Harnack's inequality and principle

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Let DD be a domain (connected open set) in C\mathbb {C} and let {un}\left\{ u_n\right\} be a sequence of harmonic functions un:D→(0,+∞)u_n : D \to (0,+\infty ). Show that if un(z0)→0u_n(z_0) \to 0 for some z0∈Dz_0 \in D, then un→0u_n \to 0 uniformly on compact subsets of DD.

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The harmonic functions

u0(x,y),u1(x,y),u2(x,y),…,un(x,y),… u_{0}(x,y), u_{1}(x,y), u_{2}(x,y), \ldots , u_{n}(x,y), \ldots

are assumed to be regular and positive in a certain open region R\mathfrak {R} of the x,yx,y-plane. If the infinite series

u0(x,y)+u1(x,y)+u2(x,y)+⋯+un(x,y)+⋯ u_{0}(x,y) + u_{1}(x,y) + u_{2}(x,y) + \cdots + u_{n}(x,y) + \cdots

converges at a single point of R\mathfrak {R} it converges everywhere in R\mathfrak {R}; in fact, it converges uniformly in any closed subdomain of R\mathfrak {R}.

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We suppose that the functions of the sequence f0(z),f1(z),f2(z),…,fn(z),…f_{0}(z),f_{1}(z),f_{2}(z),\ldots ,f_{n}(z),\ldots are analytic in the open region R\mathfrak {R} and that the sequence of their real parts converges uniformly in every closed subdomain of R\mathfrak {R}. Then the sequence of their imaginary parts either diverges at all points or it converges uniformly in any closed subdomain of R\mathfrak {R}.

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(Harnack's Inequalities) Suppose Ω\Omega is a region, KK is a compact subset of Ω,z0∈Ω\Omega , z_{0} \in \Omega. Prove that there exist positive numbers α\alpha and β\beta (depending on z0,Kz_{0}, K, and Ω\Omega) such that

αu(z0)≤u(z)≤βu(z0) \alpha u\left(z_{0}\right) \leq u(z) \leq \beta u\left(z_{0}\right)

for every positive harmonic function uu in Ω\Omega and for all z∈Kz \in K.

If {un}\left\{ u_{n}\right\} is a sequence of positive harmonic functions in Ω\Omega and if un(z0)→0u_{n}\left(z_{0}\right) \rightarrow 0, describe the behavior of {un}\left\{ u_{n}\right\} in the rest of Ω\Omega. Do the same if un(z0)→∞u_{n}\left(z_{0}\right) \rightarrow \infty. Show that the assumed positivity of {un}\left\{ u_{n}\right\} is essential for these results.

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Suppose uu is a positive harmonic function in UU and u(0)=1u(0) = 1. How large can u(12)u\left(\frac{1}{2}\right) be? How small? Get the best possible bounds.

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Problem 2.14.26 from Rudin W. "Real and Complex Analysis" [Rud87]More problems in Chapter 2.14 "Conformal Mapping"

Iteration of Step 2 in the proof of Theorem 14.8 leads to a proof (due to Koebe) of the Riemann mapping theorem which is constructive in the sense that it makes no appeal to the theory of normal families and so does not depend on the existence of some unspecified subsequence. For the final step of the proof it is convenient to assume that Ω\Omega has property (h)(h) of Theorem 13.11. Then any region conformally equivalent to Ω\Omega will satisfy (h)(h). Recall also that (h)(h) implies (j)(j), trivially.

By Step 1 in Theorem 14.8 we may assume, without loss of generality, that 0∈Ω,Ω⊂U0 \in \Omega , \Omega \subset U, and Ω≠U\Omega \neq U. Put Ω=Ω0\Omega = \Omega_{0}. The proof consists in the construction of regions Ω1,Ω2,Ω3,…\Omega_{1}, \Omega_{2}, \Omega_{3}, \ldots and of functions f1,f2,f3,…f_{1}, f_{2}, f_{3}, \ldots, so that fn(Ωn−1)=Ωnf_{n}\left(\Omega_{n-1}\right) = \Omega_{n} and so that the functions fn∘fn−1∘⋯∘f2∘f1f_{n} \circ f_{n-1} \circ \cdots \circ f_{2} \circ f_{1} converge to a conformal mapping of Ω\Omega onto UU.

Complete the details in the following outline.

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(a)

Suppose Ωn−1\Omega_{n-1} is constructed, let rnr_{n} be the largest number such that D(0;rn)⊂Ωn−1D\left(0 ; r_{n}\right) \subset \Omega_{n-1}, let αn\alpha_{n} be a boundary point of Ωn−1\Omega_{n-1} with ∣αn∣=rn\left|\alpha_{n}\right| = r_{n}, choose βn\beta_{n} so that βn2=−αn\beta_{n}^{2} = -\alpha_{n}, and put

Fn=φ−αn∘S∘φ−βn. F_{n} = \varphi _{-\alpha _{n}} \circ S \circ \varphi _{-\beta _{n}}.

(The notation is as in the proof of Theorem 14.8). Show that FnF_{n} has a holomorphic inverse GnG_{n} in Ωn−1\Omega_{n-1}, and put fn=λnGnf_{n} = \lambda_{n} G_{n}, where λn=∣c∣/c\lambda_{n} = \left|c\right| / c and c=Gn′(0)c = G_{n}^{\prime }(0). (This fnf_{n} is the Koebe mapping associated with Ωn−1\Omega_{n-1}. Note that fnf_{n} is an elementary function. It involves only two linear fractional transformations and a square root).

(b)

Compute that fn′(0)=(1+rn)/2rn>1f_{n}^{\prime }(0) = \left(1+r_{n}\right) / 2 \sqrt{r_{n}} > 1.

(c)

Put ψ0(z)=z\psi_{0}(z) = z and ψn(z)=fn(ψn−1(z))\psi_{n}(z) = f_{n}\left(\psi_{n-1}(z)\right). Show that ψn\psi_{n} is a one-to-one mapping of Ω\Omega onto a region Ωn⊂U\Omega_{n} \subset U, that {ψn′(0)}\left\{ \psi_{n}^{\prime }(0)\right\} is bounded, that

ψn′(0)=∏k=1n1+rk2rk \psi _{n}^{\prime }(0) = \prod _{k = 1}^{n} \frac{1+r_{k}}{2 \sqrt{r_{k}}}

and that therefore rn→1r_{n} \rightarrow 1 as n→∞n \rightarrow \infty.

(d)

Write ψn(z)=zhn(z)\psi_{n}(z) = z h_{n}(z), for z∈Ωz \in \Omega, show that ∣hn∣≤∣hn+1∣\left|h_{n}\right| \leq \left|h_{n+1}\right|, apply Harnack's theorem and Exercise 8 of Chap. 11 to {log⁡∣hn∣}\left\{ \log \left|h_{n}\right|\right\} to prove that {ψn}\left\{ \psi_{n}\right\} converges uniformly on compact subsets of Ω\Omega, and show that lim⁡ψn\lim \psi_{n} is a one-to-one mapping of Ω\Omega onto UU.

Let the functions un(x,y),n=1,2,…u_{n}(x, y), n = 1,2, \ldots, be harmonic in a domain DD and satisfy the conditions

un+1(x,y)⩾un(x,y)((x,y)∈D,n=1,2,…) u_{n+1}(x, y) \geqslant u_{n}(x, y) \quad ((x, y) \in D, n = 1,2, \ldots )

Prove that the sequence {un(x,y)}\left\{ u_{n}(x, y)\right\} converges uniformly on every compact subset of the domain DD (possibly to +∞+\infty).

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References

[CJP+11] Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. Problems and Solutions in Mathematics. 2nd ed., World Scientific, 2011.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.