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Homework Assigned on March 17, 2009 due March 31, 2009
Problem 1 (from Stein & Shakarchi, p.128, #7). The Poisson summation formula applied to specific examples often provides interesting identites.
(a) Let τ be fixed with Im(τ ) > 0. Apply the Poisson summation formula to
f (z) =
(τ + z)k^
where k is an integer ≥ 2, to obtain
∑^ ∞
n=−∞
(τ + n)k^
(− 2 πi)k (k − 1)!
∑^ ∞
m=
mk−^1 e^2 πimτ^.
(b) Set k = 1 in the above formula to show that if Im(τ ) > 0, then
∑^ ∞
n=−∞
(τ + n)^2
π^2 sin^2 (πτ )
(c) Can one conclude that the above formula holds true whenever τ is any complex number that is not an integer?
Hint: For (a), use residues to prove that fˆ (ξ) = 0 if ξ < 0, and
fˆ (ξ) = (−^2 πi)
k (k − 1)!
ξk−^1 e^2 πiξτ^ , when ξ > 0.
Problem 2 (from Stein & Shakarchi, p.129, #8). Suppose fˆ has compact support contained in [−M, M ] and let f (z) =
n=0 anz
n. Show that
an =
(2πi)n n!
∫ M
−M
f^ ˆ (ξ)ξndξ,
and as a result lim sup n→∞
(n!|an|) n^1 ≤ 2 πM.
In the converse direction, let f be any power series f (z) =
n=0 anz
n (^) with
lim sup n→∞
(n!|an|) n^1 ≤ 2 πM.
Then f is holomorphic in the complex plane, and for every ε > 0 there exists Aε > 0 such that |f (z)| ≤ Aεe^2 π(M^ +ε)|z|.
Problem 3 (from Stein & Shakarchi, p.129, #9). Here are further results similar to the Phragm´en-Lindel¨of theorem.
(a) Let F be a holomorphic function in the right half-plane that extends continuously to the boundary, that is, the imaginary axis. Suppose that |F (iy)| ≤ 1 for all y ∈ R, and
|F (z)| ≤ Cec|z|
γ
for some c, C > 0 and γ < 1. Prove that |F (z)| ≤ 1 for all z in the right half-plane.
(b) More generally, let S be a sector whose vertex is the origin, and forming an angle of πβ. Let F be a holomorphic function in S that is continuous on the closure of S, so that |F (z)| ≤ 1 on the boundary of S and
|F (z)| ≤ Cec|z|
α for all z ∈ S
for some c, C > 0 and 0 < α < β. Prove that |F (z)| ≤ 1 for all z ∈ S.
Problem 4 (from Stein & Shakarchi, p.131, #12). The principle that a func- tion and its Fourier transform cannot both be too small at infinity is illus- trated by the following theorem of Hardy.
If f is a function on R that satisfies
f (z) = O
e−πz
and fˆ (ξ) = O
e−πξ
then f is a constant multiple of e−πx 2
. As a result, if f (x) = O
e−πAx
and f (ξ) = O
e−πBξ
, where AB > 1 and A, B > 0, then f is identically zero.
(a) If f is even, show that fˆ extends to an even entire function. Moreover,
if g(z) = fˆ
z
, then g satisfies
|g(x)| ≤ ce−πx^ and |g(z)| ≤ ceπR^ sin
(^2) (θ 2 ) ≤ ceπ|z|
when x ∈ R and z = Reiθ^ with R ≥ 0 and θ ∈ R.
