About the sub-mean value property

Question

Let $\Omega$ be an open subset of $\mathbb{R}^N$. We say that a function $u\in C(\Omega)$ has the (surface) sub-mean value property in $\Omega$ if (1) $u(x)\leq\frac{1}{|\partial B_r(x)|}\int_{\partial B_r(x)} u(y)dSy$ for any ball $\bar{B}_r(x)\subseteq\Omega$. We say that a function $u\in C(\Omega)$ has the (volumetric) sub-mean value property in $\Omega$ if (2) $u(x)\leq \frac{1}{|B_r(x)|}\int_{B_r(x)} u(y)dy$ for any ball $\bar{B}_r(x)\subseteq\Omega$. A function that satisfies the condition (1) is called subharmonic on $\Omega$. I know that this is equivalent to (if $u\in C^2(\Omega)$) $\Delta u\geq 0$, and that (1) implies (2), but what i wish to know is if (2) implies (1). I think that is not the case but i cannot find a counterexample (The analogue version in which there is the equalities are equivalent).

Answer 1

I am not sure about the case $u \in C(\Omega)$ but if $u \in C^2(\Omega)$, then (2) does indeed implies (1). So, in all what follows, let us assume that $u \in C^2(\Omega)$.

Since you already know that $\Delta u \geq 0$ implies (2), I assume that you also know the implication $$\Delta u \leq 0\phantom{a}\text{on }\Omega \phantom{aaa}\Longrightarrow\phantom{aaa}u(x) \geq \frac{1}{|B_r(x)|}\int_{B_r(x)} u(y) d y\phantom{aa}\text{for all }\overline{B_r(x)} \subseteq \Omega$$ This follows from "$\Delta u \geq 0$ $\Rightarrow$ (2)" by multiplication with $-1$.

Now assume (2) holds. It is sufficient to show $\Delta u \geq 0$ since you already know that this is equivalent to (1).

Let's prove $\Delta u \geq 0$ by contradiction: Assume this is false. Then, by continuity of $\Delta u$, there is an open $\Omega_0 \subseteq \Omega$ such that $\Delta u < 0$ on $\Omega_0$. By my remarks above, this implies $$u(x) \geq \frac{1}{|B_r(x)|} \int_{B_r(x)} u(y) d y\phantom{aa}\text{for all }\overline{B_r(x)} \subseteq \Omega_0$$ If we combine this with (2), we get $$u(x) = \frac{1}{|B_r(x)|}\int_{B_r(x)} u(y) d y\phantom{aa}\text{for all }\overline{B_r(x)}\subseteq\Omega_0$$ But this means $\Delta u =0$ on $\Omega_0$ which contradicts our hypotheses.