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Solution of Homework Assigned on February 12, 2009 due February 17, 2009 (numbering of problems continued from the last assignment with the same due date)
Problem 8 (from Stein & Shakarchi, p.65, #6). Let Ω be an open subset of C and let T ⊂ Ω be a triangle whose interior is also contained in Ω. Suppose that f is a function holomorphic in Ω except possibly at a point w inside T. Prove that if f is bounded near w, then ∫
T
f (z)dz = 0.
Use this and Morera’s theorem to show that a function which is holomorphic in a deleted neighborhood of a point P 0 ∈ C (that is, a neighborhood of P 0 minus the point P 0 itself) and is uniformly bounded in the deleted neighbor- hood can be extended to a holomorphic function on the whole neighborhood.
Solution of Problem 8. Let Dε(w) denote the open disk of radius ε > 0 centered at w. There exists ε 0 > 0 such that the topological closure of Dε 0 (w) in C is inside Ω. For any triangle T ⊂ Ω, let Tε be the contour which is the union of T − Dε(w) and the part of the boundary of Dε(w) that is inside the solid triangle enclosed by T. Then by Cauchy’s theorem ∫
Tε
f (z)dz = 0 for 0 < ε < ε 0.
Since f is uniformly bounded in a deleted neighborhood of w, when we let ε → 0, we conclude that ∫
T
f (z)dz = lim ε→ 0
Tε
f (z)dz = 0.
First we make the additional assumption that f is continuous on Ω to con- clude that f can be made holomorphic on Ω including at the point w.
Now we look at the general case where f may not be continuous. Without loss of generality we can assume that Ω is a bounded open disk in C. Since ∫
T
f (z)dz = 0
for any triangle T inside Ω, it follows that we can define a function F on Ω by defining its value at z ∈ Ω to be the integration of f (z) from w to z through the broken line which goes horizontally and then vertically so that F ′(z) = f (z) for z ∈ Ω − {w}. Since F is continuous on Ω, by the above argument F can be made to be holomorphic on Ω. Since F ′(z) = f (z) for z ∈ Ω − {w}, it follows that f (z) can be made holomorphic on all of Ω.
Problem 9 (from Stein & Shakarchi, p.65, #8). If f is a holomorphic function on the strip − 1 < y < 1, x ∈ R with
|f (z)| ≤ A (1 + |z|)η^ , η a fixed real number
for all z in that strip, show that for each integer n ≥ 0 there exists An ≥ 0 such that (^) ∣ ∣f (n)(x)
∣ (^) ≤ An (1 + |x|)η^ , for all x ∈ R.
Hint: Use the Cauchy inequalities.
Solution of Problem 9. For x ∈ R, by applying Cauchy’s inequalities to f (z) on the open disk Dr(x) centered at x with radius 0 < r < 1, we have
∣ ∣f (n)(x)
∣ (^) ≤ n! rn^
sup |z−x|=r
|f (z)| ≤
n! rn^
A sup |z−x|=r
(1 + |z|)η^ ≤
n! rn^
A (2 + |x|)η^.
Letting r → 1, we get ∣ ∣f (n)(x)
∣ (^) ≤ n!A (2 + |x|)η^.
We need simply set
An = n! A sup x∈R
(2 + |x|)η (1 + |x|)η^
Problem 10 (from Stein & Shakarchi, p.66, #9). Let Ω be a bounded open subset of C, and ϕ : Ω → Ω a holomorphic function. Prove that if there exists a point z 0 ∈ Ω such that
ϕ (z 0 ) = z 0 and ϕ′^ (z 0 ) = 1
then ϕ is linear.
Hint: Why can one assume that z 0 = 0? Write ϕ(z) = z + anzn^ + O (zn+1) near 0, and prove that if ϕk = ϕ ◦ · · · ◦ ϕ (where ϕ appears k times), then ϕk(z) = z + kanzn^ + O (zn+1). Apply the Cauchy inqualities and let k → ∞ to conclude the proof. Here we use the standard O notation, where f (z) = O (g(z)) as z → 0 means that |f (z)| ≤ C |g(z)| for some constant C as z → 0. .
