§ 3. Some Applications of Cauchy’s Method
97
Exercise 3. Supposing that ˆ
f satisfies the conditions just proved, show directly that integral (5) is independent of y ∈ B. (Integrate along a horizontal
rectangle in B).
We could also use Dirichlet’s theorem (Chap. VII, n
◦ 30, theorem 27) and
show that
lim
N
−N
f (z)e(−tz)dz =
1
2
ˆ
f (t + 0) + ˆ
f (t − 0)
for all y ∈ I if ˆ
f is right and left differentiable everywhere; the extended
integral is over the interval |x| ≤ N of the horizontal Im(z) = y.
(ii) A Paley-Wiener theorem. One of the problems of the theory is characterizing the complex Fourier transforms of functions ˆ
f of a given category.
The simplest result is related to the space D(R) of C
∞ functions with compact support in R. In this case, f is defined for all z ∈ C, and hence is an
entire function, and if ˆ
f vanishes outside a compact interval [a, −a], then
|f (z)| ≤
a
−a
ˆ
f (t)
exp(−2πty)dt .
The exponential is bounded above for all t by exp(2πa|y|) on the integration
interval, and so
f (z) = O
e
2πa|y|
on C .
We saw that if ˆ
f is replaced by its derivative of order n, then the function
f (z) is replaced by (−2πiz)
n f (z). Thus
z
n f (z) = O
e
2πa|y|
on C
(PW)
for all n ∈ N. Conversely :
Theorem 12 (Paley-Wiener). Let ϕ be an entire function. The following
two conditions are equivalent :
(i) There is a number a > 0 such that ϕ satisfies (PW) for all n;
(ii) ϕ is the complex Fourier transform of a C
∞ function vanishing outside
[−a, a].
It suffices to prove that (i)=⇒(ii). The function z
n f (z) being bounded on
every horizontal and in particular on R, the Fourier transform
ˆ
f (t) =
f (x)e(−tx)dx
of f on R is well-defined and is C
∞ : this can be seen by differentiating under
the
sign. To show that it is zero for t ≥ a, integrate f (z)e(tz) along the
97
Exercise 3. Supposing that ˆ
f satisfies the conditions just proved, show directly that integral (5) is independent of y ∈ B. (Integrate along a horizontal
rectangle in B).
We could also use Dirichlet’s theorem (Chap. VII, n
◦ 30, theorem 27) and
show that
lim
N
−N
f (z)e(−tz)dz =
1
2
ˆ
f (t + 0) + ˆ
f (t − 0)
for all y ∈ I if ˆ
f is right and left differentiable everywhere; the extended
integral is over the interval |x| ≤ N of the horizontal Im(z) = y.
(ii) A Paley-Wiener theorem. One of the problems of the theory is characterizing the complex Fourier transforms of functions ˆ
f of a given category.
The simplest result is related to the space D(R) of C
∞ functions with compact support in R. In this case, f is defined for all z ∈ C, and hence is an
entire function, and if ˆ
f vanishes outside a compact interval [a, −a], then
|f (z)| ≤
a
−a
ˆ
f (t)
exp(−2πty)dt .
The exponential is bounded above for all t by exp(2πa|y|) on the integration
interval, and so
f (z) = O
e
2πa|y|
on C .
We saw that if ˆ
f is replaced by its derivative of order n, then the function
f (z) is replaced by (−2πiz)
n f (z). Thus
z
n f (z) = O
e
2πa|y|
on C
(PW)
for all n ∈ N. Conversely :
Theorem 12 (Paley-Wiener). Let ϕ be an entire function. The following
two conditions are equivalent :
(i) There is a number a > 0 such that ϕ satisfies (PW) for all n;
(ii) ϕ is the complex Fourier transform of a C
∞ function vanishing outside
[−a, a].
It suffices to prove that (i)=⇒(ii). The function z
n f (z) being bounded on
every horizontal and in particular on R, the Fourier transform
ˆ
f (t) =
f (x)e(−tx)dx
of f on R is well-defined and is C
∞ : this can be seen by differentiating under
the
sign. To show that it is zero for t ≥ a, integrate f (z)e(tz) along the