Let $n \in \mathbb{N}$ and $\Omega \subset \mathbb{R}^n$ be a bounded region.

Definition (subharmonic function): $u \in C(\Omega)$ is called subharmonic if for every $x \in \Omega$ there exists $r_0 \in (0, \infty)$ such that $B(x,r_0) \subset \Omega$ and for all $r \in (0,r_0)$: \begin{equation} u(x) \leq \frac{1}{n \alpha(n) r^{n-1}} \int_{\partial B(x,r)} u dS \end{equation}

Remark: From proposition 1 of my post on the mean value property it follows that $u \in C^2(\Omega)$ is subharmonic if and only if $\Delta u \geq 0$.

Proposition 1: Let $S \subset \mathbb{R}^n$ be a non-empty bounded subset, $u \in C(S)$ and $m:= \sup_{x \in S} u(x)$. Then either $m$ is attained in $S$ or there exists a sequence $(x_k)_{k \in \mathbb{N}}$ in $S$ converging to a point in $\partial S$ such that $(u(x_k))_{k \in \mathbb{N}}$ converges to $m$.

Proof: Assume that $m$ is not attained. There exists a sequence $(x_k)_{k \in \mathbb{N}}$ in $S$ such that $(u(x_k))_{k \in \mathbb{N}}$ converges to $m$: Without loss of generality (by switching to a subsequence) we may assume that $(x_k)_{k \in \mathbb{N}}$ converges in $\bar S$. This is because $\bar S$ is compact. Let $x_0 := \lim_{k \to \infty} x_k$. If $x_0 \in S$ then by continuity of $u$, $(u(x_k))_{k \in \mathbb{N}}$ converges to $u(x_0)$ and so $m = u(x_0)$ and $m$ is attained. A contradiction. Therefore $x_0 \in \partial S$. $\square$

Proposition 2 (maximum principle for subharmonic functions): Let $u \in C(\Omega)$ subharmonic. Let $m:= \sup_{x \in \Omega} u(x)$. Then:

If $u$ attains $m$, then $u$ is constant,
Let $M \in \mathbb{R}$ such that for all $x_0 \in \partial \Omega$ \begin{equation} \limsup_{\Omega \ni x \to x_0} u(x_0) \leq M. \end{equation} Then either $u<M$ or $u=M$.

Proof: By assumption (both for 1. and 2.) and the preceding proposition $m$ is finite. Let $S:= \{ x \in \Omega : u (x) = m\}$. Then the usual argument shows that either $S= \varnothing$ or $S= \Omega$. This shows 1. For 2.: Case 1: $S = \Omega$. Since $u$ is constant trivially $u \leq M$. Let $x_0 \in \Omega$ with $u(x_0) = M$. Then $u= M$, since $u$ is constant. Case 2: $S= \varnothing$, then $m \leq M$ by the previous proposition and so $u< m \leq M$. $\square$

Definition (harmonic regularisation of subharmonic functions): Let $v \in C(\Omega)$ subharmonic and $x \in \Omega$, $r \in (0, \infty)$ such that $\bar B(x,r) \subset \Omega$. Then the unique $\bar v \in C(\Omega)\cap C^2 (B(x,r))$ with $\bar v = v$ on $\Omega \setminus B(x,r)$ and $\bar v$ harmonic in $B(x,r)$ is called the harmonic regularisation of $v$ in $B(x,r)$.

Proposition 3: Let $v \in C(\Omega)$ subharmonic and $x \in \Omega$, $r \in (0, \infty)$ such that $\bar B(x,r) \subset \Omega$. Let $\bar v$ be the harmonic regularisation of $v$ in $B(x,r)$. Then $v \leq \bar v$ and $\bar v$ is subharmonic.

Proof: To show $v \leq \bar v$: We know $ v - \bar v = 0$ in $\Omega \setminus B(x,r)$ (in particular on $\partial B(x,r)$) and $ v -\bar v$ is subharmonic in $B(x,r)$. Therefore by the maximum principle for subharmonic functions $v - \bar v \leq 0$ in $B(x,r)$. To show that $\bar v$ is subharmonic: Let $y \in \partial B(x,r)$ (for all other points of $\Omega$ subharmonicity is clear). Let $\rho_0 \in (0, \infty)$ such that $B(y,\rho_0) \subset \Omega$ and for all $\rho \in (0, \rho_0)$: \begin{equation} v(y) \leq \frac{1}{\alpha(n) n r^{n-1}} \int_{\partial B(y,\rho)} v dS. \end{equation} Let $\rho \in (0, \rho_0)$. Then \begin{equation} \bar v (y) = v(y) \leq \frac{1}{\alpha(n) n r^{n-1}} \int_{\partial B(y,\rho)} v dS \leq \frac{1}{\alpha(n) n r^{n-1}} \int_{\partial B(y,\rho)} \bar v dS, \end{equation} because $v \leq \bar v$. $\square$

Definition (Perron family): A subset $P \subset C(\Omega)$ is called a Perron family if

$\forall v \in P$: $v$ is subharmonic,
$P \neq \varnothing$,
$\forall v_1, v_2 \in P: \max \{v_1, v_2 \} \in P$,
$\forall v \in P$ the harmonic regularisation of $v$ in any ball that is precompact in $\Omega$ is in $P$.

Proposition 4 (point-wise $\sup$ of Perron family is $\infty$ or harmonic): Let $P$ be a Perron family. Define $u : \Omega \to (-\infty, \infty]$ by $u(x):= \sup_{v \in P} v(x)$. Then either $u=\infty$ or $ u< \infty$ and $u$ is harmonic.

Proof: Let $S:= \{ x \in \Omega : u (x) = \infty \}$. We want to show that $S$ is open and closed (relative to $\Omega$), because then by connectedness of $\Omega$, $S= \varnothing$ or $S= \Omega$. Let $x_0 \in \Omega$ and $r \in (0, \infty)$ such that $\bar B(x_0, r) \subset \Omega$. The proof is finished if we can show the following:

If $u(x_0) = \infty$, then $u=\infty$ on $B(x_0,r)$ (this shows that $S$ is open),
If $u(x_0) <\infty$, then $u$ is equal to a harmonic function on $B(x_0,r)$ (this shows in particular that $S$ is closed relative to $\Omega$).

Let $(w_n)_{n \in \mathbb{N}}$ be a sequence in $P$ such that $\lim_{n \to \infty} w_n (x_0) = u(x_0)$. Define a new sequence $(v_n)_{n \in \mathbb{N}}$ in $P$ by \begin{equation} v_n := \max \{w_1, \dots, w_n\}. \end{equation} Then $(v_n)_{n \in \mathbb{N}}$ is monotone. Furthermore for all $n \in \mathbb{N}$: \begin{equation} w_n (x_0) \leq v_n (x_0) \leq u(x_0), \end{equation} by definition of $w_n$ and $u$. Therefore by the Sandwich lemma $\lim_{n \to \infty} v_n (x_0) = u(x_0)$. Define the sequence $(\bar v_n)_{n \in \mathbb{N}}$ in $P$ by $\bar v_n :=$ the harmonic regularisation of $v_n$ in $B(x_0, r)$. By the maximums principle for harmonic functions: for $m,n \in \mathbb{N}$ with $m\geq n$ and $x \in B (x_0,r)$: \begin{equation} \bar v_n (x) - \bar v_m (x) \leq \sup_{y \in \partial B(x_0,r)} (v_n (y) - v_m(y)) \leq 0. \end{equation} Showing that $(\bar v_n)_{n \in \mathbb{N}}$ is also monotone. Furthermore by proposition 3 for all $n \in \mathbb{N}$: \begin{equation} v_n (x_0) \leq \bar v_n (x_0) \leq u (x_0), \end{equation} which implies that $\lim_{n \to \infty} \bar v_n (x_0) = u(x_0)$. Let $v$ be the pointwise supremum of $(\bar v_n)_{n \in \mathbb{N}}$. By Harnacks convergence theorem: Either $v= \infty$ in $B(x_0,r)$ or $v<\infty$ in $B(x_0,r)$ and $v$ is harmonic in $B(x_0,r)$.

Case 1: If $u(x_0)= \infty$, then $v(x_0) =u(x_0) = \infty$ and so $v= \infty$ in $B(x_0,r)$. Now $v \leq u$ by definition of $u$ and therefore $u = \infty$ in $B(x_0,r)$.

Case 2: If $u(x_0) < \infty$, then $v<\infty$ and $v$ is harmonic in $B(x_0,r)$. We want to show that $u=v$ in $B(x_0,r)$. Let $x \in B(x_0, r)$. Let $(h_n)_{n \in \mathbb{N}}$ be a sequence in $P$ such that $\lim_{n \to \infty} h_n (x) = u(x)$. Define a new sequence $(f_n)_{n \in \mathbb{N}}$ in $P$ by \begin{equation} f_n := \max \{h_1, \dots, h_n, \bar v_n\}. \end{equation} Then $(f_n)_{n \in \mathbb{N}}$ is monotone (using that $(\bar v_n)_{n \in \mathbb{N}}$ is monotone). Furthermore for all $n \in \mathbb{N}$: \begin{equation} h_n (x) \leq f_n (x) \leq u(x), \end{equation} by definition of $f_n$ and $u$. Therefore by the Sandwich lemma $\lim_{n \to \infty} f_n (x) = u(x)$. Define the sequence $(\bar f_n)_{n \in \mathbb{N}}$ in $P$ by $\bar f_n :=$ the harmonic regularisation of $f_n$ in $B(x_0, r)$. By the maximums principle for harmonic functions: for $m,n \in \mathbb{N}$ with $m\geq n$ and $z \in B (x_0,r)$: \begin{equation} \bar f_n (z) - \bar f_m (z) \leq \sup_{y \in \partial B(x_0,r)} (f_n (y) - f_m(y)) \leq 0. \end{equation} Showing that $(\bar f_n)_{n \in \mathbb{N}}$ is also monotone. Furthermore by proposition 3 and the definition of $u$ for all $n \in \mathbb{N}$: \begin{equation} f_n (x) \leq \bar f_n (x) \leq u (x), \end{equation} which implies that $\lim_{n \to \infty} \bar f_n (x) = u(x)$. Let $f$ be the point-wise supremum ($=$ the point-wise limit) of $(\bar f_n)_{n \in \mathbb{N}}$. By Harnacks convergence theorem $f=\infty$ or $f<\infty$ in $B(x,r)$ and $f$ is harmonic in $B(x,r)$. By definition of $f$ we have $v \leq f$ on $B(x_0,r)$. Now for all $n \in \mathbb{N}$ by definition of $f_n$, proposition 3 and the definition of $u$ \begin{equation} \bar v_n (x_0) \leq f_n(x_0) \leq \bar f_n (x_0) \leq u(x_0). \end{equation} Therefore $ u(x_0) = \lim_{n \to \infty} \bar v_n (x_0) =\lim_{n \to \infty} \bar f_n (x_0) = f(x_0)$. This shows that $f\neq \infty$. Therefore $f$ is harmonic in $B(x_0,r)$. Now $v-f$ is harmonic in $B(x,r)$ and $v-f \leq 0$ in $B(x,r)$. Furthermore $(v-f)(x_0) = 0$ and therefore by the maximum principle for harmonic functions $v= f$ on $B(x,r)$. In particular $u(x) = f(x) = v(x)$. And so $v=u$ on $B(x,r)$ with $v$ harmonic. $\square$

Definition (Perron family associated to a boundary function): Let $g : \partial \Omega \to \mathbb{R}$ be a bounded function. Define the Perron family $P_g$ associated to $g$ by \begin{equation} P_g := \{ v \in C(\Omega) : \ v \ \text{subbharmonic and } \forall x_0 \in \partial \Omega : \limsup_{\Omega \ni x \to x_0} v(x_0) \leq g(x_0) \}. \end{equation}

Proposition 5 (it’s a Perron family): Let $g : \partial \Omega \to \mathbb{R}$ be a bounded function and $P_g$ the Perron family associated to $g$. Then:

$P_g \neq \varnothing$,
$\forall x \in \Omega: \sup_{ v \in P_g} v(x) <\infty$,
$\forall v_1, v_2 \in P_g: \max \{v_1, v_2 \} \in P_g$,
$\forall v \in P_g$ the harmonic regularisation of $v$ in any ball that is precompact in $\Omega$ is in $P_g$.

In particular $P_g$ is Perron family.

Proof: To 1.: $\inf_{x \in \partial \Omega} g(x))>-\infty$, because $g$ is bounded. Let $c \in (-\infty, \inf_{x \in \partial \Omega} g(x)]$, then $c \in P_g$. To 2.: Let $v \in P_g$. Then for all $x_0 \in \partial \Omega$ \begin{equation} \limsup_{ \Omega \ni x \to x_0} v(x) \leq g(x_0) \leq \sup_{x \in \partial \Omega} g(x)< \infty. \end{equation} Therefore by the maximum principle for subharmonic functions $ v\leq \sup_{x \in \partial \Omega} g(x)< \infty$. To 3.: Easy to see. To 4.: Follows from proposition 3 and the fact that the regularisation does not change the values of the function near $\partial \Omega$. $\square$

Definition (Perron solution): Let $g : \partial \Omega \to \mathbb{R}$ be a bounded function and $P_g$ the Perron family associated to $g$. The Perron solution $u_g: \Omega \to \mathbb{R}$ associated to $P_g$ is defined by \begin{equation} u_g(x) := \sup_{ v \in P_g} v(x). \end{equation}

Remark: The Perron solution is well defined ($<\infty$) by item 1 and 2 of the preceding proposition and harmonic by proposition 4.

Proposition 6 (motivation for the Perron solution): Let $g \in C(\partial \Omega)$, $P_g$ the Perron family associated to $g$ and $u_g$ the Perron solution associated to $P_g$. Let $u \in C(\bar \Omega) \cap C^2(\Omega)$ with $\Delta u =0$ on $\Omega$ and $u=g$ on $\partial \Omega$. Then $u=u_g$ on $\Omega$.

Proof: By the assumptions $u \in P_g$. Therefore $u \leq u_g$. On the other hand let $v\in P_g$. Then $v-u$ is subharmonic and for all $x_0 \in \partial \Omega$: \begin{equation} \limsup_{\Omega \ni x \to x_0} (v-u)(x) \leq \limsup_{\Omega \ni x \to x_0}v(x) + \limsup_{\Omega \ni x \to x_0}-u(x) \leq g(x_0) - u(x_0) = 0. \end{equation} Therefore by the maximum principle for subharmonic functions $v-u \leq 0$ on $\Omega$. Therefore $v \leq u$. This shows that $u_g \leq u$ and therefore $u=u_g$. $\square$

Definition (regular boundary point): Let $x_0 \in \partial \Omega$. Then $x_0$ is called regular if there exists a barrier function $b \in C(\Omega)$ at $x_0$, that is

$b$ is subharmonic,
$\lim_{x \to x_0} b(x) = 0$,
For all $y \in \partial \Omega \setminus \{ x_0\}$: $\limsup_{\Omega \ni x \to y} b(x) <0$.

Proposition 7: Let $x_0 \in \partial \Omega$ regular and Let $g : \partial \Omega \to \mathbb{R}$ be a bounded function that is continuous in $x_0$. Let $P_g$ the Perron family associated to $g$ and let $u$ be the Perron solution associated to $P_g$. Then for every $\varepsilon \in (0, \infty)$ there exists $v \in P_g$ with \begin{equation} g(x_0)- \varepsilon = \lim_{x \to x_0} v(x). \end{equation}

Proof: Let $b \in C(\Omega)$ be a barrier function at $x_0$. Let $\varepsilon \in (0, \infty)$. Let $\delta \in (0, \infty)$ such that $\forall x \in B(x_0,2 \delta) \cap \partial \Omega$: $|g(x)-g(x_0)|< \varepsilon$. Let $m := \sup_{x \in \Omega \setminus B(x_0, \delta)} b(x)$. Then $m<0$. Since $b \leq 0$ by the maximum principle for subharmonic functions we already know $m\leq 0$. Assume $m=0$. The function $b$ can not attain $0$, because then $b=0$ by the maximum principle contradicting item 3 in the definition of barrier. By proposition 1 there exists a sequence $(x_n)_{n \in \mathbb{N}}$ in $\Omega \setminus B(x_0, \delta)$ converging to a point $x \in \partial \big(\Omega \setminus B(x_0, \delta)\big)$ with $\lim_{n \to \infty} b(x_n) = 0$. Note that $\partial \big(\Omega \setminus B(x_0, \delta)\big) \subset \bar \Omega$ which implies that $ x \in \partial \Omega$, because otherwise $b$ attains $0$ (by continuity of $b$). Furthermore clearly $x \neq x_0$. But then \begin{equation} 0 = \lim_{n \to \infty} b(x_n) \leq \limsup_{\Omega \ni z \to x} b(z) <0 \end{equation} A contradiction. Let $M := \sup_{ x\in \Omega} |g(x)|$ and $c:=-2M/m>0$. Now define $v: \Omega \to \mathbb{R}$ by \begin{equation} v(x):= g(x_0) - \varepsilon + c b(x). \end{equation} Then $v$ is subharmonic, because $b$ is. Let $z \in \partial \Omega \cap B(x_0, 2 \delta)$. Then For all $x \in \Omega \cap B(x_0, 2\delta)$: \begin{equation} v(x) \leq g(z) + \underbrace{c b(x)}_{ \leq 0} \leq g(z) \end{equation} and therefore $\limsup_{\Omega \ni x \to z} v(x) \leq g(z)$. Let $z \in \partial \Omega \setminus B(x_0, 2 \delta)$. Then $\Omega \cap B(z,\delta) \subset \Omega \setminus B(x_0, \delta)$ and therefore for all $x \in \Omega \cap B(z, \delta)$: \begin{equation} v(x) \leq g(x_0) + c b(x) \leq g(x_0) + c m = g(x_0) - 2M \leq g(z), \end{equation} because $g(x_0) - g(z) \leq |g(x_0)| + |g(z)| \leq 2 M$ and by definition of $m$. This implies that \begin{equation} \limsup_{ \Omega \ni z \to x} v(x) \leq g(z). \end{equation} It follows that $v \in P_g$.

Proposition 8 (Perron’s theorem): Let $x_0 \in \partial \Omega$ regular and Let $g : \partial \Omega \to \mathbb{R}$ be a bounded function that is continuous in $x_0$. Let $P_g$ the Perron family associated to $g$ and let $u_g$ be the Perron solution associated to $P_g$. Then \begin{equation} \lim_{x \to x_0} u_g(x) = g(x_0). \end{equation}

Proof: Let $\varepsilon \in (0, \infty)$. Let $v \in P_g$ with \begin{equation} g(x_0)- \varepsilon = \lim_{x \to x_0} v(x). \end{equation} Existence of $v$ is the content of proposition 7. Consider the Perron family $P_{-g}$ associated to $-g$. The function $-g$ is also bounded and continuous in $x_0$. Let $u_{-g}$ be the Perron solution associated to $P_{-g}$. By proposition 7 there exists $w \in P_{-g}$ with \begin{equation} - g(x_0) - \varepsilon = \lim_{x \to x_0} w(x). \end{equation} Or equivalently \begin{equation} g(x_0) + \varepsilon = \lim_{x \to x_0} -w(x). \end{equation} Since $w \in P_{-g}$ for all $ z \in \partial \Omega$: \begin{equation} \limsup_{ x \to z } w(x) \leq - g(z). \end{equation} Let $v^\prime \in P_g$ arbitrary. Then for all $ z \in \partial \Omega$: \begin{equation} \limsup_{x \to z} (v^\prime+w) \leq g(x_0)- g(x_0) =0. \end{equation} By the maximum principle $v^\prime+w \leq 0$, which implies $v^\prime \leq -w$. In particular $u_g \leq -w$. Then by definition of $u_g$ for all $x \in \Omega$: \begin{equation} v(x) \leq u_g(x) \leq -w (x). \end{equation} Therefore \begin{equation} g(x_0) - \varepsilon \leq \liminf_{ x \to x_0} u_g(x) \leq g(x_0) + \varepsilon \end{equation} and \begin{equation} g(x_0) - \varepsilon \leq \limsup_{ x \to x_0} u_g(x) \leq g(x_0) + \varepsilon. \end{equation} Since this is valid for any $\varepsilon \in (0, \infty)$ we have $\lim_{x \to x_0} u_g (x) = g(x_0)$. $\square$

Proposition 9 (Dirichtlet problem solvable for all continuous boundary data $\Leftrightarrow$ boundary is regular): \begin{equation} \begin{split} &\forall g \in C(\partial \Omega) \exists u \in C(\bar \Omega) \cap C^2(\Omega): \Delta u =0 \ \text{in} \ \Omega, \ u = g \ \text{in} \ \partial \Omega \\ &\Leftrightarrow \forall x_0 \in \partial \Omega: x_0 \ \text{is regular}. \end{split} \end{equation}

Proof: $\Leftarrow$ is Perron’s theorem. To show $\Rightarrow$ let $x_0 \in \partial \Omega$ and define $g \in C(\partial \Omega)$ by $g(x):=- \| x-x_0\|$. Let $u$ in $C(\bar \Omega) \cap C^2 (\Omega)$ with $\Delta u = 0$ on $\Omega$ and $u=g$ on $\partial \Omega$. Then $u$ is a barrier at $x_0$. $\square$

Definition (exterior sphere condition): Let $x_0 \in \partial \Omega$. Then the exterior sphere condition is satisfied at $x_0$ $:\Leftrightarrow$ there exists an open ball $B \subset \mathbb{R}^n \setminus \Omega$ such that $ \bar B \cap \bar \Omega = \{x_0\}$.

Proposition 10 (exterior sphere condition implies regularity): Let $x_0 \in \partial \Omega$ such that the exterior sphere condition is satisfied at $x_0$. Then $x_0$ is regular.

Proof: Let $y \in \mathbb{R}^n \setminus \Omega$ and $\varepsilon \in (0, \infty)$ such that $\bar B(y,\varepsilon) \cap \bar \Omega = \{x_0\}$. Then $b \in C(\Omega)$ \begin{equation} b(x):= \exp ( - \alpha \| x-y \|^2) - \exp (- \alpha \varepsilon^2), \end{equation} with $\alpha \in (0, \infty)$ choosen appropriatly large (so that $\Delta b \geq 0$) is easily seen to be a barrier at $x_0$. $\square$

Remark: It is possible to show that if $\partial \Omega$ is $C^2$, then the exterior sphere condition is satisfied for all boundary points.