Key to Complex Analysis Homework 1 Spring 2012 (Thanks to Da Zheng for providing the tex-file)

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1 Key to Complex Analysis Homework 1 Spring 212 (Thanks to Da Zheng for providing the tex-file) February 9, Prove: If u is a complex-valued harmonic function, then the real and the imaginary parts of u are harmonic. Conclude that a complex-valued function u is harmonic if and only if ū is harmonic. Proof. First, we set u(x, y) = f(x, y) + ig(x, y),where f and g are both realvalued, then by the condition that u is a harmonic, it is easily seen that both f and g are in C 2. Also, taking the Laplacian of u and we obtain u = f + i g = which clearly implies that f = g =, so f and g are both harmonic. Next, if u is harmonic, then f and g are both harmonic, and it is easily seen that ū = f ig, their linear combination, is in C 2. Moreover, since ū = f i g, then f = g = implies that u =. So ū is also harmonic. Conversely, if ū is harmonic, then f and g are also harmonic, which directly means that u is harmonic. 19. Prove that there is no nonconstant harmonic functions u : C R such that u(z) for all z C. Proof. Since C is simply connected, we know that there is a harmonic function v : C R, such that f = u + iv is holomorphic on C. Thus, consider the function e f, which is also holomorphic on C, we have e f = e u+iv = e u 1 The last inequality holds, because u on C. 1

2 So, by the Liouville s theorem, e f is constant on C, which implies that u is constant. This means that there is no nonconstant function that satisfies the given condition in the problem. Remark: You can also use the Harnack inequality to prove it. 25. Compute a formula analogous to the Poisson integral formula, for the region U = {z : Iz > } (the upper half plane), by mapping U conformally to the unit disc. The formula is as follows: Suppose u is harmonic on U = {z : Iz > }, continuous and bounded on Ū = {z : Iz }, then, for z = x + yi with y >, we have u(z) = 1 + y π (x t) 2 + y 2 u(t)dt Proof. First, we set ϕ(z) = z i z+i, the Cayley transform, then ϕ maps U conformally onto the unit disc, and the boundary R to {z : z = 1} \ {z = 1}. Second, consider the mapping u ϕ 1, which is holomorphic on D(, 1), and continuous on D(, 1) since u is bounded, then we have the Poisson integral formula on the unit disc: if a D(, 1), then u ϕ 1 (a) = 1 { ζ =1}\{ζ=1} 1 a 2 ζ a 2 u ϕ 1 (ζ) dζ iζ Next, make substitution of variables by setting ζ = ϕ(t), where t R, and a = ϕ(z), we obtain u ϕ 1 (a) = u(z) = 1 1 ϕ(z) 2 (t)dt R ϕ(t) ϕ(z) 2 u(t)ϕ iϕ(t) = 1 1 z i z+i 2 2dt R t i t+i z i 2 u(t) t z+i = 1 [(y + 1) 2 (y 1) 2 ](t 2 + 1) 2 2[(t x) 2 + y 2 ] t dt = 1 π This proves the desired formula. R + y (t x) 2 + y 2 u(t)dt Remark. Here, for the desired formula we have proved, since it involves both + and as the upper and lower limit, we actually need to use a limit process to define it (e.g. C.P.V is a nice way to define, as u is bounded). Also, as is ( ) 2

3 not included in the upper half plane, the image of it under Cayley s transform omits the boundary point z = 1, so we have to integrate on { ζ = 1} \ {ζ = 1} in equality (*), which is defined via a limit process corresponding to our final formula. But this (formula involving the limit) seems a little different from averaging formula, which is purely equality. However, the following lemma will fix this difference. Lemma. Suppose that u(z) is harmonic on D(, 1), and continuous on D(, 1) except at z = 1, then for z < 1, we have that u(z) = 1 ζ z 2 u(ζ)dζ iζ = 1 lim ɛ { ζ =1}\{ζ=1} ɛ ɛ e iθ z 2 u(eiθ )dθ Proof. Fix a z D(, 1), then since u is harmonic on D(, 1), then for all < r < 1, we have Then, consider u(z) = 1 e iθ z 2 u(eiθ )dθ ɛ u(z) 1 ɛ e iθ z 2 u(eiθ )dθ = 1 ɛ ɛ e iθ z 2 [u(reiθ ) u(e iθ )]dθ ɛ + e iθ z 2 u(reiθ )dθ + e iθ z 2 u(reiθ )dθ ɛ Next, because u is uniformly continuous on {z = re iθ : r 1, ɛ θ ɛ}, and u us bounded on D(, 1) \ {1}, by letting r 1, ɛ, we can make the right hand side of the above equality arbitrarily small, that is The lemma is proved. u(z) = 1 ɛ lim ɛ ɛ e iθ z 2 u(eiθ )dθ 27. Let P (z, ζ) be the Poisson kernel for the disc. If you write z = re iθ and ζ = e iφ, then you can relate the formula for the Poisson kernel that was given in the text to the new formula P (z, ζ) = 1 z ζ 2 = 1 ζ 2 z 2 z ζ 2 3

4 Do so. Now calculate z P (z, ζ) for z D(, 1) to see that P is harmonic in z. Proof. From the expression of the Poisson formula, it is obvious that P (z, ζ) = 1 z ζ 2 = 1 ζ 2 z 2 z ζ 2 So what left to prove is that P (z, ζ) is harmonic w.r.t z. Since z = 4 2 z z, we have z P (z, ζ) = 4 2 = 2 π = 2 π = 2 π = ( 1 ζ 2 z 2 ) z ζ 2 ( ) ζ ζ z z (z ζ)(z ζ) ( ζ ζ z + ( ) ζ + 2 ζ z π This shows that P (z, ζ) is harmonic w.r.t z. ζ ) ζ z 1 2 ( ) ζ 2 z z ζ z π 2 (1) 28. If h 1, h 2, are harmonic on U C and if {h j } converges uniformly on compact subsets of U, then prove that the limit function h is harmonic. Proof. For any P U, and any δ > such that D(P, δ) U, we have, by the fact that {h n } is harmonic, for each z D(P, δ), h n (z) = 1 (z P ) δe iθ 2 h n(p + δe iθ )dδ Then, as h n h uniformly on D(P, δ), we can choose n large enough such that h n (P + δe iθ ) h (P + δe iθ ) < ɛ, for all θ [, ], so we have h n(z) 1 (z P ) δe iθ 2 h (P + δe iθ )dδ = 1 (z P ) δe iθ 2 [h n(p + δe iθ ) h (P + δe iθ )]dδ 1 ɛ 1 = ɛ, (z P ) δe iθ 2 h n (P + δe iθ ) h (P + δe iθ ) dδ (z P ) δe iθ 2 dδ 4

5 Here, we also used the fact that Thus, we know that h n (z) 1 δ 2 z P 2 (z P ) δe iθ 2 dδ = 1. δ 2 z P 2 (z P ) δe iθ 2 h (P + δe iθ )dδ. However, on the other hand, we also know that h n (z) h (z); therefore, by the uniqueness of limit, this implies that h (z) = 1 (z P ) δe iθ 2 h (P + δe iθ )dδ Finally, since our z and δ is arbitrarily chosen, we conclude that h is harmonic on U. 41. Prove that if f is C 2 on an open set U and f is subharmonic, then f on U. Proof. First, since f is C 2, for a U, and a small neighborhood around it, say D(a, R), we can expand f as second order Taylor polynomial in complex variables z and z: f(z) =f(a) + f f (a)(z a) + z + 2 f (a) z a f (z a)2 z2 z (a)( z ā) f z 2 (a)( z ā)2 + O( z a 3 ) Here, h = O( z a 3 ) means h C z a 3 on D(a, R) for some C >. Thus, set z = a + re iθ and integrate the both sides of the above equation, we obtain f(a) 1 f(a + re iθ )dθ = f(a) + 2 f (a)r2 + O(r 3 ) = f(a) f(a)r2 + O(r 3 ). ( ) By letting r, we get f, and this completes the proof. 43. (Maximum Principle for Subharmonic functions) If f is subharmonic on U, and if there is a P U such that f(p ) f(z) for any z U, then f is constant. Proof. First, we set s = f(p ), and consider the set M = {z M : f(z) = s}. We shall prove that M is both open and closed. Clearly, M is non-empty. 5

6 Since f is continuous, M is closed. On the other hand, for any w M and D(w, r) U, we have the following inequalities, by the fact that f is subharmonic: s = f(w) 1 1 = s f(w + re iθ )dθ So we actually obtain that 1 f(w+re iθ )dθ = s, and since for any θ [, ], f(w + re iθ ) s, we know that f(w + re iθ ) = s for all θ [, ]. This means that M is open. In sum, M is non-empty, open, and closed, we conclude that M = U, and therefore f is constant on U. sdθ 44. If u is harmonic on U C, then u satisfies a minimum principle as well as a maximum principle. However, subharmonic functions do not satisfy a minimum principle. Illustrate this claim. Proof. To explain this claim, we shall construct a counter-example. Set r >, and consider the following function on the closed unit disc D(, 1): Then, it is easy to check that f(z) = log(r + z 2 ) f(z) = 4 2 ( log(r + z 2 ) ) r = (r + z 2 ) 2 > Thus, f is subharmonic; however, it violates the minimum principle, since min f is attained at the origin, an interior point. Remark: Another eimple example is u(z) = z Let U C be a connected open set. Let f : U R be subharmonic. Suppose further that V is open and F : V U is holomorphic. Prove that f F is subharmonic. What happens if F is only harmonic? Proof. For any P V and D(P, r) V, and any harmonic function h on D(P, r), we need to prove that if f F h on D(P, r), then f F h on D(P, r). 6

7 To show this, we first assume that F is one-to-one, thus it is biholomorphic from V to ran(f ). We need to show that for any D(P, r) V, and any harmonic function h on D(P, r) V, such that h f F on D(, 1), then h f F on D(, 1). Consider f F h, we can rewrite it as as F is biholomorphic. f F (z) h F 1 (F (z)) Then, notice that F ( D(P, r)) is also a Jordan curve in U, which is the boundary of F (V ), hence f(f (z)) h F 1 (F (z)) means that f h F 1 on F ( D(P, r)), and by the fact that h is subharmonic, we know that f h F 1 on F (V ). This is exactly what we desire, i.e., f F (z) h(z) in D(P, r), so we proved that f F is also subharmonic. Now, suppose that F is not one-to-one, if F is constant, then proof is trivial. If F is nonconstant, then there exists a point P such that F (P ) =. In this case, we know that there is an injective holomorphic function g on D(P, r) such that F = g k + F (P ), so what we need to prove is that f(z k + F (P )) is subharmonic at z =. This can be seen from the following: 1 f(ɛe ikθ + F (P ))dθ = 1 = 1 2kπ f(f (P )) f(ɛe iζ + F (P )) dζ k f(ɛe iζ + F (P ))dζ This means f(z k + F (P )) is subharmonic at z =, i.e., f F is subharmonic at P. For a point Q such that F (Q) is not zero, we know that there is a neighborhood around Q such that F is injective on it, and the proof is the same as the above case where F is one-to-one. Thus, in sum, we conclude that f F is also subharmoic, given F is holomorphic. However, if F is only harmonic, then the claim may not be true. To see this, consider the function f(z) = 2x 2 y 2. We know that it is subharmonic on C. Since it is a composition of a subharmonic function and a increasing convex function. Next, take F (x, y) = x + xyi, then we have that f F (x, y) = 2x 2 x 2 y 2, which is not subharmonic, since f F < at some points, and this explains why harmonic functions cannot transform subharmonic functions into subharmonic functions. 7

8 69. Let f : U R be a C 2 function on an open set in U C. (a). Recall that if f > at a point P, then f cannot have a local maximum at P. Use this observation to deduce that if f > everywhere on U, then f is subharmonic. (b). If f everywhere, then, for each ɛ >, (f + ɛ z 2 ) > everywhere. Use a limiting argument and part (a) to deduce that if f everywhere, then f is subharmonic. Proof. (a). Given that f > everywhere on U, then for any neighborhood D(P, r) U, and any harmonic function h on D(P, r), such that u h on D(P, r), we need to show that u h on D(P, r). We suppose that there is a Q D(P, r) that violates our desired result, that is, u(q) h(q) > ; however, this implies that u h attains its maximum inside D(P, r), and without loss of generality, we can assume that u h attains its maximum at Q. Next, from the knowledge in mathematical analysis, we know that if u h attains maximum at Q, then its Hessian matrix at Q cannot be positive definite (indeed it can only be negative semi-definite) at Q. So we have that 2 (u h) x 2 (Q), 2 (u h) y 2 (Q) ( ) But we already have the given condition that u >, from which we know that (u h) >. This contradicts the above result (*) derived by our assumption. Hence, due to this contradiction, u h cannot attain its maximum inside D(P, r), and this implies that u h inside D(P, r). So, by definition, we conclude that u is subharmonic on U. (b). Now we only have that f, so we consider the function f + ɛ z 2. It is easy to see that (f + ɛ z 2 ) > for every ɛ >, and therefore f + ɛ z 2 is subharmonic on U. Next, we notice that for each D(P, r) U, f + ɛ z 2 converges uniformly to f on it. Moreover, for each ɛ >, we have that f(p ) + ɛ P 2 1 Then, by the uniform convergence, f(p + re iθ ) + ɛ P + re iθ 2 dθ ( ) f(p ) + ɛ P 2 f(p ), as ɛ 1 f(p + re iθ ) + ɛ P + re iθ 2 dθ 1 f(p + re iθ )dθ, as ɛ This directly implies f(p ) 1 f(p + re iθ )dθ, by the inequality (*). So we know that f is subharmonic. 8

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