mean value property for subharmonic functions (probabilistic proof)

Question

A function $u \in C^2(D)$ on a domain $D$ is called subharmonic on $D$ if $\Delta u \geq 0$. Let $B(x, r)$ be the ball of radius $r$ centered at $x$ in $\mathbb{R}^d$. Give a probabilistic proof-that is, using ideas such as Ito integration and submartingle properties - that if $u$ is subharmonic on $D$, then it satisfies the following analogue of the mean value property: If $x \in D$ is such that $B(x, r) \subseteq D$ then $$ u(x) \leq \frac{1}{|\partial B(x, r)|} \int_{\partial B(x, r)} u(y) d y $$ where $\partial B(x, r)=\{y:|x-y|=r\}$ is the sphere of radius $r$ centered at $x$ and $|\partial B(x, r)|$ denotes its surface area.

Answer 1

This is standard material in potential theory eg. see Brownian motion by Peter Mörters and Yuval Peres. Following proof modified from here Harmonic functions and Brownian motion for subharmonic

Fix a ball $B(x,r)$ and consider $\tau= \inf{ \{t>0, |B_t-x| = r} \}$.

Now compute by Itô's formula $$u(B_\tau) = u(B_0) + \int_0^\tau u'(B_s)\, dB_s + \frac{1}{2}\int_0^\tau \Delta u(B_s)\, ds $$

(since subharmonic functions are at least $C^{2}$). Take expectations: $$\Bbb{E}[u(B_\tau)] = u(B_0) + \Bbb{E}\bigg[\int_0^\tau u'(B_s)\, dB_s\bigg] + \frac{1}{2}\int_0^\tau \Delta u(B_s)\, ds $$

Since $\Bbb{E}\big[\int_0^\tau u'(B_s)\, dB_s\big] = 0$ and $\Delta u\geq 0$ (subharmonic) we obtain that

$$\Bbb{E}[u(B_\tau)] \geq u(B_0)=u(x)$$

Now it suffices to take a Brownian motion starting at $x$ and to note by rotation invariance that $$\Bbb{E}[u(B_\tau)] = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(x)\, dS$$ and therefore you obtain the mean value property

$$u(x) \leq\Bbb{E}[u(B_\tau)] = \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(x)\, dS$$

remark: It is important to note that $\tau$ is a finite almost surely stopping time and that $|B_\tau - x| = r$