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Math 246A
Homework 3 Solutions
Due Thursday April 26
Note the solutions are often just an outline, but I have tried to address all of the key points and you should be able to fill in the details. If you have questions about them feel free to ask me; also if there are any errors in my solutions please let me know. Of course there are usually many other correct ways to solve each problem as well.
2.12 Let u be a real-valued function defined on the unit disk D. Suppose that u is twice continuously differentiable and harmonic, that is, 4 u(x, y) = 0 for all (x, y) ∈ D.
(a) Prove that there exists a holomorphic function f on the unit disc such that Re (f) = u. Also show that the imaginary part of f is uniquely defined up to an additive (real) constant. [Hint: From the previous chapter we would have f′(z) = 2∂u/∂z. Therefore 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.] Following the hint, let g(z) = 2∂u/∂z = ux − iuy. The Cauchy-Riemann equations follow easily from u being harmonic, and the second partials are continuous by hypothesis, so g is holomorphic. There exists a primitive F with F ′^ = g by Theorem 2.1. Let F (z) = F (x, y) = U (x, y) + iV (x, y) where z = x + iy. Then F ′(z) = Ux − iUy = ux − iuy , and we must have Ux = ux and Uy = uy. Integrating both sides of Ux = ux with respect to x gives U = u + h(y), where h is a function of y alone. Now differentiating with respect to y we get Uy = uy + h′(y), which implies h′(y) = 0 and thus h(y) = C ∈ R, a constant. Now let f(z) = (U (x, y) − C) + iV (x, y) which is seen to be holomorphic with Re f = u. Now say that f = u + iv and g = u + iw, where f, g are holomorphic. Then ux = vy = wy, and −uy = vx = wx, and an argument as above shows that v = w + C for some constant C. (b) Deduce from this result, and from Exercise 11 (problem 9 of the last assignment), the Poisson integral representation formula from the Cauchy integral formula: If u is harmonic in the unit disc 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 (γ) is the Poisson kernel for the unit disc given by
Pr (γ) = 1 − r^2 1 − 2 r cos γ + r^2
Let f = u + iv be defined as in part (a). Then by problem 9 of the last assignment we have
f(z) =
2 π
∫ (^2) π
0
f(Reiϕ)Re
Reiϕ−θ^ + r Reiϕ−θ^ − r
dϕ
where 0 ≤ r < R < 1. Now take real parts and let R → 1 to get the result.
2.13 Suppose that f is an analytic function defined everywhere in C and such that for each z 0 ∈ C at least one coefficient in the expansion f(z) =
n=0 cn(z^ −^ z^0 ) n (^) is equal to 0. Prove that f is a polynomial.
[Hint: Use that fact that cnn! = f(n)(z 0 ) and use a countability argument.] Let En =
z ∈ C : f(n)(z) = 0
. By hypothesis ∪∞ n=0En = C, so by countability there exists some n such En is uncountable. Now any uncountable subset S ⊂ C must contain a point that is a limit point of a sequence of distinct points in S. Otherwise for every z ∈ S there would exist some z such that Dz (z) ∩ S = {z}. However every open disk contains a point with purely rational coordinates, since those points are dense in C. Thus |S| would have to be countable, a contradiction. For a different proof, let Fn =
z ∈ D¯ : f(n)(z) = 0
. Again by countability there exists some n such that Fn is uncountable. Now Fn is a closed subset of D¯ and thus compact, and so any sequence of points (make them distinct) in Fn has a convergent subsequence. Now by Theorem 4.8 we have f(n)^ (z) = 0 for all z ∈ C. Integrating n times shows that f(z) is a polynomial.
2.14 Suppose that f is holomorphic in an open set containing the closed unit disc, except for a pole at z 0 on the unit circle. Show that if
n=0 anz
n (^) denotes the power series expansion of f in the open unit disc, then limn→∞ (^) aann+1 = z 0.
It suffices to consider the case z 0 = 1 because if z 0 = eiα^ then g(z) = f(eiαz) is also holomorphic in an open set containing D¯ except for a pole at 1 , and if g(z) =
n=0 bnz
n (^) then bn = einαan limn→∞ (^) bnbn+1 = 1 implies limn→∞ (^) aann+1 = eiα.
If f has a single pole at 1 of order k and no other poles in the open set Ω containing D¯, (z − 1)k^ f(z) is holomorphic in Ω. This means that (z − 1)kf(z) can be expanded in a power series that converges in some open disk D centered at 0 of radius > 1 , which in turn means that we can write f(z) =
n=−k bn(z^ −^ 1)
n (^) in some disk D′ (^) ⊂ D centered at 1. Consider the principal part of f, that is g(z) =
n=−k bn(z^ −^ 1)
n. Now expand b− 1 1 −z =^
n=0 a−^1 z
n, and expand b−m (1−z)m^ similarly. The result will be of the form
n=0 p(n)z n (^) where p(n) is a polynomial of degree k − 1. Now f(z) − g(z) =
n=0(an^ −^ p(n))z n, and (f − g)(z) converges at z = 1, and so limn→∞ an − p(n) = 0. For some N we must have p(n) 6 = 0 for all n ≥ N and so an 6 = 0 for all n sufficiently large. Also limn→∞ (^) aann+1 = limn→∞ (^) p(pn(n+1)) = 1. The previous solution was by Prof. Duke; here is another solution that he presented in class. Let f(z) =
n=0 anz
n (^) in D. By Cauchy’s integral formulas, an = 1 2 πi
C
f (z) zn+1^ dz^ where^ C^ is a circle centered at 0 of radius less than 1. By the residue theorem, this means an is the sum of the residues of f (z) zn+1^ inside C. If D is a circle of radius greater than 1 , then (^21) πi
D
f (z) zn+1^ dz^ is the sum of the residues of^ h(z) =^
f (z) zn+ inside D, which is just an plus the residue at 1. We thus have an = (^21) πi
D h(z)^ dz^ −^ res^1 h(z). Since the radius of D is greater than 1 ,
2 πi
D h(z)^ dz
∣ (^) → 0 as n → ∞ by a simple ML bound. Therefore an → − res 1 h(z) as n → ∞. By a calculation it can be shown that res 1 h(z) = p(n), where p is a polynomial of degree k − 1. The result follows as before.
2.15 Suppose f is a non-vanishing continuous function on D that is holomorphic on D. Prove that is |f(z)| = 1 whenever |z| = 1, then f is constant.
Following the hint define F (z) =
f(z) for |z| ≤ 1 1 /f(1/z) for |z| ≥ 1
. This definition makes sense, since f 6 = 0
and if z ∈ ∂D this implies 1 /z = z and |f(z)| = 1 implies 1 /f(z) = f(z) so 1 /f(1/z) = 1/f(z) = f(z) Then F is clearly holomorphic on D and the definition on D extends continuously to the boundary. If |z| > 1 1/z ∈ D so if f(z) =
n=0 anz
n (^) on D then f(1/z) = ∑∞ n=0 an(1/z)
n (^) a Laurent series which converges absolutely and is thus holomorphic for |z| > 1. Since f is nonvanishing this implies F (z) is differentiable for |z| > 1 and this extends to the boundary continuously. Now prove a version of the symmetry principle to show that F is also holomorphic on ∂D. Use a Morera’s Theorem following the book using Problem 3 with toy contours that are triangles with one of the edges an arc instead of a straight line. Thus F is entire. Since f : D → C and D is compact this implies Im (f) is compact and therefore bounded, similarly for Im (1/f). Thus F is also a bounded function and by Liouville’s theorem F is constant which implies f is constant.
Problem 2.3 (a) Suppose that f is continuous on C, and
C f(z)dz^ = 0 for every circle^ C. Prove^ f^ is holomorphic. Following the hint if ϕ(z) is a smooth bounded function with
C ϕ(z)dxdy^ = 1^ let ϕ(z) = −^2 ϕ(z/) so ϕ(z) is an approximate identity. Define f(z) =
C f(z^ −^ w)ϕ(w)dxdy, that is, the convolution of f and ϕ. Then by Folland Prop. 8.10 f is smooth (actually smooth on compact subsets which implies smooth everywhere). By Folland Thm. 8.14 f → f uniformly on compact subset as → 0 and
C f(z)dz^ =^
C
C f(z^ −^ w)ϕ(w)dxdy dz
C ϕ(w)^
C f(z^ −^ w)dz dxdy. By a change of variables^
C f(z^ −^ w)dz^ =^
C−w f(z)dz^ = 0. Where C − w is the translate of C by w which is still a circle. So
C f^ (z)dz^ = 0. Thus if f = u + iv the partial derivatives all exist and are continuous and 0 =
C fdz^ =^
C (u^ +^ iv)dx^ + (−v^ +^ iu)dy^ =^
Interior of C
∂−v ∂x +^
∂iu ∂x −^
∂u ∂y −^
∂iv ∫^ ∂y^ dxdy^ = Interior of C −(^
∂u ∂y +^
∂v ∂x ) +^ i(^
∂u ∂x −^
∂v ∂y )dxdy.^ Since this is true for any circle (any translate or dilate) this implies ∂u ∂y = − ∂v ∂x and ∂u ∂x = ∂v ∂y the C-R equations for f so f is holomorphic and since it converges uniformly on compact subsets to f, f is also holomorphic.
