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Let $u\in C^2(\Omega)$ be such that $\Delta u \ge 0$ on $\Omega\supset \overline{B(a,r)}$. We consider the Poisson modification $U$ of $u$ for the ball $B(a,r),$ that is $U$ equals $u$ on $\Omega-B(a,r)$ and that on $B=B(a,r)$ equals the solution to Direchlet problem with boundary data $u|_{\partial B}$, which is given by the Poisson kernel classically denoted by $P(x,y)$. It is known that $U$ is subharmonic in the sense that it verifies an inequality mean property. My question is : Do we have $U\in H^2(\Omega)?$.

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I must be misunderstanding the question. Assuming $\Omega$ is intended to be an open domain, what's to prevent $u$ from blowing up fiercely near the boundary of $\Omega$? Adjusting it in a compact ball in the interior won't help, will it? –  Andreas Blass Aug 13 '12 at 3:40
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Ah, wait a minute: could you please remind us of how you are defining $H^2(\Omega)$? –  Yemon Choi Aug 13 '12 at 4:56
    
$H^2(\Omega)$ is the set of functions $u\in L^2(\Omega)$ such that $u', u"\in L^2(\Omega)$, where the derivation is in the sense of distribution. –  hardy Aug 13 '12 at 13:17
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2 Answers 2

Note that in your assumption $u$ itself need not be in $H^2(\Omega)$, nor even $L^2(\Omega)$, even if it is harmonic (so $u=U$). You are possibly interested on the local behaviour, that is whether $U\in H^2_{loc}$. But note that for $n=1$ any smooth, convex function is subharmonic; the function $U$ is affine on $(a-r,a+r)$, and in general not in $C^1$ (unless $u$ was already harmonic ), hence not $H^2_{loc}$. However, I think in your assumptions it is true that $U\in H^1_{loc}$.

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thanks. In fact I m interested with a modification $v$ of $u$ ($u$ may be taken smooth enough) such that $v\in H^2(\Omega)$ and harmonic in $\Omega-\overline{B}(a,r)$. –  hardy Aug 13 '12 at 13:53
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Subharmonicity of this modification is true and easy to prove just from the definition of a subharmonic function. Condition $C^2$ is redundant. Of course the modification will not be in $C^2$. Now what is $H(\Omega)$ ?

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Here $H^2(\Omega)$ is the Sobolev space of $u\in L^2(\Omega)$ such that its distribution derivatives $u',u"$ are in $L^2(\Omega)$. –  hardy Aug 13 '12 at 13:57
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