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Solution of Homework Assigned on February 10, 2009 due February 17, 2009 (numbering of problems continued from the last assignment with the same due date)
Problem 5 (from Stein & Shakarchi, p.65, #7). Suppose f : D → C is a holomorphic function on the open unit disk
D =
z ∈ C
∣ |z| < 1
Show that the diameter d defined by
sup z,w∈D
|f (z) − f (w)|
of the image f (D) of D under f satisfies
2 |f ′(0)| ≤ d.
Moreover, it can be shown that equality holds precisely when f is a polyno- mial of exact degree 1, f (z) = a 0 + a 1 z with a 1 6 = 0.
Hint: Use the identity
2 f ′(0) =
2 πi
|ζ|=r
f (ζ) − f (−ζ) ζ^2 dζ
for 0 < r < 1.
Solution of Problem 5. Let
g(z) =
f (z)−f (−z) z for^ z^ ∈^ D^ − {^0 }
2 f ′(0) for z = 0.
Then g(z) is holomorphic on D, because g(z) can be written as
g(z) = f (z) − f (0) z
f (−z) − f (0) −z
for z ∈ D − { 0 }. By applying Cauchy’s integral formula to g(z) on { |z| < r } for 0 < r < 1, we obtain
2 f ′(0) =
2 πi
|ζ|=r
f (ζ) − f (−ζ) ζ^2
dζ
for 0 < r < 1, from which we get the inequality
2 |f ′(0)| ≤
2 π
sup |z|=r
∣∣^ f^ (z)^ −^ f^ (−z) z^2
2 π r ≤ d r
By letting r → 1, we get the desired inequality
2 |f ′(0)| ≤ d.
We now consider the question when the equality
2 |f ′(0)| = d
holds. We rewrite the Cauchy integral formula
2 f ′(0) =
2 πi
|ζ|=r
f (ζ) − f (−ζ) ζ^2 dζ
by using the parametrization z = reiθ^ for 0 ≤ θ ≤ 2 π and get
2 f ′(0) =
2 π
∫ (^2) π
θ=
f (reiθ) − f (−reiθ)
) (^) e−iθ r dθ
for 0 < r < 1.
We now consider the question when the equality 2 |f ′(0)| = d holds. Let w = f (z). After replacing w by αw + β for some appropriate α, β ∈ C with α 6 = 0, we can assume without loss of generality that d = 2 and the image of w = f (z) is contained in D = { |w| < 1 } in the w-plane. After this coordinate change the identity 2 |f ′(0)| = d becomes |f ′(0)| = 1. Let a = f (0).
We differentiate between the two cases a = 0 and a 6 = 0. Let us first assume that a = 0. Let h(z) = f^ ( zz ) for z ∈ D − { 0 } and h(0) = f ′(0). Then h(z) is holomorphic on D and
sup |z|=r
|h(z)| ≤
r
for 0 < r < 1.
Since |h(0)| = 1, by applying the maximum principle to h(z) and the point z = 0, we conclude that h(z) is constant and we are done.
(b) Show that Re
Reiγ^ + r Reiγ^ − r
R^2 − r^2 R^2 − 2 Rr cos γ + r^2
Hint: For the first part, note that if w = R 2 ¯z , then the integral^
f (ζ) ζ−w around the circle of radius R centered at the origin is zero. Use this, together with the usual Cauchy formula, to deduce the desired identity.
Solution of Problem 6. (a) Cauchy’s integral formula yields
1 2 πi
|ζ|=R
f (ζ) ζ − z dζ = f (z),
1 2 πi
|ζ|=R
f (ζ) ζ − Rz ¯^2
dζ = 0.
Subtracting the second equation from the first and using the parametrization ζ = Reiθ, we obtain
f (z) =
2 πi
|ζ|=R
f (ζ)
ζ − z
ζ − Rz ¯^2
dζ
2 π
∫ (^2) π
θ=
f
Reiθ
Reiθ^ − z
Reiθ^ − Rz ¯^2
Reiθdθ
2 π
∫ (^2) π
θ=
f
Reiθ
) (^ Reiθ Reiθ^ − z
z¯ Re−iθ^ − z¯
dθ
2 π
∫ (^2) π
θ=
f
Reiθ
Reiθ Reiθ^ − z
z¯ Re−iθ^ − ¯z
dθ
2 π
∫ (^2) π
θ=
f
Reiθ
Reiθ^ + z Reiθ^ − z
Re−iθ^ + ¯z Re−iθ^ − ¯z
dθ
2 π
∫ (^2) π
θ=
f
Reiθ
Re
Reiθ^ + z Reiθ^ − z
dθ
(b) By multiplying both the numerator and the denominator by the complex conjugate of the denominator, we get
Reiγ^ + r Reiγ^ − r
(Reiγ^ + r) (Re−iγ^ − r) (Reiγ^ − r) (Re−iγ^ − r)
R^2 − 2 irR sin γ − r^2 R^2 − 2 Rr cos γ + r^2
which yields
Re
Reiγ^ + r Reiγ^ − r
R^2 − r^2 R^2 − 2 Rr cos γ + r^2
Problem 7 (from Stein & Shakarchi, pp.66-67, #12). Let u be a real-valued function defined on the unit disk D. Suppose that u is twice continuously differentiable and harmonic, that is
∆u = ∂^2 u ∂x^2
∂^2 u ∂y^2
for all (x, y) ∈ D.
(a) Prove that there exists a holomorphic function f on the unit disk D such that Re f = u. Also show that the imaginary part of f is uniquely defined up to an additive real constant.
Hint: For a holomorphic function f = u + iv, use the formula f ′(z) = 2 ∂u∂z from f ′(z) = ∂f∂x = ∂u∂x + i ∂v∂x and the Cauchy-Riemann equations, where (^) ∂z∂ = (^12)
∂ ∂x −^ i^
∂ ∂y
so that 2∂u∂z = ∂u∂x − i∂u∂y. Let g(z) = 2∂u∂z and prove that g is holomorphic. Why can one find F with F ′^ = g? Prove that Re (F ) differs from u by a real constant. (b) Deduce from this result, and from Problem 6 the following Poisson integral representation formula from the Cauchy integral formula. If u is harmonic in the unit disk D and continuous on its closure, then if z = reiθ^ one has
u(z) =
2 π
∫ (^2) π
0
Pr(θ − ϕ)u(ϕ)dϕ,
where Pr is the Poisson kernel for the unit disk given by
Pr(γ) = 1 − r^2 1 − 2 r cos γ + r^2
