MATH4060 Solution 1

February 2023

Exercise 1
(a) The assumption fˆ(ξ) = 0 for all ξ ∈ R implies that
Z ∞
A(ξ) − B(ξ) = e2πiξt f (x)e−2πiξx dx = 0.
−∞

(b) Consider A defined on the upper half plane. Note that for z = u + iv, v > 0, and
x ≤ t, we have
|f (x)e−2πiz(x−t) | = |f (x)|e2πv(x−t) ≤ |f (x)|.
By the moderate decrease of f , A(z) is well-defined and bounded. To see that A is
holomorphic on the upper half plane, we argue as in Theorem 3.1: define An (z) =
Rt
f (x)e−2πiz(x−t) dx and observe that An → A uniformly because |An (z) − A(z)| ≤
R−n
−n
−∞
|f (x)| dx and f has moderate decrease. Each An is holomorphic by Theorem 5.4 of
Chapter 2 and so is the uniform limit A.
Similarly, B is holomorphic and bounded on the lower half plane. Part (a) and the
symmetry principle (Theorem 5.5 of Chapter 2) imply that F is entire and bounded,
hence a constant. In fact this constant is 0, since the boundedness of f implies that (on
the upper half plane)
Z t
C
|A(u + iv)| ≤ C e2πv(x−t) dx = ,
−∞ 2πv

so that A(z) → 0 as Im(z) → ∞.

(c) The first statement follows from F (0) = 0 and the second from the continuity of f .

Exercise 3
a −2πizξ
Consider the function f (z) = a2 +z 2e having simple poles at z = ±ai with residue
±(2i)−1 e±2πaξ . When ξ ≥ 0, consider the contour from −R to R along the real axis
−
and then from R to −R along the semicircular arc CR in the lower half plane. Along
the arc z = Reiθ (with Im(z) < 0 and assume R > a),
a a
|f (z)| = e2π Im(z)ξ ≤ 2 .
|a2 + z 2 | R − a2
R
So C − f (z) dz → 0 as R → ∞. Because the contour is clockwise oriented, the residue
R
theorem implies that
Z ∞
a
2 2
e−2πixξ dx = −2πi resz=−ai f = πe−2πaξ = πe−2πa|ξ| .
−∞ a + x

The case ξ < 0 is similar and uses the semicircular contour on the upper half plane. The
second statement of the question is by direct integration.

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Exercise 7
(a) We first compute fˆ(ξ) using residue theorem. Consider the function g(z) = (τ +
z)−k e−2πiξz , with an order k pole at z = −τ (in the lower half plane) with
 k−1
1 d (−2πiξ)k−1 2πiξτ
resz=−τ g = e−2πiξz = e .
(k − 1)! dz z=−τ (k − 1)!
Similar to exercise 3, for ξ > 0, consider the semicircular
R contour in the lower half plane.
Since k ≥ 2, the same argument shows that C − g(z) dz → 0 as R → ∞, and thus the
R
residue theorem gives
Z ∞ −2πiξx
e (−2πi)k k−1 2πiξτ
fˆ(ξ) = k
dx = −2πi resz=−τ g = ξ e .
−∞ (τ + x) (k − 1)!

For ξ ≤ 0, the same argument in the upper half plane shows that fˆ(ξ) = 0 because the
contour does not enclose the pole. The desired identity is now a direct consequence of
the Poisson summation formula.
(b) Apply (a) with k = 2. Note that |e2πiτ | < 1 since Im τ > 0, so we have the following
identity (viewing as a function of τ ):
∞  ∞ 0  0
X
2πimτ 1 X 2πimτ 1
me = e =
m=1
2πi m=0 2πi(1 − e2πiτ )
e2πiτ 1 1
= = −πiτ =− .
(1 − e2πiτ )2 (e − eπiτ )2 4 sin2 (πτ )

(c) Yes, because both sides are meromorphic functions on C that have the same poles
and agree on the upper half plane.

Exercise 10
Let l > 0. First note that for z = x + it and ζ = ξ + iη ∈ Sl (i.e. |η| < l), we have
2
+2πxη bt2 +2πtξ
|f (z)e−2πizζ | = |f (x + it)|e2π(xη+tξ) ≤ ce−ax e (1a)
−ax2 +2πl|x| bt2 +2πtξ −ãx2 bt2 +2πtξ
≤ ce e ≤ c1 e e , (1b)
for any 0 < ã < a and some constant c1 independent of x and ζ ∈ Sl (but dependent on
a,
R nã, l). Similar to Theorem 3.1, observe that fˆ(ζ) is holomorphic in every Sl : let fˆn (ζ) =
−2πixζ
−n
f (x)e dx, each holomorphic by Theorem 5.4 of Chapter 2. Equation (1b) with
2 2
t = 0 and the integrability of e−ãx imply that |fˆn (ζ) − fˆ(ζ)| ≤ c1 e−ãx dx → 0
R
|x|≥n
uniformly in ζ ∈ Sl , as n → ∞. So fˆ is holomorphic by Theorem 5.2 of Chapter 2.
Next, with ζ fixed, we show that the contour of integration can be changed to {Im(z) =
y} for any fixed y ∈ R, i.e. we have
Z ∞ Z ∞
ˆ
f (ζ) = f (x)e −2πixζ
dx = f (x + iy)e−2πi(x+iy)ζ dx. (2)
−∞ −∞

(The integrals on both sides are well-defined by (1b).) To prove (2) when y 6= 0, con-
sider the entire function f (z)e−2πzζ and the rectangular contour defined by the vertices
−R, R, R + iy, −R + iy. As in the proof1 of Theorem 2.1, it suffices to show that the
integrals along the vertical segments of the contour tends to 0 as R → ∞. By (1b),
along the left vertical segment, we have
Z y
2
f (−R + it)e−2πi(−R+it)ζ dt ≤ Ce−ãR → 0
0
1 In fact, if η = 0, we could just apply the proof in Theorem 2.1.

2
as R → ∞. And the same holds for the right vertical segment. This proves (2).
Finally, we estimate |fˆ(ζ)| using the shifted contour: by (1a), we have
Z ∞
ˆ by 2 +2πyξ 2
|f (ζ)| ≤ ce e−ax +2πxη dx
−∞
Z ∞ √0
by +2πyξ b0 η 2
2 2
= ce e e−a(x− b /aη) dx
−∞
0 by 2 +2πyξ b0 η 2
≤ce e ,

where b0 > 0 is obtained by completing square (and is independent of η); while c0 is

some constant also independent of ζ = ξ + iη. Consider a sufficiently small d > 0 so that
a0 := 2πd − bd2 > 0, and then take y = −dξ in the above to obtain the desired estimate
0 2
+b0 η 2
|fˆ(ζ)| ≤ c0 e−a ξ .