Let $\Omega \subset \mathbb{R}^n$ and $f\in C^2(\Omega)$ such that $\Delta f(x)\leq 0$ for some $x\in\Omega$. Thus, can we derive: $f(x)\geq \frac{1}{|B_r(x)|}\displaystyle\int_{B_r(x)}f(y)\,dy $
where $B_r(x)\Subset \Omega$? Or is there really $\Delta f(x)\leq 0$ for all $x\in\Omega$ needed?
Thanks.
It's well known that if $f$ is subharmonic and smooth then $\Delta f\le0$. Perhaps less well known is the fact that the mean-value inequality at a single point implies the inequality for the Laplacian at that point. This is clear if you know the result expressing the Laplacian in terms of spherical averages; I don't recall the result so I'll give the proof first: If $f\in C^2(\Bbb R^2)$ then Taylor's theorem says $$f(x,y)=f(0,0)+Ax+By+Cxy+Dx^2+Ey^2+o(x^2+y^2).$$Note that $$\Delta f(0,0)=2D+2E$$while $$\frac1{\pi r^2}\int_{D(0,r)}f(x,y)\,dxdy=f(0,0)+\frac12(D+E)+o(1).$$So