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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.

HelloEveryone
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    Whether this follows from that inequality depends on what you mean by "this": The inequality certainly does not imply that $f\in C^2$, but if you assume that $f\in C^2$ then $\Delta f\le0$ does imply the inequality.. – David C. Ullrich Jul 04 '22 at 11:19
  • Thank you and sorry: Yes we know that $f\in C^2(\Omega)$. So to be clear: it suffices if $\Delta f(x) \leq 0$ only for that one point $x\in\Omega$ without knowing anything about all the other $y \in \Omega$? – HelloEveryone Jul 04 '22 at 11:24

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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

$\frac1{\pi r^2}\int_{D(0,r)}f(x,y)\,dxdy\le f(0,0)$ for all $r>0$ implies $\Delta f(0,0)\le0$.

David C. Ullrich
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  • Thanks again. But I would need the other implication. Or maybe I am misunderstanding you...But I can't see how to derive that the mean value inequality holds for $f$ only if $\Delta f(x) \leq 0$ for only " the one" $x\in\Omega$ instead of $\Delta f \leq 0$ on the whole of $\Omega$... – HelloEveryone Jul 04 '22 at 12:31
  • Maybe I misunderstood you! I guess with subharmonic you mean exactly that: that there is a single point $x\in\Omega$ with $\Delta f(x) \leq 0$ right? And then, if $f$ is smooth, it follows that $\Delta f\leq 0$ on the whole of $\Omega$? And thus one would get the above mean value inequality... If you do, I was not aware of this! – HelloEveryone Jul 04 '22 at 12:47
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    Yes, when I said subharmonic I meant exactly subharmonic. Which is not what you go on to conjecture I meant! The mean-value inequality at a single point does not imply that $f$ is subharmonic. You've had a few questions about what I meant by this or that. Fyi in each case what I meant was exactly what I wrote. – David C. Ullrich Jul 04 '22 at 19:12
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    You seem to be sort of mixing half of one sentence with half of the next sentence. The first paragraph reads "It is well known that WK, perhaps less well known is LWK", then I go on to prove LWK. Your questions sound like you think my proof of LWK has something to do with WK. It does not. LWK never mentins subharmonic functions, for example. – David C. Ullrich Jul 04 '22 at 19:15