MATH 566 LECTURE NOTES 1: HARMONIC FUNCTIONS

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1 MATH 566 LECTURE NOTES 1: HARMONIC FUNCTIONS TSOGTGEREL GANTUMUR 1. The mean value property In this set of notes, we consider real-valued functions on two-dimensional domains, although it is straightforward to generalize the results to, e.g., vector-valued functions defined on n- dimensional domains. The symbol D r (z) denotes the open disk with radius r and centre z, and we set D r = D r (). Unless otherwise specified, in what follows will be an open subset of R 2. Definition 1. Let R 2 be open. Then a continuous function u C() is said to have the mean value property and written u MVP(), if it satisfies u(z) = 1 πr 2 u(w)dw, for all D r (z). It is easy to show (by differentiating and integrating with respect to r) that the above condition is equivalent to u(z) = 1 2π u(z + r cis θ)dθ, for all D r (z), where we have used the shorthand notation cis θ = (cos θ, sin θ) R 2. Let us also introduce the notation Har() = {u C 2 () : u = } for the set of harmonic functions in. Theorem 2. Har() MVP(). Proof. Let u Har(), and let D r (z) and ρ (, r). Then obviously the following integral must vanish: u = ν udl = ρ ρ u(z + ρ cis θ)dθ = ρ ρ u(z + ρ cis θ)dθ, D ρ(z) D ρ(z) where commuting the derivative ρ with the integration over θ is justified because the functions (ρ, θ) u(z + ρ cis θ) and (ρ, θ) ρ u(z + ρ cis θ) are both continuous. Now from the fundamental theorem of calculus we have ρ u(z + ρ cis θ)dθ = u(z + r cis θ)dθ 2πu(z) =. So viewed as a tool, the mean value property can be used to prove properties of harmonic functions. The following converse shows that the mean value property can also be used to prove harmonicity. Theorem 3. MVP() Har() and MVP() C (). Date: September 27, 21. 1

2 2 TSOGTGEREL GANTUMUR Proof. Let ϕ C (D 1) be a positive function with ϕ = 1, which is also radial, i.e., ϕ(z) = ψ( z ) for some ψ. The normalization of ϕ implies 2π 1 rψ(r)dr = 1. Set ϕ ε(z) = ε 2 ϕ(z/ε) for ε >, and assume that z and ε < dist(z, ). Then u(w)ϕ ε (w z)dw = u(z + w)ϕ ε (w)dw = 1 ε 2 u(z + w)ψ( w w <ε ε )dw = 1 ε 2 u(z + r cis θ)ψ( r 2πu(z) )rdrdθ = ε ε 2 ψ( r )rdr = u(z), ε r<ε which means that u = ϕ ε u, and so u C (). Now let D r (z), and compute u(w)dw = r r u(z + r cis θ)dθ = r r (2πu(z)) =. Since D r (z) is arbitrary, we conclude that u = in. r<ε 2. Simple consequences of the mean value property Recall that an open set is connected if it cannot be decomposed as the disjoint union of two open sets. Theorem 4 (Maximum principle). Let R 2 be open, and let u C 2 () be a subharmonic function in, i.e., u in. (a) If is connected and u(z) = sup u at some z, then u is constant. (b) If is bounded and u C( ), then we have sup u max u. Proof. For part (a), let M = u(z) = sup u. Then we can write the mean value property as [u(z) u(z + r cis θ)]dθ =, for all sufficiently small r >. The integrand is a nonnegative continuous function, so it must vanish, which implies that u(w) = M in a small disk w D r (z). In other words, the set Σ = {z : u(z) = M} is open. Since u is continuous, the set {z : u(z) < M} = \ Σ is also open. This means that either Σ = or Σ =, but Σ is not empty by hypothesis. For part (b), there is z with u(z) = sup u since is bounded and u is continuous on. If z then we are done; otherwise applying part (a) to the connected component of that contains z concludes the proof. Remark 5. In part (b) of the above theorem, the condition that be bounded cannot be simply removed, as can be seen from the example u(x, y) = e x cos y on (, ) ( π 2, π 2 ). In the following theorem we use complex notation for convenience. Theorem 6 (Schwarz reflection principle). Let C be an open set symmetric with respect to the real axis, and let + = {Iz > } be the part of it in the upper half plane. Moreover, assume that u Har(), and that u(z) as z + tends to any point on the real axis {Iz = }. Then u extends to be harmonic on, and the extension satisfies u( z) = u(z), z.

3 MATH 566 LECTURE NOTES 1: HARMONIC FUNCTIONS 3 Proof. Let = { z : z + }, and extend u to a function on by u(z) = for z {Iz = } and u(z) = u( z) for z. Then the mean value property is obviously satisfied at points z +. But for (the remaining) points z with Iz =, since u(z) = u( z), the integral of u over any small circle centred at z is zero, giving the mean value property for those points. For R 2 and a function u C(), we define the uniform norm u = sup u. We say that a sequence of functions {u k } C() converges uniformly on to u C(), if u k u. Recall that continuity and the Riemann integral are preserved under uniform limits. Theorem 7. If {u k } Har() converges uniformly on to u, then u Har(). Proof. We have u C(). Let D r (z). Then we have 2πu k (z) = u k (z + r cis θ)dθ, for all k. Since the left hand side converges to 2πu(z), and the integrand converges uniformly to θ u(z + r cis θ), the mean value property over D r (z) is satisfied for u, and since D r (z) is arbitrary we have u Har(). 3. Derivative bounds Noting that partial derivatives of harmonic functions are also harmonic, and by using the mean value property for the partial derivatives, we can bound the derivatives of harmonic functions by the size of the function itself. Recall that for ν = (ν 1, ν 2 ) with ν = 1, the directional derivative along ν is defined by ν u = ν 1 x u + ν 2 y u. Theorem 8. Let u Har() and D r (z). Then for any ν R 2 with ν = 1, we have If in addition u in D r (z), we have ν u(z) 2 r max u. ν u(z) 2 r u(z). Proof. Since u = and u is smooth, we have ν u = in, i.e., ν u Har(). Using the mean value property and the divergence theorem, we get ν u(z) = 1 πr 2 ν u(w)dw = 1 πr 2 u(z + r cis θ)(ν cis θ)rdθ. Now if u, since the inner product ν cis θ 1, we infer ν u(z) 1 πr u(z + r cis θ)dθ = 1 πr 2πu(z). On the other hand, for general u, we have x u(z) 1 πr max u 2π = 2 r max u. The following can be called the Liouville theorem for harmonic functions. Corollary 9. If u Har(R 2 ) is bounded above or bounded below, then u is constant.

4 4 TSOGTGEREL GANTUMUR Proof. If u a for some constant a R, then set v = u a, or if u b for some constant b R, then set v = b u. In either case, we have v and v Har(R 2 ). Now let z R 2 and apply the above theorem to v, inferring which gives ν v in any direction ν. Now we estimate higher derivatives. ν v(z) 2 v(z), for any r >, r Corollary 1. Let u Har() and D r (z). Then for j, k and n = j + k, we have n x u(z) 2 ρ max D ρ(z) j x k y u(z) 2n n n r n max u. Proof. We prove it only for k =, since the general case is completely analogous. All the derivatives x m u are harmonic, so with ρ = r/n, we have ( 2 x n 1 u... ρ which is what we need. 4. Analyticity ) n max u, D nρ(z) Let us recall the Taylor theorem for two variables, with a somewhat simplified hypothesis on the smoothness of the function involved. Theorem 11. Let f C n ( D r (z)), and h = (h 1, h 2 ) D r (). Then we have f(z + h) = j+k<n with the following estimate on the error term x j y k f(z) h j 1 j!k! hk 2 + R j,k (z) sup w j+k=n x j y k f(w) j!k!. R j,k (z)h j 1 hk 2, Definition 12. A function f C () is called (real) analytic at z if the following Taylor series converges in a neighbourhood of z. x j y k f(z) f(z + h) = h j 1 j!k! hk 2. j= k= Moreover, f is said to be (real) analytic in if it is analytic at each z. The set of analytic functions in is denoted by C ω (). Theorem 13. Har() C ω (). Proof. Let D r (z) and ρ (, r). Then we have x j y k u R j,k (z) sup j!k! 2 n n n j!k!(r ρ) n max u D ρ(z) Using the crude bounds ( n j) 2 n and n n e n n!, we get 2 n n n ( ) n 2 n n n = 4 n e n, j!k! j n!

5 MATH 566 LECTURE NOTES 1: HARMONIC FUNCTIONS 5 and with this in mind, for h ρ the Taylor remainder term can be estimated as ( ) R j,k (z)h j 4eρ n 1 hk 2 max u. r ρ We see that the Taylor series converges in D ρ (z) whenever ρ < r 4e+1. As a consequence of analyticity, we have the following rigidity that is typical in analytic setting. What is remarkable is that there is no condition on the size of D relative to ; we only need D nonempty and open. Theorem 14 (Identity theorem). Let u Har() with a connected open set, and let u in an open subset D. Then u in. Proof. Each Σ j,k = {z : x j y k u(z) = } is closed, so the intersection Σ = j,k Σ j,k is also closed. But Σ is also open, because z Σ implies that u in a small disk centred at z by a Taylor series argument. Since Σ is nonempty by hypothesis, we conclude that Σ =. The following corollary records the fact that a harmonic function is completely determined by its restriction to any open subset of its domain of definition. In other words, if it is at all possible to extend a harmonic function (defined on an open set) to a bigger domain, then there is only one way to do the extension. Corollary 15. Let u, v Har() with a connected open set, and let u v in an open subset D. Then u v in.