§7.2 Harnack’s Principle

7.2 Harnack’s Principle

Due to the perceived similarity between properties of holomorphic functions and those of harmonic functions, we now provide an analogous statement to Montel’s Theorem (Theorem 5.2.6 $Theorem 5.2.6 Theorem 5.2.6$ ) which is commonly known Harnack’s Principle, which will be useful for the proceeding sections in solving the Dirichlet Problem for more general domains.

First we provide a result that in many ways resembles Theorem 4.1.1 $Theorem 4.1.1 Theorem 4.1.1$ .

Theorem 7.2.1.

Let

{𝑢_{𝑛}}_{𝑛 \in ℕ}

be a sequence of real-valued harmonic functions on an open set

𝑈 \subseteq ℂ

. If

{𝑢_{𝑛}}

locally uniformly converges to

𝑢

𝑈

, then

𝑢

is harmonic on

𝑈

Proof.

For any closed disk $\overline{𝐷 (𝑎, 𝑟)} ⋐ 𝑈$ and any $𝜀 > 0$ , there exists $𝑁 \in ℕ$ such that $\forall 𝑛 > 𝑁$ , $\forall 𝑧 \in 𝜕 𝐷 (𝑎, 𝑟)$ ,

| 𝑢 (𝑧) - 𝑢_{𝑛} (𝑧) | < 𝜀,

implying that

| \frac{1}{2 π} \int_{0}^{2 π} (𝑢 - 𝑢_{𝑛}) (𝑎 + 𝑟 e^{i 𝜃}) d 𝜃 | \leq 𝜀,

which is equivalent to saying

lim_{𝑛 \to \infty} \frac{1}{2 π} \int_{0}^{2 π} 𝑢_{𝑛} (𝑎 + 𝑟 e^{i 𝜃}) d 𝜃 = \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝑎 + 𝑟 e^{i 𝜃}) d 𝜃

By the global mean-value property,

\frac{1}{2 π} \int_{0}^{2 π} 𝑢_{𝑛} (𝑎 + 𝑟 e^{i 𝜃}) d 𝜃 = 𝑢_{𝑛} (𝑎) ⟹ \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝑎 + 𝑟 e^{i 𝜃}) d 𝜃 = 𝑢 (𝑎) .

Thus,

𝑢

is harmonic as it satisfies the global mean-value property on all disks in

𝑈

. ⁠

□

Remark.

We note here that the statement can be considerably relaxed by replacing locally uniform convergence with locally uniform boundedness and pointwise convergence. In this case, the exchange between the limit and integral is directly justifiable through Lebesgue’s Dominated Convergence Theorem.

Theorem 7.2.2 (Harnack Inequality).

Let $𝑢 : 𝑈 \to ℝ$ be nonnegative and harmonic on some neighborhood $𝑈 \supset \overline{𝐷 (𝑝, 𝑅)}$ for $𝑟 \in ℝ_{> 0}$ , $𝑝 \in ℂ$ . Then for any $𝑧 \in 𝐷 (𝑝, 𝑅)$ ,

\frac{𝑅 - | 𝑧 - 𝑝 |}{𝑅 + | 𝑧 - 𝑝 |} 𝑢 (𝑝) \leq 𝑢 (𝑧) \leq \frac{𝑅 + | 𝑧 - 𝑝 |}{𝑅 - | 𝑧 - 𝑝 |} 𝑢 (𝑝) (7.2 . 1)

Proof.

First assume $𝑝 = 0$ .

By the Poisson Integral Formula ((3.6.3) $(3.6.3) (3.6.3)$ ), we have that

\begin{aligned} 𝑢 (𝑧) & = \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝑅 e^{i 𝑡}) \frac{𝑅^{2} - {| 𝑧 |}^{2}}{{| 𝑅 e^{i 𝑡} - 𝑧 |}^{2}} d 𝑡 \\ (reverse triangle inequality) & \leq \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝑅 e^{i 𝑡}) \frac{(𝑅 + | 𝑧 |) (𝑅 - | 𝑧 |)}{{| 𝑅 - | 𝑧 | |}^{2}} d 𝑡 \\ \leq \frac{𝑅 + | 𝑧 |}{𝑅 - | 𝑧 |} \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝑅 e^{i 𝑡}) d 𝑡 \\ = \frac{𝑅 + | 𝑧 |}{𝑅 - | 𝑧 |} 𝑢 (0) . (global mean-value property) \end{aligned}

Similarly,

\frac{𝑅^{2} - {| 𝑧 |}^{2}}{{| 𝑅 e^{i 𝑡} - 𝑧 |}^{2}} \geq \frac{(𝑅 + | 𝑧 |) (𝑅 - | 𝑧 |)}{{(𝑅 + | 𝑧 |)}^{2}} = \frac{𝑅 - | 𝑧 |}{𝑅 + | 𝑧 |}

by the triangle inequality, so therefore

𝑢 (𝑧) \geq \frac{𝑅 - | 𝑧 |}{𝑅 + | 𝑧 |} \frac{1}{2 π} \int_{0}^{2 π} 𝑢 (𝑅 e^{i 𝑡}) d 𝑡 = \frac{𝑅 - | 𝑧 |}{𝑅 + | 𝑧 |} 𝑢 (0)

by the global mean-value property once more.

Lastly, for nonzero $𝑝$ , a translation suffices to obtain (7.2.1) $(7.2.1) (7.2.1)$ from the $𝑝 = 0$ case. ⁠ $□$

Theorem 7.2.3 (Harnack’s Principle).

Let ${𝑢_{𝑛}}_{𝑛 \in ℕ}$ be a pointwise monotonically non-decreasing sequence of harmonic functions on an open region $𝑈 \subseteq ℂ$ . Then either $𝑢_{𝑛} \to \infty$ locally uniformly:

\forall 𝑀 > 0, \forall 𝐾 ⋐ 𝑈 compact, \exists 𝑁 \in ℕ : \forall 𝑛 > 𝑁, \forall 𝑧 \in 𝐾, 𝑢_{𝑛} (𝑧) > 𝑀,

or instead,

𝑢_{𝑛} ⇉ 𝑢

locally uniformly, where

𝑢

is harmonic on

𝑈

Proof.

First assume that there exists a point $𝑝 \in 𝑈$ such that $𝑢_{𝑛} (𝑝) \to \infty$ . Then $\exists 𝑛^{'} \in ℕ$ such that $𝑢_{𝑛^{'}} (𝑝) > 0$ (and by the increasing nature of the sequence, for any $𝑛 > 𝑛^{'}$ , $𝑢_{𝑛} (𝑝) > 0$ is also necessarily true).

By the continuity of each $𝑢_{𝑛} \geq 𝑢_{𝑛^{'}}$ , there exists some closed disk $\overline{𝐷 (𝑝, 𝑟)}$ in $𝑈$ on which $𝑢_{𝑛}$ is positive. Then for any $𝑧$ within the smaller disk $𝐷 (𝑝, \frac{𝑟}{2})$ and any $𝑛 > 𝑛^{'}$ , we have that

\frac{𝑟 - \frac{𝑟}{2}}{𝑟 + \frac{𝑟}{2}} 𝑢_{𝑛} (𝑝) \leq 𝑢_{𝑛} (𝑧) ⟹ 𝑢_{𝑛} (𝑧) \geq \frac{1}{3} 𝑢_{𝑛 (𝑝)} \to \infty

by Theorem 7.2.2 $Theorem 7.2.2 Theorem 7.2.2$ . Therefore, $𝑢_{𝑛} ⇉ \infty$ uniformly on a neighborhood of $𝑝$ .

Contrarily, if the sequence converges to a finite limit at a point $𝑝^{'} \in 𝑈$ (so $𝑢_{𝑛} (𝑝^{'}) \to 𝐿$ ), let $𝑟^{'} > 0$ be chosen such that $\overline{𝐷 (𝑝^{'}, 𝑟^{'})} ⋐ 𝑈$ . Then for any $𝑧 \in 𝐷 (𝑝^{'}, \frac{𝑟^{'}}{2})$

\begin{aligned} lim_{𝑁 \to \infty} sup_{𝑛 > 𝑚 > 𝑁} | 𝑢_{𝑛} (𝑧) - 𝑢_{𝑚} (𝑧) | & = lim_{𝑁 \to \infty} sup_{𝑛 > 𝑚 > 𝑁} 𝑢_{𝑛} (𝑧) - 𝑢_{𝑚} (𝑧) \\ \leq lim_{𝑁 \to \infty} sup_{𝑛 > 𝑚 > 𝑁} \frac{𝑟^{'} + | 𝑧 - 𝑝^{'} |}{𝑟^{'} - | 𝑧 - 𝑝^{'} |} (𝑢_{𝑛} (𝑝^{'}) - 𝑢_{𝑚} (𝑝^{'})) \\ \leq lim_{𝑁 \to \infty} sup_{𝑛 > 𝑚 > 𝑁} 3 (𝑢_{𝑛} (𝑝^{'}) - 𝑢_{𝑚} (𝑝^{'})) = 0, \end{aligned}

where Theorem 7.2.2 $Theorem 7.2.2 Theorem 7.2.2$ is justified by the harmonicity of $𝑢_{𝑛} - 𝑢_{𝑚}$ and the final limit is obtained from the Cauchy Criterion (Theorem 1.2.17 $Theorem 1.2.17 Theorem 1.2.17$ ). Therefore, ${𝑢_{𝑛}}$ uniformly converges (to a harmonic function by Theorem 7.2.1 $Theorem 7.2.1 Theorem 7.2.1$ ) on $𝐷 (𝑝^{'}, \frac{𝑟^{'}}{2})$ by the Cauchy Criterion for functions (Theorem 2.3.2 $Theorem 2.3.2 Theorem 2.3.2$ ).

Therefore, we have

\begin{aligned} 𝐴 & = {𝑝 : 𝑢_{𝑛} (𝑝) \to \infty} \\ = {𝑝 : 𝑢_{𝑛} ⇉ \infty uniformly on a neighborhood of 𝑝} \end{aligned}

and

\begin{aligned} 𝐵 & = {𝑝^{'} : 𝑢_{𝑛} (𝑝^{'}) converges} \\ = {𝑝^{'} : 𝑢_{𝑛} uniformly converges on a neighborhood of 𝑝} . \end{aligned}

Moreover, $𝐴$ and $𝐵$ are disjoint by definition and $𝐴 \cup 𝐵 = 𝑈$ because any nondecreasing real sequence has an accumulation point in $ℝ \cup {+ \infty}$ by the Monotone Convergence Theorem.

Then by the connectivity argument (Theorem 3.2.1.19 $Theorem 3.2.1.19 Theorem 3.2.1.19$ ), exactly one of these two sets is $𝑈$ by the connectivity of $𝑈$ .

Then either $𝑢_{𝑛} ⇉ \infty$ locally uniformly in $𝑈$ or $𝑢_{𝑛}$ locally uniformly converges in $𝑈$ to a harmonic function by Theorem 7.2.1 $Theorem 7.2.1 Theorem 7.2.1$ . ⁠ $□$