Strong maximum principle for subharmonic functions?

Question

I think I just "proved" the Strong MP for subharmonic functions. But I don't know where things went wrong.

Suppose $U$ is a connected bounded region in $\Bbb R^n$, $u\in C^2(\bar U)$ and $\Delta u\ge 0$ in $U$. Then I try to prove in the following that if $u$ isn't constant, then $\max_{\bar U}u$ is only attained on the boundary $\partial U$.

Proof: for any $x\in U$ and $r>0$ such that $B_r(x)\subset U$, consider the spherical mean $$\phi(r):=\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)} u(y)dS(y)=\frac1{|\partial B_1(0)|}\int_{\partial B_1(0)} u(x+rz)dS(z).$$ Then $$ \begin{align} \phi'(r)&=\frac1{|\partial B_1(0)|}\int_{\partial B_1(0)} Du(x+rz)\cdot zdS(z)\\ &=\frac1{|\partial B_1(0)|}\int_{B_1(0)}\Delta u(x+rz)dz\ge 0 \end{align} $$ Hence $\phi(r)$ is monotonously increasing and in particular $\ge \phi(0^+)=u(x)$ when $r>0$. Hence, if at $x\in U$ is attained the max $M$, then the above mean value inequality forces $u=M$ on a small ball centred at $x$. So $E:=\{x\in U\mid u(x)=M\}$ is open in $U$, but is also closed by continuity. So either $E$ is empty (max is only attained on the boundary) or $E$ is all of $U$ ($u$ is constant).

Answer 1

The proof is correct. The strong maximum principle "a nonconstant subharmonic function attains its global maximum only on the boundary" does hold for subharmonic functions.

I guess your source was referring to "a nonconstant harmonic function has no local maxima", which indeed does not generalize to subharmonic functions. E.g., $u(x) = \max(x_1, 0)$ is a nonconstant subharmonic function on the unit ball which has a local maximum at every point of the left half of the ball.