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Consider a continuous $H^1$ function $u$ on a bounded open set $\Omega \subset \mathbb{R}^n$. We additionally have that $|\Delta u|^2 \leq c |\nabla u|^2$ pointwise on $\Omega \setminus \Sigma$, where $\Sigma$ is an embedded hypersurface. Also, if $X$ denotes the unit normal vector field to the hypersurface $\Sigma$, we additionally have $Xu = 0$. I am trying to find a reference that says that $u$ is locally Lipschitz. I am looking in Gilbarg-Trudinger, but am unable to find it. Thanks for your help!

Note: I edited the question after Denis Serre's answer (see comments below).

Further note: I edited the question again after Michael Renardy and Connor Mooney's comments.

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    $\begingroup$ Write the equation as $\Delta u + b \cdot \nabla u = 0$ where $b$ is bounded. If $\Sigma$ is flat one can take the even reflection and reduce to interior regularity (apply the answer of Denis Serre). If not one can take similar extensions or flatten the boundary (getting an equation of the same form with smooth leading order coefficients) and do the same. This is related to boundary regularity for the Neumann problem. $\endgroup$
    – Connor Mooney
    Commented Jan 28, 2016 at 22:57

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This is done by bootstrapping, and you'll find the ingredients in G.-T. By assumption, you know that $\Delta u\in L^2(\Omega)$. By elliptic regularity, you obtain $u\in H^2_{loc}$. Hence $\nabla u\in H^1_{loc}$. By Sobolev embedding, $\nabla u\in L^p_{loc}$ for $\frac1p=\frac12-\frac1n$. By your assumption, the same is true for $\Delta u$, and by elliptic regularity, $u\in W^{2,p}_{loc}$. And so on, your apply circularly 1) your assumption, 2) elliptic regularity, 3) Sobolev embedding. At each step, you get $\nabla u\in L^{q_k}_{loc}$, where $\frac1{q_k}=\frac12-\frac kn\,$. When $k>\frac n2$, you obtain actually that $\nabla u$ is Hölder continous. In particular, $u$ is locally Lipschitz. all

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  • $\begingroup$ Actually I wrote the question a bit wrong. Of course, statements like $|\Delta u|^2 \leq c|\nabla u|^2$ make no sense pointwise if $u$ is just continuous and $H^1$. It turns out that $u$ satisfies the above inequality pointwise everywhere except on an embedded hypersurface. But I think what you wrote still goes through? $\endgroup$
    – student
    Commented Jan 28, 2016 at 18:11
  • $\begingroup$ You do not know that $\Delta u\in L^2$. $\Delta u$ might include a surface delta function if $\nabla u$ is discontinuous across $\Sigma$. $\endgroup$
    – Michael Renardy
    Commented Jan 28, 2016 at 19:51
  • $\begingroup$ The function $r^{1/2}sin(\theta/2)$ on $\mathbb{R}^2$ satisfies the desired conditions (it's harmonic away from the positive $x$-axis, continuous and $H^1$) and is not Lipschitz. In this example the gradient is discontinuous on $\Sigma$ so (as Michael Renardy's comment points out) the Laplace concentrates there. $\endgroup$
    – Connor Mooney
    Commented Jan 28, 2016 at 20:24
  • $\begingroup$ When I answered the question, the assumption was pointwise, not away from a hypersurface. $\endgroup$
    – Denis Serre
    Commented Jan 28, 2016 at 21:03
  • $\begingroup$ @DenisSerre Sorry for changing the question on you, I have added a note and edited it further. $\endgroup$
    – student
    Commented Jan 28, 2016 at 21:28

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