Background Information:
I am studying the book: Partial Differential Equations by Walter A. Strauss. I will be trying to adapt the proof given in chapter 6 section 1 to prove the following question.
Definition:
If $U$ is a bounded, open set, we say that $v\in C^2(U)\cap C(\overline{U})$ is subharmonic if $$-\Delta v \leq 0 \ \ \ \text{in} \ U$$
This may be relevant for this particular proof from what I have read online.
Relevant proof to be considered:
a.) Prove that if $v$ is subharmonic and $B(x,r)\subseteq U$, then $$v(x) \leq ⨍_{\partial B(x,r)}v(y)dS_y$$
Proof a.) - Let $$f(r) = ⨍_{\partial B(x,r)}v(y)d S_y = ⨍_{|y-x| = r}v(y)d S_y = \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}v(y) dS_y = \frac{1}{n\alpha(n)}\int_{|z| = 1}u(x+rz) d S_z$$ where $y = x + rz$ and $\alpha(n) = \int_{|x|\leq 1}dx = $ volume of unit ball in $n$ dimensions. The outer unit normal at $y\in\partial B(x,r)$ is $\nu(y) = \frac{y-x}{r}$, thus \begin{align*} f'(r) = \frac{1}{n\alpha(n)}\int_{|z|=1}z \nabla v(x+rz) dS_z &= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y) \frac{y-x}{r}d S_y\\ &= \frac{1}{n\alpha(n)r^{n-1}}\int_{|y-x| = r}\nabla v(y)\nu(y)d S_y \end{align*} Since $-\Delta v\leq 0$ in $U$, we have $$0\leq \int_{B(x,r)}\Delta v(y) dy = \int_{\partial B(x,r)}v_{\nu}d S_y$$ thus $f'(r)\geq 0$ for $r > 0$. Then we get for $r > 0$ $$v(x) = f(0) \leq f(r) = ⨍_{\partial B(x,r)}v(y)d S_y$$ Therefore, $$v(x) \leq ⨍_{\partial B(x,r)}v(y)dS_y$$
b.) Prove that if $v$ is subharmonic and $B(x,r)\subseteq U$, then $$v(x)\leq ⨍_{B(x,r)}v(y)dy$$
Proof b.) - Following from part a.), using polar coordinates, we get \begin{align*} ⨍_{B(x,r)}v(y)dy = \frac{1}{\alpha(n)r^{n}}\int_{B(x,r)}v(y)dy &= \frac{1}{\alpha(n)r^{n}}\int_{0}^{r}d\rho \int_{\partial B(x,\rho)}v(y)dS_y\\ &\geq \frac{1}{\alpha(n)r^{n}}\int_{0}^{r}n\alpha(n)\rho^{n-1}v(x)d\rho\\ &= \frac{v(x)}{\alpha(n)r^{n}}\int_{0}^{r}n \rho^{n-1}d\rho\\ &= v(x) \end{align*} Hence, $$v(x)\leq ⨍_{B(x,r)}v(y)dy$$
Question:
Prove the weak-maximum principle for subharmonic functions: if $v$ is subharmonic, then $$\max_{x\in \bar{U}}v(x) = \max_{x\in \partial\bar{U}}v(x)$$
Attempted proof - Let $\epsilon > 0$ and $v(x) = u(x) + |x|^2$. Then still in two dimensions, say, $$\Delta v = \Delta u + \epsilon (x^2 + y^2) \geq 0 + 4\epsilon > 0 \ \ \text{in} \ U$$ But $\Delta v = v_{xx} + v_{yy} \leq 0$ at an interior maximum point, by the second derivative test in Calculus. Therefore, $v(x)$ has no interior maximum in $U$. Now, $v(x)$, being a continuous function, has to have a maximum somewhere in the closure $\bar{U} = U\cup \text{bdy} U$. Say the maximum is attained at $x_0\in \text{bdy}U$. Then fore all $x\in U$ $$u(x)\leq v(x) \leq v(x_0) = u(x_0) + \epsilon |x_0|^2\leq \max_{\text{bdy}U} u + \epsilon l^2$$ where $l$ is the greatest distance from $\text{bdy}U$ to the origin. Since this holds for all $\epsilon > 0$ we have $$u(x)\leq \max_{\text{bdy}U}u \ \forall x\in U$$
Attempted proof 2 - Assume that $v$ takes an interior maximum at some point $x_0$. Then for $r > 0$, if $B(x_0,r)\subseteq U$ by part b.) above we have $$⨍_{B(x_0,r)}v(y) - v(x_0) \geq 0$$ but this integrand is nonpositive so $v(y) = v(x_0)$ for all $y\in B(x_0,r)$
I am not sure where to go from here, any suggestions or comments are greatly appreciated.
Let us proof by contradiction. We know that $$\max_{x\in \bar{U}}v(x) \ge \max_{x\in \partial\bar{U}}v(x)$$
Suppose that $$\max_{x\in \bar{U}}v(x) > \max_{x\in \partial\bar{U}}v(x) $$
Then $v$ takes an interior maximum at some point $x_0 \in U$. So, we have that there is a point $x_0 \in U$, such that
$$v(x_0) > \max_{x\in \partial\bar{U}}v(x) \tag{1}$$
By part b.) above we have, for any $r > 0$, if $B(x_0,r)\subseteq U$, $$⨍_{ B(x_0,r)}v(y) - v(x_0) = ⨍_{ B(x_0,r)}(v(y) - v(x_0)) \geq 0$$ but the integrand $(v(y) - v(x_0))$ is nonpositive so $v(y) = v(x_0)$ for all $y\in B(x_0,r)$
Let us consider $L = \{r>0 |B(x_0,r) \subseteq U\}$ and $s=\sup L$
Since $B(x_0,s) = \bigcup_{r \in L}B(x_0,r) \subseteq U$, we have that, for all $y\in B(x_0,s)$, $v(y) = v(x_0)$
Since $v \in C(\overline{U})$, we have that, for all $y\in \overline{B(x_0,s)}$, $v(y) = v(x_0)$.
Claim: $ \overline{B(x_0,s)} \cap \partial U \neq \emptyset$.
Once this claim is proved, we immediately have that there is $y \in \partial U$, such that $v(y) = v(x_0)$, which contradicts $(1)$.
Proof of the claim:
Let us proof it by contradicion. Suppose that $ \overline{B(x_0,s)} \cap \partial U = \emptyset$.
Since $ \overline{B(x_0,s)} \subset \overline{U} $, so if $ \overline{B(x_0,s)} \cap \partial U = \emptyset$, then $\overline{B(x_0,s)} \cap (\mathbb{R}^n \setminus U) = \emptyset$.
But since $\overline{B(x_0,s)}$ is compact and $(\mathbb{R}^n \setminus U)$ is closed, if they are disjoint, then
$$ 0< d=\inf \{ |b-a| \; | \; b \in \overline{B(x_0,s)} \textrm{ and } a \in (\mathbb{R}^n \setminus U)\} $$
So $B(x_0,s+\frac{d}{2})$ and $(\mathbb{R}^n \setminus U)$ are disjoint, which means that $B(x_0,s+ \frac{d}{2}) \subseteq U$. So $s+\frac{d}{2} \in L$ and $s+\frac{d}{2} >s =\sup L$. Contradiction. So we must have $ \overline{B(x_0,s)} \cap \partial U \neq \emptyset$.