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Math 113 (Spring 2009) Yum-Tong Siu 1
Homework Assigned on March 19, 2009 due March 31, 2009 (numbering of problems continued from the last assignment with the same due date)
Problem 5 (modified from Stein & Shakarchi, p.131, #1). Suppose a > 0
and fˆ (ξ) = O
e−^
2 πap p |ξ|p^
as |ξ| → ∞, for some p > 1. Let q > 1 such that 1 p +^
1 q = 1. Then^ f^ is holomorphic for all of^ z^ and for^ b >^
2 π qaq^ there exists some Ab > 0 such that the growth condition
|f (z)| ≤ Abeb|z|
q
is satisfied for z ∈ C.
Hint: Use the inequality αβ ≤ α p p +^
βq q for^ α >^ 0 and^ β >^ 0 and^
1 p +^
1 q = 1, which follows from the geometric means no greater than the arithmetic means of n positive numbers.
Remark. Note that on the one hand, when p → ∞ then q → 1, and this limiting case can be interpreted as part of Theorem 3.3 on p.122 of Stein & Shakarchi with M = 1. On the other hand, when p → 1 then q → ∞, and this limiting case in a sense brings us back to Theorem 2.1 on p.114 of Stein & Shakarchi.
Problem 6 (from Stein & Shakarchi, p.132, #2). The problem is to solve the differential equation
an
dn dtn^
u(t) + an− 1
dn−^1 dtn−^1
u(t) + · · · + a 0 u(t) = f (t),
where a 0 , a 1 , · · · , an are complex constants, and f is a given function. Here we suppose that f has bounded support and is smooth (say of class C^2 ).
(a) Let
fˆ (z) =
−∞
f (t)e−^2 πiztdt.
Observe that fˆ is an entire function, and using integration by parts show that (^) ∣ ∣ ∣ fˆ (x + iy)
A
1 + x^2
if |y| ≤ a for any fixed a > 0.
Math 113 (Spring 2009) Yum-Tong Siu 2
(b) Write P (z) = an(2πiz)n^ + an− 1 (2πiz)n−^1 + · · · + a 0.
Find a real number c so that P (z) does not vanish on the line
L =
z ∈ C
∣ z = x + ic, x ∈ R
(c) Set
u(t) =
L
e^2 πizt P (z)
fˆ (z)dz.
Check that (^) n ∑
j=
aj
d dt
)j u(t) =
L
e^2 πizt^ fˆ (z)dz
and (^) ∫
L
e^2 πizt^ fˆ (z)dz =
−∞
e^2 πixt^ fˆ (x)dx.
Conclude by the Fourier inversion theorem that
∑^ n
j=
aj
d dt
)j u(t) = f (t).
Note that the solution u depends on the choice of c.
Problem 7 (from Stein & Shakarchi, p.133, #3). In this problem, we inves- tigate the behavior of certain bounded holomorphic functions in an infinite strip. The particular result described here is sometimes called the three-lines lemma.
(a) Suppose F (z) is holomorphic and bounded in the strip 0 < Im(z) < 1 and continuous on its closure. If |f (z)| ≤ 1 on the boundary lines, then |F (z)| ≤ 1 throughout the strip.
(b) For the more general F , let supx∈R |F (x)| = M 0 and supx∈R |F (x + i)| = M 1. Then sup x∈R
|F (x + iy)| = M 01 −yM 1 y if 0 ≤ y ≤ 1.
(c) As a consequence, prove that log supx∈R |F (x + iy)| is a convex function of y when 0 ≤ y ≤ 1.
Hint: For part (a), apply the maximum modulus principle to Fε(z) = F (z)e−εz 2 . For part (b), consider M 0 z −^1 M 1 − zF (z).
