
x
with the conditions Φ(0, y) = g∗ (y) being an even function g∗ (y) = g(y)
for y > 0 and g∗ (y) = g(−y) for y < 0, and Φx (0, y) = 0 Now we define
B p [g(x)] = Φ(x, 0).
Changing
√ the variables (x,√y) to the variables (s, t) be means of the
formulae s 2 = y + x and t 2 = y − x, we see by simple computas−t s+t
is a solution of the partial differential
,
tion that u(s, t) = Φ
2
2
equation
p + 21
p + 12
us +
ut = 0
s−t
s−t
√
with the conditions u(s, s) = g∗ (s 2)
ust −
22
(us − ut )s=t = 0
1. The operator B p for Rep > −1/2
19
1
For α = p + , Poisson’s solution has the form
2
Z 1 n
√ o
Γ(2α)
∗
2 (1 − ρ)α−1 ρα−1 dρ
g
[s
+
(t
−
s)ρ]
u(s, t) =
2
[Γ(α)] 0
1
valid for Reα > 0 (or Rep > − ). Then
2
x −x
B[g(x)] = Φ(x, 0) = u( √ , √ )
2
2
Z 1
Γ(2α)
g∗ [x(1 − 2ρ)](1 − ρ)α−1 ρα−1 dρ
=
[Γ(α)]2 0
setting 1 − 2ρ = r,
Z 1
Γ(2α) 1
B p [g(x)] =
(1 − r2 )α−1 g∗ (rx)dr
[Γ(α)]2 22α−2 0
Z 1
1
2Γ(p + 1)
(1 − t2 ) p− 2 g(tx)dt.
= √
1
πΓ(p + 2 ) 0

Γ(α)Γ(α + 12 )22α−1 
( Using Γ(2α) =
 .
√
π
Note that B p [g(0)] = 1.
Proposition
1. The mapping f →
n
o B p f is a linear continuous map of 23
F 2 = f / f ∈ E 2 (x ≥ 0), f ′ (0) = 0 into itself and satisfies
B p D2 f = L p B p f for any f ∈ F 2 .
As for Rep > − 12 , differentiation
) of integration is permissible,
( under sign
d
B p[ f ]
= f ′ (0) k (k being some
B p f ∈ E 2 (x ≥ 0). Further
dx
x=0
constant),
= 0.
Hence B p f ∈ F 2 . Evidently B p is linear. In order to prove continuity,
since F 2 has a metrizable topology induced by that of E 2 (x ≥ 0), it is
20
sufficient to show that if a sequence { fn }, n = 1, 2, . . . , fn ∈ F 2 . converges to 0 in F 2 , then B p fn converges to zero in F 2 . But fn → 0 in
F 2 implies fn → 0 uniformly on each compact set, in particular on the
compact set [0, 1] from which it follows that B p fn → 0. It remains to
2Γ(p+1)
verify that B p satisfies the given condition. Writing β p = √πΓ(p+1/2)
Z 1
o
1
1
1 n
L p B p f − B p D2 f =
t2 (1 − t2 ) p− 2 f ′′ (tx)
βp
βp 0
1
1
(2p + 1)t
(1 − t2 ) p− 2 f ′ (tx) − (1 − t2 ) p− 2 f ′′ (tx) dt
+
x
Z
Z 1
1
2p + 1 1
2 p− 12 ′
t(1 − t )
f (tx)dt −
(1 − t2 ) p+ 2 f ′′ (tx)dt.
=
x
0
0 

1
Z 1
1




2 ) p+ 12


(1
−
t
2p + 1 
1
 1  (1 − t2 ) p+ 2 ′

′′

−
f
(tx)
f
(tx)dt
=
+








1


x  2
2
x

p+
−
Z
0
2
1
0
0
1
(1 − t2 ) p+ 2 f ′′ (tx)dt (integrating the first integral by parts).
= 0.
24
Remark. Let J p (x) denote the classical Bessel function and let
j p (x) = 2 p Γ(p + 1)x−p J p (x)
25
d2 y
j p (x) is in F2 and is the unique solution of the differential equation 2 +
! dx
2p + 1 dy
dy
2
+ s y = 0 with the conditions (y)x=0 = 1 and
= 0.
x dx
dx x=0
Now cos sx ∈ F 2 . Let B p (cos sx) = g(x). Using L p B p = B p D2 , we
get
L p B p [cos sx] = L p [g(x)] = B p D2 (cos sx)
= −s2 B p [cos sx] = −s2 g(x).
i.e.,
that
L p g + s2 g = 0.
Further g(0) = cos 0 = 1 and g′ (0) = 0 since g ∈ F 2 . This shows
B p (cos sx) = g(x) j p (sx).
1. The operator B p for Rep > −1/2
21
Hence we obtain the classical formula,
Z 1
1
2Γ(p + 1)
(1 − t2 ) p− 2 cos(stx)dt.
j p (sx) = √
1
πΓ(p + 2 ) 0
The operator B p was considered by Poisson in this particular question
of the transformation of the consine into the function j p . The operator
1
B p for −1 < Rep < −
2
For f ∈ E 0 (x ≥ 0), the definition of B p is
Z x
−p−3
B p [ f (x)] = b p x
(x2 − y2 ) 2 · y2p+1 f (y)dy
0
Z 1
= bp
t2p+1 (1 − t2 )−p−3/2 f (tx)dt
0
!
1
1
where
1/b p = √ Γ(p + 1)Γ −p −
2
2 π
1
and f ∈ F 2 , f → B p f is a linear 26
2
continuous map of F 2 into itself satisfying
Proposition 2. For −1 < Rep < −
D2 B p f = B p L p f.
The proof of the fact that B p is a linear continuous map of F 2 into
itself is analogous to the one we have given in Proposition 1. We verify that B p satisfies the given condition, again as in Proposition 1, by
integration by parts
Z 1
o
1 n
2
t2p+1 (1 − t2 )−p−3/2
B p L p − D B p f (x) =
bp
0
(
)
2p + 1 ′
′′
2 ′′
f (tx) +
f (tx) − t f (tx) dt
tx
Z 1
Z
2p + 1 1 2p
2p+1
2 −p− 12 ′′
) f (tx) +
=
t
(1 − t )
t (1 − t2 )−p−3/2 ) f ′ (tx)dt
x
0
0
1 Z 1 f ′ (tx) d
1
= t2p+1 (1 − t2 )−p− 2 f ′ (tx) −
x dt
0
0
22



27

Z

2p + 1 1 2p
f ′ (tx)

dt
+
dt = 0.
t
1 
x
(1 − t2 ) p+3/2
0
(1 − t2 ) p+ 2
t2p+1
1
The Sonine operator B¯ p for −1 < Rep < .
2
B¯ p is defined for every f ∈ E 0 (x ≥ 0)by
B¯ p f = xb¯ p
Z
1
0
x
Z
t2p+1
1
(1 − t2 ) p+ 2
f (tx)dt
1
y2p+1 (x2 − y2 )−p− 2 f (y)dy
0
√
π
b¯ p =
.
Γ(p + 1)Γ(−p + 12 )
= b¯ p
where
The integral converges if −1 < Rep <
the sign fo integration
1
2
and we can differentiate under
d ¯
B p [( f (x))] = B p [ f (x)] for every f ∈ E 0 (x ≥ 0)
dx
Relation between B p and B¯ p
1
1
When − < Rep < , both B and B¯ p are defined and it is easy to
2
2
prove by direct computation Abel’s functional equation
Z x
B¯ p B p [ f (x)] =
f (y)dy.
0
In fact
Z
x
2 −p− 12
−2p
Z
y
1
βpy
f (z)(y2 − z2 ) p− 2 dz
(x − y )
0
0
Z y Z y
1
¯
= bpβp
y(x2 − y2 )−p− 2 f (z)dz
dy
0
Z0 x
Z x
1
1
= b¯ p β p
[ f (z)
(x2 − y2 )−p− 2 (y2 − z2 ) p− 2 ydy]dz
B¯ p B p [ f (x)] = b¯ p
2p+1
y
0
2
z
2. Continuation of the operator B p
28
23
Setting x2 sin2 θ + z2 cos2 θ = y2 , we have
x2 − y2 = (x2 − z2 ) cos2 θ, y2 − z2 = (x2 − z2 ) sin2 θ
ydy = 2(x2 − z2 ) sin θ cos θdθ
Z
x
z
1
1
(x2 − y2 )−p− 2 (y2 − z2 ) p− 2 ydy
Z
π
2
1
1
1
=
cos θ sin θdθ = B −p + , p +
2
2
2
0
1
1
1 Γ −p + 2 Γ p + 2
=
2
Γ(1)
Rx
Hence B¯ p B p [ f (x)] = 0 f (z)dz so that D B¯ p B p [ f (x)] = f (x).
−2p
2p
!
2 Continuation of the operator B p
For any f ∈ E (x ≥ 0) = E ∞ (x ≥ 0), we define
T 1 [ f (t, x)] = Dt [t f (tx)],
T 2 [ f (t, x)] = Dt [t3 T 1 { f (t, x)}].
In general
T n [ f (t, x)] = Dt {t3 T n−1 [ f (t, x)]}.
Lemma 1. T n [ f (t, x)] = t2n−2 gn (t, x) where gn (t, x) is an indefinitely 29
differentiable function in [0, 1] × [0, ∞).
The proof of the lemma is trivial and is based on indication on n.
For n = 1, we have only to set g1 (t, x) = Dt [t f (tx)]. Assume that the
lemma is true for n − 1 so that
T n−1 [ f (t, x)] = t2n−4 gn−1 (t, x).
Define
gn (t, x) = 3gn−1 (t, x) + (2n − 4)g(n−1) (t, x) + tDt gn−1 (t, x).
24
By definition
T n [ f (t, x)] = 3t2 T n−1 [ f (t, x)] + t3 Dt T n−1 [ f (t, x)]
n
o
= 3t2 t2n−4 gn−1 (t, x) + t3 (2n − 4)t2n−5 gn−1 (t, x) + t2n−4 Dt gn−1 (t, x) .
= t2n−2 gn (t, x).

 2n
−p+
R1
t2p−(2n−3) (1 − t2 )
1
verges for −1 < Rep < n − .
2
Corollary. The integral
30
0

− 3 

2  T n f (x, t)dt con-
The corollary is immediate since the integral can be written as
2n − 3
R1
−p+
2p+1
2
2 gn (t, x).
t
(1 − t )
0
1
Proposition 1. For −1 < Rep < − , and for f ∈ E (x ≥ 0),
2
B p [ f (x)] =
(−1)n b p
(2p + 1)(2p − 1) · · · (2p − 2n − 3)
Z 1 2p−(2n−3)
t
0
(1 − t2 ) p−
2n−3
2
T n f (t, x)dt
The proof is based on induction on n and the following formula which is
obvious:


d  t2p−λ  (2p − λ)t2p−λ−1
(1)
 =

dt (1 − t2 ) p− λ2
(1 − t2 ) p−λ/2+1
Z 1
for any
λB p [ f (x)] = b p
t2p+1 (1 − t2 )−p−3/2 f (tx)dt
0
Let
n = 1. If p +
3
λ
= p − + 1,
2
2
i.e.,
λ = −1


1
t2p
d  t2p+1 


=

(1 − t2 ) p+3/2 (2p + 1) dt (1 − t2 ) p+ 12
25
so that
bp
B p [ f (x)] =
2p + 1
Z
bp
=−
2p + 1
1
0
Z
0


d  t2p+1 

 t f (tx)dt
dt (1 − t2 ) p+ 12
1
t2p+1
(1 −
1
t2 ) p+ 2
T 1 f (t, x) dt
(integrating by parts, the integrated part being zero since Rep + 12 < 0 31
and 2 Rep + 2 > 0). Thus the formula to be proved holds for n = 1.
Assuming it for n − 1, we establish it for n
B p f (x) =
(−1)n−1 b p
(2p + 1)(2p − 1) . . . (2p − 2n + 5)
Z 1 2p−(2n−5)
t
0
(1 − t2 ) p−
2n−5
2
T n−1 [ f (t, x)]dt
2n − 5
λ
Using (1) with p −
= p − + 1 i.e., λ = 2n − 3, the integral
2
2
on the right hand side equals
1
2p − 2n + 3
Z
1
d
dt




 3
 t T n−1 [ f (t, x)]dt
2n−3 
t2p−2n+3
(1 − t2 ) p− 2

1



t2p+2
1



g (t, x)
=

2n−3 n−1


p−
2p − 2n + 3 (1 − t2 ) 2
0

Z 1
2p−2n+3


t

T n f (t, x) dt
−
2n−3


p−
2
0 (1 − t2 )
Z 1
1
t2p−2n+3
T n f (t, x) dt,
=
2n−3
2p − 2n + 3 0 (1 − t2 ) p− 2
0
1
the integrated part being zero since −1 < Rep < − . We write
2
Z 1 2p−(2n−3)
t
(n)
n
B p f (x) = b p
T n f (t, x) dt
2n−3
p−
2
0 (1 − t2 )
32
(2)
26
where
n
b(n)
p = (−1)
bp
(2p + 1)(2p − 1)(2p − 3) · · · (2p − 2n + 3)
(3)
1
The integral is convergent for −1 < Rep < n − so that under
2
this condition, we can differentiate under the sign of integration and
Bnp f ∈ E . We obtain for each n a function which assigns to each p in
1
−1 < Rep < n − , a map Bnp of E into itself which coincides with B p if
2
1
−1 < Rep < − . It is easy to see that Bnp is a linear map of E into itself.
2
In order to show that it is continuous, as E is metrizable, it is enough to
prove that if a sequence f j j=1,2,... tends to zero in E , then Bnp f j tends to
0 in E . We have
Z 1 2p−(2n−3)
r n t
(n)
D x B p f j (x) ≤ b p M(r, x, j)
dt
2n−3
0 (1 − t2 ) p− 2
h
i
where
M(r, x, j) = sup Drx T n f j (t, x) 1≤t≤1
33
Now T n f f (t, x) is a polynomial in t, x with coefficients which are
derivatives of order ≤ n of f j and f j together with all its derivatives
converge to zero on each compact subset. Hence M(r, x, j) → 0 as
j → ∞ uniformly for x on each compact subset i.e. Bnp f j → 0 in E .
1
Thus we obtain a function p → Bnp on −1 < Rep < − with values
2
in L (E , E ), the space of linear continuous maps of E into itself. We
intend to prove that this function is analytic and can be continued in the
whole complex plane into a function which ia analytic in the half plane
Rep > −1 and meromorphic in Rep < −1 with a sequence of poles
lying on the real axis. Before proving this continuation theorem we give
first the definition of a vector valued analytic function and some of its
properties which follow immediately from the definition.
Definition. Let 0 be an open subset of the complex plane and E a locally
convex vector space. A function f : 0 → E is called analytic if for every
e′ in the topological dual E ′ of E (i.e. the space of linear continuous
forms on E) the function z →< f (z), e′ > is analytic in 0 where <, >
denotes the scalar product between E and E ′ .
27
Lemma 2. If E is locally convex vector space in which closed convex
envelope of a compact set is compact, a function f : 0 → E is analytic if
f is continuous and for every e′ in a total set M ′ of E ′ , z →< f (z), e′ >
is analytic in 0.
Let C be any simple closed curve lying entirely in 0, enclosing re- 34
′
′
gion ( open connected set ) contained in 0.
R For e ∈ M , the function
′
′
z →< f (z), e > is analytic in 0, so that c < f (z), e > dz = 0 i. e.
R
R
< c f (z)dz, e′ >= 0. c f (z)dz is the integral of the continuous E-valued
function f over the compact set C and is an element of E since E has the
property that the closed convex envelope of any compact subset is compact. ( For integration of a vector valued function, refer to N.RBourbaki,
Elements de Mathematique, Integration, Chapter III). Hence c f (z)dz =
R
R
0 since M ′ is total, so that < c f (z)dz, e′ >= c < f (z), e′ > dz = 0 for
every e′ ∈ E ′ . Also as f is continuous it follows that z →< f (z), e′ >
is continuous. This proves that z →< f (z), e′ > is analytic for every
e′ ∈ E ′ since the choice of C was arbitrary.
1
Proposition 2. Suppose that −1 < Rep < n − . Then
2
a) Bnp f ∈ E (x ≥ 0) for every f ∈ E (x ≥ 0)
b) The mapping f → Bnp f is linear continuous of E into itself.
1
c) The function p → Bnp on the strip −1 < Rep < n − with val2
ues in Ls (E , E ) is analytic where Ls (E , E ) is the space of linear
continuous maps of E into E endowed with the topology of simple
convergence.
We have to prove only (c). We first observe that any linear continuous form on Ls (E, F) is given by finite linear combination of forms 35
of the type u →< ue, f ′ > with e ∈ E and f ′ ∈ F ′ . The theorem
will be proved if we show that the map p →< Bnp f, T > is analytic in
1
−1 < Rep < n − where f is any element of E and T any element of
2
E ′ ; i.e. for fixed f we have to show that the map p → Bnp f is analytic
with values in E . Now E ia a Hausdorff complete locally convex vector
28
space and therefore closed convex envelope of each compact subset of
E is compact ( refer to N. Bourbaki, Espaces Vectoriels Topologiques,
Ch.II, §4, Prop.2). Further it is easy to see that p → Bnp f is continuous
and that the set δx x≥0 , where δx is the Dirac measure with support at x,
is total in E ′ . Applying the lemma, we see that in order to prove analyticity of the function p → Bnp f we have only to prove that the function
p →< Bnp f, δx > i.e. the function p → Bnp f (x) (for f and x fixed ) is
1
analytic in −1 < Rep < n − .
2
√
2 π
Observing that b p =
is an entire function with
1
Γ(p + 1)Γ(−p − )
2
1 1 3
zeros at p = −1, −2, −3, . . . and p = − , , , . . . we see that bnp in (3)
2 2 2
is an entire function. The integral in (2) on the other hand converges for
1
−1 < Rep < n − and therefore is analytic in the same region, Hence
2
1
n
p → B p f (x) is analytic in the strip −1 < Rep < n − .
2
36
Corollary. The functions Bmp and Bnp where m and n are two distinct
positive integers are identical in the intersection of their domains of
definition.
We have in fact two analytic functions Bmp and Bnp which coincide
1
with B p in −1 < Rep < − which is common to their domains of
2
definitions and therefore the two functions coincide everywhere in the
domain which is the intersection of their domains of definition due to
analyticity. It follows from the corollary that we have a unique analytic
function B p defined for Rep > −1.
Remark. We have B−1/2 = identity.
For n = 1,
−b p
B p( f ) =
2p + 1
Z
0
1
t2p+1
(1 −
1
t2 ) p+ 2
d
t f (tx) dt
dt
29
1 −b p
= 1 and
and if p = − ,
2 2p + 1
B− 1
2
f (x) =
Z
0
1
d
t f (tx) dt = f (x).
dt
1
Continuation of B p for Rep < − .
2
We define
In general
1
U1 f (t, x) = Dt (1 − t2 ) 2 f (tx) .
U2 f (t, x) = Dt (1 − t2 )3/2 f (tx) .
Un f (t, x) = Dt (1 − t2 )3/2 Un−1 f (tx) .
Lemma 3. For every f in E(x ≥ 0) = E we have
1
Un f (t, x) = (1 − t2 )3/2 Dt Un−1 f (t, x) − 3t(1 − t2 ) 2 Un−1 f (t, x) (4)
n−2
(5)
and Un f (t, x) = (1 − t2 ) 2 hn (t, x)
where hn (t, x) is indefinitely differentiable in 0, 1 × 0, ∞ .
Relation (4) is evident. (5) can be proved by induction on n. It is
true for n = 1 if we set
h1 (t, x) = −t f (tx) + x(1 − t2 ) f ′ (t, x).
Assuming it for n − 1, it is easy to verify that (5) holds for n if
hn (t, x) = −nthn−1 (t, x) + (1 − t2 )Dt hn−1 (t, x)
R 1 2p+n+1
Corollary. The integral 0 t p+ n+1 Un f (x, t) dt converges for −1 −
(1−t2 )
n
2
< Rep <
− 12 .
1
Proposition 3. If −1 < Rep < − ,
2
2
37
30
B p f (x) =
(−1)n b p
(2p + 2)(2p + 3) · · · (2p + n + 1)
Z 1
t2p+n+1
0
(1 −
n+1
t2 ) p+ 2
Un f (t, x) dt
The proof of this proposition is analogous to that of Proposition 1.
We prove it by induction on n and by using formula (1).
1
For −1 < Rep < − ,
2
Z 1
3
t2p+1 (1 − t2 )−p− 2 f (tx)dt
B p f (x) = b p
0
39
Using (1) with 2p + 1 = 2p − λ − 1 i. e. λ = −2, we have,
1 d
t2p+2
t2p+1
=
)
(
(1 − t2 ) p+2 2p + 2 dt (1 − t2 ) p+1
so that
Z 1
bp
t2p+2
d
2 12
)(1
−
t
) f (tx)dt
(
2p + 2 0 dt (1 − t2 ) p+1
Z 1
−b p
t2p+2
=
U1 f (t, x) dt,
2
p+1
(2p + 2) 0 (1 − t )
B p f (x) =
1
the integrated part being zero since −1 < Rep < . The formula is
2
proved for n = 1. We assume it for n − 1
B p f (x) =
(−1)n−1 b p
(2p + 2)(2p + 3) · · · (2p + n)
Z 1
0
t2p+n
Un−1 [ f (t, x)]dt
(1 − t2 ) p+n/2
Using (1) with 2p + n = 2p − λ − 1 i. e. λ = −(n + 1), we get


t2p+n+2
d  t2p+n+1 


=
(2p
+
n
+
1)
,

n+3
dt (1 − t2 ) p+ n+1
2
(1 − t2 ) p+ 2
38
31
so that
B p f (x) =
−(−1)n−1 b p
(2p + 2) · · · (2p + n + 1)
Z
0
1
t2p+n+1
n+1
(1 − t2 ) p+ 2
2 32
Dt (1 − t ) Un−2 f (t, x) dt
we shall now set
nB p f (x) = (n)b p
(n)b p
t2p+n+1
p+ n+1
2
Un f (t, x) dt
(1 − t2 )
(−1)n b p
=
(2p + 2)(2p + 3) · · · (2p + n + 1)
(6)
(7)
(n)b p is a meromorphic function of p, with poles at the points
−3 −5
,
, . . ..
p=
2
2
Proposition 4. Suppose that p satisfies
−1−
n
1
< Rep < −
2
2
(8)
and does not assume any of the values
3 −5
, ...
− ,
2 2
(9)
Then
a) n B p f ∈ E for each f ∈ E
b) The mapping f → n B p f is linear continuous from E into E .
c) The function p → n B p is meromorphic in the strip defined by (8)
with poles situated at the points given by (9).
We omit the proof of a) and b) since it is exactly similar to that of
Proposition 2. The proof of c) reduces as in Proposition 2, to showing
that the function
Z 1
t2p+n+1
(n)
p
bp
Un f (t, x) dt
n+1
p+
2
0 (1 − t2 )
40
32
is meromorphic in the strip (8) with poles at the points (9), which is
obvious since the integral converges in (8) and is therefore analytic in
(8) and nb p is meromorphic in (8) with poles given by (9).
Corollary. If m, n are two distinct positive integers, the two functions
mB p , nB p coincide in the common part of their domains of definition.
41
This corollary is immediate like the corollary of Proposition 2. Con1
sequently we have a unique function B p defined for Rep < − .
2
Propositions (2) and (4) give finally the continuation theorem for the
operator B p .
1
Theorem. The function p−B p defined initially in −1 < − Rep < − with
2
values in L (E , E ) endowed with the topology of simple convergence
can be continued in the whole plane into a meromorphic function. The
−3 −5 −7
,
,
, . . ..
poles of this function are situated at the points
2 2 2
Remark. The notion of an analytic function with values in a locally
convex vector space E depends only on the system of bounded subsets
of E. E being complete it is easy to see (in view of Theorem 1, page 21,
Ch.III, Espaces vectorieles Topologiques by N. Bourbaki) that the space
L (E , E ) when furnished with the topology of simple convergence has
the same system of bounded sets as when furnished with the topology
of uniform convergence on the system of bounded sets of E . Hence in
the theorem we can replace the topology of simple convergence by the
topology of uniform convergence on bounded subsets of E or by any
other locally convex topology which lies between these two topologies.
Let E∗ and D◦ be subspaces of E(x≥0) defined by
n
o
E∗ = f | f ∈ E (x ≥ 0), f 2n+1 (0) = 0 for n ≥ 0
D◦ = f | f ∈ (x ≥ 0), f n (0) = 0 for n ≥ 0
42
1
When −1 < Rep < − , we have
2
Z 1
−p− 3
t2p+1+r (1 − t2 ) 2 f r (tx)dt
Dr B p f (x) = b p
0
so that
where
33
Dr B p f (0) = b p,r f r (0)
(10)
r
πΓ p + + 1
= 2
Γ r+1
2 Γ(p + 1)
√
bq,r
This shows that when r is even, p → b p,r is an entire function
of p and when r is odd it is a meromorphic function with poles at
−3 −5
2 , 2 , . . . (10) is therefore true for all p not equal to the exceptional
−5
values −3
2 , 2 , . . .. Then B p [ f ] is in E∗ or in D◦ according as f is in E∗
−5
or D◦ . Therefore for p , −3
2 , 2 , . . . , B p ∈ L (ε∗ , ε∗ )( or ∈ L (Do , Do ))
−3
and p → B p is meromorphic with poles at p =
, . . .. Actually the
2
following stronger result holds.
Theorem. The function p → B p is an entire analytic function with values in L (ε∗ , ε∗ ) (also in (Do , Do )).
We have seen more than once, that in order to investigate the analyticity of the function p → B p , it is sufficient to do the same for the
function p → B p [ f (x)], where f ∈ E and x ≥ 0 are arbitrarily chosen and are fixed. In this case we study the behaviour of the function 43
p − B p [ f (x)] where f ∈ E ∗ and x ≥ 0, in the neighbourhood of the point
2m + 1
, (a supposed singularity of the function).
po = −
2
By Taylor’s formula,
Z
N
X
xn tn n
xN+1 t
f (tx) =
f (0) +
f (tξ) f N+1 (ξx)dξ,
n!
N
o
0
N being an arbitrary integer.
1
Suppose first that −1 < Rep < − . Then
2
√
n
N
π XΓ p+ 2 +1 n n
B p [ f (x)] =
x f (0)
Γ(p + 1) o Γ n + 1 n!
2
2
34
+ bp
44
xN+1
N!
Z
o
1
3
t2p+1 (1 − t2 )−p− 2 dt
Z
1
o
(t − ξ)N f N+1 (ξx)dξ.
N +3
<
The right hand side of this formula is well defined when −
2
1
Rep < − and depends analytically on p and therefore coincides with
2
N+3
<
the continuation of B already obtained. Choosing N such that −
2
po , the integral on the right is analytic at the point po . Hence we have
n
only to consider the finite sum at po . The function Γ(p+ + 1) has poles
2
n
at the points p such that p + + 1 = −µ, µ a positive integer. It has a
2
n
1
pole at p0 if = −µ − 1 − p0 = m − µ − i.e. n = 2(m − µ) − 1. But when
2
2
n is odd, f n (0) = 0 since f ∈ E∗ . The finite sum is therefore analytic
2m + 1
at p0 and B p [ f (x)] has false singularity at −
and the proof of the
2
theorem is complete.
Theorem. The formula
D2 B p f = B p L p f holds for every f ∈ E ∗
and for every complex number p. L p f is in E∗ if f ∈ E∗ so that L p ∈
L (E∗ , E∗ ) and it is easy to verify that p → L p is an entire function with
value is L (E∗ , E∗ ). The two entire functions p → B p L p and p → D2 B p
1
coincide by Proposition 2, §1 in −1 < Rep < − and are therefore
2
identical in the whole plane.
Remark. It is necessary to suppose that f ∈ E∗ . If f is only in E , L p f is
not in E (always).
3 Continuation of B p
We now consider the extension of the operator B p initially defined for
1
Rep > − by
2
Z 1
1
B p [ f (x)] = β p
(1 − t2 ) p− 2 f (tx)dt.
(I)
o
3. Continuation of B p
where
35
2Γ(p + 1)
β p = √Q
Γ(p + 12 )
Changing the variable t = sin Θ,
B p [ f (x)] = β p
Z
Q
2
f (x sin θ) cos2p θdθ.
o
Let M 1p [ f (x, θ)] = f (x sin θ)
)
(
d
d
1
sin θ [sin θ f (x sin θ)]
+
(2p + 1)(2p + 2) dθ
dθ
and
N 1p [ f (x, θ)] =
1
d
{sin θ f (x sin θ)}
2p + 1 dθ
We determine by induction, the functions
d
1
M np [ f (x, θ)] = M n−1
p [ f (x, θ)] +
2p + 2n dθ
(
)
d
d
1
n−1
sin θ sin θN p [ f (x, θ] +
dθ
(2p + 2n − 1)(2p + 2n) dθ
)
(
d
n−1
sin θ [sin θM p ( f (x, θ))
dθ
and
N np [ f (x, θ)] = N n−1
p [ f (x, θ)] +
o
1
d n
sin θM n−1
p [ f (x, θ)]
2p + 2n − 1 dθ
1
Proposition 1. For Rep > − , f ∈ E (x ≥ 0) and for any n,
2
B p [ f (x)] = β p
Z
o
π
2
cos2p+2n θM np [ f (x, θ)]dθ
Z π
2
+ βp
cos2p+2n+1 θN np f (x, θ)dθ (1)
0
45
36
The proof of this proposition is elementary and is based on induction 46
on n and the process of integration by parts.
B p f (x) = β p
Z
π
2
f (x sin θ) cos2p+2 θdθ
0
+ βp
Z
π
2
f (x sin θ) sin2 θ cos2p θdθ.
o
But as sin θ cos2p θ =
integral, we get,
B p f (x) = β p
Z
π
2
d cos2p+1 θ
(
) integrating by parts the second
dθ −(2p + 1)
f (x sin θ) cos2p+2 θdθ
o
+ βp
Z
0
π
2
1
d
(sin θ f (x sin θ)) cos2p+1 θdθ
2p + 1 dθ
Now the second integral in this equation equals
Z
47
π
2
d
1
2p
+
1
dθ
0
Z π
2
1
d
+
(sin θ f (x sin θ)) cos2p+1 θ sin2 θdθ
2p
+
1
dθ
0
Z π
2
1
d
=
2p
+
1
dθ
0
Z π
2
1
d
d
+
sin θ (sin θ f (x sin θ)) dθ
cos2p+2 θ
dθ
dθ
0 (2p + 1)(2p + 2)
so that
B p f (x) = β p
Z
0
π
2
cos2p+2 θ f (x sin θ)
d
d
1
(sin θ (sin θ f (x sin θ))) dθ
+
(2p + 1)(2p + 2) dθ
dθ
Z π
2
d
1
+ βp
(sin θ f (x sin θ))dθ
cos2p+3 θ
2p
+
1
dθ
0
37
Hence formula (1) holds for n = 1. Assuming it for n − 1, it can be
verified for n by integration by parts.
We define for every integer n > 0,
B np f (x) = β p
Z
π
2
0
cos2p+2n θM np f (x, θ)dθ+
βp
Z
π
2
o
cos2p+2n+1 θN np f (x, θ)dθ
1
The two integrals converge for Rep > − − n.
2
Proposition 2. If p satisfies
48
1
Rep > − − n
2
(2)
p , −1, −2, −3, . . .
(3)
and
Then
a) For f ∈ E (x ≥ 0) = E , B np f ∈ E .
b) The mapping f → B np f is linear continuous of E into itself.
c) The function p → B np is meromorphic in the half plane defined by
(2) with values in L (E , E ) the poles being situated at the points
(3). We shall give only the outline of the proof since it is analo2Γ(p + 1)
gous to that of Proposition (2), §2. The function β p =
√
1
πΓ(p + )
2
1 3
is a meromorphic function of p which vanishes at − , − , . . . , and
2 2
poles at −1, −2, −3, . . . and for x, θ fixed, M np f (x, θ), and N np f (x, θ)
−3
1
, −2, . . .. Hence
are meromorphic function with poles at − , −1,
2
2
for p fixed and , −1, −2, . . . , the function (x, θ) → β p M np (x, θ) and
π
(x, θ) → β p N np f (x, θ) are indefinitely differentiable in [0, ∞) × (0, )
2
38
49
so that (a) is true. The proof for (b) is similar to that of (b) of proposition 2, §2. For f, x and θ fixed, M np f (x, θ) and N np f (x, θ) are meromorphic functions with poles at −1, −2, . . . so that p → B p f (x) is
1
meromorphic Rep > − − n with poles at (3). Further since p → B p
2
is continuous in the region given by (2) and (3), (c) follows.
Corollary. If m, n are two distinct positive integers, the functions B m
p
and B m
p coincide in the intersection of their domains of definitions.
In fact the two meromorphic functions coincide with B p in the half
1
plane Rep > − which is common to their domains of definitions and
2
hence they coincide everywhere in the intersection of their domains of
definition.
We have proved the following
1
Theorem. The function p → B p defined in Rep > − by (I) with values
2
in L (E , E ) can be continued analytically into a function meromorphic
in the whole plane with poles at −1, −2, −3, . . ..
Remark 1. B− 1 = identity.
2
1
Remark 2. For Rep > − , we have
2
r
D B p f (x) = β p
and
50
r
Z
1
◦
r
1
t (1 − t ) 2 f r (tx)dt
r
[D B p f (x)]x=0 = β p,r f (0)
2
p−
(4)
!
r+1
Γ(p + 1)Γ
2
is a meromorphic function of p with
where β p,r = √ r
πΓ + p + 1
2
poles at −1, −2, . . .. By analytic continuation, the equation (I) holds for
p , −1, −2, . . .. Then if f ∈ E∗ (resp. Do ), B p f ∈ E∗ (resp. Do ) and
p → B p is meromorphic with values in L (E∗ , E∗ )(resp. L (Do , Do )).
39
Theorem. For any p , −1, −2, . . . we have
B p D2 f = D2 B p f, f ∈ E∗ .
1
By Proposition 1, §1, the given equation holds for Rep > − and
2
hence for all p , −1, −2, . . . by analytic continuation.
Theorem. For any complex p , −1, −2, . . . the operators B p and B p
are isomorphisms of E∗ (resp Do ) onto itself which are inverses of each
other.
1
1
< Rep < . Also
2
2
1
1
¯
for p in the same region D B p = B p so that B p B p = I for − < Rep <
2
2
and the same holds for all p , −1, −2, −3, by analytic continuation.
Similarly B p B p = I for p , −1, −2, . . ..
We have seen that D B¯ p B p = I(identity) for −
Chapter 4
Transmutation in the
Irregular Case
!
2p + 3
Case of the general operator : D +
+ M(x)D + N(x)
x
2
1
Let M and N be two indefinitely differentiable functions, M odd (i.e., 51
M (2n) (0) = 0 for every n ≥ 0) and N even (i.e., N 2n+1 (0) = 0 for every
n ≥ 0 i.e., N ∈ E∗ ).
We consider the two following problems.
Problem 1. To find u(x, y), indefinitely differentiable even in x and y
which is a solution of
!
∂u
∂2 u
2p + 1
∂2 u
+
M(x)
+
N(x)u
−
+
=0
(1)
x
∂x
∂x2
∂y2
with
u(0, y) = g(y), g ∈ E∗
(2)
Problem 2. To find v(x, y) indefinitely differentiable, even in x and y,
solution of the same equation (1) with
41
42
v(x, 0) = f (x), f ∈ E∗
(3)
We intend to prove the following
Theorem 1. For p , −1, −2, . . ., each of the two problems stated above
admits a unique solution, which depends continuously on g (or f ).
52
Once we prove this theorem, we can define Operators X p and X p .
Definition 1. For g ∈ E∗ , and p , −1, −2, . . .
X p [g(x)] = u(x, 0),
where u is the solution of the Problem 1.
Definition 2. For f ∈ E∗ , and p , −1, −2, . . .
X p [ f (y)] = v(0, y)
where v is the solution of the second problem. Assuming Theorem 1,
definitions (1) and (2) give immediately
Theorem 2. For p , −1, −2, . . . , X p and X p are in L(E∗ , E∗ ).
We now prove
Theorem 3. For p , −1, −2, . . ., we have
D2 X p = X p Λ p
2
X p D = Λ pX p
where
2p + 1
Λ p = D2 +
D + M(x)D + N(x)
x
∂2
to the two sides of
∂y2
!
∂v
∂2 v
∂2 v
2p + 1
+
M(x)
+
N(x)v
−
+
=0
x
∂x
∂x2
∂y2
Applying the operator
(4)
(5)
(6)
1.
53
43
and setting
∂2 v
= V, we have
∂y2
(Λ p ) x V −
∂2 V
= 0,
∂y2
∂2 v
(x, 0) = (Λ p ) x (v(x, 0)) = Λ p [ f (x)]; and by the defini∂y2
tion of X p , V(0, y) = X p Λ p [ f (x)].
and V(x, 0) =
∂2 V
(0, y) = D2y [X p f (y)]. Thus (4) is proved. Sim∂y2
ilarly (5) can be proved. But (5) follows from (4) and the
But we have also
Theorem 4. For p , −1, −2, . . . , the operators X p , X p are isomorphisms of E∗ onto itself and X p X p = X p X p = the identity.
Let g ∈ E∗ and u be the solution of Problem 1. Then u(x, 0) =
X p [g(x)]. In order to find X p X p [g(x)] it is sufficient to determine the
solution v(x, y) of
2p + 1
∂2 v
∂v
∂2 v
+
(
=0
+
M(x))
+
N(x)v
−
x
∂x
∂y2
∂y2
with v(x, 0) = X p [g(x)] = u(x, 0).
Then, in view of Theorem 1, u(x, y) = v(x, y). Hence X p X p [g(y)] =
u(0, y) = g(y) similarly X p X p = the identity.
54
The Theorem 1. We now proceed to solve Problem 1. Let (B p )x
[u(x, y)] = w(x, y) where (B p )x denotes the operator B p relative to the
variable x. Applying (B p )x to the two sides of equation (1), i. e., of
(L p ) x [u(x, y)] + M(x)
∂2 u
∂u
+ N(x)u − 2 = 0
∂x
∂y
and using D2 B p = B p L p , we obtain
∂2 w ∂2 w
∂u
− 2 + (B p )x [M(x) + N(x)u] = 0
∂x
∂x2
∂y
and we have also w(0, y) = g(y) = u(0, y) since B p f (0) = f (0).
We intend to put this equation which is equivalent to equation (1) in
a proper form, We first prove
44
Rx
Proposition 1. B p AB p f (x) = A(x) f (x) + x o T p [A](x, y) f (y)dy where
A(x) and f (x) ∈ E∗ and T p is a map defined in E∗ with values in the
space of even indefinitely differentiable functions of the variables x and
y.
For −1 < Rep <
1
,
2
B¯ p f (x) = b¯ p
where
and for
b¯ p =
Z
x
√o
π
1
(x2 − y2 )−p− 2 y2p+1 f (y)dy
1
Γ(p + 1)Γ(−p + )
2
, B¯ p f (0) = 0, D B¯ p = B p
1
Rep > − ,
2
Z x
1
−2p
2
B p g(x) = β p x
(x − y2 ) p− 2 f (y)dy
o
55
2Γ(p + 1)
.
√
1
πΓ(p + )
2
1
1
Hence for − < Rep < , we compute
2
2
Z x
1
B¯ p AB p f (x) = b¯ p β p
(x2 − z2 )−p− 2 z2p+1 A(z)
o
"
#
Z z
−2p
2
2 p− 12
z
(z − y )
f (y)dy dz
o
Z x
Z x
2
2 −p− 12 2
2 p− 12
¯
f (y)
= bpβp
(x − z )
(z − y ) A(z)zdz dy.
where
βp =
o
y
Setting z2 = x2 sin2 θ + y2 cos2 θ, we get
Z
B¯ p AB p f (x) = b¯ p β p φ p (x, y) f (y)dy
where,
φ p (x, y) =
Z
o
o
π
2
sin
2p
θ cos
−2p
θA
q
x2 sin2 θ + y2 cos2 θ dθ
1.
45
1
which converges for − < Rep < 12 . It follows that
2
Z x
∂
¯
¯
φ p (x, y) f (y)dy
D B¯ p AB p f (x) = b p β p φ p (x, x) f (x) + b p β p
0 ∂x
But we have
56
1
1
1
φ p (x, x) = A(x) Γ(p + )Γ −p +
2
2
2
¯b p β p φ p (x, x) = A(x).
so that
Hence B p AB p f (x) = A(x) f (x) + γ p
γp =
R
x
0
∂
φ p (x, y) f (y)dy where
∂x
2
Γ(p +
1
2 )Γ(−p
!
+ 21 )
.
π
R
∂
A′ (z)
xdθ so that if we give
Now φ p (x, y) = 02 sin2p+2 θ cos−2p θ
∂x
z
Definition 3. For f ∈ E∗
T p [ f ](x, y) = γ p
Z
π
2
sin2p+2 θ cos−2p θ f1 (z)d
0
where f1 (z) =
can write
h
i1
f ′ (z)
, z = x2 sin2 θ + y2 cos2 θ 2 , γ p = b¯ p β p , then we
z
B p AB f (x) = A(x) f (x) + x
(7) is valid for | Rep | <
Z
x
T p [A](x, y) f (y)dy
(7)
o
1
and f ∈ E∗ .
2
1 3
Remark. We have γ p = 0 for p = ± , ± , . . . so that T − 1 = 0. Again 57
2
2 2
for p = − 12 , B p = B p = the identity. Hence (7) is valid for p =
− 21 . Now w(x, y) = (B p ) x u(x, y) so that u(x, y) = (B p ) x w(x, y) and
46
∂u
= (B p )x M(x)D x (B) x w(x, y). We shall denote by φ(x), the
(B p ) x M(x)
∂x
function w(x, y) considered as a function of x; let (B p ) x w(x, y) = B p φ =
ψ. Then B p ψ = φ, B p MDB p φ = B p MDψ = B p (D(Mψ))B p (M ′ ψ) =
M(x)
B p [D(xM ∗ ψ)] − B p (M ′ ψ) where M ∗(x) =
∈ E∗ and B p [MDψ] =
x
1
B p [xDM ∗ ψ] + B p [(M ∗ − M ′ )ψ]. But for −1 < Rep < − ,
2
Z 1
B p [xDF] = b p x
t2p+2 (1 − t2 )−p−3/2 F ′ (tx)dt = xD(B p [F])
o
Hence B p (MDψ) = xDB p M ∗ ψ + B p [(M ∗ − M)ψ]
Now by (7),
)
T p [M ](x, y)φ(y)dy
B p [MDψ] = xD M φ + x
o
Z x
∗
′
+ (M − M )φ + x
T p [M ∗ − M](x, y)φ(y)dy.
o
Z x
2
∗
= x T p [M ](x, x)φ(x) + x
S p [M](x, y)φ(y)dy + M(x)φ′ (x)
(
Z
∗
x
∗
o
58
where S p [M](x, y) = T p [2M ∗ − M](x, y) + x
∂
T p [M ∗](x, y)
∂x
As
Z π
2 sin2p+2 θ cos−2p θdθ
o
!
′
1 M ∗ (x)
,
= p+
2
x
!
1
′
B p MDB p [φ] = M(x)φ (x) + p + [M ′ (x) − M ∗(x)]φ(x)
2
Z x
S p [M](x, y)φ(y)dy
+x
′
M ∗ (x)
T p [M ](x, x) = γ p
x
∗
o
Also B p [N(x)ψ] = B p NB p φ = N(x)φ(x) + x
R
x
T p [N](x, y)φ(y)dy
o
3. Continuation of T p
47
Hence the differential equation in w has the form
!
1
∂w
∂2 w ∂2 w
+ P + [M ′ (x) − M ∗ (x)]w
− 2 + M(x)
∂x
2
∂x2
∂y
Z xn
o
+ N(x)w + x
S p [M](x, ξ) + T p [N](x, ξ) w(ξ, y)dξ = 0
o
!
!
1
M(x)
′
Let Q p (x) = N(x) + p +
M (x) −
∈ ξ∗ and R p (x, ξ) =
2
x
S p [M](x, ξ) + T p [N](x, ξ). We see that Problem 1 is equivalent to the
determination of the indefinitely differentiable solution even in x and y
the integro differentiable equation
∂w
∂2 w ∂2 w
− 2 M(x)
+ Q p (x)w + x
2
∂x
∂x
∂y
Z
x
R p (x, ξ, )w(ξ, y)dξ = 0
o
with the condition w(0, y) = g(y).
It is easy and classical to transform this problem to a purely integral 59
equation of Valterra type and the solution is obtained by the method of
successive approximation. It can be verified that all the conditions of
w are verified . A process completely analogous gives the solution of
Problem 2.
3 Continuation of T p
For f ∈ E∗ fixed we define by induction the functions gn (x, y, θ) as follows
f ′ (z)
where z =
go (x, y, θ) = f1 (z) =
z
q
x2 sin2 θ + y2 cos2 θ
sin2p θ cos θg1 (x, y, θ) =
i
∂ h 2p+1
sin
θgo (x, y, θ)
∂θ
In general
sin2p−2n+2 θ cos θ gn (x, y, θ) =
∂ h 2p−2n+3
sin
θgn−1 (x, y, θ)
∂θ
48
i.e.,
where
g1 (x, y, θ) = (2p + 1) f1 (z) + (x2 − y2 ) sin2 θ f2 (z)
f ′ (z)
f2 (z) = 1
∈ E∗ .
z
In fact we can prove immediately by induction on n the following
60
Lemma 1. gn (x, y, θ) is a linear combination of f1 , f2 , . . . , fn+1 with
coefficients which are polynomials in x2 , y2 and sin2 θ where fn (z) =
1
fn−1 (z) ∈ E∗ .
z
Corollary. The functions gn (x, y, θ) are indefinitely differentiable in x, y,
π
θ and even in x and y for (x, y) ∈ R2 and θ ∈ [0, ].
2
Definition.
T p(n) [ f ](x, y)
Z π
2
(−1)n γ p
=
(2p − 1)(2p − 3) · · · (2p − 2n + 1) 0
cos−2p+2n θ sin2p−2n+2 θgn (x, y, θ)dθ
2n − 3
2n + 1
The integral converges for
< Rep <
and T p(n) ∈
2
2
L (E∗ , E) where E is the space of indefinitely differentiable functions
of two variables x, y which are also even in x and y with the usual topology of uniform convergence on every compact subset of R2 of functions
together with their derivatives. It can be verified that p → T p(n) is an
analytic function for p in the strip in view of the fact that
γp
is an entire function.
p→
(2p − 1) · · · (2p − 2n + 1)
Lemma 2. In each strip
T p(n) = T p(n−1)
T p(n−1) [ f ](x, y) =
2n − 3
2n − 1
< Rep <
, we have
2
2
(−1)n−1 γ p
(2p − 1)(2p − 3)˙(2p − 2n + 3)
Z π
2
cos−2p+2n−2 θ sin2p−2n+4 θgn−1 (x, y, θ)dθ
0
3. Continuation of T p
61
49
The integral can be written as
Z
π
2
cos−2p+2n−2 θ sin θ sin2p−2n+3 θgn−1 (x, y, θ)dθ
0
−1
=
(2p − 2n + 1)
Z
π
2
cos−2p+2n−1 θ
0
o
∂ n +2p−2n+3
sin
θgn−1 (x, y, θ) dθ
∂θ
2n − 3
< Rep <
(integrating by parts, the integrated part being zero for
2
2n − 1
). The lemma is now evident by the recurrence formula for gn .
2
Thus we have
3
1
< Rep < admits
2
2
−3
, with values in
an analytic continuation in the half plane Rep >
2
L (ε∗ , E). The explicit definition of T p is given by the formula for T P(n)
for suitable n.
Proposition. The function p → T p defined for −
−3
Analytic continuation of T p in the half plane Rep <
is obtained
2
by exactly similar process. We introduce functions
1
ho (x, y, θ) = f1 (z), z = (x2 sin2 θ + y2 cos2 θ) 2 ,
∂ −2p−1
cos
θho (x, y, θ) ,
cos−2p−2 θ sin θh1 (x, y, θ) =
∂θ
and by induction,
cos−2p−2n θ sin θhn (x, y, θ) =
∂
cos−2p−2n+1 θhn−1 (x, y, θ) ,
∂θ
It is easy to see by induction on n that hn (x, y, θ) is a linear com- 62
bination of f1 , f2 , . . . , fn+1 , with coefficients which are polynomials in
x2 , y2 , cos2 θ so that the functions hn (x, y, θ) are indefinitely differentiable in x, y, θ and are even in x and y. If we set
(n)T p f (x, y) =
(−1)n γ p
(2p + 3)(2p + 5) · · · (2p + 2n + 1)
50
π
Z2
sin2p+2n+2 θ cos−2p−2n θhn (x, y, θ)dθ
0
1
3
< Rep < −n + and
2
2
p →(n) T p is analytic in this strip with values in L (ε∗ , E) and that for
1
1
−n − < Rep < −n + , (n−1) T p =(n) T p . Finally we have the
2
2
for n ≥ 1, then
(n) T
p
∈ L (ε∗ , E) for −n −
1
3
Theorem. The function p → T p initially defined for − < Rep < ,
2
2
admits an analytic continuation into an entire function with values in
L (ε∗ , E) and we have the explicit formula for this continuation.
In particular, we have, by analytic continuation, T p [ f ](x, x) = (p +
1
) f1 (x) for any p.
2
Corollary. For p , −1, −2, . . . , the formula
Z x
T p [A](x, y) f (y)dy
B p AB p f (x) = A(x) f (x) + x
o
is valid.
63
The first member B p AB p f (x) is defined and is analytic except for
p = −1, −2, . . .. The second member is an entire function of p, and the
1
two members are equal for |Rep| < so that the corollary follows.
2
Remark. If A(x) is a constant, we have T p [A](x, y) = 0 and the formula
of the corollary is equivalent to B p B p = the identity.
Part II
Topics In Mean-Periodic
Functions
51
Chapter 1
Expansion of a
Mean-periodic Function in
Series
Introduction.
There is no connection between Part I and Part II of this lecture 64
course. But both these parts are essential for the two-radius theorem
which will be proved in the last part.
The theory of Mean periodic functions was founded in 1935 (refer
to “Functions Moyenne-periodiques” by J. Delsarte in Journal de Mathematique pures et appliques, Vol. 14, 1935). A mean periodic function
was then defined as the solution f of the integral equation
Z b
K(ξ) f (x + ξ)dξ = 0
(1)
a
where K is “ density ” given by a continuous function K(x), given on
a bounded interval [a, b]. It was obvious that the study of an ordinary
differential equation with constant coefficients or of periodic functions
with period b − a was a problem exactly of the same type as the study of
equation (1). It was proved in the paper mentioned above that if K satisfied certain conditions and if f were a solution of (1) i.e. f were mean
periodic with respect to K then f is developable in a series of exponential functions which converges towards f uniformly on each interval on 65
53
1. Expansion of a Mean-periodic Function in Series
54
which f is continuous. Equation (1) in modern notation is essentially a
convolution equation
K∗ f =0
∨ ∨
if we replace f by f ( f (x) = f (−x)).
L.Schwartz in 1947 applied the theory of distributions to the theory of Mean periodic functions. He calls a mean periodic function a
continuous function f which is a solution of
µ∗ f =0
where µ is measure with compact support. More generally a continuous
function f is mean periodic with respect to the distribution T with compact support if T ∗ f = 0. He has also given a new definition of mean
periodic function which is important from the topological point of view
and the theory developed in his paper (Annals of Mathematics, 1947)
is complete in the case of one variable. J.P. Kahane has given a special
and delicate development of the theory which gives connection between
Mean periodicity and almost periodicity. The extension to the case of
several variables is certainly difficult and the first known result in Rn is
due to B. Malgrange.
For x ∈ R1 , λ ∈ C we have,
∨
T ∗ (eλx ) = e−λx M(λ)
66
where M(λ) = hT, eλξ i is the Fourier - Laplace transform of the distribution T . When T is a density K or a measure, µ,
M(λ) =
Z
b
a
K(ξ)eλξ dξ or M(λ) =
Z
b
eλξ dµ(ξ)
a
respectively and for periodic function of period a, we have M(λ) =
eaλ − 1. In the case of the differential operator with constant coefficients, M(λ) is the characteristic polynomial of the operator. In all these
cases M(λ) is an entire function of λ, of exponential type and behaves
like a polynomial if Re λ is bounded and the (simple) zeros of this function give the exponential functions (or the exponential polynomials in
1. Determination of the coefficients in the formal series
55
the case of multiple zeros) which are mean periodic with respect to the
distribution T . It is clear that any linear combination of mean periodic
exponentials is also mean periodic and the first problem is to determine,
for any mean - periodic function f , the coefficients of the mean periodic
exponentials in the expansion of the function.
1 Determination of the coefficients in the formal series
Let T be a distribution with compact support and M(λ) its FourierLaplace transform. The set of zeros of M(λ) will be called the spectrum
of M(λ) and will be denoted by (σ).
For the sake of simplicity we suppose that these zeros are simple. 67
Let F be a function sufficiently regular for the validity of the scalar
product
Z
x
hT,
eλ(x−ξ) F(ξ)dξ
o
If T is a measure with compact support, F need be only an integrable
function and if T is a distribution with compact support and order m, F
can be any (m − 1) times continuously differentiable function.
Let F(x) = eαx where α ∈ C is fixed
Z x
eλx − eαx
eλ(x−ξ) eαξ dξ =
λ−α
Z x o
M(λ) − M(α)
so that
hT,
eλ(x−ξ) eαξ dξi =
= τα (λ) · τα (λ)
λ−α
o
is an integral function of exponential type and for α, β ∈ (σ),
τα (β) = 0 if β , α
= M ′ (α) if α = β.
1
= 1if α = β
= 0 if α , β
Then tα (λ) is an entire function of exponential type and therefore by
the theorem of Paley-Wiener, there exists a distribution T α with compact 68
Let tα (λ) =
τα (λ)
M ′ (α)
β
so that tα (β) = δα
56
support whose Fourier-Laplace transform is
tα (λ) i.e. hT α , eλx i = tα (λ).
Thus we see that the system of distributions {T α }α∈(σ) and the functions {eαx }α∈(σ) form a biorthogonal system relative to the distribution
T.
If
F(x) =
X
α∈(σ)
=
cα eαx then cα = hT α , Fi
1
hT,
M ′ (α)
Z
x
eα(x−ξ) F(ξ)dξi.
o
For any F which satisfies suitable regularity conditions (stated at the
beginning) we consider the formal expansion
F(x) ∼
X
α∈(σ)
1
cα = ′ hT,
M (α)
where
cα eαx
Z
x
eα(x−ξ) F(ξ)dξi
o
and we have immediately two problems.
Problem 1. If F is mean periodic relatively to a distribution T i.e. if
T ∗ F = 0 and if we compute the coefficients (cα ) and construct the
P
series
cα eαx .
α∈(σ)
69
Then what is the significance of this series? If the series converges
(in some sense) what is the connection between the sum of the series
and the given function F? In particular does there exist a one-to-one
correspondence between the mean periodic function F and the system
of coefficients (cα )α∈σ ?
Problem 2. If F is given one the smallest closed interval which contains
the support of T it is possible to compute the cα by the formula. In this
case is it possible to extend F into a mean periodic function on R′ ?
2. Examples
57
2 Examples
Example 1. We consider a periodic function F(x) with [0, 1] as the interval of periodicity. In this case T = δ1 −δ0 (δa being the Direct measure
at a) and T ∗ F = F(x + 1) − F(x) = 0 and M(λ) = eλ − 1 so that the
spectrum (σ) is given by = 2kπi, k an integer; M ′ (λ) = eλ , M ′ (α) = 1
for every α ∈ (σ). We have
Z x
eα(x−ξ) F(ξ)dξ
cα = hT α , Fi = hT,
o
Z 1
Z 1
α(1−ξ)
=
e
F(ξ)dξ =
e−2ikπξ F(ξ)dξ.
o
o
This is the classical formula for the coefficients of the Fourier series
for F(x) on the interval [0, 1] and the answers to Problem 1 and Problem
2 are classical.
Example 2. Let T be a distribution in R1 with compact support which
has the property
T ∗ F = F(x + 1) − kF(x) −
Z
1
K(ξ)F(x + ξ)dξ
(2)
0
where k is a constant , 0 and K(x) is continuously differentiable and 70
T ∗ F is a function defined by
T ∗ F(x) = hT ξ , F(x + ξ)i = T ξ .F(x + ξ).
We shall study in this case the spectrum and the formal development
in series of exponentials for a function F mean periodic with respect to
T.
a) Let K be continuous in [0, 1].
The definition T ∗ eλx = e+λx M(λ) gives
Z 1
λ
K(ξ)eλξ dξ
(3)
M(λ) = e − k −
o
Z 1
or
e−λ M(λ) − 1 = −ke−λ −
K(ξ)eλ(ξ−1) dξ
o
58
and if λ = λo + iλ1 ,
−λo Z
−λ
e M(λ) − 1 ≤ ke +
71
o
K(ξ)eλo (ξ−1) dξ
1
The second member of this inequality → 0 when λ0 → +∞ so that
the real parts of the zeros of M(λ)
are bounded
above.
R 1 λo
Similarly from M(λ) + k ≤ e + o K(ξ)eλoξ dξ, we see that the
real parts of the zeros are bounded below. Thus the spectrum lies in a
vertical band of complex numbers with real parts bounded.
b) Let K be once continuously differentiable in [0, 1]. Integrating
by parts the integral in (3),
Z
K(1)eλ − K(0) 1 1 ′
M1 (λ)
λ
M(λ) = e − k −
+
K (ξ)eλ dξ = eλ − k −
λ
λ o
λ
where M1 (λ) is an entire function of exponential type which remains
bounded when the real part λ0 of λ remains bounded. The zeros of M(λ)
are therefore asymptotic with the zeros of eλ −k i.e. with α+2πih, where
α is a determination of log α and h = 0, ±1, ±2, . . .. Let αh be the zero of
+∞
P
1
M(λ) near α + 2πih. Then the convergence of the series
2
h=−∞ |α + 2πih|
+∞
P 1
implies the convergence of
.
2
h=−∞ |αh |
c) Let F be (C, 2) in [0, 1] and K be (C, 1). We consider the ex
pansion of F(x) in terms of the exponentials eαh x h=0,±1,±2,... (αh ∈ (σ)),
mean periodic with respect to T .
F(x) ∼
where
72
+∞
X
Ah eαn
x
(4)
h−−∞
1
hT,
Ah = ′
M (αh )
Z
x
eαh (x−ξ) F(ξ)dξ
R1
We have
M ′ (λ) = eλ − o K(ξ)eλξ dξ
since M(αh ) = 0,
Z 1
eαh = k +
K(ξ)eαh ξ dξ
o
(5)
o
(6)
2. Examples
and
59
M ′ (αn ) = k +
Z
o
1
(1 − ξ)K(ξ)eαhξ dξ
Integrating by parts integral and observing that K ′ (ξ) is bounded in
M2 (αh )
[0, 1] since it is continuous, it is clear that M ′ (αh ) = k +
where
αh
M2 (αh ) is bounded when Reαh is bounded. Also from (a), we know that
if |h| → ∞, |αh | → ∞ with |Reαh | bounded. Hence
1
Bh
∼ Bh as |h| → ∞ · · ·
(7)
M ′ (αh ) K
Z x
where
Bh = hT, Gh (x)i, Gh (x) =
eαh (x−ξ) F(ξ)dξ
o
Z 1
hT, Gh (x)i = T ∗ Gh (0) = Gh (1) − kGh (0) −
K(ξ)Gh (ξ)dξ
o
Z 1
Z 1Z ξ
αh (1−ξ)
Bh =
e
F(ξ)dξ −
K(ξ)eαh (ξ−η) F(η)dηdξ
Ah =
o
Now
Z
o
and
o
o
73
1
Z 1
1
F(1) eαh F(o)
+
+
eαh (1−ξ) F ′ (ξ)dξ
αh
αh
αh o
1
1
=−
F(1) − eαh F(0) − 2 F ′ (1) − eαh F ′ (0)
αh
αh
Z 1
1
+ 2
eαh (1−ξ) F ′′ (ξ)dξ
αh o
Z 1Z ξ
−
K(ξ)eαh (ξ−η) F(η)dηdξ
o
o
Z 1
Z
1
F(0) 1 αh ξ
=
K(ξ)F(ξ)dξ −
e K(ξ)dξ
αh o
αh o
Z 1Z ξ
1
eαh (ξ−η) K(ξ)F ′ (η)dηdξ
−
αh o o
Z 1
Z
1
F(0) 1 αh ξ
=
K(ξ)F(ξ)dξ −
e K(ξ)dξ
αh o
αh o
eαh (1−ξ) F(ξ)dξ = −
60
+
−
74
1
α2h
1
α2h
Z
o
Z
o
1
K(ξ)F ′ (ξ)dξ −
1Z ξ
F ′ (0)
α2h
Z
1
eαh ξ K(ξ)dξ
o
eαh (ξ−η) K(ξ)F ′′ (η)dηdξ
o
whence
(
Z 1
Z1
1
αh
K(ξ)F(ξ)dξ − F(0) eαh ξ K(ξ)dξ
− F(1) + e F(0) +
Bh =
αh
o
o
(
Z1
1
− F ′ (1) + eαh F ′ (0) + eαh (1−ξ) F ′′ (ξ)dξ
αh 2
o
Z 1
Z 1
′
′
+
K(ξ)F (ξ)dξ − F (0)
eαh ξ K(ξ)dξ
o
o
)
Z 1Z ξ
αh (ξ−η)
′′
e
K(ξ)F (η)dηdξ
−
o
o
substituting for eαh from (6) the first bracket above becomes
R1
R1
−F(1) + kF(0) + F(0) o K(ξ)eαh ξ dξ + o K(ξ)F(ξ)dξ
R1
−F(0) o eαh ξ K(ξ)dξ = −(T ∗ F)(0) = 0 since F is assumed to be
 
 1 
mean periodic with respect to T . Hence Bh = o  2  provided F ′ and
αh
F ′′ are bounded in [0, 1]. As |Reαh | remains bounded when |h| → ∞,
+∞
P 1
from (7) it is immediate that the series (4) is comparable with
2
h=−∞ αh
and therefore converges uniformly and absolutely on each compact subset of R1 to a continuous function F1 (x)H(x) = F(x) − F1 (x) is mean
periodic with respect to T and the coefficients cαh (αh ∈ σ) in the expansion for H(x) in mean periodic exponentials are all zero. Hence H = 0
by the uniqueness of development. (Problem 1).
Chapter 2
Mean Periodic Function in
R2
We shall study function of two variables mean periodic with respect 75
to two distributions, and discuss problems 1 and 2 stated in Chapter 1
(page 56). The general case being difficult we shall give solutions of the
problems in some special cases when the distributions are of particular
type with the spectrum satisfying certain conditions.
Definition. A continuous function F on R2 is said to be mean periodic
with respect to T 1 , T 2 ∈ ε′ (R2 ) if it verifies simultaneously the convolution equations
T 1 ∗ F = 0, T 2 ∗ F = 0.
The Fourier- Laplace transforms
Mi(λ) = hT i , e<λ,x> i
where x = (x1 , x2 ) R2 , λ = (λ1 , λ2 ) ∈ C 2 , < λ, x >= λ1 x1 + λ2 x2 of T i (i =
1, 2) are entire functions of λ1 , λ2 of exponential type. The spectrum
(σ) is defined by Mi (λ) = 0, i = 1, 2,. In all that follows we shall restrict
ourselves to the case when i) (σ) is countable and (ii) (σ) is simple i.e.
α ∈ (σ) is a simple zero of M1 (λ) and M2 (λ) and the Jacobian
76
61
2. Mean Periodic Function in R2
62
∂M1
∂λ1
D(λ) = ∂M2
∂λ
∂M1
∂λ2
does not vanish at any α ∈ (σ). We define
∂M2
∂λ2
1
J(M, M)
1
tα (λ) =
D(α)(λ1 − α1 )(λ2 − α2 )
J(λ, α)
where
M (λ , λ ) − M1 (α1 , λ2 ) M1 (α1 , λ2 ) − M1 (α1 , α2 )
J(M, M) 1 1 2
= J(λ, α)
M (λ , λ ) − M (α , λ ) M (α , λ ) − M (α , α )
2 1 2
2 1 2
2 1 2
2 1 2
is the determinant of Jacobi. It is clear that
tα (β) = 0
for
=1
for
α, β ∈ (σ), α , β
α, β ∈ (σ), α = β
For α∈ (σ), tα (λ) is an entire function of λ of exponential type and by
Paley-Wiener theorem, tα (λ) is the Fourier Laplace transform of a distri1 2
nbutiono T α ∈ E (R ). {T α }α∈(σ) together with the exponentials
mean periodic with respect to T 1 , T 2 form a biorthogonal
ehα,x i
α∈(σ)
system, i. e.
hT α , e<β,x> i = 0 for α, β ∈ (σ), α , β
(1)
hT α , e<α,x> i = 1 for α ∈ (σ)
77
If F is mean periodic with respect to T 1 , T 2 and if we suppose the existence of an expansion of F in a series of mean periodic exponentials
X
F(x) ∼
cα e<α,x>
(2)
α∈(σ)
we have formally
cα = hT α , Fi
Let sα (λ) =
1 J(M, M)
so that
D(α J(λ, α)
sα (λ) = (λ1 − α1 )(λ2 − α2 )tα (λ)
(3)
63
For α fixed in (σ), sα (λ) is an entire function of λ of exponential type.
Let S α be the distribution in E ′ (R2 ) whose Fourier-Laplace image is
∂
by Di (i = i, 2), we have
sα (λ). Denoting
∂xi
h(D1 + α1 )(D2 + α2 )T α , eλx i = (D1 D2 + α1 D2 + α2 D1 + α1 α2 )T α , eλx i
= hT α , (D1 D2 − α1 D2 − α2 D1 + α1 α2 )eλx i
(by the definition of the derivative of a distribution hD p T , φi = (−1) p 78
hT, D p φi, refer to ‘theory des distributions’ by L. Schwartz)
= hT α , (λ1 − α)(λ2 − α2 )eλx i = (λ1 − α1 )(λ2 − α2 )tα (λ)
Hence
S α = (D1 + α1 )(D2 + α2 )T α
(4)
e−hα,xi D1 D2 e<α,x> T α = S α
(5)
(4) is equivalent to
For
i
h
he−hα,xi D1 D2 ehα,xi T α , φi = hT α , e<α,x> D1 D2 e−hα,xi φi
(by the definition of the multiplicative product of a distribution by a
function and the derivative of a distribution; refer to ‘Theory des distributions’)
= hT α , (α1 α2 − α2 D1 − α1 D2 + D1 D2 )φi
= h(D1 + α1 )(D2 + α2 )T α , φ >=< S α , φi
Let G(x) be any solution of the partial differential equation
h
i
ehα,xi D1 D2 e−hα,xi G(x) = F(x)
or
where D =
DG = F
(6)
(6′ )
∂
∂
∂2
− α1
− α2
+ α1 α2 is a differential operator 79
∂x1 ∂x2
∂x2
∂x1
2. Mean Periodic Function in R2
64
with constant coefficients. For F ∈ E , we can choose G ∈ E . We can
actually take G to be
Z x1 Z x2
G(x1 , x2 ) =
eα1 (x1 −ξ1 )+α2 (x2 −ξ2 ) F(ξ1 , ξ2 )dξ1 dξ2
(7)
a1
a2
where a1 , a2 are arbitrary.
Thus we obtain from
the formula
h
i
hT α , Fi = hT α , ehα,xi D1 D2 e−hα,xi G(x) i
i
h
= e−hα,xi D1 D2 ehα,xi T α , Gi = hS α , Gi,
cα =< S α ,
Z
x1
a1
Z
x2
eα1 (x1 −ξ1 )+α2 (x2 −ξ2 ) F(ξ1 , ξ2 )dξ1 dξ2 >
(8)
a2
which is a natural generalization of the formula know in the case of R1 .
80
Remark. The distributions S are completely explicit. We have
S α (λ) =
1
M1 (λ1 , λ2 )M2 (α1 , λ2 ) − M2 (λ1 , λ2 )M1 (α1 , λ2 )
D(α)
P P
1 T 1 ∗ 2 −T 2 ∗ 1 where T 1 and T 2 are the given disD(α)P
tributions and 2 for instance is a distribution in the variable x2 determined by its Fourier-Laplace image M2 (α1 , λ2 ).
Theorem. The mean periodic exponentials e<α,x> α∈(σ) form a free system of functions in E .
and S α =
In fact they form a biorthogonal system with the distributions
{T α }α∈(σ) in E ′ and it nis clearo from (1) that no ehα,xi is in the closed
subspace generated by e<β,x> β∈(σ) .
β,α
Remark. It is possible to choose the solution Iα ( f ) = G of (6) such that
for F ∈ E , the function λ → Iα (F) from C 2 to E is an entire function.
In fact it suffices to take G as in (7). In this case
cα
ehα,xi =
ehα,xi
hS α , Iα (F)i
D(α)
65
81
λ → S λ is an entire function of λ with values in E ′ and λ → Iλ (F) is an
entire function of λ with values in E . Hence
F (x, λ) =
ehλ,xi hS λ , Iλ (F)i
M1 (λ)M2 (λ)
is a meromorphic function of λ in which the ‘local residues’ in the sense
of Poincare are precisely cα e<gα,x> .
The Problem is now as follows.
If F is given in E mean periodic with respect to T 1 and T 2 then is F
P
then sum of the series
eα e<α,x> ?
α∈(σ)
If for x fixed, the integrals of F (x, λ) over a system of varieties in C 2
tending to infinity have for limit F(x) and if the global Cauchy theorem
were true in C 2 , then the answer to this question would be in affirmative.
In this manner, the problem is equivalent to the global Cauchy theorem
in C 2 and the answer is completely unknown.
Chapter 3
The Heuristic Method
1
We consider the problem of continuation of a function given on a subset 82
of R1 into a function defined on the whole of the real line, mean periodic
with respect to a distribution with compact support (Problem 2, page 56)
T ∈ E ′ (R1 ) has its support contained in a finite closed interval say [0,
1]. Then the distributions {T α }, for α of T have their supports contained
in [0, 1]. Let F be a function given in [0, 1] so that the computation of
the coefficients cα = hT α , Fi is possible. The heuristic point of view
is the following. We suppose that problem 2 is completely solved for
the distribution T and for the function F i.e. any function F given on
the minimal set [0, 1] admits of an extension F˜ in the whole of R1 as a
function mean periodic with respect to T . F is not necessarily infinitely
differentiable, even if F may be in [0, 1] but it is probably sufficiently
regular piecewise. We have T ∗ F = 0 and in a certain sense F˜ =
P
cα eαx . In particular, for x ∈ (0, 1) we have
α∈(σ)
˜
F(x) = F(x)
=
X
cα eαx
(1)
α∈(σ)
Now we suppose the possibility of the change of the distribution T , 83
with its compact support keeping in a fixed interval [0, 1] and we shall
obtain several representations of F in series of mean periodic exponentials (relative to several distributions). For instance we suppose that T
67
3. The Heuristic Method
68
depends on a parameter ε which tends to zero, but that the support of T
is contained in [0, 1]. In such a situation, the spectrum (σ) and therefore
each term in the right hand member of (1) depends on ε whereas the left
hand member is fixed whatever ε. Thus we have


 X

d 
cα eαx  = 0 x ∈ (0, 1)
α∈(σ)
where the differential d is related to the variation of the distribution T .
Formal computation gives
X
X
d[cα ]eαx +
x eαx cα dα = 0
α∈(σ)
α∈(σ)
But applying our heuristic concept to the distribution T and the function
xeαx we obtain
X
xeαx =
kαβ eβx , for x ∈ (0, 1),
β∈(σ)
which gives
)cαx
X (
X
dcα +
cβ kβα dβ
=0
α∈(σ)
84
β∈(σ)
so that
dcα +
X
cβ kβα dβ = 0
β∈(σ)
But
˜ where
cα = hT α , Fi
M(λ)
hT α , eλx i = tα (λ) =
(λ − α)M ′ (α)
As support T α ⊂ [0, 1], cα = hT α , Ri
so that
dcα = hdT α , Fi
(2)
1.
69
since F is independent of ε and (2) becomes
X
hdT α +
kβα T β dβ, Fi = 0
(3)
β∈(σ)
(3) holds for any arbitrarily chosen F on [0, 1] and for the distribution
P
dT α +
kβα T β dβ with support contained in [0, 1] whenever the
β∈(σ)
n
scalar product is meaningful whenever F is sufficiently regular in [0, 1]
o
in order that the scalar product be defined . Hence
85
P
dT α +
kβα T β dβ = 0 (in the sense of distributions)
β∈(σ)
Taking Fourier-Laplace image,
X
dT α (λ) +
kβα T β (λ) dβ = 0
β∈(σ)
Now
kβα
1
= ′ hT,
M (α)
Z
x
eα(x−ξ) ξ eβξ dξi
0
But
αx
e
Z
x
(β−α)ξ
e
0
xeβx
− e−αx
dξ =
(β − α)
kβα
eβx
o
eαx
e(β−α)ξ
dξ
(β − α)
−
−
β − α (β − α)2
)
(
1
M(β) − M(α)
1
βx
= ′
hT, xe i −
M (α) (β − α)
(β − α)2
)
( ′
1
M (β) M(β) − M(α)
= ′
−
M (α) β − α
(β − α)2
=
and
xeβx
Zx
Hence
86
kβα =
and
1 M ′ (β)
for β , α
β − α M ′ (α)
1 M ′′ (α)
kαα =
2 M ′ (α)
70
We have also
dT α (λ) = d
=
M(λ)
(λ − α)M ′ (α)
M(λ)
d[M(λ)]
M(λ)
+
dα −
d[M ′ (α)]
(λ − α)M ′ (α) (λ − α)2 M ′ (α)
(λ − α)[M ′ (α)]2
and
d[M ′ (α)] = d[M ′ (λ)]λ=α + M ′′ (α)dα
This gives
d[T α (λ)]+
X
1 M ′′ (α)
M(λ)
M(λ)
dα+
dβ = 0
′ (α)
2 M ′(α) (λ − α)M ′ (α)
(λ
−
β)(β
−
α)M
β∈(σ)
β,α
or
M(λ)(d[M ′ (λ)])
d[M(λ)]
λ
−
(λ − α)M ′ (α) (λ − α)[M ′ (α)]2
)
(
M(λ)M ′′ (α)
M(λ)
−
=α+
(λ − α)2 M ′ (α) (λ − α)[M ′ (α)]2
X
1 M ′′ (α)
M(λ)
M(λ)
dα +
dα +
dβ = 0
′
′
2 M (α) (λ − α)M (α)
(λ − β)(β − α)M ′ (α)
β∈(σ)
β,α
87
After simplification we finally obtain the formula
(
)
1
1 M ′′ (α)
d[M(λ)] (d[M ′ (λ)])λ=α
−
+
−
M(λ)
M ′ (α)
λ−α
2 M ′(α)
X
λ−α
dβ = 0 . . . . . .
dα +
(λ − β)(β − α)
β∈(σ)
(F )
β,α
2
The formula (F ) is obtained by a purely heuristic process and the following example will serve as a partial verification of the computation.
2.
71
Suppose that M(λ) is a polynomial. (In this case T is a differential
operator with constant coefficients)
n
Q
M(λ) = (λ − βi ). Let βi , β j for i , j
i=1
X 1
dM(λ)
=−
dβ j
M(λ)
λ − βj
j=1
X
M ′ (λ) =
(λ − β1 )(λ − β2 ) . . . (λ d
− βn ) . . . (λ − βn )
n
j
and
(d[M ′ (λ)])λ=α=βi = −
(λ − βn )dβk
!
X
j,k
λ=βi
(λ − β1 ) . . . (λd
− β j ) . . . (λd
− βk ) . . .
−
X
(βi − β1 ) . . . (βid
− β j ) . . . (βi − βk )dβi
j
But
88
M ′ (βi ) = M ′ (α) = (βi − β1 ) . . . (βid
− βi ) . . . (βi − βn )
X
hX 1 i
(d[M ′ (λ)])λ=α
1
=−
dβk −
dβi
M ′ (α)
β
−
β
β
−
β
i
k
i
j
j,i
k,i
so that
M ′′ (λ) = (λ − β1 ) . . . (λd
− β j ) . . . (λd
− βk ) . . . (λ − βn )k, j
X
′′
M (λ) = 2 (βi − β j ) . . . (βid
− β j ) . . . (βid
− βk ) . . . (βi − βn )
k,i
and
X
1
M ′′ (α)
=
2
.
′
M (α)
βi − βk
k,i
Thus we can write
X 1
d[M(λ)]
=−
dβ
M(λ)
λ−β
β∈(σ)
X 1
1 M ′′ (α)
(d[M ′ (λ)])λ=α
=
−
dβ
−
dα
M ′ (α)
α−β
2 M ′ (α)
β
β,α
72
and
89
d[M(λ)]
−
M(λ)
(d[M ′ (λ)])λ=α
M ′ (α)
M ′′ (α)
1
dα
2 M ′(α)
X 1
X 1
1
dβ +
dβ −
dα
=−
λ−β
α−β
λ−α
β,α
β,α
X
1
λ−α
=−
dα +
dβ
λ−α
(λ − β)(α − β)
β
−
β,α
=−
1
dα −
λ−α
X
β
β,α
λ−α
dβ
(λ − β)(β − α)
This is exactly the formula (F ).
Remark. It is important to note that in this case, the support of T is only
{0}.
3 The general formula in R2 by the heuristic process
Let T 1 , T 2 ∈ E ′ (R2 ) with their spectrum (σ) simple. Suppose that F is
a function given on a subset E of R2 , which depends on the supports of
T 1 and T 2 so that the computation of cα =< T α , F >, α ∈ (σ) is possible
(page 64). We can write formally
X
F≈
cα e<α,x>
α∈(σ)
90
suppose that the distributions vary in certain family which depends on a
parameter their supports fixed. Then dF = 0 gives
X
α∈(σ)
dcα e<α,x> +
2
XX
cβ xi dβi e<β,x = 0 β = (β1 , β2 )
β∈(σ) i=1
But
xi ehβ,xi =
X
β∈(σ)
kβαi e<α,x> on E,
4.
73
and so
dcα +
2
XX
kβαi dβi cβ = 0.
β∈(σ) i=1
The coefficients cα are given by cα = hF, T α i and
dT α +
2
XX
kβαi T β dβi = 0
β∈(σ) i=1
Taking Fourier-Laplace transform,
d(tα (λ)) +
2
XX
kβαi tβ (λ)dβi = 0
β∈(σ) i=1
Now kβαi = hxi
e<β,xi , T
the general formula is
#
"
i
∂
∂ h
<λ,x>
tα (λ)
and
< Tα, e
=
αi =
λ=β
∂λi
∂λi
λ=β
d[tα (λ)] +
2
XX
β∈(σ) i=1
n
o
where tα (λ)
tβ (λ)
∂
tα (λ)λ−β dβi = 0
∂λi
(G )
is a biorthogonalising system of functions on the 91


0, β , α
spectrum: tα (β) = 

1, β = α
α∈(σ) 
4
We now again consider the formula (F ) and give another interpretation
of the formula by making precise the variation of T . Let UεE ′ (R1 ) be a
distribution with support contained in [0, 1] and ε a parameter infinitely
small. The distribution T − εU has Fourier-Laplace transform M(λ) −
εA(λ) where A(λ) is the Fourier-Laplace transform of U and support
(T − εU) ⊂ [0, 1]. If {α} is the spectrum of T, {α + dα} is the spectrum
of T − εU. M(α + dα) − εA(α + dα) = 0 give
M(α) + dαM ′ (α) + · · · − ∈ {A(α) + dαA′ (α) + · · · } = 0
74
using M(α) = 0 and neglecting terms of higher order, we have
M ′ (α)dα− ∈ A(α) = 0 so that dα =
∈ A(α)
M ′ (α)
substituting in (F ) for dα,
!
X
A(λ)
A(λ)
1
1
A(β)
=
+
+
M(λ) (λ − α)M ′ (α) βε(σ) λ − β β − α M ′ (β)
β,α
(
)
1 M ′′ (α)A(α)
1 d[M ′ (λ)]λ=α
−
− lim
ε→0 2
2 [M ′ (α)]2
M ′ (α)
92
But M ′ (λ) + dM ′ (λ) = M ′ − εA′ (λ) and d[M ′ (λ)] = −εA′ (λ) so that
finally
!
X
1
A(α)
1
A(β)
A(λ)
=
+
+
′
M(λ) (λ − α)M (α) β∈(σ) λ − β β − α M ′ (β)
β,α
1 M ′′ (α)A(α) A′ (α)
+ ′
· · · · · · (F1 )
2 [M ′ (α)]2
M (α)
!
A(β)
1
1
+
= 0 for λ = α, and it
λ − β β − α M ′ (β)
−
Remark. The series
P
βε(σ)
β,α
is easy to prove that
"
#
A(λ)
A(α)
A′ (α) 1 M ′′ (α)A(α)
lim
−
−
=
λ→α M(λ)
(λ − α)M ′ (α)
M ′ (α) 2 [M ′ (α)]2
We have in fact, in the neighbourhood of α
A(λ) = A(α) + (λ − α)A′ (α) + · · · · · ·
1
M(λ) = (λ − α)M ′ (α) + (λ − α)2 M ′′ (α) + · · · · · ·
2
A(α) + (λ − α)A′ (α) + · · · · · ·
1
A(λ)
=
M(λ) (λ − α) M ′ (α) + 21 (λ − α)M ′′ (α) + · · · · · ·
4.
75
(
1
A(α)
1
A(λ)
−
=
[A(α) + (λ − α)A′ (α) + · · · ]
′
′
M(λ) (λ − α)M (α) λ − α M (α)
)
M ′′ (α)
A(α)
1
+ ···]− ′
[1 − (λ − α) ′
2
M (α)
M (α)
(
1
1
1
=
[A(α) + (λ − α)A′ (α)][1 − (λ − α)
′
λ − α M (α)
2
)
′′
A(α)
M (α)
+ ···] − ′
M ′(α)
M (α)
(
)
′′
′
h
i
A (α) 1 A(α)M (α)
1
−
=
(λ − α) ′
+ ···
λ−α
M (α) 2 [M ′ (α)]2
n o
A′ (α) 1 A(α)M ′′ (α)
+ (λ − α) · · ·
−
= ′
′
2
M (α) 2 [M (α)]
which gives the required result.
93
Remark 2. The formula (F ) is exactly a formula of Mittag-Leffler; the
A(λ)
principal part of
in the neighbourhood of the simple pole β is
M(λ)
A(β)
A(β)
; the term (β−α)M
′ (β) is a corrective term which gives the
(λ − β)M ′ (β)
P 1
P
convergence of the series , by reason of the convergence of
|β|2
β
(which itself is a consequence of the fact that M(λ) is of exponential
type) only if A(β) is bounded on (σ). We can also write
!
X
1
M(λ)A(α)
1
M(λ)
A(λ) =
+
+
A(β)
′
(λ − α)M (α) βε(σ) λ − β β − α M ′ (β)
β,α
and we consider this formula as an interpolation formula for A(λ), with 94
the interpolation function
tβ (λ) =
M(λ)
(λ − β)M ′ (β)
Now we see that the computation of §1 is in a certain sense the
converse of the theorem of the Mean periodic functions, viz a mean
76
periodic function admits of an expansion in mean periodic exponentials (the spectrum being simple otherwise mean periodic exponentialmonomials) (Problem 1). Mittag-Leffler’s theorem is used in the proof
of this theorem. Conversely problem (2) if solved gives the MittagLeffler’s theorem.
5
We shall now consider the formula (G ), in the case of R2 and C 2 . For
convenience we shall first fix the following notation.
Let S denote the function from C 2 into C 2 given by (λ1 , λ2 ) − − −
−(S 1 (λ1 , λ2 ), S 2 (λ1 , λ2 )). S is an entire analytic function on C 2 into
itself since S 1 and S 2 are so. For λ, α, βεC 2 let [λ−α] = (λ1 −α1 )(λ2 −α2 )
and
[α − β] = (α1 − β1 )(α2 − β2 ).
Let
and
95
∂S 1 ∂S 1 ∂(S 1 , S 2 ) ∂α1 ∂α2 = D(α) =
∂(α1 , α2 ) ∂S 2 ∂S 2 ∂α1
∂αn
J(S , A) S 1 (λ1 , λ2 ) A1 (α1 , λ2 )
=
J(λ, α) S 2 (λ1 , λ2 ) A2 (λ1 , λ2 ) where S and A are two entire functions of C 2 into itself and λ, α are two
points of C 2 . If S = A then
J(S , A) S 1 (λ1 , λ2 ) S 1 (α1 , λ2 )
=
J(λ, α) S 2 (λ1 , λ2 ) S 2 (λ1 , λ2 ) The point α is a zero of the function S if and only if S 1 (α1 , α2 ) =
0 = S 2 (α1 , α2 ) i.e., if αε(σ) and
D(α) = lim
λ→α
96
1 J(S , S )
(λ − α) J(λ, α)
The set (α) is countable and the zeros αε(σ) are simple (i.e. D(α) ,
0 for αε(σ)).
We shall now compute
5.
77
a) dβi (i = 1, 2),
b)
∂
(tα (λ))λ=β ,
∂λi
c) d[tα (λ)] which occur in (G ).
a) Let ε be a small parameter which has 0 for limit. We substitute for
the function S , a neighbouring function S − εA where A = (A1 , A2 )
is another entire analytic function of C 2 into itself and we compute
the variations of the spectrum (σ).
S 1 (α1 + dα1 , α2 + dα2 )− ∈ A1 (α1 + dα1 , α2 + dα2 ) = 0
S 2 (α1 + dα1 , α2 + dα2 )− ∈ A2 (α1 + dα1 , α2 + dα2 ) = 0
But (α) is a simple zero of the functions S . Hence neglecting the
terms of the second order in ε, we have,
∂S 2
∂S 1
dα1 +
=∈ A1 (α1 , α2 )
∂α1
∂α2
∂S 2
∂S 2
dα1 +
dα2 =∈ A2 (α1 , α2 )
∂α1
∂α2
and by the Cramer’s rule,
dα1 =
∂S
∂S
∈ J(A, ∂λ2 )
∈ J( ∂λ1 , A)
, dα2 =
D(α) J(α, α)
D(α) J(α, α)
b) As β is a zero of the
! the function S , it is obvious that, for β , α, the 97
∂
tα (λ)
which are different from zero arise from the
terms of
∂λi
λ=β
differentiation of S 1 and S 2 and we have immediately,
!
∂S
,
S
J
∂
1
∂λi
tα (λ)
=
, i = 1, 2
∂λi
D(α)[β
−
α]
(β,
α)
λ=β
substituting for dβ, we obtain
78
2
X
i=1
!
∈
∂
tα (λ)
dβi =
∂λi
D(α)D(β)[β − α]
λ=β
∂S
∂S
∂S
∂S
J ∂λ
J ∂λ
, S J A, ∂λ
, S J ∂λ
,A
1
2
2
1
+
J(β, α) J(β, β)
J(β, α) J(β, β)
The curly bracket equals
#
"
∂
∂
S 1 (β1 , β2 )S 2 (α1 , β2 ) −
S 2 (β1 , β2 )S 1 (α1 , β2 ) ×
∂β1
∂β1
"
#
∂
∂
A1 (β1 , β2 )
S 2 (β1 , β2 ) − A2 (β1 , β2 )
S 1 (β1 , β2 )
∂β2
∂β2
#
"
∂
∂
S 1 (β1 , β2 )S 2 (α1 , β2 ) −
S 2 (β1 , β2 )S 1 (α1 , β2 ) ×
+
∂β2
∂β2
∂
∂
S 1 (β1 , β2 )A2 (β1 , β2 ) −
S 2 (β1 , β2 )A1 (β1 , β2 )
β1
∂β1
∂
∂
S 1 (β1 , β2 )
S 2 (β1 , β2 )A1 (β1 , β2 )S 2 (α1 , β2 )
=
∂β1
∂β2
∂
∂
−
S 1 (β1 , β2 )
S 2 (β1 , β2 )A2 (β1 , β2 )S 1 (α1 , β2 )
∂β1
∂β2
∂
∂
S 1 (β1 , β2 )
S 2 (β1 , β2 )A2 (β1 , β2 )S 1 (α1 , β2 )
+
∂β2
∂β2
∂
∂
S 1 (β1 , β2 )
S 1 (β1 , β2 )A1 (β1 , β2 )S 2 (α1 , β2 )
−
∂β2
∂β2
J(A, S )
= D(β)
.
J(β, α)
98
Finally,
2
X
∂
tα (λ)dβi
tβ (λ)
∂λi
i=1
1
J(S , S )
∈
J(A, S )
D(β)
D(β)[λ − β] J(λ, β) D(α)D(β)[β − α]
J(β, α)
1 J(S , S ) J(A, S )
∈
=
D(α)D(β)[β − α] λ − β J(λ, β) J(β, α)
=
5.
79
99
c)
J(S , S )
1
D(α)(λ − α) J(λ, α)
(
)
(
)
J(S , S )
1
1
J(S , S )
d tα (λ) =
d
+d
D(α)(λ − α)
J(λ, α)
D(α)(λ − α) J(λ, α)
tα (λ) =
The differential of
S 1 (λ1 , λ2 ) S 1 (α1 , λ2 )
J(S , S ) =
J(λ, α) S 2 (λ1 , λ2 ) S 2 (α1 , λ2 )
is the sum of two determinants, one of which is obtained from the differentiation of the first column and the second by the differentiation of
the second column. The first determinant gives in d{tα (λ)} the term
J(A, S )
−∈
D(α)(λ − α) J(λ, α)
and this term is, in a certain sense, a principal term, because it is composed of variations A1 and A2 of S 1 and S 2 respectively, in the generic
sense, that is which depend on the two independent variables : λ1 , λ2 . In
the second determinant, the second column has elements
d[S 1 (α1 , α2 )] and d[S 2 (α1 , λ2 )]
which are equal to
100
− ∈ A1 (α1 , λ2 ) +
and −εA2 (α1 , λ2 ) +
∂
S 1 (α, λ2 )dα1
∂α1
∂
∂α1 S 1 (α, λ2 )dα1
respectively, in which
∂S
J(A, ∂λ
,A)
1
ε
D(α)
∂S
J(A, ∂λ
)
2
J(α,α)
ε
has to be substituted for dα1 and D(α)
J(α,α) has to be substituted for
dα2 . Other terms is d[tα (λ)] artist from the denominator D(α)[λ − α]
in which again we have to substitute for dα1 and dα2 . For us, it is
sufficient to write the principal term and to write R for the others which
80
are, in fact, of irregular formation, depend on only, finite in number and
contain A1 and A2 (variations of S 1 and S 2 ) in a non-generic manner.
Then the formula is reduced to
J(S , S ) J(A, S )
[λ − α]
J(A, S ) X
=
+ (R)
(G1 )
J(λ, α)
D(β)[λ − β][β − α] J(λ, β) J(β, α)
β,α
βε(σ)
101
in which the terms in R are i) finite in number, ii) non-generic in A, iii)
depend only on λ and α. It is good to note that in dimension 1, formula
G1 becomes
X
λ−α
A(λ) =
S (λ)A(β) + · · · · · · · · ·
D(β)(λ − β)(β − α)
β,α
β∈(σ)
where the terms not written are exactly
S (λ)A′ (α) 1 S (λ)A(α)D′ (α)
S (λ)A(α)
+
−
D(α)(λ − α)
D(α)
2
D2 (α)
That is the formula (F1 ) of §4 which is analogous to G1 .
It is natural to say that G1 is a generalisation, in C 2 , and for the
function S of the formula of Mittag-Leffler in C 1 .
6
We now proceed to give an example in which the preceding results,
which are purely formal at present, are correct.
Let T 1 , T 2 in E ′ (R2 ) be finite linear combinations of Dirac measures,
situated at rational points.
Their Fourier-Laplace transforms will be of the form
pλ + qλ X
1
2
(1)
S 1 (λ1 , λ2 ) =
a pq exp
N
pλ + qλ X
1
2
(2)
S 2 (λ1 , λ2 ) =
b pq exp
N
102
where N is a fixed integer, and where p, q are also integers and the to
6.
81
summations are finite.
λ1
When (λ1 , λ2 ) is in spectrum, the point (X1 , X2 ) with X1 = exp ,
N
λ2
X2 = exp , have to satisfy the two algebraic equations
N
X
p q
a pq X1 X2 = 0
(3)
X
p q
and
b pq X1 X2 = 0
(4)
The two algebraic curves represented by (3) and (4) intersect in a
finite number, say M, of points. We suppose that all these points (ξ1 , ξ2 )
are simple and finite.
α1
Let ξ1 = exp , ξ2 = exp αN2 . The spectrum (σ) is defined by
N
β1 = α1 + 2hnπi , β2 = α2 + 2knπi
where h and k are integers and the general solutions of
T1 ∗ F = 0 = T2 ∗ F
can be formally expressed as
X
F(x1 , x2 ) =
exp(α1 x1 + α2 x2 ) fα1 α2 (x1 , x2 )
α
where the M functions fα1 α2 (x1 , x2 ) are periodic in x1 and x2 with pe- 103
1
riod . The developments of the fα1 α2 (x1 , x2 ) in Fourier series give
N
immediately the development of F in mean periodic exponentials of the
spectrum (σ) and the computation of the coefficients by the use of the
distributions T α is perfectly correct in this case.
Let the convex envelope of the supports of T 1 and T 2 be the rectangle
0 ≤ x1 ≤ a , 0 ≤ x2 ≤ b
and let T 1 , T 2 have Fourier-Laplace transforms
S 1 (λ1 , λ2 ) =
X
paλ1 qbλ2
a pq exp
+
m
n
!
(1)′
82
S 2 (λ1 , λ2 ) =
X
paλ1 qbλ2
+
b pq exp
m
n
!
(2)′
respectively, where m, n are positive integers and p, q are integers satisfying
0≤ p≤m, 0≤q≤n
104
aλ1
bλ2
setting exp
= X1 , exp
= X2 , we obtain (3) and (4).
m
n
If the coefficients a, b are generic, the equations (3), (4) have degree
n in X2 , the coefficients relatively to X2 being the polynomial in X1 of
degree m. We eliminate X2 by Sylvester’s resultant
0
0 · · · 0
A0 A1 · · · An
0 A0 · · · An−1 An 0 · · · 0
···
···
· · · · · ·
· · ·
···
···
· · · 0 0 · · · A1 A2 · · · An = 0
B0 B1 · · · B1
0 ···
0 0 0 B0 · · · Bn−1 Bn · · ·
· · ·
···
···
· · · 0 0 ···
B1 · · · Bn The determinant is given by the elimination of X20 , X21 , . . . , X22n−1 between the 2n equations
A0 X22n−1 + A1 X22n−2 · · · + An X2n−1 = 0
A0 X22n−2 + · · · + An X2n−2 = 0
··· ··· ··· ··· ···
A0 X2n + A1 X2n−1 + · · · + An = 0
B0 X22n−1 + · · · · · · + Bn X2n−1 = 0
···
B0 X2n
105
···
···
+ · · · + Bn = 0
in which the Ai and the Bi are polynomials in X1 of degree m so that the
determinant of Sylvestor is a polynomial in X1 of degree 2mn and the
6.
83
number M of common points of (3) and (4) is 2mn. Any point (ξ1 , ξ2 )
aα1
bα2
is given by ξ1 = exp
, ξ2 = exp
and the spectrum (σ) consists
m
n
2knπi
2hmπi
, λ2 = α2 +
and F(x, y) =
of (λ1 , λ2 ) where λ1 = α1 +
a
b
P
exp(α1 x1 + α2 x2 ) fα1 α2 (x1 , x2 ) with fα1 ,α2 (x1 , x2 ) periodic in x1 of
α1 ,α2
b
a
period and in x2 of period .
m
n
We know that for αε(σ),
Z x1 Z x2
eα1 (x1 −ξ1 )+α2 (x2 −ξ2 ) F(ξ1 , ξ2 )dξ1 dξ2 i
cα = hS α ,
P
where S α =
1
D(α)
Q
P
∗ 1 } and if
X
ρi F(ai , bi )
hT 1 , Fi =
{T 1 ∗
P
2 −T 2
hT 2 , Fi =
P P
1 , 2 are defined by
i
X
σ j F(c j , d j )
then,
j
106
hΣ1 , Fi = Σi ρi eα2 bi F(ai , 0)
hΣ2 , Fi = Σ j σ j eα2 d j F(c j , 0)
so that
hS α , Fi =
X
i, j
ρi σ j eα2 d j F(ai + c j , bi ) − eα2 bi F(ai + c j , d j )
As the points (ai , bi ) and (c j , d j ) are in the rectangle 0 ≤ x1 ≤ a , 0 ≤
x2 ≤ b, it is clear that in the computation of cα1 ,α2 , the values of the
function F in the rectangle
0 ≤ x1 ≤ 2a , 0 ≤ x2 ≤ b
(5)
are used. This rectangle can be divided in M = 2mn small rectangles
ap
a(p + 1) bq
b(q + 1)
≤ x1 ≤
,
≤ x2 ≤
m
m
n
n
84
p = 0, 1, 2, . . . , (2m − 1)
for
q = 0, 1, 2, . . . , (n − 1)
setting
107
ap
bq
+ x1 ,
+ x2
m
n
p = 0, 1, . . . , 2m − 1
φ pq (x1 , x2 ) = F
!
q = 0, 1, . . . , n − 1,
where 0 ≤ x1 ≤
b
a
, 0 ≤ x2 ≤ we obtain
m
n
φ pq (x1 , x2 ) =
X
(α1 ,α2
aα1 p bα2 q
+
exp
m
n
)
!
exp (α1 x1 + α2 x2 ) fα1 α2 (x1 , x2 )
(6)
p = 0, 1, . . . , 2m − 1 ; q = 0, 1, . . . , (n − 1)
bq
(since fα1 ,α2 ap
= fα1 ,α2 (x1 , x2 )).
+
x
,
+
x
1
2
m
n
The numbers of equations is (6) is equal to the number of unknowns
exp(α1 x1 + α2 x2 ) fα1 ,α2 (x1 , x2 )
and the solution of the linear system gives periodic functions in the recta
b
angle 0 ≤ x1 ≤ , 0 ≤ x2 ≤ . Thus we see that
m
n
108
i) The continuation of F in all the plane is completely known.
ii) The coefficients of the mean periodic exponentials in the development of F are determined by the process of the Jacobi determinant.
iii) The expansion converges as the Fourier series of the function
fα1 ,α2 (x1 , x2 ).
For instance if F(x1 , x2 ) is given and continuous in the rectangle
(5) together with their derivatives of order 1, 2, 3, 4, we have uniform
convergence in any compact set contained in the interior of the rectangle
(5).
7. The formula (F ) for a polynomial
85
The heuristic computation has given us the following important result:
The formula of the type of Mittag-Leffler in C 2 , is valid if S 1 and
S 2 are two linear combinations of exponentials
X
paλ1 qbλ2
+
)
m
n
X
paλ1 qbλ2
+
)
S 2 (λ1 , λ2 ) =
b pq exp(
m
n
p = 0, 1, . . . , m − 1
S 1 (λ1 , λ2 ) =
a pq exp(
q = 0, 1, . . . , n − 1
where the coefficients a pq , b pq are generic
7 The formula (F ) for a polynomial
We shall now give the formula F in the case of a polynomial. It is possible to establish the formula for more than one variable but the proof and
computation will be very long and therefore for the sake of simplicity
we consider only the case of one variable.
Let M(λ) be a polynomial in eλ . M(λ) = P(X) = a0 +a1 X+· · ·+an X n
with X = eλ . Let A1 , A2 , . . . , An be n distinct roots of P(X) = 0. The
spectrum (σ) is in this case composed of n arithmetical progressions
β j (h) = α j + 2hπi where eα j = A j , j = 1, 2, . . . , n, h an integer. In the
formula F :
"
#
1
1 M ′′ (α)
dM(λ) d[M ′ (λ)]λ=α
−
+
−
dα
M(λ)
M ′ (λ)
1 − λ 2 M ′ (α)
X
β−α
+
dβ = 0,
(λ − β)(β − α)
β,α
the summation
P
can be divided into n summations with respect to h,
β,α
corresponding to n arithmetic progressions which constitute (σ). Let
α = α j , be fixed. We first consider the summation for β(h) = αk +
109
86
2hπi, k , j, i.e.,
+∞
X
1
(λ
−
α
−
2hπi)(α
k
k − α j − 2hπi)
h=−∞
(1)
The well known classical formula (Mittag-Laffler) give
110
!
h=+∞
X∗
1
1 1
1
1
=− + +
+
;
eu − 1
2 u h=−∞ u − 2hπi 2hπi
setting u = λ − αk and u = α j − αk in this, we have
1
eλ−αk
and
1
X
1
1
1
1
+
+
=− +
2 λ − αk
λ − αk − 2hπi 2hπi
−1
h
∗
= − 21 +
eα j −αk − 1
By subtraction,
1
eλ−αk
−1
−
1
eα j −αk
1
α j −αk
+
∗ P
h
1
α j −αk −2hπi
+
1
2hπi
!
αj − λ
− 1 (λ − αk )(α j − αk )
)
∗ (
X
1
1
−
+
λ − αk − 2πi α j − αk − 2hπi
+∞
X
1
= (α j − λ)
(λ − αk − 2πi)(α j − αk − 2hπi)
h=−∞
=
Hence
∞
X
1
(λ − αk − 2πi)(αh − α j − 2hπi)
h=−∞
#
"
#
"
1
Ak
1
1
Ak
1
=
−1 =
−
(λ − α j ) eλ−αk − 1 eα j −αk
λ − α j X − Ak A j − Ak
Ak [X − A j ]
.
=
(λ − α j )(X − Ak )(Ak − A j )
111
+∞
But d βk (h) = dαk is independent of h and the part of the summation
7. The formula (F ) for a polynomial
(λ − α j )
87
P
dβ
which is under consideration is
β,α (λ − α)(β − α)
(X − A j )Ak dαk
(X − Ak )(Ak − A j )
For the part of the summation corresponding to the arithmetical
progression β j (h) = α j + 2hgπi, h , 0, it is necessary to compute
∗
P
1
setting z = λ − α j , we have,
h (λ − α j − 2hπi)2hπi
∗
X
h
#
∗ "
1
1X
1
1
=
+
(z − 2hπi)(2hπi) z h z − 2hπi 2hπi
!
1 1
1
1
=
+
−
2z z ez − 1 z
#
"
Aj
1
1
1
−
+
=
(λ − α j ) X − A j λ − α j
2(λ − α j )
Hence the second part of the summation in gives
A j dα j
d αj
d αj
−
+
.
X − Aj λ − αj
2
For any h, dAk = eαk dαk = Ak dαk . (T ) now becomes
′
d M(λ) (d M (λ))λ=α j 1 M ′′ (α j )
−
−
d αj
M(λ)
M ′ (α j )
2 M ′ (α j )
X
dα j
(X − A j )dAk
dA j
+
+
= 0 (1)
+
2
X − A j k, j (X − Ak )(Ak − A j )
Now M(λ) = P(X)
M ′ (λ) = XP′ (X); M ′′ (λ) = X 2 P′′ (X) + XP′ (X).
We have
d[A j P′ (A j )] = (d[XP′ (X)])X=A j + [A j P′′ (A j ) + P′ (A j )]dA j
112
88
and also,
d[A j P′ (A j )] = dA j P′ (A j ) + A j d[P′ (A j )]
= dA j P′ (A j ) + A j (d P′ (X))X=A j + A j P′′ (A j )dA j
Hence
also
(d[XP′ (X)])X=A j = A j (d[P′ (X)])X=A j ;
(dM ′ (λ))λ=α j
(d[XP′ (X)])X=A j (d P′ (X))X=A j
=
=
M ′ (α j )
A j P′ (A j )
P′ (A j )
1 M ′′ (α j )
1 P′′ (A j )
1 dA j
=
−
A j dα j −
′
′
2 M (α)
2 P (A j )
2 Aj
′′
1 P (A j )
1
=−
dA j − dα.
2 P′ (A j )
2
From (1) we obtain,
and
113
−
"
#
′
1
1 A′′ (A j )
d P(X) d P (X)X=A j
−
+
−
dA j
P(X)
P′ (A j )
X − A j 2 P′ (A j )
X (X − A j )dAk
=0
+
(X
−
A
)(A
−
A
)
k
k
j
k, j
which is the formula (F ) for the polynomial P(X).
8
We have seen that the formula (G1 ) holds for the Fourier - Laplace transforms of distributions which are finite linear combinations of Dirac measures placed at rational points in a rectangle in R2 such that the convex
envelops of the supports of these distributions (i.e. the set of rational
points) is precisely the rectangle. We shall now prove that (G1 ) holds
for Fourier - Laplace transforms of two distributions T 1 and T 2 in ε′ (R2 )
which differ slightly (in fact by a measure defined by a density) from a
finite linear combination of Dirac measures. Let T 1 and T 2 be defined
by
T 1 ∗ F = a1 F(x, y) + b1 F(x + 1, y) + c1 F(x, y + 1) + d1 F(x + 1, y + 1)
8.
89
+
Z1 Z1
o
k1 (ξ, η)F(x + ξ, y + η)d ξ d η
o
T 2 ∗ F = a2 F(x, y) + b2 F(x + 1, y) + c2 F(x, y + 1) + d2 F(x + 1, y + 1)
+
Z1 Z1
o
k2 (ξ, η)F(x + ξ, y + η)d ξ dη
o
where
114
i) ai , bi , ci , di (i = 1, 2) are all not zero
ii) ki (i = 1, 2) are continuous.
Let F L T i = Mi (λ, µ) be Fourier - Laplace transforms of
T i (i = 1, 2)
x
M1 (λ, µ) = a1 + b1 eλ + c1 eµ + d1 eλ+µ +
k1 (ξ, η)eλξ+µη d ξ d η
x
M2 (λ, µ) = a2 + b2 eλ + c2 eµ + d2 eλ+µ +
k2 (ξ, η)eλξ+µη d ξ d η
Let (σ) denote the spectrum.
Properties of the spectrum (σ) of T 1 and T 2 .
We have
M1 (λ, µ) e−λ−µ − d1 = a1 e−λ−µ + b1 e−µ + c1 e−λ
x
+
k1 (ξ, η)eλ(ξ−1)+µ(η−1) dξdη
Let
115
(λ, µ) ∈ (σ), λ = λo +i λi , µ = µo +i µ1 |M1 (λ, µ)e−λ−µ −d1 | ≤ |a1 |e−λo −µo
−µo
+ |b1 |e
−λo
+ |c1 | e
Z1 Z1
o
o
|k1 (ξ, η)|eλo (ξ−1)+µo (η−1) d ξ d η
For (λ, µ) ∈ (σ), the left hand member is −d1 and as λo , µo both
tend to +∞, the right hand member tends to 0. This cannot happen as
90
d1 , 0. Hence for (λ, µ) ∈ (σ), λo , µo both cannot tend to +∞. Similarly
the pairs (λo , −µo ), (−λo , µo ) and (−λo , −µo ) cannot each tend to positive
infinity. We have only to consider for i = 1, 2,
Mi (λ, µ)e−λ − bi = ai e−λ + di eµ +
Z1 Z1
ki (ξ, η)eλ(ξ−1)+µη d ξ d η
o
o
Mi (λ, µ)e−µ − ci = ai e−µ + bi eλ−µ + di eλ
+
Z1 Z1
o
−µ
Mi (λ, µ) e
116
λ
µ
o
λ+µ
− di = bi e + ci e + di e
ki (ξ, η)eλξ+µ(η−1) d ξ d η
+
Z1 Z1
o
ki (ξ, η)eλξ+µη d ξ d η.
o
Proposition 1. If each of b2 d1 −b1 d2 , c1 d2 −c2 d1 , c1 a2 −a1 c1 , b1 a2 −b2 a1
is distinct from zero, then the spectrum (σ) of T 1 , T 2 is contained in a
vertical band in C 2 (i.e. the projection (λo , µo ) of (λ, µ) ∈ (σ) in the real
plane remains bounded).
In view of the observations made in the preceding paragraph we
have only to show that (λ, µ) ∈ (σ), λ = λo + iλ1 , µ = µo + i µ1 , one of
λo , µo cannot tend to infinity while the other remains bounded. Eliminating e−µ from the equation
M1 (λ, µ)e−λ−µ − d1 = a1 e−λ−µ + b1 e−µ + c1 e−λ
+
−λ−µ
M2 (λ, µ)e
Z1 Z1
o o
−λ−µ
− d2 = a2 e
+
+ b2 e−µ + c2 e−λ
Z1 Z1
o
we have
k1 (ξ, η)eλ(ξ−1)+µ(η−1) d ξ d η
o
k2 (ξ, η)eλ(ξ−1)+µ(η−1) d ξ d η
8.
91
b2 M1 (λ, µ)e−λ−µ − b1 M2 (λ, µ)e−λ−µ − b2 d1 + b1 d2
= (b2 a1 − b1 a2 )e−λ−µ + (b2 c1 − b1 c2 )e−λ
+
Z1 Z1
o
[b2 k1 (ξ, η) − b1 k2 (ξ, η)]eλ(ξ−1)+µ(η−1) dξdη
o
Hence
|b2 M1 (λ, µ)e−λ−µ − b1 M2 (λ, µ)e−λ−µ − b2 d1 + b1 d2 |
≤ |b2 a1 − b1 a2 |e−λo −µo + |b2 c1 − b1 c2 |e−λo
+
Z1 Z1
o
|b2 k1 (ξ, η) − b1 k2 (ξ, η)|eλo (ξ−1)+µo (η−1) d ξ d η
o
For (λ, µ) ∈ (σ), Mi (λ, µ) = 0 so that passing to the limit as λo → 117
+∞ on (σ) while µo remains bounded, we obtain b2 d1 − b1 d2 = 0 which
is supposed to be not true.
Similarly we can show that
µo → +∞ with λo remaining bounded implies that c1 d2 − c2 d1 = 0;
λo → −∞ with µo remaining bounded implies that c1 a2 − c2 a1 = 0;
and µo → −∞ with λo remaining bounded implies that a1 b2 − a2 b1 = 0.
Remark. For the hyperbolas
a1 + b1 X + c1 Y + d1 XY = 0
(H1 )
a2 + b2 X + c2 Y + d2 XY = 0
(H2 )
and
The condition a) b2 d1 − b1 d2 = 0 expresses that (H1 ) and (H2 ) have
the same horizontal asymptote b) c2 d1 − c1 d2 = 0 implies that (H1 ) and
(H2 ) have the same vertical asymptote; c) c2 a1 − c1 a2 = 0 implies that
(H1 ) and (H2 ) have a common point on o Y; d) b2 a1 − b1 a2 = 0 implies
that (H1 ) and (H2 ) have a common point on oX.
We assume that the conditions a), b), c), d) are not valid. Besides we
suppose that the hyperbolas (H1 ) and (H2 ) are not tangent to each other.
92
Proposition 2. If ki (i = 1, 2) are twice continuously differentiable in 118
0 ≤ x ≤ 1, 0 ≤ y ≤ 1, then the spectrum (σ) is asymptotic to (λ, µ)
where


λ = α1 + 2hπi



µ = β1 + 2kπi
λ = α2 + 2h′ πi
and
µ = β2 + 2k′ πi
where h, k, h′ , k′ are integers and (ξi , ηi ) with eαi = ξi , eβi = ηi (i, 1, 2)
are the distinct common points of (H1 ) and (H2 )
λ
µ
λ+µ
M1 (λ, µ) = α1 + β1 e + γ1 e + δ1 e
1
+
λµ
Z1
o
Z1
o
k1 (ξ, η)
∂2 h λξ+µη i
e
dξdη
∂ξ∂η
setting H(x, y) = eλx+µy and applying Green’s formula,
Z1 Z1
o
o
k1 (ξ, η)
∂2
∂2
H(ξ, η) − H(ξ, η)
k1 (ξ, η) d ξ d η
∂ξ∂η
∂ξ∂η
Z h
i
∂H
∂k1
dξ + k1 (ξ, η)
dη
=
H(ξ, η)
∂ξ
∂η
Q
8.
93
where Q↑ denotes the perimeter of the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
in the senseROABC. The integral over the vertical sides AB, CD will be
denoted by ↑ and it equals
Z
Z
Z
Z
∂H
k1 (ξ, η)
dη =
k1 dH1 =
d(k1 H) − Hdk1
∂η
↑
↑
↑
↑
Z
∂k1
dη + (k1 H)(1,1) − (k1 H)(1,0)
=−
H
∂η
↑
− (k1 H)(0,1) + (k1 H)(0,0)
Finally
119
M1 (λ, µ) = a1 + b1 eλ + c1 eµ + d1 eλ+µ
1h
+
k1 (1, 1)eλ+µ − k1 (1, 0)eλ − k1 (0, 1)eµ + k1 (0, 0)
λµ
# Z1 Z1
"
Z
∂k
∂k
∂k
1
1
dξ −
dη +
eλξ+µη 1 d ξ d η
eλξ+µη
∂ξ
∂η
∂ξ∂η
Q↑
o
o
and we have a similar expression for M2 (λ, µ) i.e.
¯ 1 (λ, µ)
M
λµ
¯
M
2 (λµ)
M2 (λ, µ) = a2 + b2 eλ + c2 eµ + d2 eλ+µ +
λµ
M1 (λ, µ) = a1 + b1 eλ + c1 eµ + d1 eλ+µ +
¯ 1 (λ, µ) and M
¯ 2 (λ, µ) are entire functions of expowhere the functions M
nential type which remains bounded when (λ, µ) lie in a vertical band
M1 (λ, µ)
| → 0. as λ, µ → ∞ in a vertical band and in
of C 2 so that |
λµ
particular when (λ, µ) ∈ (σ) by proposition 1. Thus the spectrum (σ) is
asymptotic with the solutions of
120
φ1 (λ, µ) = a1 + b1 eλ + c1 eµ + d1 eλ+µ = 0
φ2 (λ, µ) = a2 + b2 eλ + c2 eµ + d2 eλ,µ = 0
Setting eλ = X and eµ = Y we obtain (H1 ) and (H2 ) and the required
result follows.
94
Corollary. |D(λ, µ)|, D(λ, µ) being the Jacobian of M1 (λ, µ), M2 (λ, µ)
possesses a positive lower bound on (σ). D(λ, µ) it asymptotic with
the Jacobian of φ1 (λ, µ) and φ2 (λ, µ) which is
∂2 (11 , ϕ2 ) b1 eλ + d1 eλ+µ
= λ
b2 e + d2 eλ+µ
(λ, µ)
c1 eµ + d1 eλ+µ c2 eµ + d2 eλ+µ When (λ, µ) ∈ (σ), (eλ , eµ ) = (ξi , ηi ) i = 1, 2, (ξi , ηi ) being the common points of (H1 ) and (H2 ). Then
i
h
h ∂(ϕ1 , φ2 ) i
= ξi ηi (b1 c2 −b2 c1 )+(b1 d2 −b2 d1 )ξi +(d1 c2 −d2 c1 )ηi
∂(λ, µ) (λ,µ)∈(σ)
121
for i = 1, 2.
But the expression in the square bracket is precisely the Jacobian
of (H1 ) and (H2 ) at the common point (xi , ηi ) which is distinct from
zero since the two hyperbolas do not touch each other. In view of the
conditions of proposition 1 and the remarks following the proposition
ξi ηi , 0. Thus |D(λ, µ)| for (λ, µ) ∈ (σ) is asymptotic with two nonzero
values. Further by the usual hypothesis that the spectrum (σ) is ‘simple’
i.e. D(λ, µ) , 0 for any (λ, µ) ∈ (σ) it follows that |D(λ, µ)| possesses a
positive (strictly) lower bound.
Let A1 and A2 be Fourier- Laplace transforms of two densities U1
and U2 with supports in the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
x
U1 ∗ F =
U1 (ξ, η)F(x + ξ, y + η)d ξ dη
x
U2 ∗ F =
U2 (ξ, η)F(x + ξ, y + η)dξ dη
We wish to prove the following
Theorem. The formula
X
J(M, M) J(A, M)
[λ − α]
J(A, M)
=
+R
J(λ, α)
D(β)[λ
−
β][β
−
α]
J(λ,
β)
J(β,
α)
β,α
β∈(σ)
holds if M1 and M2 are Fourier- Laplace transforms of distributions T 1
and T 2 given by
8.
95
T i ∗ F = ai F(x, y) + bi F(x + 1, y) + ci F(x, y + 1)
x
+ di F(x + 1, y + 1) +
ki (ξ, η)F(x + ξ, y + η)dξη i = 1, 2
(provided the functions ki (ξ, η) and the coefficients ai , bi , ci , di , i = 1, 2, 122
satisfy the required conditions in order that the spectrum (σ) should
have the desirable properties; cf. propositions proved above).
If T 1 , T 2 , U1 , U2 be replaced by T 1 (η), T 2(n) , U1(n) , U2(n) respectively
where
T in ∗ F = ai F(x, y) + bi F(x + 1, y) + ci F(x, y + 1) + di F(x + 1, y + 1)
n−1
X
p
q
1 p q F
x
+
, i = 1, 2; and
,
,
y
+
k
+
i
n n
n
n
n2
p,q=0
Uin
n−1
p q X
1
p
q
∗F =
U1 , F x + , y + ,
n n
n
n
n2
p,q=0
then we know that the theorem holds for M1n , M2n , An1 , An2 (with the obvious notation: F L T 1n = M1n · · · etc). Hence the first step in the proof
is to study the spectrum (σn ) of T 1n and T 2n and its relation to (σ) as
n → ∞.
Proposition 3. (σn ) is contained in a vertical band which is independent
of n.
The proof of the proposition is analogous to that of Proposition 1 123
and we shall give only a partial verification. For instance we prove that
if (λ, µ) ∈ (σn ) and λ = λo + iλ, µ = µo + i µ1 , then i) λo , µo cannot
both tend to +∞; ii) λo cannot tend to +∞ when µo → −∞, iii) when
|λo | < m < ∞, then µo cannot tend to +∞.
i) For (λ, µ) ∈ (σn ), we have Min (λ, µ) = 0, i = 1, 2. If λo →
+∞, µo → ∞, we write
X 1 p q
λp + µq
k1 , exp
−d1 = a1 e−λ−µ + b1 e−µ + c1 e−λ + e−λ−µ
2
n n
n
n
Z1 Z1
|d1 | ≤ |a1 |e−λo −µo + |b1 |e−λo + |c1 |e−λo + e−λo −µo L1
eλo ξ+µo η dξdη
o
o
96
L1 = sup k1 (ξ, η)
where
0≤ξ≤1
0≤η≤1
since (x, y) → eλo x+µo y is an increasing function of (x, y) when λo , µo > 0
which may be assumed to be the case since λo , µo both tend to +∞.
Or
|d1 | ≤ |a1 |e−λo −µo + |b1 |e−µo |c1 |e−λo + e−λo −µ L1
(eλo −1 )(eµo −1 )
λo µo
→ 0 as λo , µo → +∞.
This contradicts the hypothesis that d1 , 0.
ii) If λo → +∞, µo → −∞, we write
124
−b1 = a1 e−λ + c1 eµ−λ + d1 eµ + e−λ
XX 1
p q
λp + µq
k1 ( , ) exp
2
n n
n
n
p
q
|b1 | ≤ |a1 |e−λo + |c1 |eµo −λo + |d1 |eµo + e−λo
n−1
n−1 X
X
λo p
1
p q
|k ( , )| exp
2 1 n n
n
n
p=0 q=0
since exp µno q ≤ 1
i.e.,
−λo
|b1 | ≤ |a1 |e
µo −λo
+ |c1 |e
µo
−λo
+ |d1 |e + e
L1
Z1 Z1
o
o
= |a1 |e−λo + |c1 |eµo −λo + |d1 |eµo + e−λo L1
eλo ξ dηdξ
eλo
−1
λo
→ 0 as λo → +∞ and µo → −∞ which is not possible since b1 , 0.
iii) Suppose that |λo | < m1 and µo → +∞. We write
−µ
0 = a1 e
λ−µ
+ b1 e
−µ
+ c1 + d1 e
−µ
n−1 X
n−1
X
−µ
n−1 X
n−1
X
+e
p=0 q=0
−µ
0 = a2 e
λ−µ
+ b2 e
−µ
+ c2 + d2 e
+e
k1
p q
λp + µq
, exp
n n
n
p q
λp + µq
k2 , exp
n
n
n
p=0 q=0
8.
125
97
Eliminating eλ from these we get
|c1 d2 − c2 d1 | ≤ |a1 d2 − a2 d1 |e−µo + |b1 d2 − b2 d1 |
λo −µo
e
−µo
+e
L2
Z1 Z1
o
eµo η dξdη
o
where L2 depends upon the supremum of |d1 k2 (x, y) − d2 k1 (x, y)|eλo x for
(eµo − 1)
.
0 ≤ x ≤ 1, 0 ≤ y ≤ 1. i.e. the last term is majorised by e−µo L2
µo
Hence right hand side of the above inequality → 0 when |λo | < m and
µo → +∞. But this contradicts the assumption that c1 d2 − c2 d1 , 0.
Summation of the series S n (λ, αn ).
S n (λ, αn ) =
X
[λ − αn ]
J(M n , M n ) J(An , M n )
+ Rn
n )[λ − βn ][βn − αn ] J(λ, βn )
n , αn )
D(β
J(β
n
n
β ∈(σ )
(where R n can be got from R by replacing M, A by M n , An respectively), where λ = (λ1 , λ2 ) denotes a point in C 2 . The functions
M1n (λ), M2n (λ), An1 (λ), An2 (λ) are periodic in λ1 , λ2 with periods 2πni. The
same statement holds for the Jacobian Dn (λ) of the functions M n (λ) as
also the two determinants of Jacobi:
J(An , M n )
J(M n , M n )
and
J(λ, α)
J(β, α)
considered as functions of couples of points of C 2 , (λ, α) and (β, α), 126
λ, α, β ∈ C 2 . These are therefore periodic functions of the four complex variables with the same period 2nπi. Moreover M1n , and M2n are
λ1
λ2
polynomials in exp , exp
so that the zeros βn in the spectrum (σn )
n
n
can be arranged in 2n2 classes, each of these classes being situated in a
plane parallel to the purely imaginary plane of C 2 and forming in this
plane a network of squares of sides 2πn. By virtue of this remark, the
series S n (λ, αn ), which is absolutely convergent can be broken up in 2n2
98
partial sums corresponding to 2n2 classes in which the spectrum (σn ) is
divided. In each of these partial sums the factors
1 J(M n , M n ) J(An , M n )
Dn (βn ) J(λ, βn ) J(βn , αn )
has the same value for all the terms of the partial sum (λ, αn are fixed)
due to the periodicity of the functions Dn , M1n , M2n , An1 , M2n . We shall
then choose a representative βn in each class which will be fixed (for the
class λ, σn , we choose αn itself as its representative). It is necessary to
calculate, for each class
+∞
+∞ X
X
λ1 − αn1
(λ1 − βn1 − 2nhπi)(βn1 + 2nhπi − αn1 )
h=−∞ k=−∞
λ2 − αn2
(λ2 − βn2 − 2nkπi)(βn2 + 2nkπi − αn2 )
127
if αn does not belong to the class considered and
+∞
X′
X′ +∞
λ1 − αn1
λ2 − αn2
(λ1 − αn1 − 2nh′ πi)(2nhπi) (λ1 − αn1 − 2nkπi)(2nkπi)
h=−∞ k=−∞
P P
if αn belongs to the class considered. ( ′ ′ denotes that that value
h = 0, k = 0 is excluded in the summation). The calculation of these
two sums is classical and gives



 


 

1
1
1
1
1 

 


−
−
n −βn
n −βn
n
n
2
α
α
λ
−β
λ
−β



n  exp 1 1 − 1 exp 1 1 − 1   exp 2 2 − 1 exp 2 2 − 1 
n
n
n
n
in the first case and






1 
n
1  
1 
n
1
1

−
+  
+ 
−
n
n2  exp λ1 −βn1 − 1 λ1 − αn1 2   exp λ2 −α2 − 1 λ2 − αn2 2 
n
n
is the second.
8.
99
J(An , M n )
= 0 since αn ∈ e(σn ), we see that the series S n (λ, αn )
J(λn , αn )
can now be put in the form of a finite sum (with 2n2 − 1 terms) each
of these terms corresponding to classes in which the spectrum (σn ) is
divided. Denoting the classes by capital letters,
As
an = class of αn , Bn = class of βn ,
the representatives αn , βn being fixed in their class, we can write




X


1
1
1

n

n n
−
S n (λ, α ) =

n
n2 n n  exp λ1 −β1 − 1 exp α1 −β1 − 1 
B ,a
n
n





1
1


n n
−

n
λ
−β
α
−β
 exp 2 2 − 1 exp 2 2 − 1 
n
n
1
J(M n , M n ) J(An , M n )
128
(1)
Behaviour of S n (λ, αn ) for n large.
z
which is holoApplying Taylor’s formula for the function z
e −1
morphic in the neighbourhood of the origin, we have
Z
x
x
x2
dz
=
1
−
+
x
e −1
2 2πi 0 z(z − x)(ez − 1)
where C is the circumference of a circle with centre origin and radius <
2π (2π is the radius of convergence of the Taylor’s series of the function
λ
about the origin). Let x = with |λ| < πn so that |x| < π. Dividing by
n
λ, we have
Z
1
dz
1
1
λ
−
,λ , 0
+
=
λ
1/n
2
n(e − 1) λ 2π 2π i n
z(z − n )(ez − 1)
C
similarly for |µ| < π n,
129
100
1
µ
1
=
− +
µ/n
n(e − 1) µ 2n 2π i n2
1
Z
C
dz
z(z −
µ
z
n )(e
− 1)
,µ , 0
subtracting,
!
Z " λ
µ #
1 1
1
n
−
−
− nλ
−
=
λ
λ µ
2πin
n(eλ/n − 1) n(eµ/n − 1)
z− n z− n
Z C
λ−µ
dz
dz
=
λ
z(ez − 1) 2πin2
z − n z − µn (ez − 1)
1
1
C
µ
λ
λ
The length of C is < 4π2 , | | < π and | | < π. Hence |z − | ≥
n
n
n
µ
1
π, |z − | ≥ π; let M denote max | z
|, (R < 2π). Then
n
|z|=R e − 1
!
1 1 1
|λ − µ|
1
−
−
−
≤ Co
λ/n
µ/n
λ µ
n2
n(e − 1) n(e − 1)
2M
and |λ| < πn, |µ| ≤ πn. Changing λ into λ1 − βn1 , µ into
with Co =
π
αn1 − βn1 or λ into λ2 − βn2 , µ into αn2 − βn2 , we have the majorisation





λ1 − αn1

1 
1
1
 −
n n
−

n
n
n
n
n  exp λ1 −β1 − 1 exp α1 −β1 − 1  (λ1 − β1 )(β1 − α1 ) n
n
c0
|λ − αn1 |
2 1
n





λ2 − αn2

1 
1
1
 −
n n
−

n
n
n
n
n  exp λ2 −β2 − 1 exp α2 −β2 − 1  (λ2 − β2 )(β2 − α2 ) n
n
≤
≤
130
provided that


|λ1 − βn1 | ≤ πn , |λ2 − βn2 | ≤ πn 


n
n
n

|β1 − α1 | ≤ πn , |β2 − α2 | ≤ πn
c0
|λ2 − αn2 |
n2
(2)
8.
101
[ Note that |λ1 − α1 | and |λ2 − α2 | are independent of ρ1 , β2 and therefore
fixed in the summation (1) ].
Using the majorisation (2), we shall study (1) and compare it with a
finite sum



1
1
1 X 


λ −β α −β −
τ(λ, α) = 2
n β,α exp 1 1 − 1 exp 1 1 − 1 
n
n




 1 J(M, M) J(A, M)
1
1


λ −β λ −ρ −
.

2
2
2
2
exp
− 1 exp
− 1 D(β) J(λ, β) J(β, α)
n
n
The summation in β is made for β ∈ (σ) which are “near” to those βn 131
which are the chosen representatives of the 2n2 − 1 classes Bn , an . It
is necessary for this to compare the functions M and M n , A and An , (σ)
and (σn ).
Comparison of M with M n and of A with An .


M1 (λ) = a1 + b1 eλ1 + c1 eλ2 + d1 eλ1 +λ2 + N1 (λ)


λ1
λ2
λ1 +λ2

M2 (λ) = a2 + b2 e + c2 e + d2 e
+ N2 (λ)


M1n (λ) = a1 + b1 eλ1 + c1 eλ2 + d1 eλ1 +λ2 + N1n (λ)


λ1
λ2
λ1 +λ2
n
n

+ N2 (λ)
M2 (λ) = a2 + b2 e + c2 e + d2 e
(3)
(4)
Hence it is sufficient to compare N and N n . The calculation will be
similar for A and An .
We shall now suppose (for simplifying the proof) that the functions
k1 , k2 , a1 , a2 in R2 are indefinitely differentiable with compact support 132
contained in the square 0 ≤ x1 ≤ 1, 0 ≤ x2 ≤ 1. Then the functions
A(λ), N(λ) decrease rapidly when λ recedes to infinity keeping itself in
a vertical plane. Hence the functions such as
λ y + λ y 1 y1 y2 1 1
2 2
exp
= K1n (y1 , y2 )
,
k
1
2
n n
n
n
are indefinitely differentiable with compact support contained in the
square 0 ≤ y1 ≤ n, 0 ≤ y2 ≤ n and we can write for example
X
K1n (p, q)
N1n (λ) =
p,q
102
where the summation is made over all the couples of integers (p, q). In
order to evaluate this sum it suffices to apply Poisson’s formula. The
Fourier transform of K1n (y1 , y2 ) is (µ1 , µ2 being real):
F [K1n ] =
x
K1n (y1 , y2 ) exp[−2πi(µ1 + µ2 y2 )]dy1 dy2
λ
λ
1 x
2
1
= 2
− 2πµ1 y1 +
− 2πiµ2 y2 ]dy1 dy2
K1 (y1 , y2 ) exp[
n
n
n
x
=
k1 (u) exphλ − 2πiµn, uidu = N1 (λ − 2πiµn)
Hence by Poisson’s formula,



N(λ1 − 2πihn, λ2 − 2πikn)





h,k
X



n

A (λ) =
A(λ1 − 2πihn, λ2 − 2πikn) 



N n (λ) =
X
(5)
h,k




N (λ) − N(λ) =
N(λ1 − 2πihn, λ2 − 2πikn) = M (λ) − M(λ)





h,k

(6)


′
X



n


A (λ) − A(λ) =
A(λ1 − 2πihn, λ2 − 2πikn)



n
′
X
n
h,k
133
where the accent indicates that in the summation the couple h = 0, k = 0
is excluded.
Majorisation of the difference N n − N and An − A.
As N decreases rapidly in the vertical bound, we have
|N(λ1 , λ2 )| ≤
c1 (r)
|λ1 λ2 |r
in a vertical band where r is a positive integer, arbitrarily large and where
c1 (r) is a constant which depends on r and the function k1 (x1 , x2 ). Hence
|M1n (λ) − M1 (λ)| ≤ c1 (r)
′
X
h,k
1
|λ1 −
2πinh|r |λ2
− 2πink|r
8.
103
(for λ1 and λ2 different from the multiples of 2πin; this restriction is
artificial).
Majorising the second member for r ≥ 3, it is easy to verify that
c2 (r)
(2πn)r
(7)
|Imλ1 | ≤ πn, |Imλ2 | ≤ πn
(8)
|M1n (λ) − M1 (λ)| ≤
if
and we have analogous inequalities for the functions M2 , A1 , A2 under 134
the same conditions (8). We can always denote by c2 (r) the positive
constant figuring in the numerator of the second member of (1) by taking
the same constant for the functions.
The volume Vn ; zeros of M(λ) in the interior of Vn .
We know that (σ) and (σn ) are in a fixed vertical band B independent of n. We shall intersect the vertical band by a horizontal band in
C 2 defined by (8). Its section by a vertical plane is a square of side 2πn.
Such a square contains one and only one point of each of the 2n2 classes
in which (σn ) is decomposed, since each of these 2n2 classes form, in its
plane, which is vertical, a network of squares of side 2πn. It follows that
the volume Vn in C 2 contains exactly 2n2 points of the spectrum (σn ).
We wish to find the points of (σ) which are also in Vn . We first recall a
classical result due to Kronecker.
Let f, g, h be three functions continuously differentiable in a region
in R3 . Let V be a volume contained in the region with boundary S . Then
1 x
(A cos λ + B cos µ + C cos ν)dS
4π
S
#
"
1
D(h, f )
D( f, g)
D(g, h)
+g
+h
A= f
2
2
D(y, z)
D(y, z)
D(y, z) [ f + g + h2 ]3/2
m=−
where
B, C being analogously defined and cos λ, cos µ, cos, γ are the direction 135
cosines of the interior normal to S and m equals the difference between
the number of solutions lying in V of the systems f = g = h = 0 for
104
D( f, g, h)
D( f, g, h)
> 0 and the number of solutions for which
<
D(x, y, z)
D(x, y, z)
0. We shall use the analogue of this proposition in R4 = C 2 i.e.,
If f1 , f2 , f3 , f4 are four functions which are (C, 1) in a region of R4 ,
then
1 y
m=− 2
(A1 cos λ1 + A2 cos λ2 + A3 + A4 cos λ4 )dS
 2π

X D( f , f , f ) 
2
3
4
 " 1 #
with
A1 = 
2
D(x2 , x3 , x4 )  P
4
f1
fi2
which
1=1
and A2 , A3 , A4 similarly defined, where m is defined as above for the
D( f1 , f2 , f3 , f4 )
.
system of equations f1 = f2 = f3 = f4 = 0 and the Jacobian D(x
1 ,x2 ,x3 ,x4 )
The integral on the right hand side will be called the Kronecker integral.
We consider the following analytic transformation of C 2 into itself,
(λ1 , λ2 ) → (M1 (λ1 , λ2 ), M2 (λ1 , λ2 ))
136
This can be considered as a transformation of R4 into itself. Instead of
the four variables which are the real and imaginary parts of both λ1 , λ2 ,
we take λ1 , λ¯1 , λ2 , λ¯2 . Then
∂M1
∂M1
dλ1 +
dλ2 and
∂λ1
∂λ2
∂M2
∂M2
dM2 =
dλ1 +
dλ2 ,
∂λ1
∂λ2
dM1 =
as M1 , M2 are analytic. The volume elements which correspond to each
other by this transformation are proportional to
dλ1 ∧ dλ¯1 ∧ dλ2 ∧ dλ¯2 and dM1 ∧ d M¯ 1 ∧ dM2 ∧ d M¯ 2
Now
¯1
∂M
dλ¯1 +
∂λ1
¯2
∂M
dλ¯1 +
d M¯ 2 =
∂λ1
d M¯ 1 =
¯1
∂M
dλ¯ 2 and
∂λ2
¯2
∂M
dλ¯ 2
∂λ2
8.
105
Hence
¯2
dM1 ∧ d M¯ 1 ∧ dM2 ∧ d M

 ∂M1 ∂M1 ∂M2 ∂M2 ∂M1 ∂M1 ∂M2 ∂M2 ∂M1 ∂M1 ∂M2
−
−
= 
∂λ1 ∂λ1 ∂λ2 ∂λ2
∂λ1 ∂λ2 ∂λ2 ∂λ1
∂λ2 ∂λ1 ∂λ1

∂M2 ∂M1 ∂M1 ∂M2 ∂M2 
 dλ1 ∧ dλ¯1 ∧ λ2 ∧ dλ¯2
+
∂λ2
∂λ2 ∂λ2 ∂λ1 ∂λ1
=
D(M1 , M2 ) D(M1 , M2 )
dλ1 ∧ dλ¯1 ∧ dλ2 ∧ dλ¯2
D(λ1 , λ2 ) D(λ1 , λ2 )
Thus the Jacobian of the transformation under consideration is al- 137
ways real and ≥ 0. Taking real and imaginary parts of M1 and M2 the
system M1 = 0, M2 = 0 is equivalent to the four equations fi = 0 i =
1
1, 2, 3, 4. The Kronecker integral which can be briefly denoted by − 2
2π
t
K(M1 , M2 )dS in this case gives m exactly equal to the number of
solutions of the system fi = 0 in the volume V enclosed by S (since the
Jacobian does not change sign) i.e., the number of elements of (σ)inV.
Proposition 4. The volume Vn contains 2n2 points of σ for n sufficiently
large.
We know that Vn contains 2n2 points of (σn ). Using Kronecker’s
result it follows that
Z
2
2n =
K(M1n , M2n )dS
∂Vn
where ∂Vn denotes the boundary of Vn whose measure is of the form
c3 n2 , where c3 is a fixed constant (viz. the product of 4π2 by the length
of the parameter of the right section of the vertical band B) and the
proposition will be proved if we know that
Z
2
K(M1 , M2 )dS
(i)
2n =
∂Vn
Consider the difference K(M1 , M2 ) − K(M1n , M2n ) in ∂Vn . Let λ1 = 138
x1 + ix2 , λ2 = x3 + ix4 , M1 = f1 + i f2 , M2 = f3 + i f4 , M1n = f1n + i f2n , M2n =
106
f3n + i f4n where the f ′ s are real valued functions of the four real variables
c2 (r)
x1 , x2 , x3 , x4 . For (λ1 , λ2 ) ∈ Vn , |Mi − Min | ≤
i = 1, 2. Hence
(2πn)r
| fi − fin | ≤ |
c2 (r)
i = 1, 2, 3, 4.
(2πn)r
c′ (r)
∂ fi ∂ fin
−
| ≤ 2 r for (λ1 , λ2 ) ∈ Vn , since the partial deriva∂x j ∂x j
(2πn)
tives of the fi with respect to x j are exactly of the same form as the fi .
We consider the difference K(M1 , M2 ) − K(M1n , M1n ) on ∂vn . In
∂ fi
may be regarded as a finite set of variables
K(M1 , M2 ) the fi and
∂x j
u1 , u2 , . . . and K(M1n , M2n ) is the same function with the variables fi and
∂fn
∂ fi
replaced by fin and i respectively or the variables u1 , u2 , . . . re∂x j
∂x j
placed by un1 , un2 , . . . respectively. Hence applying mean value theorem
of differential calculus,
Similarly |
|K(M1 , M2 ) − K(M1n , M2n )| ≤
139
c3 (r)
L on ∂Vn
(2πn)r
(ii)
where L depends on the maximum modulus of the derivatives of K
with respect to u1 , u2 , . . . over a region which contains (u1 , u2 , . . .) as
also (un1 , un2 , . . .) while (λ1 , λ2 ) = (x1 , x2 , x3 , x4 ) varies in ∂Vn . Since
1
|ui − uni | = 0( r ) in Vn and therefore on ∂Vn , for n ≥ N1 , and since
n
∂K
are continuous functions of (u1 , u2 , . . .) in estimating L it is enough
∂ui
∂K
when (λ1 , λ2 ) ∈ ∂Vn . Now
to consider the maximum modulus of
∂ui
the numerator of K(u1 , u2 , . . .) is a homogeneous polynomial in all the
u’s of total degree 4 with coefficients which are functions of λ1 , λ2 with
maximum moduls ≤ 1 and the denominator is 2π2 (u21 + u22 + u23 + u24 )
or 2π2 [|M1 (λ)|2 + |M2 (λ)|2 ]. Both the numerator and denominator as
also their partial derivatives with respect to ui are uniformly bounded
on ∂Vn since they are uniformly bounded on B and Vn is a closed subset of B. Hence L can be found to be a fixed positive number which
8.
107
does not depend on n if we prove that the denominator of K i. e.
2π2 [|M1 (λ)|2 + |M2 (λ)|2 ] is bounded below uniformly for (λ1 , λ2 ) ∈ ∂Vn
by a fixed number > 0 which does not depend on n.
First we consider two vertical parts of ∂Vn , denoted by (∂Vn )1 i. e.
parts contained in ∂B. On ∂B real parts of λ1 , λ2 are constant and the
imaginary parts vary from −∞ to +∞. We can suppose that the principal
parts
φ1 (λ1 , λ2 ) = a1 + b1 eλ1 + c1 eλ2 + d1 eλ1 +λ2 and
φ2 (λ1 , λ2 ) = a2 + b2 eλ1 + c2 eλ2 + d2 eλ1 +λ2
of M1 (λ) and M2 (λ) respectively do not vanish on ∂B (we have only to 140
choose B suitably) and have their moduli bounded below by m1 > 0
if (λ1 , λ2 ) ∈ ∂B and |Imλ1 | ≤ π and |Imλ2 | ≤ π. But φ1 and φ2 are
periodic in λ1 and λ2 with periods 2πi so that |φ1 | > m1 and |φ2 | > m1
m1
> 0, we can find a compact set K1 such that
on ∂B. Now given
2
m
Mi (λ1 , λ2 ) − φi (λ1 , λ2 ) < 1 , i = 1, 2 for (λ1 , λ2 ) < K1 by Proposition
2
m1
2. Hence for (λ1 , λ2 ) ∈ (∂Vn )1 and (λ1 , λ2 ) < K1 , |Mi (λ1 , λ2 )| >
,i =
2
1, 2. As M1 , M2 do not vanish on ∂B and therefore on (∂Vn )1 ∩ K1 , |M1 |
m1
and |M2 | > m2 > 0 on (∂Vn )1 ∩ K1 so that if m = Min( , m2 ) >
2
0, 2π2 (|M1 (λ)|2 + |M2 (λ)|2 ) > 2π2 m2 > 0 on (∂Vn )1 . On the horizontal
part (∂Vn )2 of ∂Vn i. e. where |Imλ1 | = πn and |Imλ2 | = πn , since the
a′ s, b′ s · · · etc. are generic, we may suppose that φ1 and φ2 do not vanish
on (∂Vn )2 so that φ1 and φ2 have a lower bound m′ > 0 on (∂Vn )2 (since
(∂Vn )n is compact) and m′ is independent of n because of the periodicity 141
m′
, there exists a compact set outside which
of φ1 and φ2 . Now given
2
′
m
|Mi − φi | <
, i = 1, 2. Also for n > N2 , (∂Vn )2 lies out side this
2
m′
compact set so that |M1 | and |M2 | ≥
> 0 on (∂Vn )2 for n ≥ N2 . Thus
2
12
m
(|M1 (λ)|2 + |M2 (λ)|2 ≥ Min(m2 ,
> 0) on ∂Vn for n ≥ N2 .
2
Thus the moduli of the denominator and numerator of K(u1 , u2 , . . .)
as also their partial derivatives are bounded above uniformly on ∂Vn and
108
the modulus of the denominator is bounded below by a strictly posi∂K
tive number on ∂Vn for n ≥ N2 , so that | | is bounded above by a
∂ui
fixed number on ∂Vn independent of n ≥ N2 . Hence for all n ≥ N3 =
Max(N1 , N2 ) there exists a fixed positive L satisfying (ii) independent
of n ≥ N3 . Hence
Z
Z
c3 (r)
K(M1 , M2 )dS −
Lc3 n2
K(M1n , M2n )dS ≤
(2πn)r
∂Vn
∂Vn
c4 (r)
c4 (r)
for n ≥ N3 , and r−2 < 1 for n ≥ N4 .
r−2
n
n
But each of the integrals equals an integerR and their difference has
R
to be zero for n ≥ N = Max{N3 , N4 }, and ∂V K(M1 , M2 )dS =
=
n
K(M1n ,
142
M2n )dS
=
∂Vn
2n2 .
Proposition 5. For n sufficiently large (n ≥ no fixed), there exists a
one-to-one correspondence between the 2n2 points of (σ) and of (σn )
contained in the volume Vn such that the distance between the correg5 (r)
sponding points of (σ) and of (σn ) is uniformly majorised in Vn by r/2
n
where c5 (r) > 0 depends only on the maximum modulus of the real and
imaginary parts of M1 and M2 as also their partial derivatives with respect to real coordinates.
As M n converges to M uniformly on each compact subset of C 2 ,
it is easy to see using Kronecker’s integral that in any arbitrary neighbourhood of a point of (σ), there exists a point of (σn ) for n sufficiently
large. Also by Proposition 4, for n sufficiently large, both (σ) and (σn )
have the same number (= 2n2 ) of points in Vn . Now we show that if we
c5 (r)
describe a sphere of radius r/2 about each of the 2n2 points of (σ) in
n
Vn , then there exists in the interior of each of these spheres a point of
(σn ) lying in Vn .
Let S (α, ∈) denote the sphere of centre α and radius ε. The points
of (σ) are asymptotic with (σ′ ) which consist of points
(λ1 λ2 ), λ1 = α′1 + 2hπi
λ2 = α′2 + 2kπi
and
λ1 = β′1 + 2h′ πi
λ2 = β′2 + 2kπi
8.
109
(by Proposition 2). It follows therefore that for ε sufficiently small,Rthe
143
sphere S (α, ε) does not contain any other point of (σ) so that
∂S (α,ε)
K(M1 , M2 )dS = 1∀αε(σ).R We shall establish the proposition by comK(M1n , M2n )dS .
paring this integral with
∂S (α,ε)
The denominator of K(M1 , M2 ) which is 2π2 [|M1 (λ)|2 +|M2 (λ)|2 ]2 is
infinitely small of fourth order in ε on ∂S (α, ε). We have λ = (λ1 , λ2 ) =
(x1 , x2 , x3 , x4 ) = x and
M1 (λ) = f1 (x) + i f2 (x), M2 (λ) = f3 (x) + i f4 (x).
As x = α is a zero of each of f j ,
X xi − xα ∂ f j
X (xi − xα ) (xk − xα )  ∂2 f j 
i
i
k 
2


+ε
f j (x) = ε
ε ∂xαi
ε
ε
∂xi ∂xk x=x′
where x′ is some point of S (α, ε)
|M1 (λ)|2 + |M2 (λ)|2 = f12 + f22 + f32 + f42

4 
X
X xi − xαi ∂ f j 2
2
 + ε3 ψ(x, α) · · · (i)

=ε
α
ε
∂x
i
i
j=1
For αε(σ) and xεS (α, ε), |ψ(x, α)| is bounded above by M say, uniformly
for αε(σ) since the f j as also their partial derivatives are uniformly
bounded in the vertical band B. As α is a simple zero M1 (α) and M2 (α),
!2
P P xi − xαi α f j
is positive definite for xε∂S (α, ε) and has a 144
j
ε ∂xαi
i
strictly positive minimum A(α) depending α. We know that at a great
distance in B, M1 and M2 behave as their principal parts φ1 and φ2 respectively and the same is true for their corresponding real and imaginary parts as also their first partial derivatives with respect to real coor"
#2
P P xi − xαi αg j
′
′
dinates. We observe that for xε∂S (α , ε), j
(where
ε ∂xαi
i
g j are the real and imaginary parts of φ1 and φ2 and other obvious notation) is a positive definite quadratic form with strictly positive minimum
110
B(α′ ). But g′j s are periodic in x3 and x4 as also their partial derivatives
and therefore B(α′ ) has a lower bound m′ > 0 independent of α′ ε(σ′ ).
Let α, α′ denote points of (σ) and (σ′ ) respectively which are very near
to each other (at a great distance in B). We have
145
Let


X X xi − xα ∂ f j 2
i



A(α) = min
α 
xε∂S (α,ε)
ε
∂x
i
j
i


X X xi − xα ∂g j 2
i


′
b(α ) = min

α′ 
xε∂S ′ (α′ ,ε′ )
ε
∂x
i
i
j
 
2 


α


X
X


x
−
x
∂g
i


 
i  
i
.
B(α) = min 



α 



xε∂S (α,ε) 
ε
∂x

 j
i
i
The g′j s are periodic functions of x3 and x4 and hence are uniformly
∂g j
continuous in B; so are
. Hence B(α′ ) is a uniformly continuous
∂xi
m′
function of α′ εB; i.e. given
> 0, there exists a δ > 0 such that
4
m′
|α − α′ | < δ implies that |B(α) − B(α′ )| <
. Now given δ, there exists
4
m′
, there
a compact set K1 such that |α − α′ | < δ for α < K1 and given
4
′
m
exists a compact set K2 such that |A(α) − B(β)| <
for α < K2 . Hence
4
for α < K1 ∪ K2 ,
A(α) > B(α′ ) − |B(α) − B(α′ )| − |A(α) − B(α)|
m′ m′ m′
−
=
> 0.
> m′ −
4
4
2
Also K1 ∪ K2 contain only a finite number of αε(σ) lying in Vn .
Hence A(α) > M ′′ > 0 for αεK1 ∪ K2 , αεVn . From (i),.
|M1 (λ)|2 + || |M2 (λ)|2 > m1 ε2 − ε3 M
m′
m1 = min
, m′′
2
where
8.
146
111
m1
, so that |M1 (λ)|2 + |M2 (λ)|2
2
has a strictly positive lower bound mε2 for λε∂S (α, ε) which does not
depend on αε(σ).
R
K(M1 , M2 )dS and
In order to compare the integrals
∂S
(α,ε)
R
K(M1n , M2n )dS for αεVn , we adopt the procedure in proposition
Choosing ε small enough, ε3 M < ε2
∂S (α,ε)
4 in which integrand K is treated as a function of u1 , u2 , . . . which are fi
∂ fi
and apply the mean value theorem for differential calculus to
and
∂x j
the difference K(u1 , u2 , . . .) − K(un1 , un2 , . . .) = K(M1 , M2 ) − K(M1n , M2n ).
We suppose that S (α, ε) ⊂ Vn for αεVn which is possible if ε is sufficiently small.
X
∂K
K(M1 , M2 ) − K(M1n , M2n ) =
(ui − uni ) ,
∂ui
!
1
n
since |ui − ui | = 0 r ,
n
c′ (r)
|K(M1 , M2 ) − K(M1n , M2n )| <
L
(2πn)r
∂K
where L depends on the maximum modules of
where (λ1 , λ2 ) =
∂ui
xε∂S (α, ε). The derivatives of the numerator of K are uniformly bounded in B and therefore on ∂S (α, ε). In the derivatives of the denomih
i−3
nator appears the term |M1 (λ)|2 + |M2 (λ)|2 , partially compensated in
the numerator by terms which involve derivatives of |M1 (λ)|2 + |M2 (λ)|2
or u21 + u22 + u23 + u34 . These letter term are uniformly majorised in Vn 147
and on ∂S (α, ε) by quantities which are of the first order in ε. Thus the
term in the denominator are uniformly bounded in Vn and therefore on
∂S (α, ε) by a quantity of order ε−5 since [|M1 (λ)|2 + |M2 (λ)|2 ] is uniformly bounded below on all ∂S (α, ε), αεVn, by mε2 with m > 0. The
measure of ∂S (α, ε) being proportional to ε3 , we have for all the spheres
S (α, ε) situated in Vn ,
Z
Z
′′
< c (r)
n
n
K(M
,
M
)ds
−
K(M
,
M
)ds
1
2
1
2
ε2 nr
∂S (α,ε)
σS (α,ε)
112
#1
c′′ (r) 2
c5 (r)
If ε =
= r/2 and n > n0 in order that be sufficiently
r
n
n
small we shall have the right hand side of the above inequality < 1 and
hence equal to zero as it is an integer and
Z
Z
n
n
K(M1 , M2 )ds = 1.
K(M1 , M2 )dS =
"
∂S (α,ε)
∂S (α,ε)
The passage to the limit.
We now compare the series
S (λ, α) =
X
β,α
148
[λ − α]
J(M, M) J(A, M)
D(β)[λ − β][β − α] J(λ, β) J(β, α)
with
S n (λ, αn ) =
X
βn ,αn
βn ε(σn )
[λ − αn ]
J(An , M n )J(An , M n )
Dn (βn )[λ − βn ][βn − αn ] J(λ, αn )J(βn , αn )
using the finite form of S n (λ, αn ) given by (1 ). We first remark that the
preceding properties permit on e to establish a sequence of one-to-one
correspondences
an ↔ αn ↔ α
among the set of 2n2 classes of zeros of M n , the set of M n which are in
Vn and the set of zeros of M which are in the same volume, the distance
between the two zeros of M and M n being estimated in Proposition 5.
Let Wn be the volume which consists of points of B satisfying
1
1
|Imλ1 | ≤ πn 4 , |Imλ2 | ≤ πn 4
λ being fixed, we can suppose that Wn contains the point λ for n sufficiently large (since one can always suppose that for λ fixed, B contains
λ). We choose an , α, αn such that α, αn εWn . Then
|λ1 − α1 |, |λ2 − α2 |, |λ1 − αn1 |, |λ2 − αn2 |
8.
149
113
1
are majorised by c6 n 4 .|D(β)| is bounded away from zero for βε(σ) and
therefore |D(βn )| by proposition 5 for βn εVn and therefore |Dn (βn )| for n
sufficiently large and βn εVn . Also An , M n , A, M as also their derivatives
are uniformly bounded so that by (7) and (8),
!
1 J(M n , M n ) J(An , M n )
1
1 J(M, M) J(A, M)
−
=0 r
D(βn ) J(λ, βn ) J(βn , αn )
n
for β, βn εVn .
Similarly by Proposition 4 and 5, it is clear that
1 J(M, M) J(A, M)
1 J(M n , M n ) J(An , M n )
=
+ An
n
n
n
n
Dn (β ) J(λ, α ) J(α , β )
J(βn ) J(λ, β) J(β, α)
(9)
with
c7 (r)
.
(10)
nr/2
The constant c7 (r) being the same for all the terms of (1).
The second factor



 

1
1
1 
1
1
 


−
−

n
n
n
n
n
n
n2  exp λ1 −β1 − 1 exp α1 −β1 − 1   exp λ2 −β2 − 1 exp α2 −β2 − 1 
n
n
n
n
(11)
of the general term in (1) can be written as (because of (2))
|A| <
λ1 − αn1
λ2 − αn2
+
B
+
C
n
n
(λ1 − βn1 )(βn1 − αn1 )
(λ2 − βn2 )(βn2 − αn2 )
with
|Bn | ≤
c0
c0
|λ1 − αn1 |, |Cn | ≤ 2 |λ2 − αn2 |
2
n
n
(12)
150
(13)
In (12), the term
λ1 − αn1
λ1 − α1
may be replaced by
n
n
n
(λ1 − β1 )(β1 − α1 )
(λ1 − β1 )(β1 − α1 )
=
1
1
+
···
λ1 − β1 β1 − α1
(14)
114
But when β describes (α),
λ1 − β1 , λ2 − β2 , β1 − α1 , β2 − α2
have a strictly positive minimum. Hence (11) can be written as
"
#"
#
λ1 − α1
λ2 − α2
+ Bn + B′n
+ Cn + Cn′ (15)
(λ1 − β1 )(β1 − α1 )
(λ2 − β2 )(β2 − α2 )
with the conditions (13) and
|B′n | <
151
cg (r)
cg (r) ′
, |Cn | < r/2
r/2
n
n
We shall now put the second member of (9) as also (15) in place of
(11) in the general term of (1). Then the principle term is evidently
X 1 J(M, M) J(A, M)
[λ − α]
J(β) J(λ, β) J(β, α) [λ − β][β − α]
β,α
βεVn
where the summation is extended to 2n2 − 1 points of β contained in Vn
(and distinct from α). The corrective terms are of different kinds. There
are two terms of type
)(
)
X(
1 J(M, M) J(A, M)
λ1 − α1
Cn
(16)
(λ1 − β1 )(β1 − α1 )
D(β) J(λ1 , β1 ) J(β, α)
βεV
n
β,α
The second bracket is uniformly bounded in B and
|Cn (λ1 − α1 )| ≤
c9
c0
|λ1 − α1 ||λ2 − α2 | ≤ 3/2
2
n
n
Now we prove that
X
βεVn
β,α
1
= 0(n)
(λ1 − β1 )(β1 − α1 )
(16)′
8.
115
c10 (λ, α)
where c10 (λ, α) de√
n
pends upon the shortest distance of λ1 from the set of β1 and the shortest
distance of α1 from the set of β1 , α1 . Let n = m4 . The number of terms
√
of (16)′ in Wn = Vm is 2m2 = 2 n. Let Un = Vm2 . Let d denote the
minimum if the shortest distances of λ from α and of β , α from α.
Then
X
1
≤ 1 2n.
(λ1 − β1 )(β1 − α1 ) d2
βεU
such that (16) will have the majorisation
152
n
β,α
√
There are 2n2 − 2n points of (σ) in Vn − Un and |β| > π n for
√
βεVn −Un . Also |λ1 | < c6 n1/4 , |α1 | <6 n1/4 gives |λ1 −β1 | > π n−c6 n1/4
√
and |β1 − α1 | > π n − c6 n1/4 so that
1
1
(λ1 − β1 )(β1 − α1 ) ≤ (π √n − c n1/4 )(π √n − c n1/4 )
6
6
!
1
1
=0
or
for βεVn − Un
(λ1 − β1 )(β1 − α1 )
n
Hence
X
βεVn −U n
!
1
1
2
= (2n − 2n) 0
= 0(n)
(λ1 − β1 )(β1 − α1 )
n
and
X
βεVn
β,α
X
1
1
=
(λ1 − β1 )(β1 − α1 ) βεU (λ1 − β1 )(β1 − α1 )
n
β,α
+
X
βεVn −U n
1
(λ1 − β1 )(β1 − α1 )
= 0(n).
The terms of the type
λ1 − α1
1 J(M, M) J(A, M) ′
C
(λ1 − β1 )(β1 − α1 ) D(β) J(λ, β) J(β, α) n
153
(17)
116
have majorisation of the form
c11 (λ, α)
and the terms such as
2r−5
n 4
1 J(M, M) J(A, M) ′ ′
BC
D(β) J(λ, β) J(β, α) n n
1 J(M, M) J(A, M)
BnCn′
D(β) J(λ, β) J(β, α)
1 J(M, M) J(A, M) ′ ′
BC
D(β) J(λ, β) J(β, α) n n
(18)
(19)
(20)
have respectively the evident majorisation
c12 c13 (r) c14 (r)
,
, r
n
n7/2 n 2r + 74
c15 (r, λ, α)
[λ − α]
An is majorised by
where the
r−1
[λ − β][β − α]
n 2
constant c15 depends on λ, α and contains in the denominator the shortest distance of λ1 , λ2 from the set of β1 , β2 respectively and the shortest
distance of α1 , α2 from the set of β1 , α1 , β2 , α2 respectively.
Similarly the terms
Then term
154
λ1 − α1
λ1 − α1
An C n ,
AnCn′
(λ1 − β1 )(β1 − α1 )
(λ1 − β1 )(β1 − α1 )
An BnCn , An BnCn′ , An B′nCn′
give by summation, the majorisation of the form
c15 (r, λ, α) c16 (r, λ, α) c17 (r) c18 (r)
,
, r+7 , 3r .
r+1
5
n 2
nr− 2
n2
n2
Conclusion. For r sufficiently large, and for λ, α fixed, we have












X


J(M,
M)
J(A,
M)
[λ
−
α]
Lt


n
n → ∞
S
(λ,
α
)
−
=0

n




[λ
−
β][β
−
α]
J(λ,
β)
J(β,
α)




β,α




βεV
n
Now the same summation, extended to β exterior to the volume Vn is
c19 (λ, α)
(this is obvious if we consider the asymptotic
majorised by
n
behaviour of σ described by Proposition 2).
9. The fundamental theorem of Mean...
117
Hence for λ, α fixed and r sufficiently large
Lt
n → ∞S n (λ, α) =
X
β,α
βεVn
[λ − α]
J(M, M) J(A, M)
1
.
D(β) [λ − β][β − α] J(λ, β) J(β, α)
J(An , M n )
J(A, M)
− R n which tends to
− R as 155
n
J(λ, α )
J(λ, α)
n → ∞ for λ, α fixed. Hence
But S n (λ, αn ) =
J(M, M) J(A, M)
[λ − α]
J(A, M) X
=
+R
J(λ, α)
D(β)[λ − β][β − α] J(β, α) J(λ, α)
β,α
βε(σ)
and we have proved the formula (G1 ) for the distributions T 1 and T 2 .
9 The fundamental theorem of Mean periodic functions in the case of two variables
The fundamental theorem for Mean periodic functions, Viz. expansion
of a mean periodic function in term of mean periodic exponentials in
the case of one variable is well-known. But its analogue in Rn is not
know. We shall prove it over for a function mean periodic relative to
two special kinds of distributions in R2 . Even as in the case of R1 , the
theorem is proved by making use of Mittag-Leffler theorem in C 1 , the
proof given here depends upon the formula (G1 ) which may be considered as an analogue of Mittage-Leffler theorem in C 2 .
Let T 1 , T 2 εE ′ (R2 ) be defined as in the preceding article by
T i ∗ F = ai F(x, y) + b1 F(x + 1, y) + c1 F(x, y + 1) + d1 F(x + 1, y + 1)
+
R1 R1
o 0
angle R1
ki (ξ, η)F(x + ξ, y + η)dξdη, i = 1, 2, with supports in the rect-
: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 where the densities ki (x, y) and the 156
coefficients ai , bi , . . . (i = 1, 2) satisfy all the necessary conditions in order that all the results of §8 should hold. Let u1 , u2 εD(R1 ) (the space of
indefinitely differentiable functions with compact support) so that their
118
Fourier-Laplace transforms A1 (λ1 , λ2 ), A2 (λ1 , λ2 ) are functions of expo1
nential type which decrease more rapidly than any power of
,
|λ1 | + |λ2 |
as |λ1 |, |λ2 | → ∞ in any vertical band; then we have
[λ − α]
J(S , S ) J(A, S )
J(A, S ) X
=
+R
J(A, α) βεσ D(β)[λ − β][β − α] J(λ, β) J(λ, β)
(1)
β,α
where S i (λ) = F L T i , i = 1, 2, and (σ) is the spectrum. The series (1)
converges uniformly on each compact set. Let ϕ(x)εDΦ(λ)F L φ is a
function rapidly decreasing in any vertical band.
Setting A¯ i (λ) = Φ(λ)A(λ), S¯ i (λ) = Φ(λ)S (λ), i = 1, 2, we obtain,
(after multiplying both sides of (1) by Φ(λ)),
¯ S) X
J(A,
[λ − α]
J(S¯ , S ) J(A, S )
=
+ R¯
J(λ, α)
D(β)[λ − β][β − α] J(λ, β) J(β, α)
(2)
The series (2), like (1), converges uniformly on each compact set. Moreover we can prove the following
157
Proposition 1. The series (2) converges in the sense of L1 (R2 ) in every
plane of C 2 for which the real parts of λ1 , λ2 are fixed (i.e. for (λ1 , λ2 )
in a vertical plane).
As A1 (λ), A2 (λ) decrease rapidly and S 1 (λ), S 2 (λ) are bounded in a
vertical band and (σ) is contained in one such band,
J(A, S ) < χα (β)
J(α, β)
J(S¯ , S )
| < c1 |Φ(λ)|
J(λ, β)
where c1 > 0 is a constant which depends only on two vertical bands,
one containing (σ) and the other containing λ. Further by the corollary
of proposition 2, §8, |D(β)| ≥ k > 0 on (σ). Hence
1 J(S¯ , S ) J(A, S ) < 1 χ (β)|Φ(λ)|
D(β) J(λ, β) J(β, α) k α
where χα (β) decreases rapidly in βε(σ). Similarly |
119
We now consider the factor
1
1
[λ − α]
=
+
[λ − β][β − α] (λ1 − β1 )(λ1 − β2 ) (λ1 − β1 )(β2 − α2 )
1
1
+
+
(λ2 − β2 )(β1 − α1 ) (λ1 − β1 )(β2 − α2 )
(3)
For (λ1 , λ2 ) fixed, we investigate the term of (2) in the four cases
158
i) |λ1 − β1 | ≥ 1, |λ2 − β2 | ≥ 1
ii) |λ1 − β1 | < 1, |λ2 − β2 | < 1
iii) |λ1 − β1 | < 1, |λ2 − β2 | ≥ 1
iv) |λ1 − β1 | ≥ 1, |λ2 − β2 | < 1
For βε(σ) verifying (i),
|
[λ − α]
1
1
1
|≤1+
+
+
[λ − β][β − α]
|β1 − α1 | |β2 − α2 | |β1 − α1 ||β2 − α2 |
The general term of the series (2) corresponding to such a β is majorised by
c′1 (1 +
1
1
)(1 +
)χα (β)|Φ(λ)|
|β1 − α1 |
|β2 − α2 |
Summing for β since χα (β) decrease rapidly on (σ), this part of the
series is majorised in modulus by c3 (α)|Φ(λ)|. (ii):- For λ fixed, there
are only a finite number of terms verifying this condition.
Now
159
J(S¯ 1 , S ) S¯ 1 (λ1 , λ2 ) − S¯ 1 (β1 , λ2 ), S 1 (β1 , λ2 ) − S 1 (β1 , β2 )
= ¯
S 2 (λ1 , λ2 ) − S¯ 2 (β1 , λ2 ), S 2 (β1 , λ2 ) − S 2 (β1 , β2 )
J(λ, β)
and
S¯ 1 (λ1 , λ2 ) − S¯ 1 (β1 , λ2 ) =
Zλ1
β1
∂ ¯
[S 1 (ρ1 , λ2 )]dρ1
∂ρ1
120
Then
J(S¯ 1 , S )
=
J(λ, β)
Zλ1 Zλ2
β1 β2
∂
∂ ¯
S 1 (ρ1 , λ2 )
S 2 (β1 , ρ2 )
∂ρ1
∂ρ2
−
∂ ¯
∂
S 1 (β1 , ρ2 ) dρ1 dρ2
S 2 (ρ1 , λ2 )
∂ρ1
∂ρ2
Let ψ(λ) denote the maximum modulus of the functions
∂ ¯
S 1 (ρ1 , λ2 ),
∂ρ1
∂ ¯
S 2 (ρ1 , λ2 ) when (ρ1 , ρ2 ) is such that |λ1 −ρ1 | < 1, |λ2 −ρ2 | < 1. Then
∂ρ1
ψ(λ) is a rapidly decreasing function in a vertical band, and
¯
J(S 1 , S ) ≤ c |ψ(λ)||λ − β ||λ − β |
2
1
1 2
2
J(λ, β) 160
and the terms of (2) for which (ii) holds are majorised by c3 (α)ψ1 (λ)
where ψ1 (λ) is a rapidly decreasing function in a vertical band.
J(S¯ , S )
is the form
(iii) : - Writing
J(λ, β)
Zλ1 (
β1
)
∂ ¯
∂ ¯
S 1 (ρ1 , λ2 )S 2 (β1 , λ2 ) −
S 2 (ρ1 , λ2 )S 1 (β1 , λ2 ) dρ1
∂ρ1
∂ρ1
[λ − α] we have in this case the terms of (2) majorised by cχα (β)ψ2 (λ)
[β − α]
where ψ2 (λ) decreases rapidly in a vertical band and the sum of the
series of this part is majorised by c(α)ψ3 (λ) where c(α) is a constant
which depends on α and ψ3 (λ) decreases rapidly in a vertical band and
we have similar majorisation in case (iv). Thus the sum of (2) being
majorised by a rapidly decreasing function of λ in a vertical plane, the
series converges in the sense of L1 (R2 ) in a vertical plane.
In view of Proposition 1, we can apply the inverse of Fourier Laplace
transformation to both sides of (2) and we obtain
X J(A, S )
u¯ =
s¯αβ (x, y) + ν¯
(4)
J(β, α)
β ∈σ
β,α
121
¯ S)
J(A,
,
J(λ, α)
[λ − α]
J(S¯ , S )
=
= S¯αβ (λ),
D(β)[λ − β][β − α] J(λ, β)
F L¯u =
where
F L s¯αβ
F LR¯ = ν¯ and the series converges uniformly on each compact sub- 161
set of R2 . Then for any continuous function F on R
X J(A, S )
u¯ ∗ F =
s¯αβ ∗ F + ν¯ ∗ F
(5)
J(β,
α)
β,α
β∈(σ)
Using (3), we write
4
P
[λ − α]
j
J(S¯ ,S )
=
S¯αβ (λ) where
J(λ,β)
[λ − β][β − α]
j=1
1
J(S¯ , S )
D(β)(λ1 − β1 )(λ2 − β2 ) J(λ, β)
J(S¯ , S )
1
2
S¯αβ
(λ) =
D(β)(β1 − α1 )(λ2 − β2 ) J(λ, β)
J(S¯ , S )
1
3
S¯αβ
(λ) =
D(β)(β2 − α2 )(λ1 − β1 ) J(λ, β)
J(S¯ , S )
1
4
S¯αβ
(λ) =
D(β)(β1 − α1 )(β2 − α2 ) J(λ, β)
1
S¯αβ
(λ) =
−j
j
Let F Lsαβ = S¯αβ (λ) j = 1, 2, 3, 4. Then
s¯αβ ∗ F =
4
X
j=1
−j
sαβ ∗ F
1
Φ(λ) S 1 (λ1 , λ2 )S 2 (β1 , β2 )
D(β)(λ1 − β1 )(λ2 − β2 )
− S 2 (λ1 , λ2 )S 1 (β1 , λ2 )
1
= S¯αβ
(λ) =
= Φ(λ)tβ(λ).
Let F LT β = tβ (λ). Then s−1 αβ ∗ F = ϕ ∗ T β ∗ F
S 2 (β1 , λ2 ) − S 2 (β1 , β2 )
1 S 1 (λ1 , λ2 )
(λ1 − β1 )tβ (λ) =
D(β)
λ2 − β2
162
122
S 1 (β1 , λ2 ) − S 1 (β1 , β2 ) S 2 (λ1 , λ2 )
λ2 − β2
S 2 (β1 , λ2 ) − S 2 (β1 , β2 ) S 1 (β1 , λ2 ) − S 1 (β1 , β2 )
,
λ2 − β2
λ2 − β2
are entire functions of λ2 of exponential type and are Fourier Laplace
transforms of distributions s1 and s2 in the variable x2 with support compact in which β1 appears as a parameter. Then
1
(λ1 − β1 )tβ (λ) = F L
T 1 ∗ (δx1 ⊗ s2 ) − T 2 ∗ (δx1 ⊗ s1 )
D(β)
where δx1 is the Dirac measure in the space of x1 . Setting T β ∗ F = G,
we have
∂G
1 T 1 ∗ (δx1 ⊗ s2 ) ∗ F − T 2 ∗ (δx1 ⊗ s1 ) ∗ F = 0
− β1 G =
∂x1
D(β)
Similarly
(λ2 − β2 )tβ (λ) =
"
S 1 (λ1 , λ2 ) − S 1 (β1 , λ2 )
1
S 2 (λ1 , λ2 )
D(β)
λ1 − β1
S 1 (λ1 , λ2 )
163
S 2 (λ1 , λ2 ) − S 2 (β1 , λ2 )
λ2 − β2
gives
∂G
− β2 G = 0
∂x2
Hence
G = k exp < β, x >
For x = 0, G(0) = T β ∗ F(0) = hT β , Fi so that
T β ∗ F = hT β , Fi exphβ, xi
s−1
αβ
and
∗ F = φ ∗ T β ∗ FhT β , Fiφ ∗ exphβ, xi
= Φ(β)hT β , Fi exphβ, xi
1
2
Sαβ
(λ) =
D(β)(λ2 − β2 )(β1 − α1 )
#
(3.1)
123
S¯ 1 (λ1 , λ2 )S 2 (β1 , λ2 ) − S¯ 2 (λ1 , λ2 )S 1 (β1 , λ2 )
S 1 (β1 , λ2 )
S 2 (β1 , λ2 )
and
are entire functions of exponential type and
λ2 − β2
λ2 − β2
the distributions S 1 and S 2 with compact support in the variable x2 have
these functions as Fourier Laplace transforms
s−2
αβ
1
=
φ ∗ T 1 ∗ (δx1 ⊗ s2 ) − φ ∗ T 2 ∗ (δx1 ⊗ S 1 )
D(β)(β1 − α1 )
and s¯2αβ ∗ F = 0.
Similarly s¯3αβ ∗ F = 0, s¯4αβ ∗ F = 0 and we obtain finally,
u¯ ∗ F =
=
X
Φ(β)
β∈(σ)
X
J(A, S )
hT β , Fi exphβ, xi + ν¯ ∗ F
J(β, α)
hT β , Fi exphβ, xi + ν¯ ∗ F.
β∈(σ)
¯ = Φ(λ)R where R 164
We shall now verify that ν¯ ∗ F = 0. F L¯ν = R
contains only a finite number of terms of ‘irregular type’ in the formula
G1 , (refer to page 80). The terms in R come from
J(S , S )
1
a) D(α)[λ − α]d
and
D(α)[λ − α] J(λ, α)
S 1 (λ1 , λ2 ), −εA1 (α1 , λ2 ) + ∂α∂ 1 S 1 (α1 , λ2 )dα1 b) the determinant S 2 (λ1 , λ2 ), −εA2 (λ1 , λ2 ) + ∂α∂ S 1 (α1 , λ2 )dα2 1
multiplied by D(α)[λ −! α] in which for dα1 and
! dα2 we have to substi∂S
∂S
J A,
J
,A
∂α2
α1
ε
ε
tute
and
.(b) gives in R a term of
D(α)
J(α, α)
D(α)
J(α, α)
J(S , S )
where kα depends only in α and this functhe form Kα [λ − α]
J(λ, α)
tion in the ideal generated by S 1 (λ1 , λ2 ) and S 2 (λ1 , λ2 ) in the ring of
d[D(α)]
entire functions. Similarly in (a) the term
multiplied by
[D(α)]2 [λ − α]
124
D(α)[λ − α]
J(S , S )
gives a function in the same ideal. The other term 165
J(λ, α)
in (a) viz.
d[λ − α]
J(S , S )
equals
J(λ, α) D(α)[λ − α]2
J(S , S )
[(λ1 − α1 )dα2 + (λ2 − α2 )dα].Now
J(λ, α)
1
1
J(S , S )
and
belong to the ideal. This
λ1 − α1
J(λ, α) λ2 − α2
¯ lies in the
as in the study of S¯2 (λ) and S¯3 (λ). Thus R
D(α)[λ − α]
1
[λ − α]
J(S , S )
J(λ, α)
can be seen
ideal so that
αβ
αβ
γ¯ ∗ F = 0
We have
u¯ ∗ F =
X J(A,
¯ S)
hT β , Fi exphβ, xi
J(β, α)
β∈(σ)
(6)
in which the series converges uniformly on compact of R2 and where
¯ S)
J(A,
.
F L¯u =
J(λ, α)
We shall now prove that the continuous function F mean periodic
with respect to T 1 and T 2 a uniquely determined by the system of coefficients cβ = hT β , Fi corresponding to the mean periodic exponentials
(e<β,x> )βεσ by establishing the following
Representation theorem. If F is mean periodic with respect to T 1 and T 2
and φεD(R 2 ) and u ∈ D(R1 ) then
X
φ∗u∗F =
ψ(β) < T β , F > e<β,x>
(7)
β∈(σ)
166
where ψ(λ) = F Lφ ∗ u and the series converges uniformly in every
compact subset of R2 .
For φεD(R2 ) and u1 , u2 εD(R1 ) we had F Lφ ∗ ui = A¯ i (λ1 , λ2 )i =
1, 2 and
¯ S)
J(A,
= A¯ 1 (λ1 , λ2 )S 2 (α1 , λ2 ) − A¯ 2 (λ1 , λ2 )S 1 (α1 , λ2 )
J(λ, α)
125
Let φ ∗ u1 ∗ F = G1 , φ ∗ u2 ∗ F = G2 . Gi are in E (R2 ) and are mean
periodic with respect to T 1 , T 2 . Let F Lµi = Mi (α1 , λ2 ). Then
X J(A,
¯ S)
hT β , Fiehβ,xi
µ2 ∗ G 1 − µ1 ∗ G 2 =
J(β,
α)
β∈σ
Setting in this equation first u1 = uεD(R1 ), u2 = 0 and then u1 = 0,
u2 = −u, and writing φ ∗ u ∗ F = G we obtain
X
Ψ(β)S 1 (α1 , β2 )hT β , Fiehβ,xi
(8)
µ1 ∗ G =
β∈σ
µ2 ∗ G =
X
Ψ(β)S 2 (α1 , β2 )hT β , Fiehβ,xi
(9)
β∈σ
Let H denote the sum of the series on the right hand side of (7). This
series converges uniform;y on every compact set since ψ(λ) is a rapidly
decreasing function in any vertical band and (σ) is contained in such 167
a band. Hence convolution with H is obtained by convoling with each
term of the series then summing. But µ1 ∗ H and µ2 ∗ H so obtained are
nothing but the series (8) and (9) respectively. Therefore µ1 ∗G = µ1 and
µ2 ∗ G = µ2 ∗ H. Hence G = H if we show that the two homogeneous
equations µ1 ∗ L = 0 = µ2 ∗ L = 0 have L = 0 for the unique solution in
E (R2 ).
(2)
1
2
Now µ1 = δx1 ⊗ s(1)
x2 µ2 = δ x1 ⊗ s x2 where s and s are distributions
in the variable x2 having for Fourier Laplace transforms S 1 (α1 , λ2 ) and
S 2 (α1 , λ2 ) respectively in which α1 is a parameter. Hence L(X1 , x2 ) considered as a function of x2 for x1 fixed is means periodic with respect to
s1x2 and s2x2 . The spectrum (σα1 ) in C for the variable λ2 defined by the
equations S 1 (α1 , λ2 ) = S 2 (α1 , λ2 ) = 0
X
cα2 (x1 )eα2 x2
L(x1 , x2 ) =
α2 ∈σα1
where σα1 is the set of α2 such that (α1 , α2 )ε(σ) by the fundamental
theorem of mean periodic functions in R1 .
Taking β = (β1 , β2 )ε(σ) sum that α1 , β1 we obtain again
X
dβ2 (x1 )eβ2 x2 .
L(x1 , x2 ) =
β2 ∈σβ 1
126
The spectrum (σ) is therefore decomposed into a countable union of 168
subset each subset consisting of all αε(σ) with the same first co-ordinate
α1 and corresponding to each such subset we have the expansion of
L(x1 , x2 ) in mean periodic exponentials in x2 with coefficients in which
the fixed co-ordinate x1 occurs as a parameter. But all these expansions
have to be the same and therefore there will be at leat one α2 common
to all σα1 i.e σ is decomposed in countable subset of elements with the
same first co-ordinate and we can choose one element from each of these
such that the second co-ordinates of all these chosen elements are the
same. But this is impossible for (σ). For (σ) is indefinitely near to (σ◦ )
which is the set of common zeros of the principal parts of S 1 (λ1 , λ2 ) and
S 2 (λ1 , λ2 ) { refer to §8 } and the above type of decomposition of (σ0 ) is
not possible for the following reason.
Setting Xi = eλi in these principal parts of S 1 and S 2 , (σ0 ) is given
in terms of the points of intersection of two rectangular hyperbolas
a1 + b1 X1 + c1 X2 + d1 X1 X2 = 0
a2 + b2 X1 + c2 X2 + d2 X1 X2 = 0
169
These have two distinct points of intersection (X1′ , X2′ ) and (Y1′ , Y2′ )
′
′
′
′
let X1′ = eα1 , X2′ = eα2 ; Y1′ = eβ1 Y2′ = eβ2 . Then (σ0 ) is defined by
(α′1 + 2hπi,2 +2kπi) and (β′1 + 2h′ πi, β2 + 2k′ πi)
where h, k, h′ , k′ very in the set of rational integers Z. If the situation
described for (σ) exists in the case of (σ◦ ) then for all distinct first coordinates
α′1 + 2hπi, β′1 + 2h′ πi h, h′ εZ,
there will exist a single 2nd co-ordinate such that the points formed will
lie in (σ0 )i.e there exist integers k, k′ such that
α2 + 2kπi = β + 2k′ πi
This gives Y2′ = Y2′ . But it is clear that the two points of intersection (of
the two coines ) which are assumed to be distinct (see §8) cannot have
the same ordinates.
127
Thus we have proved that the only solution for the con-volution
equation (8) µ1 ∗ L = µ2 ∗ L = 0 is L ≡ 0 and the theorem is proved.
The Uniqueness Theorem.
If all the coefficients cβ = hT β , Fi are zero then F = 0.
170
cβ = 0 imply that φ ∗ u ∗ F = 0 ∀φεD(R2 ), ∀uεD(R∞ ). Letting
φ → δ the Dirac measure u ∗ F = 0 ∀uεD(R1 ). Letting u tend towards
the Dirac at a point x0 in the interior of the rectangle R1 , we have
F(x + x0 ) = 0 i.e F = 0.
Part III
The Two-Radius Theorem
129
Chapter 1
1
The subject of this part is the two-radius theorem, which is the converse 171
of the classical theorem of Gauss on the spherical mean of the harmonic
functions in Rn
P
P
Let kn denote the area of the sphere n−1 of radius 1 in Rn ( n−1 =
P
{x/x = (x1 , x2 , . . . xn )εRn , x2i = 1). Let f (x) be a function which is
(C, 2) in Rn . The spherical mean of f (x) on the surface of the sphere
with center x and radius r is by definition
Z
1
f (ξ)dσ
(1)
M(x, r) =
kn rn−1 S rn−1 (x)
where S rn−1 (x) is the sphere of center x and radius r in Rn , ξ is the generic
point of the sphere and dσ the element of area of the sphere. We have
also
Z
1
−
M(x, r) =
f (x + r →
u )dω
(2)
kn Pn−1
P
−u is the unit vector at 172
where dω is the element of the sphere n−1 and →
P
the origin, whose other extremity describes n−1 .
Proposition 1 (Poisson). The function M(r, x) is a solution of the partial
differential equation
∂2 M n − 1
+
△ x M(x, r) =
r
∂r2
131
∂M
∂r
(3)
1.
132
As △ is a convolution operator in Rn , we have
Z
1
△ξ f (ξ) dσ
△ x M(x, r) =
kn rn−1 S rn−1 (x)
From 2,
Z
∂M
1
∂
−u )dω
=
f (x + r→
P
∂r
kn
∂r
Z n−1
d
1
f (ξ) dσ
n−1
n−1
dν
kn r
S r (x)
where ν is the exterior normal to S rn−1 (x) at the point ξ. By Green’s
formula, we also have,
Z Z
1
∂M
1
= r n−1
△ξ f (ξ) dV = n−1 J
∂r
kn
r
173
1 R R △ξ f (ξ) dV where the integral is taken over the volkn
ume of the solid sphere in Rn with S rn−1 (x) as boundary. Then
where J =
(n − 1)
1 ∂J
∂2 M
= − n J + n−1
2
∂r
∂r
r
Zr
1
∂J
=
△ξ [ f (ξ)]dσ
∂r kn S rn−1 (x)
But
since dV = dσdr.
Thus
1
∂2 M n − 1 ∂M
+
=
2
r ∂r
∂r
kn rn−1
Remark. For r = 0, obviously
And
∂M ∂r r=0
Z
△ξ [ f (ξ)]dσ
S rn−1
= △ x M(x, r) .
M(x, 0) = f (x).
1 R R
△ f (ξ) dV
= limr=0
S rn−1 ξ
kn rn−1
2. Study of certain Cauchy-Problems
133
The integral in this expression is of order rn
#
"
∂M
=0
∂r r=0
Hence the solution M(x, r) of (3), verifies the Cauchy conditions
M(x, 0) = f (x),
174
∂
M(x, r) r=0 = 0
∂r
2 Study of certain Cauchy-Problems
Let E∗ denote the vector space of even functions indefinitely differentiable in R with the usual topology (that induced by E ). Let
L = D2 +
q(x)
D
x
d
n+1
be a differential operator where D =
, qεE∗ (with q(0) , −
,n
dx
2
integer ≥ 1). Then (by the results obtained in part I) there exists an
isomorphism B of E∗ onto itself the property
D2 B = BL and B0 f (ξ) = f (0)
Problem 1. To find a function F(x, y) which is twice differentiable in R2
and which is a solution of the Cauchy-problem
with
and
∂2 F q(x) ∂F ∂2 F q(y) ∂F
+
= 2 +
x ∂x
y ∂y
∂x2
∂y
F(x, 0) = f (x),
f ∈ E∗
" #
∂F
=0
∂y y=0
Suppose that there exists a solution F(x, y) which is an even function 175
of x for y fixed and also an even function of y for x fixed.
Let
G(x, y) = B x By[F(ξ, η)]
1.
134
where B x operates on ξ and By on η. G(x, y) is an element in E∗ as a
function of y for x fixed and of x for y fixed. We have
∂r G
= D2x Bξ By [F(ξ1 , η)] = By D2x Bξ [F(ξ1 , η)]
∂x2
= By B x Lξ [F(ξ1 , η)] = By B x Lη [F(ξ, η1 )]
= B x By Lη [F(ξ, η1 )] = B x D2y Bη [F(ξ, η1 )]
∂2G
= D2y B x Bη[F(ξ, η1 )] = 2 .
∂y
Also
G(x, 0) = B x Bo [F(ξ, η)] = B x [F(ξ, η)] = B x [ f (ξ)] = g(x).
"
#
∂G(x, y)
= 0.
∂y
y=0
176
Hence Problem 1 is reduced to the problem of finding a solution of
the following Cauchy problem
∂2 G ∂2 G
= 2
∂x2
∂y
G(x, 0) = g(x) ∈ E∗
" #
∂G
= 0.
∂y y=0
It is wellknown that this problem has the unique solution
1
g(x + y) + g(x − y)
2
(4)
1
B x By [g(ξ + η) + g(ξ − η)]
2
(5)
G(x, y) =
Then there exists a unique solution of Problem (1), defined by
F(x, y) =
where B = B−1 is an isomorphism of E∗ .
Remark. As g is even, G(x, y) is symmetric in x and y. Then F(x, y) is
also symmetric in x and y.
135
Definition 1. For any f ∈ E∗ let, M x,y f (ξ) denotes the solution of
problem (1):
M xy [ f (ξ)] =
n
o
1
B x By Bξ+η [ f ] + Bξ−η [ f ]
2
(6)
Problem 2. To determine F(x, y), (r ∈ R1 , r ≥ 0) which satisfies the 177
differential equation
A x [F(ξ, r)] = Lr [F(x, ρ)],
where A is an elliptic differential operator in Rn with indefinitely differentiable coefficients, and the Cauchy conditions
" #
∂F
F(x, 0) = f (x);
= 0.
∂r r=0
The existence and uniqueness of the solution of Problem 2 defines
the operator Mr by
Mr [ f (ξ)] = F(x, r).
Remark. By Poisson’s theorem, when A = ∆ and q(r) = n − 1, then the
solution of Problem 2 is given by the spherical mean.
Proposition 2. The operator M commutes with A.
Let
G(x, r) = A x [F(ξ, r)] = A x Mr [ f (ξ)].
We have
A x [G(ξ, r)] = A x Aξ [F(ξ1 , r)]
Moreover,
and
"
#
= A x Lr [F(ξ, ρ)] = Lr A x [F(ξ, ρ)] = Lr G(x, ρ)
G(x, 0) = A x [F(ξ, 0] = A x [ f (ξ)]
∂G
= 0 since G(x, r) ∈ E∗ , as a function of r for x fixed.
∂r r=0
Hence
G(x, r) = Mr Aξ [ f (ξ1 )]
178
1.
136
and the proposition is proved.
Iteration of the operator
For f ∈ E (Rn ), let F(x, r) = Mr [ f (ξ)] be the solution of Problem 2.
By iteration we consider the Cauchy problem,
A x [F (ξ, s)] = L s [F (x, σ)]
with
F (x, 0) = F(x, r)
"
#
∂F
=0
∂s s=0
in with r is a positive parameter. The solution F is a function of x, r, s,
F (x|r, s) = M s F(ξ, s) = M s Mr [ f (ξ)]
(7)
Proposition 3. For x fixed in Rn , F (x|r, s) is a solution of Problem 1, i.
e.
Lr F (x|ρ, s) = L s [F (x|r, σ)]
F (x|r, 0) = F(x, r)
#
"
∂F
= 0.
∂s s=0
and
179
We compute
n Lr F (x|ρ, s) = M s Lr F(ξ, ρ)
n o
= M s Aξ F(ξ1 , r) = A x M s [F(ξ, r)]
(by Proposition 2)
= A x F (ξ|r, s) = L s [F (x|r, σ)] .
By Definition 1, we have
F (x|r, s) = Mr,s [F(x, θ)].
137
Hence we obtain the formula
M s Mr = Mr M s = Mr,s [M0 ]
(8)
We now consider the solution
F(x, r) = Mr [ f (ξ)]
of Problem 2 and suppose that the it satisfies for a > 0 fixed the condi- 180
tion
F(x, a) = f (x)
(9)
for any x.
Condition (9) expresses the fact for r = a fixed, the value F(x, a)
reproduces the initial value f (x) of the function for r = 0 in the space
Rn .
n−1
D then F(x, r) = M(x, r) and the
Remark. If A = ∆, and L = D2 +
r
condition (9) is ’Gauss’s condition’ for the fixed radius a.
In view of (9) and (7) we have
F (x, a, s) = F(x, s)
where x and s are arbitrary.
By definition (1),
Ma,s [F(x, θ)] = F(x, s)
(10)
where the left-hand side the operator M operates on the variable θ (Equation (10) thus gives a ’transposition’ from Rn to R1 ). Using (6), (10)
becomes,
2F(x, s) = Ba B s {Bα+σ F(x, θ) + Bα−σ [F(x, θ)]}
which gives
2B s [F(x, σ)] = Ba {Bα+s F(x, θ) + Bα−s F(x, θ)}
since B is the inverse of B.
181
1.
138
Setting
K(x, s) = B s [F(x, σ)].
we have
2K(x, s) = Ba {K(x, α + s) + K(x, α − s)}
(11)
in which a is fixed.
Since K(x, s) is even in s, (11) can be written as
2K(x, s) = Ba {K(x, α + s) + K(x, s − α)}
(11)′
so that the function K(x, s) is mean periodic in s as the equation (11)′ is
clearly an equation of convolution in s.
Remark. In order to obtain the spectrum, we have to substitute eλs in
place of K(x, s) in (11)′ which leads to
1 = Ba [cos hα]
3 The generalized two-radius theorem
Let a, b ∈ R1 ; a, b, > 0; a , b
182
Definition 2. A function f (x) which is (C, 2) in Rn possesses the two
radius property with respect to the elliptic operator A and the singular
operator L, if
F(x, a) = F(x, b) = f (x)
(12)
for x ∈ Rn , where
F(x, r) = Mr [ f (ξ)]
is the solution of Problem 2.
n−1
D. Condition (12) is the ’Gauss’s
Remark. If A = ∆ and L = D2 +
r
condition’ for two fixed radii a and b.
Now in place of (11)′ we have two equations
2K(x, s) = Ba {K(x, s + α) + K(x, S − α)}
3. The generalized two-radius theorem
139
2K(x, s) = Bb {K(x, s + α) + K(x, s − α)}
so that K(x, s) is mean periodic in s with respect to two distribution,
and by the classical result of the theory of mean periodicity in R1 , the
elements λ in the spectrum σ(a, b) have to satisfy two equations


1 = Ba [cos hλα]

(13)


1 = Bb [cos hλα]
Now we show that λ = 0 is a double solution of (13).
By definition D2 B = BL.
183
2
If B s [1] = ϕ(s) , D [ϕ(s)] = 0 since L s (1) = 0. As ϕ(s) is even, ϕ(s)
has to be a constant equal to ϕ(0). But ϕ(0) = B0 [1] = 1 so that ϕ(s) ≡ 1.
Since B is the inverse of B, and B s [1] = 1, we have B s [1] ≡ 1.
As Ba [cos hλα] is an even function of λ, (13) possesses the double
solution λ = 0.
We shall hereafter restrict ourselves to the following hypothesis
Hypothesis (H)- The equations
1 = Ba [cos hλα] = Bb [cos hλα]
have the only double solution λ = 0. In this case K(x, s) is necessarily
of the form
K(x, s) = k1 (x) + sk2 (x)
(14)
by the fundamental theorem of the mean periodic functions in R1 . But
K(x, s) is even in s so that k2 (x) ≡ 0 in Rn and we have
K(x, s) = k1 (x)
(15)
By inversion, K(x, s) = B s [F(x, σ)] gives
F(x, s) = k1 (x)B s [1] = k1 (x)
and for s = 0, F(x, s) = f (x). Hence
F(x, s) = f (x)
184
(16)
1.
140
for x ∈ Rn and s > 0.
But F(x, s) is a solution of
A x F(ξ, s) = L s [F(x, σ)]
and F(x, s) = f (x) gives necessarily
A x f (ξ) = 0
(17)
Thus we have in conclusion the
Theorem. The hypothesis (H) and the condition
F(x, a) = F(x, b) = F(x),
gives
F(x, s) = f (x) for any s ≥ 0
and
A x f (ξ) = 0.
Corollary. A is the Laplacian ∆, then f is a harmonic function.
4 Discussion of the hypothesis H
In general, the hypothesis (H) is satisfied because the equations
1 = Ba [cos hλα] = Bb [cos hλα]
185
are a system of two equations with only one unknown λ. But it is necessary to investigate certain exceptional values of a, b(a , b, a > 0, b > 0)
for which the two-radius Theorem is false. The question of existence of
such exceptional couples (a, b) is difficult in the general case but in the
2
case of A = ∆ of R3 and L = D2 + D, (n = 3), we can assert that there
r
do not exists any such exceptional couples and the two-radius is always
true in R3 .
4. Discussion of the hypothesis H
141
Remark. This discussion is completely independent of the function f
and consequently the couples of exceptional values (a, b) are also independent of f (x).
Results of the discussion in the case A = ∆.
n−2
In this case, B = B p in the notation of part I with p =
.
2p
We know that B [cosh λx] = j p (λix) where j p (z) = 2 Γ(p + 1)
z−p J p (z). The function which assumes the value 1 for z = 0. In this case
the equations under consideration are
jP (λia) = j p (λib) = 1
(18)
Thus it is sufficient to consider in C 1 , the equation
j p (z) = 1
(19)
and to examine whether there exists two roots of (19) with the same 186
argument.
It is easy to see that the set of points ζ in C 1 which are roots of (19)
have for axes of symmetry the two axes 0ξ and 0η if z = ξ + iη. This
set is countable and contains the origin (ξ = η = 0). By an intricate
discussion based on the asymptotic expansion of the Bessel functions, it
is even possible to prove that for a given p (i.e. for a given dimension n
of the space) the number of couples of roots of (19) which have the same
argument is necessarily finite. Hence for any dimension n the number of
a
exceptional ratios is necessarily finite.
b
1
In the case n = 3 i.e. p = ,
2
j p (z) =
sin z
z
(for any odd n , j p (z) has an expression which depends algebraically
on z, sin z and cos z).
Thus sin z = z gives
sin ξ cosh η = ξ
1.
142
cos ξ sinh η = η.
Eliminating the hyperbolic functions and the circular functions, we
obtain respectively
ξ2
sin2 ξ
ξ2
and
so that
and
187
−
cosh2 η
η2
=1
cos2 ξ
+
η2
sinh2 η
=1
"
#1
sin2 ξ 2
η = ±ξ cot ξ 1 − 2
ξ
(A)
"
(B)
# 21
sinh2 η
ξ = ±η coth η
−1
η2
The equations define two real curves in the plane (ξ, η), and the roots
z = ξ + iη of (19) are a subset of the set points of intersection of two
curves.
As 0ξ, 0η are the axes symmetry of the two curves it is sufficient to
examine the situation for ξ ≥ 0, η ≥ 0 (A) can be written as
(B′ )
ξ = f1 (η) f2 (η)
1
#
sinh η
2
−1 .
where f1 (η) = η coth η, f2 (η) =
2
η
We have,
"
f1′ (η) = coth η −
η
2
2
=
sinh η
=
sinh η cosh η − η
1
2
sinh η
1
2
2 sinh η
sinh 2η − 2η > 0
and
f2′ (η)
"
#
sinh η cosh η sinh2 η sinh η cosh η 1
=
−
η − tanh η > 0.
2
3
3
f2 (η)
η
η
η f2 (η)
4. Discussion of the hypothesis H
188
143
Thus f1 (η) and f2 (η) are increasing functions of η > 0 and so is their
product f1 (η) f2 (η).
1
dξ
sinh η cosh η
(η − tanh η) + f2 (η)
(sinh 2η − 2η)
= f1 (η)
dη
η3 f2 (η)
2 sinh2 η
and ξ/η = coth η f2 (η).
Hence
1
dξ ξ
sinh 2η − 2η
/ =
sinh η cosh η(η − tanh η) +
2
dη η [η f2 (η)]
sinh 2η
But η f2 (η) 2 = sinh2 η − η2 and finally
dξ ξ
η
η − tanh
−
/ − 1 = sinh η cosh η
2
2
dη η
sinh
η
cosh η
sinh η − η
=
η3 + η sinh4 η − sinh3 η cosh η
sinh η cosh η(sinh2 η − η2 )
.
d
(η sinh η − cosh η) = η cosh η > 0, η sinh η − cosh η and theredη
fore F(η) = η3 + sinh3 η(η sinh η − cosh η) is an increasing function of η. 189
But F(0) = 0 so that F(η) > 0 for η > 0. Hence
As
dξ ξ
> on B for ξ > 0, η > 0
dη η
from which it is clear that there does not exist any point (ξ, η) on B,
ξ > 0, η > 0, the tangent at which passes through the origin. Then any
chord through the origin can cut the curve only in one point which is not
the origin. But the roots of sinz = z lie on the curve and it is impossible
to find out distinct roots other than zero which have the same argument.
a
Finally, for n = 3 there are no exceptional ratios and the two-radius
b
theorem is completely proved in R3 .
Bibliography
[1] N. Bourbaki, Espaces vectoriels topogiques; Paris, Hermann.
[2] N. Bourbaki, Integration, Paris, Hermann 1952.
[3] J. Delsarte, ’Les fonctions moyenne-periodiques’, J. Math. Pures
et Appl. 14 (1935), 403-453
[4] L. Ehrenpreis, Mean periodic functions, Amer. J.Math., 77 (1995),
293-326.
[5] J.P. Kahane, ’Sur quelques problems d’unicite et de prolongement relatifs aux fonctions approchables par des sommes
d’exponentilles’, Ann. Inst. Fourier, 5(1954),39-130.
[6]
’Sur les fonctions moyenne-periodiques borne’, Ann. Inst.
Fourier, 7(1957), 293-314.
[7]
Lectures on mean periodic functions, Tata Institute of Fundamental Research, (1958).
[8] B. Malgrange, ’Existence et approximations des solutions des
equations aux derivees partielles et des ’equations deconvolutions’, Ann. Inst. Fourier, (1956), 271-356.
[9] L. Schwartz, ’Theorie generale des fonctions moyenneperiodiques’, Ann. of Math. 48(1947), 857-925.
[10]
Theorie des distributions, Tome I et lI, Paris Hermann.
145
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Lectures on Topics in Mean Periodic Functions And The Two

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