§7.4 The Perron Method and the Solution to the Dirichlet Problem

7.4 The Perron Method and the Solution to the Dirichlet Problem

The method we provide here is due to Oskar Perron, which will subsequently apply to a large variety of curves (although it is not of highest possible generality, it will suffice for our general purposes).

Similar to the Riemann Mapping Theorem, the solution here will also pertain largely to solving an extremal problem. Moreover, as a corollary to the problem, one also will obtain a proof of the Riemann Mapping Theorem (which in spirit, is much closer to the original proof of the mapping theorem).

Proposition 7.4.1 (Boundary Maximum Principle).

Let $𝑈 \subset ℂ$ be a bounded region and $ℎ : 𝑈 \to ℝ$ be subharmonic. Suppose that there is some $𝐴 \in ℝ$ such that

\underset{\begin{matrix} 𝑧 \to 𝑞 \\ 𝑧 \in 𝑈 \end{matrix}}{lim sup} ℎ (𝑧) \leq 𝐴

for every

𝑞 \in 𝜕 𝑈

. Then

ℎ \leq 𝐴

throughout

𝑈

Proof.

For each $𝜀 > 0$ , define

𝐸_{𝜀} = {𝑧 \in 𝑈 : ℎ (𝑧) \geq 𝐴 + 𝜀} \subseteq 𝑈 .

We aim to show that $𝐸_{𝜀} = ⌀$ , thus for contradiction, first assume otherwise.

Let $𝑧 \in ℂ ∖ 𝐸_{𝜀}$ . If we are able to show that a neighborhood of $𝑧$ lies in $ℂ ∖ 𝐸_{𝜀}$ , then $ℂ ∖ 𝐸_{𝜀}$ is open and $𝐸_{𝜀}$ is then closed.

If $𝑧 \in ℂ ∖ \overline{𝑈}$ , such a neighborhood holds trivially, in particular $\exists 𝑉 ∋ 𝑧$ open such that $𝑉 \subseteq ℂ ∖ \overline{𝑈} \subseteq ℂ ∖ 𝐸_{𝜀}$ .
If $𝑧 \in 𝜕 𝑈$ , then $\exists 𝛿 > 0$ such that $\forall 𝜁 \in 𝑈$ with $| 𝑧 - 𝜁 | < 𝛿$ ,
$ℎ (𝜁) < \underset{\begin{matrix} 𝜁 \to 𝑧 \\ 𝜁 \in 𝑈 \end{matrix}}{lim sup} ℎ (𝜁) + 𝜀 \leq 𝐴 + 𝜀$
by the definition of $lim sup$ , meaning that $𝑈 \cap 𝐷 (𝑧, 𝛿) \subseteq ℂ ∖ 𝐸_{𝜀}$ . Moreover, $𝐷 (𝑧, 𝛿) ∖ 𝑈 \subseteq ℂ ∖ 𝐸_{𝜀}$ as well since $𝐸_{𝜀} \subseteq 𝑈$ . Therefore, $𝐷 (𝑧, 𝛿) \subseteq ℂ ∖ 𝐸_{𝜀}$ .
If $𝑧 \in 𝑈 ∖ 𝐸_{𝜀}$ , then $ℎ (𝑧) < 𝐴 + 𝜀$ . By continuity, there exists a neighborhood $𝑉 ∋ 𝑧$ on which $ℎ < 𝐴 + 𝜀$ holds.

Therefore, $ℂ ∖ 𝐸_{𝜀}$ is open; then $𝐸_{𝜀}$ is closed in $ℂ$ . Since $𝑈$ is bounded, $𝐸_{𝜀}$ is compact. Then $ℎ$ attains a finite maximum $𝐴^{'}$ within $𝐸_{𝜀}$ by continuity (by the Extreme Value Theorem, Theorem 1.2.23 $Theorem 1.2.23 Theorem 1.2.23$ ) at some point $𝑧$ .

By definition of $𝐸_{𝜀}$ , $𝐴^{'} \geq ℎ (𝑧) \geq 𝐴 + 𝜀$ for $𝑧 \in 𝐸_{𝜀}$ . Outside of $𝐸_{𝜀}$ , $𝐴^{'} \geq ℎ$ continues to hold as $ℎ < 𝐴 + 𝜀$ . Therefore $𝐴^{'}$ is an absolute maximum of $ℎ$ on all of $𝑈$ . By the Maximum Principle for subharmonic functions (Theorem 7.3.10 $Theorem 7.3.10 Theorem 7.3.10$ ), $ℎ \equiv 𝐴^{'} \geq 𝐴 + 𝜀$ must be constant on $𝑈$ .

Therefore $\forall 𝑞 \in 𝜕 𝑈$ , ${lim sup}_{\begin{matrix} 𝜁 \to 𝑞 \\ 𝜁 \in 𝑈 \end{matrix}} ℎ (𝜁) = 𝐴^{'} > 𝐴$ , which contradicts the hypothesis. Therefore, $𝐸_{𝜀}$ must be the empty set.

Hence, for any $𝜀 > 0$ , $𝐸_{𝜀} = ⌀$ ; which implies that $\forall 𝑧 \in 𝑈$ , $ℎ (𝑧) \leq 𝐴$ . ⁠ $□$

Theorem 7.4.2 (Solution to the Dirichlet Problem).

Let

𝑈

be a bounded open region such that for each point

𝑝 \in 𝜕 𝑈

, there is a barrier

𝑏_{𝑝} : \overline{𝑈} \to ℝ

. Then

\forall 𝑓 \in 𝐶 (𝜕 𝑈, ℝ)

\exists! 𝑢 \in 𝐶 (\overline{𝑈}, ℝ)

harmonic in

𝑈

such that

𝑢 \equiv 𝑓

𝜕 𝑈

Proof.

Let

𝑆 = {𝜓 : 𝜓 is subharmonic on 𝑈 and \forall 𝑝 \in 𝜕 𝑈, \underset{\begin{matrix} 𝑧 \to 𝑝 \\ 𝑧 \in 𝑈 \end{matrix}}{lim sup} 𝜓 (𝑧) \leq 𝑓 (𝑝)} (7.4 . 1)

By the compactness of $𝜕 𝑈$ , let $𝑚 = {min}_{𝑧 \in 𝜕 𝑈} 𝑓 (𝑧)$ and $𝑀 = {max}_{𝑧 \in 𝜕 𝑈} 𝑓 (𝑧)$ . Then $(𝑧 \mapsto 𝑚) \in 𝑆$ , meaning $𝑆 \neq ⌀$ . Let $𝑢 : 𝑈 \to ℝ$ be defined by

𝑢 (𝑧) = sup_{𝜓 \in 𝑆} 𝜓 (𝑧) .

It will turn out to be the case that $𝑢$ is the precise solution for which we are searching, it remains to verify its harmonicity and extension to $𝑓$ .

Claim.

{sup}_{𝑧 \in 𝑈} 𝑢 (𝑧) \leq 𝑀

Proof of the claim.

Let $𝜓 \in 𝑆$ . Then for every $𝑞 \in 𝜕 𝑈$ ,

\underset{\begin{matrix} 𝑧 \to 𝑞 \\ 𝑧 \in 𝑈 \end{matrix}}{lim sup} 𝜓 (𝑧) \leq 𝑓 (𝑞) \leq 𝑀 .

By Proposition 7.4.1

Proposition 7.4.1 Proposition 7.4.1

𝜓 \leq 𝑀

𝑈

. Taking the supremum over

𝜓 \in 𝑆

gives

𝑢 \leq 𝑀

. ⁠

□

Claim.

The function

𝑢

is harmonic in

𝑈

Proof of the claim.

Let $\overline{𝐷 (𝑎, 𝑟)} ⋐ 𝑈$ be arbitrary. Then for each $𝑝 \in 𝐷 (𝑎, 𝑟)$ , since $𝑢 (𝑝) = {sup}_{𝜓 \in 𝑆} 𝜓 (𝑝)$ , there is some sequence of functions ${𝜓_{𝑝, 𝑛}}_{𝑛}$ such that ${lim}_{𝑛 \to \infty} 𝜓_{𝑝, 𝑛} (𝑝) = 𝑢 (𝑝)$ .

For $𝑧 \in 𝑈$ , define the function sequence ${Ψ_{𝑝, 𝑛}}_{𝑛}$ of partial maximums

Ψ_{𝑝, 𝑛} (𝑧) = {max}_{𝑘 = 1}^{𝑛} 𝜓_{𝑝, 𝑘} (𝑧) .

By Proposition 7.3.11 $Proposition 7.3.11 Proposition 7.3.11$ , ${Ψ_{𝑝, 𝑛}}_{𝑛}$ is a pointwise-increasing sequence of subharmonic functions on $𝑈$ .

For each $𝑛$ , the function $Ψ_{𝑝, 𝑛}$ is continuous on $𝜕 𝐷 (𝑎, 𝑟)$ . Then by the Poisson Integral Formula ((3.6.5) $(3.6.5) (3.6.5)$ ), there is a continuous function $𝜑_{𝑝, 𝑛}$ on $\overline{𝐷 (𝑎, 𝑟)}$ which is identical to $Ψ_{𝑝, 𝑛}$ on the boundary and harmonic on the interior. Now let

Φ_{𝑝, 𝑛} (𝑧) = {\begin{cases} Ψ_{𝑝, 𝑛} (𝑧) & if 𝑧 \notin 𝐷 (𝑎, 𝑟) \\ 𝜑_{𝑝, 𝑛} (𝑧) & if 𝑧 \in 𝐷 (𝑎, 𝑟), \end{cases}

of which we aim to show subharmonicity (continuity follows since the pieces agree on the circle). By Theorem 7.3.5 $Theorem 7.3.5 Theorem 7.3.5$ , it suffices to show the sub-mean-value property holds locally at each point.

If $𝑧 \in 𝑈 ∖ \overline{𝐷 (𝑎, 𝑟)}$ , then by openness, there exists a disk $\overline{𝐷 (𝑧, 𝜀^{'})} ⋐ 𝑈 ∖ \overline{𝐷 (𝑎, 𝑟)}$ . By the subharmonicity of $Ψ_{𝑝, 𝑛} \equiv Φ_{𝑝, 𝑛}$ , the local sub-mean-value property holds therein.
If $𝑧 \in 𝐷 (𝑎, 𝑟)$ , then there is a disk centered at $𝑧$ therein such that $Φ_{𝑝, 𝑛} \equiv 𝜑_{𝑝, 𝑛}$ thereon. Then in this smaller disk the sub-mean-value property holds by harmonicity.
Lastly, if $𝑧 \in 𝜕 𝐷 (𝑎, 𝑟)$ , observe that by construction, $Φ_{𝑝, 𝑛} (𝑧) = Ψ_{𝑝, 𝑛} (𝑧)$ . Moreover, by Proposition 7.3.3 $Proposition 7.3.3 Proposition 7.3.3$ , for $𝜁 \in 𝐷 (𝑎, 𝑟)$ , $Ψ_{𝑝, 𝑛} (𝜁) \leq 𝜑_{𝑝, 𝑛} (𝜁)$ . Moreover, for all $𝜁 \in 𝑈$ , this inequality continues to hold. Let $\overline{𝐷 (𝑧, 𝜀)} ⋐ 𝑈$ be arbitrary. Therefore by Theorem 7.3.5 $Theorem 7.3.5 Theorem 7.3.5$ , $Ψ_{𝑝, 𝑛}$ satisfies the global sub-mean-value property, and

Φ_{𝑝, 𝑛} (𝑧) = Ψ_{𝑝, 𝑛} (𝑧) \leq \frac{1}{2 π} \int_{0}^{2 π} Ψ_{𝑝, 𝑛} (𝑧 + 𝜀 e^{i 𝜃}) d 𝜃 \leq \frac{1}{2 π} \int_{0}^{2 π} Φ_{𝑝, 𝑛} (𝑧 + 𝜀 e^{i 𝜃}) d 𝜃 .

Then at $𝑧$ , $Φ_{𝑝, 𝑛}$ also satisfies the local sub-mean-value property.

By Theorem 7.3.5 $Theorem 7.3.5 Theorem 7.3.5$ , $Φ_{𝑝, 𝑛}$ is subharmonic. Moreover, by construction, both $Φ_{𝑝, 𝑛}$ and $Ψ_{𝑝, 𝑛}$ are contained within $𝑆$ (defined in (7.4.1) $(7.4.1) (7.4.1)$ ). Since $Ψ_{𝑝, 1} \leq Ψ_{𝑝, 2} \leq \dots$ , for $𝑧 \in 𝑈 ∖ 𝐷 (𝑎, 𝑟)$ , $Φ_{𝑝, 1} (𝑧) \leq Φ_{𝑝, 2} (𝑧) \leq \dots$ .

Moreover, in $𝐷 (𝑎, 𝑟)$ , $\forall 𝑛 \in ℕ$ , $Φ_{𝑝, 𝑛} - Φ_{𝑝, 𝑛 + 1}$ is harmonic (as it is equal to $𝜑_{𝑝, 𝑛} - 𝜑_{𝑝, 𝑛 + 1}$ ) and is non-positive on the boundary. By the Maximum Principle (Theorem 3.4.5 $Theorem 3.4.5 Theorem 3.4.5$ ), it is non-positive everywhere in $\overline{𝐷 (𝑎, 𝑟)}$ . Therefore,

Φ_{𝑝, 1} \leq Φ_{𝑝, 2} \leq Φ_{𝑝, 3} \leq \dots

throughout $𝑈$ . Since for each $𝑛$ ,

𝜓_{𝑝, 𝑛} (𝑝) \leq Ψ_{𝑝, 𝑛} (𝑝) \leq Φ_{𝑝, 𝑛} (𝑝) \leq 𝑢 (𝑝) (since Φ_{𝑝, 𝑛} \in 𝑆)

and ${lim}_{𝑛 \to \infty} 𝜓_{𝑝, 𝑛} (𝑝) = 𝑢 (𝑝)$ , it follows that ${lim}_{𝑛 \to \infty} Φ_{𝑝, 𝑛} (𝑝) = 𝑢 (𝑝)$ . We now aim to show that this limit holds not only at $𝑝$ but also any arbitrary point $𝑞 \in 𝐷 (𝑎, 𝑟)$ .

Fix such $𝑞$ and define similarly the sequence ${𝜓_{𝑞, 𝑛}} \subseteq 𝑆$ such that ${lim}_{𝑛 \to \infty} 𝜓_{𝑞, 𝑛} (𝑞) = 𝑢 (𝑞)$ . Now for each $𝑛$ , $𝑧 \in 𝑈$ , define

Λ_{𝑛} (𝑧) = max ({𝜓_{𝑞, 𝑘} (𝑧) : 𝑘 \in ℕ_{\leq 𝑛}} \cup {𝜓_{𝑝, 𝑘} (𝑧) : 𝑘 \in ℕ_{\leq 𝑛}}) .

Then $Λ_{𝑛}$ is the maximum of finitely many subharmonic functions and hence subharmonic itself (Proposition 7.3.11 $Proposition 7.3.11 Proposition 7.3.11$ ). Moreover, ((7.4.1) $(7.4.1) (7.4.1)$ ) $Λ_{𝑛} \in 𝑆$ , ${Λ_{𝑛}}$ is pointwise-increasing in $𝑈$ , and for fixed $𝜀$ and sufficiently large $𝑛$ ,

𝜓_{𝑞, 𝑛} (𝑞) \leq Λ_{𝑛} (𝑞) < 𝑢 (𝑞) + 𝜀 .

Taking $𝜀 \to 0$ gives

lim_{𝑛 \to \infty} 𝜓_{𝑞, 𝑛} (𝑞) = 𝑢 (𝑞) \leq lim_{𝑛 \to \infty} Λ_{𝑛} (𝑞) \leq 𝑢 (𝑞),

meaning ${lim}_{𝑛 \to \infty} Λ_{𝑛} (𝑞) = 𝑢 (𝑞)$ .

On $\overline{𝐷 (𝑎, 𝑟)}$ , the Poisson Integral Formula gives a solution $𝜆_{𝑛} \in 𝐶 (\overline{𝐷 (𝑎, 𝑟)})$ to the Dirichlet problem with ${Λ_{𝑛} |}_{𝜕 𝐷 (𝑎, 𝑟)}$ such that $𝜆_{𝑛}$ is harmonic in $𝐷 (𝑎, 𝑟)$ . Then defining

Η_{𝑛} (𝑧) = {\begin{cases} Λ_{𝑛} (𝑧) & if 𝑧 \notin 𝐷 (𝑎, 𝑟) \\ 𝜆_{𝑛} (𝑧) & if 𝑧 \in 𝐷 (𝑎, 𝑟), \end{cases}

$Η_{𝑛}$ is continuous and subsequently subharmonic by justification identical to that of $Φ_{𝑝, 𝑛}$ (via the sub-mean-value property). Moreover, similar to the case of $Φ_{𝑝, 𝑛}$ , we have $Η_{𝑛} \in 𝑆$ and $Η_{𝑛} \leq 𝑢$ , and

𝑢 (𝑞) = lim_{𝑛 \to \infty} 𝜓_{𝑞, 𝑛} (𝑞) \leq lim_{𝑛 \to \infty} Λ_{𝑛} (𝑞) \leq lim_{𝑛 \to \infty} Η_{𝑛} (𝑞) \leq 𝑢 (𝑞) .

By the Maximum Principle (Theorem 3.4.5 $Theorem 3.4.5 Theorem 3.4.5$ , as with previously), $Η_{𝑛} - Η_{𝑛 + 1}$ is harmonic in $𝐷 (𝑎, 𝑟)$ and non-positive on the boundary and therefore non-positive throughout. Then ${Η_{𝑛}}_{𝑛}$ is pointwise-increasing everywhere in $𝑈$ .

Both $Η_{𝑛}$ and $Φ_{𝑝, 𝑛}$ are harmonic in $𝐷 (𝑎, 𝑟)$ and by Harnack’s Principle (Theorem 7.2.3 $Theorem 7.2.3 Theorem 7.2.3$ ), $Φ_{𝑝, 𝑛} ⇉ Φ_{𝑝}$ and $Η_{𝑛} ⇉ Η$ locally uniformly in this disk, where $Φ_{𝑝}$ and $Η$ are each either $\infty$ or harmonic in the disk (the former is an impossibility since $Φ_{𝑝, 𝑛}, Η_{𝑛} \leq 𝑢 \leq 𝑀$ ). Therefore the two sequences have harmonic limits in $𝐷 (𝑎, 𝑟)$ . By the discussion above, $Η (𝑞) = 𝑢 (𝑞)$ and $Φ_{𝑝} (𝑝) = 𝑢 (𝑝)$ .

For every $𝑛$ , we have $Ψ_{𝑝, 𝑛} \leq Λ_{𝑛}$ on $𝑈$ . On $𝜕 𝐷 (𝑎, 𝑟)$ , this gives

𝜑_{𝑝, 𝑛} = Ψ_{𝑝, 𝑛} \leq Λ_{𝑛} = 𝜆_{𝑛} .

Since $𝜑_{𝑝, 𝑛}$ and $𝜆_{𝑛}$ are harmonic on $𝐷 (𝑎, 𝑟)$ , the Maximum Principle (Theorem 3.4.5 $Theorem 3.4.5 Theorem 3.4.5$ ) gives

𝜑_{𝑝, 𝑛} \leq 𝜆_{𝑛} ⟹ Φ_{𝑝, 𝑛} \leq Η_{𝑛} ⟹ Φ_{𝑝} \leq Η

throughout $𝐷 (𝑎, 𝑟)$ . Moreover, $Η_{𝑛} \in 𝑆$ , so $Η_{𝑛} \leq 𝑢$ for every $𝑛$ ; hence $Η \leq 𝑢$ . Therefore, $Φ_{𝑝} \leq Η \leq 𝑢$ on $𝐷 (𝑎, 𝑟)$ . Then $Φ_{𝑝} - Η \leq 0$ throughout $𝐷 (𝑎, 𝑟)$ and

𝑢 (𝑝) = Φ_{𝑝} (𝑝) \leq Η (𝑝) \leq 𝑢 (𝑝) ⟹ Η (𝑝) = 𝑢 (𝑝),

meaning $(Φ_{𝑝} - Η) (𝑝) = 0$ . By the Maximum Principle once more, $Φ_{𝑝} \equiv Η$ in $𝐷 (𝑎, 𝑟)$ . Then since $Φ_{𝑝} (𝑞) = Η (𝑞) = 𝑢 (𝑞)$ and $𝑞$ was arbitrary, $Φ_{𝑝} \equiv 𝑢$ throughout $𝐷 (𝑎, 𝑟)$ . Therefore, $𝑢$ is harmonic on $𝐷 (𝑎, 𝑟)$ . Since $𝐷 (𝑎, 𝑟)$ was arbitrary, $𝑢$ is harmonic in $𝑈$ . ⁠ $□$

Claim.

The function

𝑢

extends continuously to

𝑓

: for each

𝑝 \in 𝜕 𝑈

{lim}_{\begin{matrix} 𝑧 \to 𝑝 \\ 𝑧 \in 𝑈 \end{matrix}} 𝑢 (𝑧) = 𝑓 (𝑝)

Proof of the claim.

Fix $𝑝 \in 𝜕 𝑈$ and $𝜀 > 0$ , and let $𝑏 = 𝑏_{𝑝}$ be a barrier at $𝑝$ .

By continuity of $𝑓$ , choose a relatively open neighborhood $𝑉$ of $𝑝$ in $𝜕 𝑈$ such that

| 𝑓 (𝑝) - 𝑓 (𝑞) | < 𝜀

for every $𝑞 \in 𝑉$ and $𝐾 = 𝜕 𝑈 ∖ 𝑉$ is nonempty. Since $𝑏 < 0$ on the compact set $𝐾$ , let $𝜂 = - {max}_{𝑞 \in 𝐾} 𝑏 (𝑞) > 0$ . Thus $𝑏 (𝑞) \leq - 𝜂$ for every $𝑞 \in 𝐾$ . Set

𝐶 = max {1, \frac{𝑀 - 𝑓 (𝑝) - 𝜀}{𝜂}} and 𝐷 = max {1, \frac{𝑓 (𝑝) - 𝜀 - 𝑚}{𝜂}} .

Then, for every $𝑞 \in 𝐾$ , if $\frac{𝑀 - 𝑓 (𝑝) - 𝜀}{𝜂} > 1$ ,

𝑀 + 𝐶 𝑏 (𝑞) \leq 𝑀 - 𝐶 𝜂 \leq 𝑓 (𝑝) + 𝜀,

and if $𝐶 = 1$ , we must have $\frac{𝑀 - 𝑓 (𝑝) - 𝜀}{𝜂} \leq 1$ , implying

𝑀 - 𝑓 (𝑝) - 𝜀 \leq 𝜂 \leq - 𝑏 (𝑞) ⟹ 𝑀 + 𝐶 𝑏 (𝑞) = 𝑀 + 𝑏 (𝑞) \leq 𝑓 (𝑝) + 𝜀 .

Similarly, one derives $𝑓 (𝑝) - 𝜀 \leq 𝑚 - 𝐷 𝑏 (𝑞)$ .

Let $𝜓 \in 𝑆$ be arbitrary. The function $𝜓 + 𝐶 𝑏$ is subharmonic. For $𝑞 \in 𝑉$ ,

\underset{\begin{matrix} 𝜁 \to 𝑞 \\ 𝜁 \in 𝑈 \end{matrix}}{lim sup} (𝜓 (𝜁) + 𝐶 𝑏 (𝜁)) \leq 𝑓 (𝑞) + 𝐶 𝑏 (𝑞) \leq 𝑓 (𝑞) < 𝑓 (𝑝) + 𝜀 .

For $𝑞 \in 𝐾$ , we instead have

\underset{\begin{matrix} 𝜁 \to 𝑞 \\ 𝜁 \in 𝑈 \end{matrix}}{lim sup} (𝜓 (𝜁) + 𝐶 𝑏 (𝜁)) \leq 𝑓 (𝑞) + 𝐶 𝑏 (𝑞) \leq 𝑀 + 𝐶 𝑏 (𝑞) \leq 𝑓 (𝑝) + 𝜀 .

Hence, by the Boundary Maximum Principle (Proposition 7.4.1 $Proposition 7.4.1 Proposition 7.4.1$ ), $𝜓 + 𝐶 𝑏 \leq 𝑓 (𝑝) + 𝜀$ throughout $𝑈$ . Taking the supremum over $𝜓 \in 𝑆$ ,

sup_{𝜓 \in 𝑆} 𝜓 (𝑧) + 𝐶 𝑏 \leq 𝑓 (𝑝) + 𝜀 ⟹ 𝑢 (𝑧) \leq 𝑓 (𝑝) + 𝜀 - 𝐶 𝑏 (𝑧)

for every $𝑧 \in 𝑈$ . As $𝑧 \to 𝑝$ within $𝑈$ , $𝑏 (𝑧) \to 𝑏 (𝑝) = 0$ , so

\underset{\begin{matrix} 𝑧 \to 𝑝 \\ 𝑧 \in 𝑈 \end{matrix}}{lim sup} 𝑢 (𝑧) \leq 𝑓 (𝑝) + 𝜀 \to 𝑓 (𝑝) .

as $𝜀 \to 0^{+}$ . Notice that $𝑧 \mapsto 𝑓 (𝑝) - 𝜀 + 𝐷 𝑏 (𝑧)$ is subharmonic, and on $𝑉$ , we have

𝑓 (𝑝) - 𝜀 + 𝐷 𝑏 (𝑞) \leq 𝑓 (𝑝) - 𝜀 < 𝑓 (𝑞),

while on $𝐾$ , we have $𝑓 (𝑝) - 𝜀 + 𝐷 𝑏 (𝑞) \leq 𝑚 \leq 𝑓 (𝑞)$ . Thus $𝑓 (𝑝) - 𝜀 + 𝐷 𝑏 \in 𝑆$ , and hence

𝑢 (𝑧) \geq 𝑓 (𝑝) - 𝜀 + 𝐷 𝑏 (𝑧) = 𝑓 (𝑝) - 𝜀 + 𝐷 𝑏 (𝑧) .

Letting $𝑧 \to 𝑝$ within $𝑈$ gives

\underset{\begin{matrix} 𝑧 \to 𝑝 \\ 𝑧 \in 𝑈 \end{matrix}}{lim inf} 𝑢 (𝑧) \geq 𝑓 (𝑝) - 𝜀 \to 𝑓 (𝑝) .

as $𝜀 \to 0^{+}$ . Therefore,

lim_{\begin{matrix} 𝑧 \to 𝑝 \\ 𝑧 \in 𝑈 \end{matrix}} 𝑢 (𝑧) = 𝑓 (𝑝) .

Since $𝑝 \in 𝜕 𝑈$ was arbitrary, $𝑢$ extends continuously to $\overline{𝑈}$ by setting $𝑢 \equiv 𝑓$ on $𝜕 𝑈$ . ⁠ $□$

This completes the proof. ⁠

□