Convolution of nonnegative bounded subharmonic function

Question

Let $u:\mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous subharmonic function with $n>2$. Now, suppose that $0 \leq u \leq 1$ and $\sup_{\mathbb{R}^n}u = 1$. I would like to ask whether $$\lim_{r\rightarrow \infty} u \ast \rho_r(z) = 1$$ where $u \ast \rho_r(z) := \int_{\mathbb{R}^n} u(y) \rho(\frac{y-z}{r})r^{-n}dy$ and $\rho$ is a smooth mollifier with $\rho|_{B_1(0)} > 0$, $\rho|_{\mathbb{R}^n \backslash B_1(0)} = 0$ and $\int_{R^n} \rho(y)dy=1$.

Answer 1

In fact this is true, but in a close to vacuuous way: Any function subharmonic and bounded above in $\Bbb R^n$ is constant. See here for a proof in the plane that can be generalized to $\Bbb R^n$ (replacing $\log(r)$ by an appropriate power of $r$...)

(In an earlier version of this answer I said "No, the limit is $u(x)$. Yes, of course the limit is $u(x)$, but curiously that actually say "yes", not "no".)