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Let $\Omega$ open bounded be given and $G(x,y)$ denote the Green function in Evans setting. That is, we have $$ \Delta_yG(x,y)=\delta_{x-y} $$ and $$ G(x,y)=0 $$ if $y\in\partial\Omega$.

Now for $u$ we define $$ u(x):=\int_\Omega G(x,y)f(y)dy $$ where $f(y)$ is bounded and integrable. I want to prove that $$ u(x)\to0 \,\,\,\text{ as }x\to \partial\Omega$$

This is intuitively right but I can not prove it in details... Please help me with that. Thank you!

spatially
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1 Answers1

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Here's an outline

First of all note that, by the maximum principle, there is a constant $c=c_n$ (the normalization constant in the definition of the fundamental solution for the laplacian), so that $|G(x,y)|\leq c|x-y|^{2-n}$. Therefore, by a variant of the dominated convergence theorem, it's enough to show that $$ \int_\Omega \frac{|f(y)|}{|x-y|^{n-2}} dy \to \int_\Omega \frac{|f(y)|}{|z-y|^{n-2}}dy, \qquad \text{ when } x\to z\in \partial \Omega. $$ This is easy to see: Clearly the integrands converge pointwise a.e. in $\Omega$, and moreover the functions $f_x(y)=|x-y|^{2-n}$ are uniformly bounded (in $x$) in $L^p(\Omega)$ for $p<n/(n-2)$, and thus they're weakly convergent as $x\to z$ by uniqueness of the pointwise and weak limits. But since $f\in L^\infty(\Omega) \subset L^{p'}(\Omega)$, we conclude the desired limit.

Jose27
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  • I understand your answer but could you explain me more that why the weak limit and a.e. limit has to agree? – spatially Dec 10 '14 at 03:08
  • @wisher: It's a rather technical point: Look for Mazur's lemma, which says that if a sequence converges weakly, then a sequence of convex combinations of the original sequence converge in norm. This limit therefore coincides with the weak and pointwise limit (since convex combinations dont' affect the limit, and a subsequence converges pointwise). – Jose27 Dec 10 '14 at 04:09
  • Sorry I still confused. I checked Mazur's lemma. Yes, I can have a sequence of convex combinations and it convergence strongly. But I don't get "since convex combinations dont' affect the limit" this part. In my opinion it does affect the limit point-wisely... Or I am making some mistake. Could you please write some details to explain this sentence in your answer? Thx a lot!! – spatially Dec 10 '14 at 22:19
  • If a sequence $f_n$ converges (pointwise or weakly) to $f$ then, if $\sum_i a_if_{j_i}$ is a convex combination, meaning $\sum_i a_i=1$, then you can estimate the distance from this to the function $f$ by writing $f=\sum_i a_i f$. From this, it's clear that a sequence of convex combination can't affect either a weak or pointwise limit. – Jose27 Dec 10 '14 at 23:44
  • Oh I see. You are keep taking the tail terms! Sorry in H.B. book he is keep taking terms from $i=1\ldots, n$ to make the convex combination so that makes me very confused. Thx! – spatially Dec 11 '14 at 00:31
  • Actually a convex combination is necessarily a finite sum. But this is not a problem since, for instance in the case of convergence a.e. $$\sum_i a_i|f_{j_i}(x)-f(x)|\to 0$$ as $j_i\to \infty$. – Jose27 Dec 11 '14 at 02:26