A subharmonic function in a ball

Question

This question is related to this (A nonnegative harmonic function in a ball). Let $B(a,r)$ be an open ball in $\mathbb{R}^m$, ($m\geq2$). Is there a function $u$, $u\not\equiv0$, that is subharmonic on $B(a,r)$, harmonic on a neighborhood of $0$, and $u(a)=0$?

A function $u$ defined on a domain $D$ of $\mathbb{R}^m$ with $m\geq 2$ and with values in $[-\infty,\infty)$ is called subharmonic, if

1) $u$ is upper semi continuous, meaning that for each $x\in D$, $$\limsup_{y\to x}u(x)\leq u(x),$$ (limisup is taken from inside $D$)and

2. $$u(x)\leq \frac{1}{d_mr^m}\int_{B(x,r)}u(y)dy, $$ where $d_mr^m$ is the volume of the ball $B(x,r)$.

Let me explain what is the use of such construction. This way from a function that is subharmonic locally, we can construct a function that is subharmonic on a neighborhood of the infinity, including at infinity. This a well-know result that if $u$ is subharmonic on a ball $B(a,r)$, harmonic on a neighborhood of $a$ and $u(a)=0$, then the function defined by $$u^*(x^*)=\Big(\frac{r}{|x^*-a|}\Big)^{m-2}u(x) $$ for $|x^*|\geq r$ and $0$ at infinity, is subharmonic for $|x^*|\geq r$ (see Lester L. Helms, Potential Theory, pg 208)

Answer 1

Let $v(x)=|x|$ where $|.|$ designates the Euclidean norm in $\mathbb{R}^m$. This is a subharmonic function. We take $0<r'<r$ and replace $v$ by its Poisson's integral on $B(0,r')$ and leave as $v$ on the rest of $B(0,r)$. Then the function we obtain, say $v'$, is harmonic on a neighborhood of $0$. Finally to make it $0$ at the origin we set $u(x)=v'(x)-v'(0).$ I think it works.