Let $u \in C^2(\Omega)$, with $\Omega \subset \mathbb{R}^2$ open. How do I prove that if $u$ is subharmonic, that is $\Delta u \ge 0$, then for all $P=(x,y) \in \Omega$, for all $r>0$ such that $\overline{B}_r(P) \subset \Omega$, we have $$ u(P) \leq \frac{1}{|\partial B_r(P)|}\int_{\partial B_r(P)}u(s)ds, $$ without using the divergence theorem or other similar results from vector calculus?
Prove $u$ subharmonic $\implies$ $ u(P) \leq \frac{1}{|\partial B_r(P)|}\int_{\partial B_r(P)}u(s)ds$ (in $\mathbb{R}^2$ without divergence theorem)
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real-analysis
complex-analysis
ordinary-differential-equations
pde
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0By $f(x)$ do you mean $u(P)$? – 2017-01-17
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0@ByronSchmuland Yes of course. Thanks for the correction. – 2017-01-17
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0Let $h$ be the solution of the Dirichlet problem on $B_r(P)$ with boundary values $u$. What do you know about the relation betwen $h$ and $u$? – 2017-01-17
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0@DanielFischer I'd like to prove this without using PDE arguments (Dirichlet problems, etc.) However, I also want to try this way. We know that $\max_{\partial B_r(P)} u = \max_{B_r(P)} h$. What follows from this? – 2017-01-17
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0Can you use Brownian motion and submartingales? – 2017-01-17
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0@ByronSchmuland No, I cannot. I can use definitions and the mean value property for harmonic functions and the maximum principle for harmonic functions. – 2017-01-17
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0We know $u - h \leqslant 0$ on $\partial B_r(P)$. And $u - h$ is subharmonic. If you have some maximum principle that applies to subharmonic functions, that will tell you something useful about $u(P) - h(P)$. – 2017-01-17
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0@DanielFischer I do not have a maximum principle for subharmonic functions. I have one for functions that satisfy the mean value inequality. In fact, this is why I'm posing this question. – 2017-01-17
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0Okay. Then start with proving that a _strictly_ subharmonic ($C^2$) function (that is, $\Delta f > 0$ everywhere) cannot have a local maximum. After that, consider $u_{\varepsilon} \colon (\xi,\eta) \mapsto u(\xi,\eta) + \varepsilon((\xi - x)^2 + (\eta - y)^2)$ for $\varepsilon > 0$. If $h_{\varepsilon}$ is the solution to the Dirichlet problem with boundary values $u_{\varepsilon}$, then since $u_{\varepsilon} - h_{\varepsilon}$ is strictly subharmonic, it follows that $u_{\varepsilon}(P) - h_{\varepsilon}(P) < 0$. – 2017-01-17
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0See this link https://www.math.ucdavis.edu/~hunter/pdes/ch2.pdf for the answer. – 2017-01-17