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Let

  • $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open

Let $u\in H_0^1(\Lambda)$ admit a weak Laplacian $\Delta u\in L^2(\Lambda)$. Are we able to show that $$\nabla u\in L^p(\Lambda,\mathbb R^d)\tag1$$ for some $p>2$?

If

  • $\partial\Lambda\in C^2$ or
  • $\Lambda$ is convex,

then we can even show that $u\in H^2(\Lambda)$ and the desired conclusion is possible (for small $d$) by the Sobolev inequality. However, I wondered whether we can show the weaker result $(1)$ without one of these assumptions. It would be fine for me, if we assume

  • $d\le 3$,
  • $\Lambda$ is connected or
  • $\partial\Lambda$ is Lipschitz
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    What did you get exactly for $d\le 3$ ? – 2017-02-03
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    By the Gagliardo-Niremberg interpolation inequality (https://en.wikipedia.org/wiki/Gagliardo%E2%80%93Nirenberg_interpolation_inequality) the supremum of the set of $p$s such that $\Delta u\in L^p$ should depend on $d$, too. – 2017-02-03
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    @user1952009 I'm $d\le3$, then $H_0^1(\Lambda)\subseteq L^6(\Lambda)$ by the Sobolev inequality. But that doesn't help. Maybe I've missed your point. – 2017-02-03
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    @JackD'Aurizio I'm sorry, but I think that I don't understand what you're telling me. I don't what that $\Delta u\in L^p(\Lambda)$ for some $p>2$. I want that $\Delta u\in L^2(\Lambda)$ implies $\nabla u\in L^p(\Lambda,\mathbb R^d)$ for some (preferably large) $p>2$. In fact, I hope for $p=4$ at least in the case $d\le3$. – 2017-02-03

1 Answers 1

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Look at Theorem 3.7 in http://www.math.lsu.edu/%7Epcnguyen/papers_html/Adi_Phuc_10_07_14.pdf for a result in this direction.

If $-\Delta u = f \in L^{2+\epsilon}$, then the answer to your question is true and this is given in the link I posted.

If you only take $-\Delta u =f \in L^2$, then I am not aware of higher integrability results. I would probably expect the higher integrability you are asking for to be false.