§7.3 Subharmonic Functions

7.3 Subharmonic Functions

Our goal of introducing subharmonic functions is principally to solve the Dirichlet Problem. In a previous section, it was demonstrated that a solution exists for all disks ((3.6.5) $(3.6.5) (3.6.5)$ ). Subsequently, one can deduce that a solution in a general domain, if one exists, is unique by the Maximum Principle (Theorem 3.4.5 $Theorem 3.4.5 Theorem 3.4.5$ ): if two solutions $𝑢, 𝑣$ exist, $𝑢 - 𝑣$ vanishes on the boundary and thus vanishes everywhere.

Historically, the problem itself was heavily motivated by physics. Consider a thin simply connected heat-conducting two-dimensional plate with a smooth boundary that is in thermal equilibrium; as time passes, the temperature remains constant at each point. Consider $𝑢 (𝑥, 𝑦, 𝑡)$ to be the temperature at each point $𝑧 = 𝑥 + i 𝑦$ of the plate at time $𝑡$ . The heat equation states that

\frac{𝜕 𝑢}{𝜕 𝑡} \equiv 𝑘 \nabla_{𝑥 𝑦}^{2} 𝑢 .

If the plate is in thermal equilibrium, we have $\frac{𝜕 𝑢}{𝜕 𝑡} \equiv \nabla^{2} 𝑢 \equiv 0$ , or that its temperature is modeled by a harmonic function. Now artificially fix temperatures at each point along its boundary. Eventually, as time passes, the points within the material will reach thermal equilibrium, and at the same time, there is a harmonic temperature function solving the Dirichlet problem. Therefore, physically, it is very plausible for the solution to exist (and therefore also be unique from the Maximum Principle). However, the situation is considerably more complicated from a rigorous mathematical standpoint.

In fact, the solution will not always exist.

Example 7.3.1.

Let $𝑈 = 𝔻^{*} = 𝔻 ∖ {0}$ ,

𝑓 (𝑧) = {\begin{cases} 1 & if | 𝑧 | = 1 \\ 0 & if 𝑧 = 0 \end{cases} .

Then there exists no harmonic function in

𝔻^{*}

with a continuous extension to

𝑓

Proof.

For the sake of contradiction, assume the existence of a solution $𝑢$ .

If we arbitrarily choose $𝑧_{1}, 𝑧_{2} \in 𝜕 𝔻$ , then $𝑣 = 𝑢 \circ (𝑧 \mapsto 𝑧 \frac{𝑧_{1}}{𝑧_{2}})$ is also a solution to the particular problem. We already know solutions are unique from the Maximum Principle, thus $𝑢 (𝑧_{2}) = 𝑣 (𝑧_{2}) = 𝑢 (𝑧_{1})$ . Therefore the unique solution $𝑢$ is radial ( $\frac{𝜕 𝑢}{𝜕 𝜃} \equiv 0$ ). Then applying ((2.2.3) $(2.2.3) (2.2.3)$ ),

\frac{𝜕^{2} 𝑢}{𝜕 𝑟^{2}} + \frac{1}{𝑟} \frac{𝜕 𝑢}{𝜕 𝑟} + \frac{1}{𝑟^{2}} \frac{𝜕^{2} 𝑢}{𝜕 𝜃^{2}} \equiv \frac{1}{𝑟} (𝑟 \frac{𝜕^{2} 𝑢}{𝜕 𝑟^{2}} + \frac{𝜕 𝑢}{𝜕 𝑟}) \equiv 0,

which implies

\frac{𝜕}{𝜕 𝑟} (𝑟 \frac{𝜕 𝑢}{𝜕 𝑟}) \equiv 0 ⟹ 𝑟 \frac{𝜕 𝑢}{𝜕 𝑟} \equiv 𝑐 ⟹ \int d 𝑢 \equiv \int \frac{𝑐}{𝑟} d 𝑟,

where $𝑐 \in ℝ$ is a constant. Then there exists some second constant $𝑐^{'} \in ℝ$ such that $𝑢 (𝑟 e^{i 𝜃}) = 𝑐 log 𝑟 + 𝑐^{'}$ . Setting $𝑟 = 0$ shows that no choice of $𝑐, 𝑐^{'}$ allows an extension to $𝑓$ . ⁠ $□$

So clearly, there are necessary conditions on $𝑓$ and/or the domain $𝑈$ for a solution to exist. Physically, forcibly fixing the temperature of a single isolated point carries too little weight to influence the nearby points. Furthermore, we found that the radial solution would attain an infinity at the $0$ point (or otherwise constancy), similar to how point masses have infinite density (colloquially modeled by the Dirac delta function).

The following section proves solvability when $𝜕 𝑈$ consists of sufficiently smooth curves (consisting of points which will be deemed to be “regular”). In fact, this hypothesis is much stronger than necessary: for planar domains, the Dirichlet problem is solvable for every continuous boundary function whenever every connected component of $𝜕 𝑈$ contains at least two points. The punctured disk from the previous example fails this condition, since the isolated point 0 forms a singleton boundary component.

The remainder of this section will be dedicated to introducing the concepts and characterizations that are precursory to the proof.

Definition 7.3.2 (Subharmonicity).

Let $𝑈 \subseteq ℂ$ be an open region and let $𝑓 : 𝑈 \to ℝ$ be continuous.

We say that $𝑓$ is subharmonic if, whenever $\overline{𝐷 (𝑝, 𝑟)} ⋐ 𝑈$ and $𝑢 : \overline{𝐷 (𝑝, 𝑟)} \to ℝ$ is harmonic (on a neighborhood) with

𝑓 \leq 𝑢 on 𝜕 𝐷 (𝑝, 𝑟),

it follows that

𝑓 \leq 𝑢 throughout 𝐷 (𝑝, 𝑟) .

At first glance, the definition may seem a little bloated. However, we aim to provide sufficient intuition. Firstly, the one-dimensional analog of a harmonic function $𝑢$ is one satisfying $\frac{d^{2} 𝑢}{d 𝑥^{2}} \equiv 0 ⟹ \frac{d 𝑢}{d 𝑥} \equiv 𝑎 ⟹ 𝑢 (𝑥) = 𝑎 𝑥 + 𝑏$ . Consider an arbitrary continuous function $𝑓 : 𝐼 \to ℝ$ , where $𝐼 \subseteq ℝ$ is an open connected set. Then the analogous scenario is: $𝑓$ is subharmonic iff: for any $𝑎, 𝑏 \in 𝐼$ and any harmonic (affine) function such that $𝑓 (𝑎) \leq 𝑢 (𝑎)$ and $𝑓 (𝑏) \leq 𝑢 (𝑏)$ , it follows that $𝑓 (𝑥) \leq 𝑢 (𝑥)$ for all $𝑥 \in (𝑎, 𝑏)$ . This is the criteria for convexity. Therefore, subharmonicity is a higher-dimension generalization of convexity.

Additionally, we note that the harmonicity of $𝑢 : \overline{𝐷 (𝑝, 𝑟)} \to ℝ$ can be weakened to simply $𝐷 (𝑝, 𝑟)$ with continuity up to $𝜕 𝐷 (𝑝, 𝑟)$ :

Proposition 7.3.3.

Let $𝑈 \subseteq ℂ$ be an open region and let $𝑓 : 𝑈 \to ℝ$ be continuous. Then $𝑓$ is subharmonic iff: whenever $\overline{𝐷 (𝑝, 𝑟)} ⋐ 𝑈$ and $𝑢 \in 𝐶 (\overline{𝐷 (𝑝, 𝑟)})$ is real-harmonic in $𝐷 (𝑝, 𝑟)$ with

𝑓 \leq 𝑢 on 𝜕 𝐷 (𝑝, 𝑟),

it follows that

𝑓 \leq 𝑢 throughout 𝐷 (𝑝, 𝑟) .

Proof.

The reverse implication (stated condition implies Definition 7.3.2 $Definition 7.3.2 Definition 7.3.2$ ) is immediate: if the stated condition holds for $𝑓$ and $𝑢$ is harmonic on a neighborhood of $\overline{𝐷 (𝑝, 𝑟)}$ , then $𝑢$ is also continuous on $\overline{𝐷 (𝑝, 𝑟)}$ and harmonic in $𝐷 (𝑝, 𝑟)$ and therefore the stated inequality can be obtained. Thus the stated condition implies the definition of subharmonicity.

Conversely, suppose that $𝑓$ is subharmonic in the sense of Definition 7.3.2 $Definition 7.3.2 Definition 7.3.2$ . Let $\overline{𝐷 (𝑝, 𝑟)} ⋐ 𝑈$ , and let $𝑢 \in 𝐶 (\overline{𝐷 (𝑝, 𝑟)})$ be real-harmonic in $𝐷 (𝑝, 𝑟)$ such that $𝑓 \leq 𝑢$ on $𝜕 𝐷 (𝑝, 𝑟)$ . We must show that the stated condition holds, or that $𝑓 \leq 𝑢$ throughout $𝐷 (𝑝, 𝑟)$ .

Fix $𝜀 > 0$ . Since $𝑓 - 𝑢$ is uniformly continuous on $\overline{𝐷 (𝑝, 𝑟)}$ and satisfies $𝑓 - 𝑢 \leq 0$ on $𝜕 𝐷 (𝑝, 𝑟)$ , there exists $𝜌 \in (0, 𝑟)$ , sufficiently close to $𝑟$ , such that $𝑓 - 𝑢 \leq 𝜀$ on $𝜕 𝐷 (𝑝, 𝜌)$ . Equivalently, $𝑓 \leq 𝑢 + 𝜀$ on $𝜕 𝐷 (𝑝, 𝜌)$ .

Since $𝑢$ is harmonic in $𝐷 (𝑝, 𝑟)$ , the function $𝑢 + 𝜀$ is harmonic on a neighborhood of $\overline{𝐷 (𝑝, 𝜌)}$ . By the subharmonicity of $𝑓$ , $𝑓 \leq 𝑢 + 𝜀$ throughout $𝐷 (𝑝, 𝜌)$ .

Now let $𝑧 \in 𝐷 (𝑝, 𝑟)$ be arbitrary. Choosing $𝜌$ above so that $| 𝑧 - 𝑝 | < 𝜌 < 𝑟$ , we obtain $𝑓 (𝑧) \leq 𝑢 (𝑧) + 𝜀$ . As this holds for every $𝜀 > 0$ , it follows that $𝑓 (𝑧) \leq 𝑢 (𝑧)$ . Since $𝑧$ was arbitrary, $𝑓 \leq 𝑢$ throughout $𝐷 (𝑝, 𝑟)$ . ⁠ $□$

By the Maximum Principle, we have the following characterization:

Proposition 7.3.4.

A harmonic function

𝑓 : 𝑈 \to ℝ

(where

𝑈 \subseteq ℂ

is an open region) is subharmonic.

Proof.

For any $\overline{𝐷 (𝑝, 𝑟)} ⋐ 𝑈$ and $𝑢 : \overline{𝐷 (𝑝, 𝑟)} \to ℝ$ harmonic such that $𝑓 \leq 𝑢$ on $𝜕 𝐷 (𝑝, 𝑟)$ , $𝑓 - 𝑢$ is harmonic on $\overline{𝐷 (𝑝, 𝑟)}$ and non-positive on the boundary circle. By the Maximum Principle, for any $𝑧 \in 𝐷 (𝑝, 𝑟)$ ,

(𝑓 - 𝑢) (𝑧) \leq sup_{𝜁 \in 𝜕 𝐷 (𝑝, 𝑟)} (𝑓 - 𝑢) (𝜁) \leq 0 ⟹ 𝑓 (𝑧) \leq 𝑢 (𝑧) .

Then

𝑓

is subharmonic. ⁠

□

Theorem 7.3.5.

Let $𝑓 : 𝑈 \to ℝ$ be continuous, where $𝑈 \subseteq ℂ$ is an open region. The following are equivalent:

$𝑓$ is subharmonic on $𝑈$ .
$𝑓$ satisfies the global sub-mean-value property: for each $\overline{𝐷 (𝑝, 𝑟)} ⋐ 𝑈$ ,
$𝑓 (𝑝) \leq \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑝 + 𝑟 e^{i 𝜃}) d 𝜃 . (7.3 . 1)$
$𝑓$ satisfies the local sub-mean-value property: for each $𝑝 \in 𝑈$ , there exists $𝑟_{𝑝} > 0$ such that $\overline{𝐷 (𝑝, 𝑟_{𝑝})} ⋐ 𝑈$ and, for every $0 < 𝑟 < 𝑟_{𝑝}$ ,
$𝑓 (𝑝) \leq \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑝 + 𝑟 e^{i 𝜃}) d 𝜃 .$

Figure 7.1: The construction of the arc subtending a positive angle on which

𝑔 (𝑧) < 𝑀

. The striped region is

𝐾

Proof.

Assume first that $𝑓$ is subharmonic. Let $\overline{𝐷 (𝑝, 𝑟)} ⋐ 𝑈$ be arbitrary. By the Poisson Integral Formula, there exists a function $ℎ \in 𝐶 (\overline{𝐷 (𝑝, 𝑟)})$ harmonic on $𝐷 (𝑝, 𝑟)$ with $ℎ \equiv 𝑓$ on $𝜕 𝐷 (𝑝, 𝑟)$ . By Proposition 7.3.3 $Proposition 7.3.3 Proposition 7.3.3$ , $𝑓 \leq ℎ$ on $𝐷 (𝑝, 𝑟)$ . Therefore,
$𝑓 (𝑝) \leq ℎ (𝑝) = \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑝 + 𝑟 e^{i 𝜃}) d 𝜃 .$
Since the disk was arbitrary, the second condition holds for every valid radius, not merely for sufficiently small radii.
The global sub-mean-value property trivially implies the local sub-mean-value property: for each $𝑝 \in 𝑈$ , choose any $𝑟_{𝑝} > 0$ such that $\overline{𝐷 (𝑝, 𝑟_{𝑝})} ⋐ 𝑈$ .
Conversely, assume the local sub-mean-value property holds and that $𝑓$ is not subharmonic. Then there exists a disk $\overline{𝐷 (𝑝, 𝑟)} ⋐ 𝑈$ and a harmonic function $𝑢 : \overline{𝐷 (𝑝, 𝑟)} \to ℝ$ such that $𝑢 \geq 𝑓$ on $𝜕 𝐷 (𝑝, 𝑟)$ but $𝑢 (𝑧_{0}) < 𝑓 (𝑧_{0})$ for some $𝑧_{0} \in 𝐷 (𝑝, 𝑟)$ . Let $𝑔 \equiv 𝑓 - 𝑢$ .
Observe that ${𝑔 |}_{𝜕 𝐷 (𝑝, 𝑟)} \leq 0$ whilst $𝑔 (𝑧_{0}) > 0$ . Let $𝑀 = {max}_{𝑧 \in \overline{𝐷 (𝑝, 𝑟)}} 𝑔 (𝑧)$ , which exists by Theorem 1.2.22 $Theorem 1.2.22 Theorem 1.2.22$ and the Extreme Value Theorem (Theorem 1.2.23 $Theorem 1.2.23 Theorem 1.2.23$ ). Moreover, $𝐾 = {𝑧 \in \overline{𝐷 (𝑝, 𝑟)} : 𝑔 (𝑧) = 𝑀}$ is not all of $𝐷 (𝑝, 𝑟)$ , because by continuity there exists some $𝛿$ such that for any $𝑧 \in 𝐷 (𝑝, 𝑟) ∖ \overline{𝐷 (𝑝, 𝑟 - 𝛿)}$ , $𝑔 (𝑧) \leq \frac{𝑀}{2}$ , since $𝑔$ is non-positive on the boundary.
Notice that $𝐾$ is compact; hence $𝐷 (𝑝, 𝑟) ∖ 𝐾$ is open and non-empty. Fix $𝑧_{𝐾} \in 𝜕 𝐾$ . Then $𝑧_{𝐾}$ is an accumulation point of $𝐷 (𝑝, 𝑟) ∖ 𝐾$ , otherwise $𝐾$ would contain a disk centered at $𝑧_{𝐾}$ and $𝑧_{𝐾}$ would not be a boundary point. Let $𝜂^{'} = dist (𝑧_{𝐾}, 𝜕 𝐷 (𝑝, 𝑟))$ . By the local sub-mean-value property at $𝑧_{𝐾}$ , choose $𝑟_{𝑧_{𝐾}} > 0$ such that the sub-mean inequality holds at $𝑧_{𝐾}$ for every radius less than $𝑟_{𝑧_{𝐾}}$ . Choose
$𝑎 \in 𝐷 (𝑧_{𝐾}, min {𝜂^{'}, 𝑟_{𝑧_{𝐾}}}) \cap (𝐷 (𝑝, 𝑟) ∖ 𝐾) .$
Letting $𝜂 = | 𝑎 - 𝑧_{𝐾} |$ , $\overline{𝐷 (𝑧_{𝐾}, 𝜂)}$ lies within $𝐷 (𝑝, 𝑟)$ , $𝜂 < 𝑟_{𝑧_{𝐾}}$ , and $𝜕 𝐷 (𝑧_{𝐾}, 𝜂) \cap (𝐷 (𝑝, 𝑟) ∖ 𝐾)$ contains a connected open arc $𝐶$ containing $𝑎$ subtending an angle $𝜗$ , on which $𝑔 (𝑧) < 𝑀$ (as described by Figure 7.1 $Figure 7.1 Figure 7.1$ ). Then by the local sub-mean-value property,
$\begin{aligned} 𝑀 & = 𝑓 (𝑧_{𝐾}) - 𝑢 (𝑧_{𝐾}) \leq \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧_{𝐾} + 𝜂 e^{i 𝜃}) d 𝜃 - 𝑢 (𝑧_{𝐾}) \\ = \frac{1}{2 π} \int_{0}^{2 π} [𝑓 (𝑧_{𝐾} + 𝜂 e^{i 𝜃}) - 𝑢 (𝑧_{𝐾} + 𝜂 e^{i 𝜃})] d 𝜃 (by the harmonicity of 𝑢) \\ = \frac{1}{2 π} \int_{0}^{2 π} 𝑔 (𝑧_{𝐾} + 𝜂 e^{i 𝜃}) d 𝜃 \\ = \frac{1}{2 π} [\int_{[0, 2 π] ∖ 𝐶} 𝑔 (𝑧_{𝐾} + 𝜂 e^{i 𝜃}) d 𝜃 + \int_{𝐶} 𝑔 (𝑧_{𝐾} + 𝜂 e^{i 𝜃}) d 𝜃] \\ < \frac{1}{2 π} [(2 π - 𝜗) 𝑀 + 𝜗 𝑀] = 𝑀, \end{aligned}$
which is impossible (the purpose of extracting a Jordan arc was to ensure a strict inequality). Therefore, $𝑓$ is subharmonic. $□$

Corollary 7.3.5.1.

Let

𝑈 \subseteq ℂ

be an open region and let

𝑓 : 𝑈 \to ℝ

be holomorphic. Then

| 𝑓 |

is subharmonic.

Proof.

Since $𝑓$ satisfies the global mean-value property on any disk $\overline{𝐷 (𝑧, 𝜀)}$ by Lemma 3.4.1 $Lemma 3.4.1 Lemma 3.4.1$ , it follows that

| 𝑓 (𝑧) | = | \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝜀 e^{i 𝜃}) d 𝜃 | \leq \frac{1}{2 π} \int_{0}^{2 π} | 𝑓 (𝑧 + 𝜀 e^{i 𝜃}) | d 𝜃 .

Therefore,

| 𝑓 |

satisfies the global sub-mean-value property and is subharmonic by Theorem 7.3.5

Theorem 7.3.5 Theorem 7.3.5

. ⁠

□

Remark.

This in no way implies

| 𝑓 |

is harmonic, clearly

𝑓 (𝑧) = 𝑧

| 𝑓 (𝑧) | = | 𝑧 |

attains an average value on the boundary of any disk centered at

0

that is strictly greater than its value at the center. (This applies to a wide variety of holomorphic functions.)

Theorem 7.3.6.

A real-valued function

𝑓 \in 𝐶^{2} (𝑈)

(where

𝑈 \subseteq ℂ

is an open region) is subharmonic iff

\nabla^{2} 𝑓 \geq 0

𝑈

Proof.

Assume $𝑓$ is subharmonic, we first show the nonnegativity of its Laplacian.
For the sake of contradiction, assume $\exists 𝑧_{0} \in 𝑈$ such that $\nabla^{2} 𝑓 (𝑧_{0}) < 0$ ; then $\exists 𝜀^{'} > 0$ such that $\overline{𝐷 (𝑧_{0}, 𝜀^{'})} ⋐ 𝑈$ and $\nabla^{2} 𝑓 < 0$ on $\overline{𝐷 (𝑧_{0}, 𝜀^{'})}$ (by continuity). By Theorem 7.3.5 $Theorem 7.3.5 Theorem 7.3.5$ , for each $0 < 𝜀 < 𝜀^{'}$ ,
$𝑓 (𝑧_{0}) \leq \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧_{0} + 𝜀 e^{i 𝜃}) d 𝜃 .$
Define the right-hand side to be a function of $𝜀$ , namely $𝑀 (𝜀)$ ; and let $𝑀 (0) = {lim}_{𝜀 \to 0^{+}} 𝑀 (𝜀) = 𝑓 (𝑧_{0})$ . Then
$𝑓 (𝑧_{0}) \leq 𝑀 (𝜀) (7.3 . 2)$
for any $𝜀 \in (0, 𝜀^{'})$ . Then
$\begin{aligned} 𝑀^{'} (𝜀) & = \frac{𝜕}{𝜕 𝜀} \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧_{0} + 𝜀 e^{i 𝜃}) d 𝜃 \\ = \frac{1}{2 π} ∳_{𝜕 𝐷 (𝑧_{0}, 𝜀)} \nabla_{\hat{𝐧}} 𝑓 (𝜁) \frac{d 𝜁}{i (𝜁 - 𝑧_{0})}, \end{aligned}$
where $\nabla_{\hat{𝐧}}$ is the normal derivative, or the directional derivative along the exterior-pointing normal vector. Since $d 𝜁$ is tangent to the circle and points $\frac{π}{2}$ counter-clockwise of the radial vector $𝜁 - 𝑧_{0}$ , $d 𝜁 = | d 𝜁 | exp [i arg (𝜁 - 𝑧_{0})] i$ , implying that $| 𝜁 - 𝑧_{0} | d 𝜁 = (𝜁 - 𝑧_{0}) i | d 𝜁 |$ . Hence,
$𝑀^{'} (𝜀) = \frac{1}{2 π 𝜀} ∳_{𝜕 𝐷 (𝑧_{0}, 𝜀)} \nabla_{\hat{𝐧}} 𝑓 (𝜁) | d 𝜁 |$
Then letting $𝜁 = 𝜉 + i 𝜂$ , using
$\nabla_{\hat{𝐧}} 𝑓 (𝜁) | d 𝜁 | = 𝛁 𝑓 (𝜁) \cdot \hat{𝐧} | d 𝜁 | = 𝛁 𝑓 (𝜁) \cdot (\begin{matrix} d 𝜂 \\ - d 𝜉 \end{matrix})$
(since $\hat{𝐧} | d 𝜁 |$ points in the direction $\frac{π}{2}$ clockwise of $d 𝜁$ ), we obtain
$\begin{aligned} 2 π 𝜀 𝑀^{'} (𝜀) & = ∳_{𝜕 𝐷 (𝑧_{0}, 𝜀)} 𝑓_{𝜉}^{'} (𝜁) d 𝜂 - 𝑓_{𝜂}^{'} (𝜁) d 𝜉 \\ = \int_{𝐷 (𝑧_{0}, 𝜀)} d (𝑓_{𝜉}^{'} (𝜁) d 𝜂 - 𝑓_{𝜂}^{'} (𝜁) d 𝜉) \\ = \int_{𝐷 (𝑧_{0}, 𝜀)} 𝑓_{𝜉 𝜉}^{″} (𝜁) d 𝜉 \land d 𝜂 \\ \begin{matrix} \end{matrix} - 𝑓_{𝜂 𝜂}^{″} (𝜁) d 𝜂 \land d 𝜉 (by Theorem 3.1.2 Theorem 3.1.2 Theorem 3.1.2) \\ = \iint_{𝐷 (𝑧_{0}, 𝜀)} \nabla_{𝜁}^{2} 𝑓 (𝜁) d 𝐴_{𝜁} . \end{aligned}$
Since $\nabla^{2} 𝑓 < 0$ on $\overline{𝐷 (𝑧_{0}, 𝜀^{'})}$ , we obtain that for each $𝜀 \in (0, 𝜀^{'})$ , $𝑀^{'} (𝜀) < 0$ , meaning that $𝑀$ is a strictly decreasing function of $𝜀$ . Therefore, for any such $𝜀$ ,
$𝑓 (𝑧_{0}) = 𝑀 (0) > 𝑀 (𝜀),$
which is impossible since we previously have asserted that $𝑓 (𝑧_{0}) \leq 𝑀 (𝜀)$ in (7.3.2) $(7.3.2) (7.3.2)$ . Thus, there exists no point $𝑧_{0} \in 𝑈$ at which the Laplacian of $𝑓$ is negative.
Now assume $\nabla^{2} 𝑓 \geq 0$ everywhere. We aim to show subharmonicity.
Let $\overline{𝐷 (𝑧_{0}, 𝑟)} ⋐ 𝑈$ be arbitrary, and let $𝑢 : \overline{𝐷 (𝑧_{0}, 𝑟)} \to ℝ$ be harmonic such that $𝑢 \geq 𝑓$ on $𝜕 𝐷 (𝑧_{0}, 𝑟)$ . Then letting $𝑣 = 𝑓 - 𝑢$ , it follows that $\nabla^{2} 𝑣 \equiv \nabla^{2} 𝑓 \geq 0$ everywhere, while $𝑣 \leq 0$ everywhere on $𝜕 𝐷 (𝑧_{0}, 𝑟)$ .
For each $𝜀 > 0$ , let $𝑣_{𝜀} (𝑧) = 𝑣 (𝑧) + 𝜀 ({| 𝑧 - 𝑧_{0} |}^{2} - 𝑟^{2})$ . Then it follows that for $𝑧 \in 𝜕 𝐷 (𝑧_{0}, 𝑟)$ , $𝑣_{𝜀} (𝑧) = 𝑣 (𝑧) \leq 0,$ while $\forall 𝑧 \in 𝐷 (𝑧_{0}, 𝑟)$ ,
$\nabla^{2} 𝑣_{𝜀} = \nabla^{2} 𝑣 + 4 𝜀 \frac{𝜕^{2}}{𝜕 𝑧 𝜕 \overline{𝑧}} [(𝑧 - 𝑧_{0}) (\overline{𝑧} - \overline{𝑧_{0}})] = \nabla^{2} 𝑣 + 4 𝜀 > 0 (strictly),$
which means that $𝑣_{𝜀}$ cannot attain a local maximum within $𝐷 (𝑧_{0}, 𝑟)$ (as one would require $\nabla^{2} 𝑣_{𝜀} \leq 0$ ). Therefore, we have $𝑣_{𝜀} (𝑧) \leq 0$ for all $𝑧 \in \overline{𝐷 (𝑧_{0}, 𝑟)}$ , and letting $𝜀 \to 0^{+}$ gives $𝑣 \leq 0 \Rightarrow 𝑓 \leq 𝑢$ thereon as well. $□$
Remark.
The purpose here of defining $𝑣_{𝜀}$ as opposed to directly using $𝑣$ is that without adding a small subharmonic (convex) paraboloid to $𝑣$ , $\nabla^{2} 𝑣$ could still vanish at some point in the interior (in which case we cannot directly determine that it is not a maximum).

(Notice that convexity in the one-dimensional case for $𝐶^{2}$ functions is equivalent to the condition that $𝑓^{″} \geq 0$ everywhere.)

Here, we must alert the reader to an important difference between harmonic and subharmonic functions. The local mean-value property implies harmonicity and therefore smoothness. The local sub-mean-value property likewise implies subharmonicity by Theorem 7.3.5 $Theorem 7.3.5 Theorem 7.3.5$ , and consequently the global sub-mean inequality, but it does not imply that the function is $𝐶^{2}$ . Thus the statement and application in Theorem 7.3.6 $Theorem 7.3.6 Theorem 7.3.6$ do not hold, and there is no analogous elliptic regularity for general subharmonic functions.

Theorem 7.3.7 (Jensen’s Inequality).

Let $[𝑎, 𝑏] \subseteq ℝ$ be an interval. If $𝜙 : ℝ \to ℝ$ is convex and $𝑓 : [𝑎, 𝑏] \to ℝ$ is continuous, then

𝜙 (\frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝑓 (𝑡) d 𝑡) \leq \frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝜙 \circ 𝑓 (𝑡) d 𝑡 .

In other words,

𝜙 ([average of 𝑓 on [𝑎, 𝑏]]) \leq [average of 𝜙 \circ 𝑓 on [𝑎, 𝑏]] .

Proof.

Let the two points $𝛼, 𝛽 \in ℝ$ be arbitrary.

First assume that $𝜙 (𝑥) = 𝛼 𝑥 + 𝛽$ is linear, then

\begin{aligned} 𝜙 (\frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝑓 (𝑡) d 𝑡) & = \frac{𝛼}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝑓 (𝑡) d 𝑡 + 𝛽 \\ = \frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝛼 𝑓 (𝑡) d 𝑡 + 𝛽 \frac{\int_{𝑎}^{𝑏} d 𝑡}{𝑏 - 𝑎} \\ = \frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} (𝛼 𝑓 (𝑡) + 𝛽) d 𝑡 \\ = \frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝜙 \circ 𝑓 (𝑡) d 𝑡 . \end{aligned}

Now assume the general case for $𝜙$ .

For any $𝑥^{'} \in ℝ$ , there exists an affine function $𝜙_{𝑥^{'}}$ such that $𝜙_{𝑥^{'}} (𝑥) \leq 𝜙 (𝑥)$ on $ℝ$ and $𝜙_{𝑥^{'}} (𝑥^{'}) = 𝜙 (𝑥^{'})$ . Then for simplicity define

𝑓_{avg} = \frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝑓 (𝑡) d 𝑡 .

Then letting $𝑥^{'} = 𝑓_{avg}$ ,

\begin{aligned} 𝜙 (\frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝑓 (𝑡) d 𝑡) = 𝜙 (𝑓_{avg}) & = 𝜙_{𝑓_{avg}} (𝑓_{avg}) = \frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝜙_{𝑓_{avg}} \circ 𝑓 (𝑡) d 𝑡 \\ \leq \frac{1}{𝑏 - 𝑎} \int_{𝑎}^{𝑏} 𝜙 \circ 𝑓 (𝑡) d 𝑡 . □ \end{aligned}

Furthermore, we have the rather predictable generalization

Theorem 7.3.8.

Let

𝑈 \subseteq ℂ

be a region and suppose

𝑓 : 𝑈 \to ℝ

is subharmonic. Let

𝜙

be non-decreasing and convex on a real connected superset of

𝑓 (𝑈)

. Then

𝜙 \circ 𝑓

is subharmonic.

Proof.

Observe that by the sub-mean-value property, for any closed disk $\overline{𝐷 (𝑧, 𝑟)} ⋐ 𝑈$ ,

(𝜙 \circ 𝑓) (𝑧) \leq 𝜙 (\frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝑟 e^{i 𝜃}) d 𝜃) .

Consider the function $ℎ (𝜃) = 𝑓 (𝑧 + 𝑟 e^{i 𝜃})$ where $ℎ : [0, 2 π] \to ℝ$ is continuous. Then by Theorem 7.3.7 $Theorem 7.3.7 Theorem 7.3.7$ ,

𝜙 ([average of ℎ on [0, 2 π]]) \leq [average of 𝜙 \circ ℎ on [0, 2 π]],

giving that

\begin{aligned} (𝜙 \circ 𝑓) (𝑧) & \leq 𝜙 ([average of ℎ on [0, 2 π]]) \\ = [average of 𝜙 \circ ℎ on [0, 2 π]] \\ \leq \frac{1}{2 π} \int_{0}^{2 π} 𝜙 \circ 𝑓 (𝑧 + 𝑟 e^{i 𝜃}) d 𝜃, \end{aligned}

hence

𝜙 \circ 𝑓

satisfies the sub-mean-value property on each compact disk in

𝑈

. Then

𝜙 \circ 𝑓

is subharmonic by Theorem 7.3.5

Theorem 7.3.5 Theorem 7.3.5

. ⁠

□

Theorem 7.3.9.

Let

𝑈 \subseteq ℂ

be a region. If

𝜙 : ℝ \to ℝ

is convex and non-decreasing and

𝑓 : 𝑈 \to ℝ

is subharmonic, then

𝜙 \circ 𝑓

is subharmonic.

Proof.

Let $\overline{𝐷 (𝑝, 𝑟)} ⋐ 𝑈$ be arbitrary. It suffices from Theorem 7.3.5 $Theorem 7.3.5 Theorem 7.3.5$ to show that the global sub-mean-value property is satisfied thereon. Observe that

𝑓 (𝑝) \leq \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑝 + 𝑟 e^{i 𝜃}) d 𝜃 .

Then, since $𝜙$ is non-decreasing,

\begin{aligned} 𝜙 \circ 𝑓 (𝑝) & \leq 𝜙 (\frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑝 + 𝑟 e^{i 𝜃}) d 𝜃) \\ \leq \frac{1}{2 π} \int_{0}^{2 π} 𝜙 \circ 𝑓 (𝑝 + 𝑟 e^{i 𝜃}) d 𝜃 . (by Theorem 7.3.7 Theorem 7.3.7 Theorem 7.3.7) \end{aligned}

Then the global sub-mean-value property holds on an arbitrary disk and hence

𝜙 \circ 𝑓

is subharmonic. ⁠

□

Theorem 7.3.10 (Maximum Principle for Subharmonic Functions).

Let $𝑈 \subseteq ℂ$ be an open region and let $𝑓 \in 𝐶 (𝑈, ℝ)$ be subharmonic. Then if $\exists 𝑧_{0} \in 𝑈$ such that for all $𝑧 \in 𝑈$ ,

𝑓 (𝑧_{0}) \geq 𝑓 (𝑧)

holds, then

𝑓

is the constant function on

𝑈

Our proof will remain the same as Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ , aside from the sole difference that the local sub-mean-value property will be used in place of the local mean-value property.

Proof.

Assume that $𝑧_{0}$ exists. We aim to show that the set

𝑆 = {𝑧 : 𝑓 (𝑧) = 𝑓 (𝑧_{0}) \land 𝑧 \in 𝑈}

is all of $𝑈$ . This is equivalent to proving that $𝑆$ is nonempty, open, and relatively closed in $𝑈$ by the connectivity argument.

Nonemptiness follows directly from $𝑧_{0} \in 𝑆$ .

For any sequence ${𝑧_{𝑛}} \in 𝑆$ converging to some $𝑧_{\infty} \in 𝑈$ , by the continuity of $𝑓$ ,

lim_{𝑛 \to \infty} 𝑓 (𝑧_{𝑛}) = 𝑓 (lim_{𝑛 \to \infty} 𝑧_{𝑛}) = 𝑓 (𝑧_{\infty}) = 𝑓 (𝑧_{0}),

and $𝑧_{\infty} \in 𝑆$ . Thus, $𝑆$ contains all of its accumulation points in $𝑈$ and is closed therein.

Since $𝑈$ is open, $\forall 𝑧 \in 𝑆$ , $\exists 𝜆 > 0$ such that $𝐷 (𝑧, 𝜆) \subseteq 𝑈$ . By the local sub-mean-value property, $\forall 0 < 𝜀 < 𝜆$ ,

𝑓 (𝑧) \leq \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝜀 e^{i 𝜃}) d 𝜃 \leq \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧) d 𝑡 = 𝑓 (𝑧),

forcing all inequalities to become inequalities:

𝑓 (𝑧) = \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝜀 e^{i 𝜃}) d 𝜃 = \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧) d 𝑡 = 𝑓 (𝑧)

From the equality of the last two integrals,

\int_{0}^{2 π} [𝑓 (𝑧) - 𝑓 (𝑧 + 𝜀 e^{i 𝑡})] d 𝑡 = 0 .

Since this integrand is strictly non-negative and continuous, we have equality everywhere. Indeed, if not, there exists a $𝑡_{0}$ at which the integrand is positive. Then the integrand is bounded below by a positive constant on a neighborhood of $𝑡_{0}$ by continuity, giving a positive integral: a contradiction. Thus for any $𝑡$ , $𝑓 (𝑧) = 𝑓 (𝑧 + 𝜀 e^{i 𝑡})$ .

Thus, $\forall 𝑧 \in 𝑆$ , $\exists 𝜆 > 0$ such that $𝐷 (𝑧, 𝜆) \subseteq 𝑆$ . In other words, every $𝑧 \in 𝑆$ has an open neighborhood that also lies in $𝑆$ , giving the openness of $𝑆$ in $𝑈$ . Therefore, $𝑆 = 𝑈$ by the connectivity argument (Theorem 3.2.1.19 $Theorem 3.2.1.19 Theorem 3.2.1.19$ ). ⁠ $□$

Remark.

Note that there is no corresponding “minimum” principle as with harmonic functions: subharmonicity is sufficiently restrictive in a “one-sided” way. If one were to instead define a “superharmonic function,” there would not be a maximum principle but instead a sole minimum principle. Although this is a widely recognized definition, it is somewhat unnecessary for our purposes as any superharmonic function could be trivially made into a subharmonic function by negation (from

𝑓

𝑧 \mapsto - 𝑓 (𝑧)

The utility of using subharmonic functions, as opposed to harmonic or holomorphic functions is due to its flexibility as we now show:

Proposition 7.3.11.

Suppose $𝑈 \subseteq ℂ$ is an open region and let $𝑓, 𝑔 : 𝑈 \to ℝ$ be two subharmonic functions. Then the function $𝜑$ defined by

𝜑 (𝑧) = max {𝑓 (𝑧), 𝑔 (𝑧)} \forall 𝑧 \in 𝑈

is subharmonic on

𝑈

as well.

Proof.

By a simple application of the global sub-mean-value property on any disk $\overline{𝐷 (𝑧, 𝑟)} ⋐ 𝑈$ :

\begin{aligned} 𝜑 (𝑧) & \leq max {\frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝑟 e^{i 𝜃}) d 𝜃, \frac{1}{2 π} \int_{0}^{2 π} 𝑔 (𝑧 + 𝑟 e^{i 𝜃}) d 𝜃} \\ \leq \frac{1}{2 π} \int_{0}^{2 π} max {𝑓 (𝑧 + 𝑟 e^{i 𝜃}) d 𝜃}, 𝑔 (𝑧 + 𝑟 e^{i 𝜃}) d 𝜃}} d 𝜃 \\ = \frac{1}{2 π} \int_{0}^{2 π} 𝜑 (𝑧 + 𝑟 e^{i 𝜃}) d 𝜃 . \end{aligned}

Applying Theorem 7.3.5

Theorem 7.3.5 Theorem 7.3.5

gives the subharmonicity of

𝜑

. ⁠

□

Before venturing into the the proof-specific definitions of the Perron Method, we remark that subharmonicity is also preserved under addition and scaling by a positive constant.

We next introduce the concept of a barrier in the establishing of a condition on the boundary of a domain which will help us avoid the situation previously described in Example 7.3.1 $Example 7.3.1 Example 7.3.1$ (a situation leading to the insolvability of the Dirichlet problem).

Definition 7.3.12 (Barrier).

Let $𝑈 ⋐ ℂ$ be an open region and let $𝑝 \in 𝜕 𝑈$ be arbitrary. Then a function $𝑏 : \overline{𝑈} \to ℝ$ is said to be a barrier iff the following conditions are satisfied:

$𝑏 \in 𝐶 (\overline{𝑈})$ .
$𝑏$ is subharmonic in $𝑈$ .
$𝑏 (𝑝) = 0$ and $\forall 𝑧 \in 𝜕 𝑈 ∖ {𝑝}$ , $𝑏 (𝑧) \neq 0$ .
$𝑏$ is non-positive in $\overline{𝑈}$ .

Although the definition of a barrier here is somewhat technical, it is the precise definition that will later allow for the construction of a solution to the Dirichlet problem.

The way in which this helps rule out the case of Example 7.3.1 $Example 7.3.1 Example 7.3.1$ is by analyzing the existence of these barriers.

We provide several important examples of barriers:

Theorem 7.3.13.

A barrier exists at every point $𝑝 \in 𝜕 𝔻$ on $𝔻$ by

𝑏 : 𝑧 \mapsto ℜ𝔢 (\frac{𝑧}{𝑝}) - 1 .

Proof.

Obviously Part 1

Part 1 Part 1

, Part 3

Part 3 Part 3

, and Part 4

Part 4 Part 4

are true. Moreover,

𝑏

is harmonic (real part of a holomorphic function) and thus subharmonic (proving Part 2

Part 2 Part 2

). ⁠

□

Theorem 7.3.14.

Let $𝑈 \subset ℂ$ be a bounded region and let $𝑝 \in 𝜕 𝑈$ be a point with greatest modulus. Then $\overline{𝑈} \subseteq \overline{𝐷 (0, | 𝑝 |)}$ . Let $𝜗 = Arg 𝑝$ , where $Arg$ is the principal branch (restricted to $[0, 2 π)$ ). Then we have that

𝑓 : 𝑧 \mapsto ℜ𝔢 (e^{- i 𝜗} 𝑧) - | 𝑝 |

serves as a barrier for

𝑈

𝑝

Proof.

By the condition on $𝑝$ , $| e^{- i 𝜗} 𝑧 | \leq | 𝑝 |$ must hold for all $𝑧 \in \overline{𝑈}$ , which implies Part 4 $Part 4 Part 4$ . For $𝑓$ to vanish at $𝑧$ , we must have

ℜ𝔢 (e^{- i 𝜗} 𝑧) = | 𝑝 | .

Therefore,

ℑ𝔪 (e^{- i 𝜗} 𝑧) = 0

, which implies that

𝑧 = 𝑝

. Thus Part 3

Part 3 Part 3

is satisfied. Moreover, Part 1

Part 1 Part 1

trivially holds and Part 2

Part 2 Part 2

holds by harmonicity (real part of a holomorphic function). ⁠

□

Theorem 7.3.15.

Let $𝑈 \subseteq ℂ$ be open and let $𝑝 \in 𝜕 𝑈$ . Assume there exists $𝑞 \in ℂ ∖ 𝑈$ and a closed line segment $𝐿$ of positive length joining $𝑝$ and $𝑞$ such that

𝐿 \cap \overline{𝑈} = {𝑝} .

Then there exists a barrier for $𝑈$ at $𝑝$ , given by

𝑏 : 𝑧 \mapsto ℜ𝔢 [- 𝜑 \circ (i 𝜓 (𝑧))] - 1,

where

𝜑 (𝑧) = \frac{𝑧 - i}{𝑧 + i}

is the Cayley transform and

𝜓 (𝑧) = \sqrt{\frac{𝑧 - 𝑝}{𝑧 - 𝑞}}

utilizes the principal branch square root.

Proof.

Let $𝜙 : 𝑧 \mapsto \frac{𝑧 - 𝑝}{𝑧 - 𝑞}$ be a Möbius transformation mapping $𝑝$ to $0$ and $𝑞$ to $\infty$ . Note that by the theorem hypotheses, $𝑞 \notin \overline{𝑈}$ . Each $𝑧 \in 𝐿$ can be written as $𝑧 = 𝑝 (1 - 𝑡) + 𝑞 𝑡$ . Then for $𝑧 \in 𝐿$ , $𝜙 (𝑧) = \frac{𝑡 (𝑞 - 𝑝)}{(1 - 𝑡) (𝑝 - 𝑞)} = \frac{𝑡}{𝑡 - 1} \leq 0$ . Therefore, $𝜙$ maps $𝐿$ to $ℝ_{\leq 0} \cup {\infty}$ and maps $ℂ ∖ 𝐿$ to $ℂ ∖ (ℝ_{\leq 0} \cup {1})$ . The the branch cut of the square root coincides to the subtracted ray, then $i 𝜓$ biholomorphically maps $ℂ ∖ 𝐿$ to

i \sqrt{ℂ ∖ (ℝ_{\leq 0} \cup {1})} = i ({𝑧 \in ℂ : ℜ𝔢 𝑧 > 0} ∖ {1}) = ℍ^{+} ∖ {i} .

Then $𝜑$ maps this region to $𝔻^{*} = 𝔻 ∖ {0}$ . It follows that

- 𝜑 \circ (i 𝜓 (\overline{𝑈} ∖ {𝑝})) \subseteq 𝔻^{*}

and extends to $𝑝$ continuously by

- 𝜑 \circ (i 𝜓 (𝑝)) = - 𝜑 (0) = 1 .

Then by Theorem 7.3.14

Theorem 7.3.14 Theorem 7.3.14

, we find that

𝑏_{𝔻 (𝑤)} = ℜ𝔢 𝑤 - 1

is a barrier on

𝔻

, thus set

𝑏 : 𝑧 \mapsto 𝑏_{𝔻} \circ (- 𝜑 \circ (i 𝜓 (𝑧)))

. By injectivity, we have Part 3

Part 3 Part 3

; Part 1

Part 1 Part 1

and Part 4

Part 4 Part 4

hold trivially. Lastly, subharmonicity (Part 2

Part 2 Part 2

) is satisfied as

𝑏

is the real part of some holomorphic function. ⁠

□

The exterior segment condition in Theorem 7.3.15 $Theorem 7.3.15 Theorem 7.3.15$ yields barriers for a broad class of domains. In particular, any domain bounded by a finite disjoint union of regular $𝐶^{1}$ Jordan curves admits a barrier at every point of its boundary (see below), though this regularity is sufficient but far from necessary. Elementary examples with corners, such as squares and rectangles, likewise admit a barrier at every boundary point, as they too satisfy the exterior segment condition.

Theorem 7.3.16.

Let

𝑈 \subset ℂ

be a bounded region such that

𝜕 𝑈

is a finite disjoint union of

𝐶^{1}

Jordan (simple, closed) curves. Then

𝑈

satisfies the exterior segment condition at every point of

𝜕 𝑈

. Consequently,

𝑈

admits a barrier at every point of its boundary.

Proof.

Fix $𝑝 \in 𝜕 𝑈$ , and let $𝛾$ be a $𝐶^{1}$ parameterization of the boundary component containing $𝑝$ (with a nowhere-vanishing derivative), with $𝛾 (0) = 𝑝$ . Define the transformation

𝜙 (𝑤) = \frac{\overline{𝛾^{'} (0)}}{| 𝛾^{'} (0) |} (𝑤 - 𝑝) .

Choose $𝜑$ to be either $𝜙$ or $𝑤 \mapsto \overline{𝜙 (𝑤)}$ , so that $𝜑 (𝑈)$ lies locally above its boundary curve at $0$ . Set $\tilde{𝑈} = 𝜑 (𝑈)$ and $\tilde{𝛾} = 𝜑 \circ 𝛾$ . Then $\tilde{𝛾} (0) = 0$ and ${\tilde{𝛾}}^{'} (0) = | 𝛾^{'} (0) | \in ℝ_{> 0}$ . Write $\tilde{𝛾} (𝑡) = 𝑢 (𝑡) + i 𝑣 (𝑡)$ . Thus $𝑢^{'} (0) > 0$ and $𝑣^{'} (0) = 0$ .

By the inverse function theorem applied to $𝑢$ , there exists an open interval $(- 𝛿, 𝛿)$ centered at $0$ and an open interval $𝐼$ such that $𝑢 : (- 𝛿, 𝛿) \to 𝐼$ is a $𝐶^{1}$ bijection with a $𝐶^{1}$ inverse $𝑢^{- 1} : 𝐼 \to (- 𝛿, 𝛿)$ . Choose $𝑎 > 0$ such that $(- 𝑎, 𝑎) \subset 𝐼$ , and define

ℎ (𝑥) = 𝑣 (𝑢^{- 1} (𝑥)) (𝑥 \in (- 𝑎, 𝑎)) .

Then

\tilde{𝛾} (𝑢^{- 1} (𝑥)) = 𝑥 + i ℎ (𝑥),

while $ℎ (0) = 0$ and $ℎ^{'} (0) = \frac{𝑣^{'} (0)}{𝑢^{'} (0)} = 0$ . Let

𝐾 = 𝜕 \tilde{𝑈} ∖ \tilde{𝛾} ((- 𝛿, 𝛿)) .

Hence $𝐾$ is a closed subset of the compact set $𝜕 \tilde{𝑈}$ , and is therefore compact. Since $0 \notin 𝐾$ , the continuous function $𝑧 \mapsto | 𝑧 |$ attains a strictly positive minimum on $𝐾$ :

𝑑 = min_{𝑧 \in 𝐾} | 𝑧 | > 0 .

Set $𝜀 = \frac{1}{2} min {𝑎, 𝑑}$ . Then $𝐷 (0, 𝜀)$ cannot meet $𝐾$ , and every point of this disk has real part in $(- 𝑎, 𝑎)$ . Consequently,

𝜕 \tilde{𝑈} \cap 𝐷 (0, 𝜀) = {𝑥 + i ℎ (𝑥) : 𝑥 \in (- 𝑎, 𝑎)} \cap 𝐷 (0, 𝜀) .

Thus

\overline{\tilde{𝑈}} \cap 𝐷 (0, 𝜀) \subseteq {𝑥 + i 𝑦 : 𝑦 \geq ℎ (𝑥)} .

Choose $0 < 𝑠 < 𝜀$ . Since $ℎ (0) = 0$ , the line segment

𝐿 = {- i 𝑡 : 0 \leq 𝑡 \leq 𝑠}

satisfies

𝐿 \cap \overline{\tilde{𝑈}} = {0} .

Applying $𝜑^{- 1}$ to $𝐿$ yields a closed line segment of positive length through $𝑝$ whose remaining points lie in $ℂ ∖ \overline{𝑈}$ . Hence the exterior segment condition holds at $𝑝$ . Since $𝑝$ was arbitrary, Theorem 7.3.15 $Theorem 7.3.15 Theorem 7.3.15$ yields a barrier at every point of $𝜕 𝑈$ . ⁠ $□$

Conversely, it is immediate that Theorem 7.3.15 $Theorem 7.3.15 Theorem 7.3.15$ does not apply to $𝔻^{*}$ : at the origin there is no segment emanating from $0$ that remains disjoint from $𝔻^{*}$ . We now provide a proof that there is no barrier for $𝔻^{*}$ at the origin.

Proposition 7.3.17.

There exists no real-valued function

𝑏 \in 𝐶 (\overline{𝔻})

that serves as a barrier on

𝔻^{*} = 𝔻 ∖ {0}

at the origin.

Proof.

For contradiction, assume that such a barrier $𝑏 \in 𝐶 (\overline{𝔻})$ exists. Let

\hat{𝑏} (𝑧) = \frac{1}{2 π} \int_{0}^{2 π} 𝑏 (𝑧 e^{i 𝜃}) d 𝜃,

which is a radial function as

\hat{𝑏} (𝑧 e^{i 𝜗}) = \frac{1}{2 π} \int_{0}^{2 π} 𝑏 (𝑧 e^{i 𝜃 + 𝜗}) d 𝜃 = \frac{1}{2 π} \int_{𝜗}^{2 π + 𝜗} 𝑏 (𝑧 e^{i 𝜃}) d 𝜃 = \hat{𝑏} (𝑧) .

Moreover, $\hat{𝑏} (0) = 𝑏 (0) = 0$ and nowhere else does $\hat{𝑏}$ vanish as $𝑏 < 0$ on each circle (proving Part 3 $Part 3 Part 3$ of Definition 7.3.12 $Definition 7.3.12 Definition 7.3.12$ ). Continuity (Part 1 $Part 1 Part 1$ ) and non-positivity (Part 4 $Part 4 Part 4$ ) are trivial. Lastly, to demonstrate the subharmonicity of $\hat{𝑏}$ , consider an arbitrary point $𝑧_{0} \in 𝔻^{*}$ . Then $\exists 𝑟 > 0$ such that $\overline{𝐷 (𝑧_{0}, 𝑟)} ⋐ 𝔻^{*}$ (and therefore for any $𝜃$ , $\overline{𝐷 (𝑧_{0} e^{i 𝜃}, 𝑟)} ⋐ 𝔻^{*}$ ),

\begin{aligned} \hat{𝑏} (𝑧_{0}) & = \frac{1}{2 π} \int_{0}^{2 π} 𝑏 (𝑧_{0} e^{i 𝜃}) d 𝜃 \\ \leq \frac{1}{2 π} \int_{0}^{2 π} \frac{1}{2 π} \int_{0}^{2 π} 𝑏 (𝑧_{0} e^{i 𝜃} + 𝑟 e^{i 𝜗}) d 𝜗 d 𝜃 . \end{aligned}

Let $𝜏 = 𝜗 - 𝜃$ , $d 𝜗 = d 𝜏$ . Then

\begin{aligned} \hat{𝑏} (𝑧_{0}) & \leq \frac{1}{2 π} \int_{0}^{2 π} \frac{1}{2 π} \int_{- 𝜃}^{2 π - 𝜃} 𝑏 ((𝑧_{0} + 𝑟 e^{i 𝜏}) e^{i 𝜃}) d 𝜏 d 𝜃 \\ = \frac{1}{2 π} \int_{0}^{2 π} [\frac{1}{2 π} \int_{0}^{2 π} 𝑏 ((𝑧_{0} + 𝑟 e^{i 𝜏}) e^{i 𝜃}) d 𝜃] d 𝜏 (by Fubini ’ s Theorem) \\ = \frac{1}{2 π} \int_{0}^{2 π} \hat{𝑏} (𝑧_{0} + 𝑟 e^{i 𝜏}) d 𝜏 . (by definition of \hat{𝑏}) \end{aligned}

This shows that $\hat{𝑏}$ satisfies the global sub-mean-value property Theorem 7.3.5 $Theorem 7.3.5 Theorem 7.3.5$ and is subharmonic. Therefore it is a barrier at the origin for $𝔻^{*}$ .

Without loss of generality, assume $\hat{𝑏} \equiv - 1$ on $𝜕 𝔻$ (this can easily be achieved by scaling). We now show that such a rotationally-invariant barrier $\hat{𝑏}$ cannot exist.

For all $𝑟 \in (0, 1)$ consider the functions on $𝔻^{*}$ in the form of

𝜑_{𝑟} (𝑧) = \frac{\hat{𝑏} (𝑟) + 1}{log 𝑟} log | 𝑧 | - 1 .

Moreover, since $log | 𝑧 |$ is harmonic, it follows that $𝜑_{𝑟}$ is harmonic. Therefore, let $𝜓 \equiv \hat{𝑏} - 𝜑_{𝑟}$ , which is subharmonic. For $| 𝑧 | = 𝑟$ ,

𝜑_{𝑟} (𝑧) = (\hat{𝑏} (𝑟) + 1) - 1 = \hat{𝑏} (𝑟) = \hat{𝑏} (𝑧) ⟹ 𝜓 (𝜕 𝐷 (0, 𝑟)) = {0}

by rotational invariance. For $| 𝑧 | = 1$ ,

𝜑_{𝑟} (𝑧) = \frac{- 1 + 1}{log 𝑟} \cdot 0 - 1 = - 1 = \hat{𝑏} (𝑧) ⟹ 𝜓 (𝜕 𝔻) = {0} .

By the Maximum Principle (Theorem 7.3.10 $Theorem 7.3.10 Theorem 7.3.10$ ),

max_{| 𝑧 | \in (𝑟, 1)} 𝜓_{𝑟} (𝑧) = max_{| 𝑧 | \in {𝑟, 1}} 𝜓_{𝑟} (𝑧) = 0 .

Therefore, for any $𝑧 \in \overline{𝔻} ∖ 𝐷 (0, 𝑟)$ , $𝜓 (𝑧) \leq 0$ , meaning that $\hat{𝑏} (𝑧) \leq 𝜑_{𝑟} (𝑧)$ .

Let $𝑟 \to 0^{+}$ . Then for any $𝑧 \in {\overline{𝔻}}^{*}$ ,

\hat{𝑏} (𝑧) \leq lim_{𝑟 \to 0^{+}} 𝜑_{𝑟} (𝑧) = - 1 .

Now let $𝑧 \to 0$ , which gives that

\hat{𝑏} (0) \leq - 1

by continuity, which contradicts $\hat{𝑏}$ being a barrier (which would otherwise require $\hat{𝑏} (0) = 0$ ). ⁠ $□$