98
VIII – Cauchy Theory
path consisting of the interval (−R, R) followed by the upper half-circle; the
result is zero since the function integrated is holomorphic everywhere. By
(PW), for all n, there is an upper bound
|f (z)e(tz)| = |f (z)| exp(−2πty) ≤ c n R
−n exp [2π(a − t)y)]
over the half-circle for y > 0. For t ≥ a, the exponential is ≤ 1; the integral
along the half-circle is then O(R
1−n ) for all n and tends to 0. For t ≤ −a,
replace the upper half-circle by the lower one. Then ˆ
f (t) = 0 for |t| ≥ a.
It remains to check that f (z) =
ˆ
f (t)e(−tz)dt. As the Fourier transform
of f (x) is in D(R), the inversion formula shows that f is the inverse Fourier
transform of ˆ
f on R. The complex Fourier transform of ˆ
f being holomorphic
everywhere and equal to F on R, the analytic extension principle (Chap. II,
n
◦ 20) shows they are both identical on C, qed.
Exercise 4. Let S be the Schwartz space, i.e. the set of C
∞ functions
ϕ(t), t ∈ R, such that all functions t
p ϕ
(q) (t) are bounded on R (Chap. VII,
§ 6, n
◦ 31). Let S + be the set of the ϕ ∈ S that are zero for t ≤ 0, so that
ϕ
(n) (0) = 0 for all n. (i) Show that the complex Fourier transform f of any
ϕ ∈ S + is defined and holomorphic for y > 0 and that, for all p, p ∈ N, the
function z
p f
(q) (z) is bounded on the half-plane Im(z) ≥ 0. (ii) Conversely,
let f be a holomorphic function on y > 0 satisfying this condition. Show
that, for any y > 0, the function x → f (x + iy) is in the space S and the the
integral of f (z)e(−tz) along the horizontal Im(z) = y does not depend on
y. Denoting this integral ϕ(t), show that ϕ ∈ S + and that f is its complex
Fourier transform.
(iii) Holomorphic functions integrable over a strip. Let
I =]a, b[⊂ R
be an open interval, ρ(y) a continuous function with values > 0 on I, and B
the open horizontal strip Im(z) ∈ I. Let f be a holomorphic function on B;
suppose that
B
|f (z)| ρ(y)dm(z) < +∞
(12.6)
for any closed horizontal strip of finite width B
⊂ B.
59 We show that, under
these conditions, f is a complex Fourier transform. The proof uses the relation
shown in n
◦ 4, (iv) between compact convergence and mean convergence, i.e
in the sense of the L
1 norm. In what follows, we will use the notation
ρ(y)dm(z) = dμ(z) .
59 As inequality 0 < m ≤ ρ(y) ≤ M < +∞ holds in every compact set contained
in I, condition (6) does not really involve ρ. We will see in the next section that
this is no longer the case if B
is replaced in (6) by the strip B. This is why I
introduce a seemingly superfluous function ρ(y) here.
VIII – Cauchy Theory
path consisting of the interval (−R, R) followed by the upper half-circle; the
result is zero since the function integrated is holomorphic everywhere. By
(PW), for all n, there is an upper bound
|f (z)e(tz)| = |f (z)| exp(−2πty) ≤ c n R
−n exp [2π(a − t)y)]
over the half-circle for y > 0. For t ≥ a, the exponential is ≤ 1; the integral
along the half-circle is then O(R
1−n ) for all n and tends to 0. For t ≤ −a,
replace the upper half-circle by the lower one. Then ˆ
f (t) = 0 for |t| ≥ a.
It remains to check that f (z) =
ˆ
f (t)e(−tz)dt. As the Fourier transform
of f (x) is in D(R), the inversion formula shows that f is the inverse Fourier
transform of ˆ
f on R. The complex Fourier transform of ˆ
f being holomorphic
everywhere and equal to F on R, the analytic extension principle (Chap. II,
n
◦ 20) shows they are both identical on C, qed.
Exercise 4. Let S be the Schwartz space, i.e. the set of C
∞ functions
ϕ(t), t ∈ R, such that all functions t
p ϕ
(q) (t) are bounded on R (Chap. VII,
§ 6, n
◦ 31). Let S + be the set of the ϕ ∈ S that are zero for t ≤ 0, so that
ϕ
(n) (0) = 0 for all n. (i) Show that the complex Fourier transform f of any
ϕ ∈ S + is defined and holomorphic for y > 0 and that, for all p, p ∈ N, the
function z
p f
(q) (z) is bounded on the half-plane Im(z) ≥ 0. (ii) Conversely,
let f be a holomorphic function on y > 0 satisfying this condition. Show
that, for any y > 0, the function x → f (x + iy) is in the space S and the the
integral of f (z)e(−tz) along the horizontal Im(z) = y does not depend on
y. Denoting this integral ϕ(t), show that ϕ ∈ S + and that f is its complex
Fourier transform.
(iii) Holomorphic functions integrable over a strip. Let
I =]a, b[⊂ R
be an open interval, ρ(y) a continuous function with values > 0 on I, and B
the open horizontal strip Im(z) ∈ I. Let f be a holomorphic function on B;
suppose that
B
|f (z)| ρ(y)dm(z) < +∞
(12.6)
for any closed horizontal strip of finite width B
⊂ B.
59 We show that, under
these conditions, f is a complex Fourier transform. The proof uses the relation
shown in n
◦ 4, (iv) between compact convergence and mean convergence, i.e
in the sense of the L
1 norm. In what follows, we will use the notation
ρ(y)dm(z) = dμ(z) .
59 As inequality 0 < m ≤ ρ(y) ≤ M < +∞ holds in every compact set contained
in I, condition (6) does not really involve ρ. We will see in the next section that
this is no longer the case if B
is replaced in (6) by the strip B. This is why I
introduce a seemingly superfluous function ρ(y) here.