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The monograph "Non-Homogeneous Boundary Value Problems and Applications" by Lions and Magenes is infamous for developing a truly extensive regularity theory for elliptic problems on domains, and for doing so under amazingly restrictive assumptions on the regularity of the boundary - basically, the domain is always assumed to be $C^\infty$.

Here comes one of their classical results, obtained by combining Theorems 2.7.3 and 2.7.4 in their first volume, when specialized to the case I am interested in:

Theorem. The trace operator $\gamma$ is bounded from $$ > D^\frac{1}{2}(\Omega):=\left\lbrace > u\in H^\frac{1}{2}(\Omega): \Delta > u\in > \Xi^{-\frac{3}{2}}(\Omega)\right\rbrace > $$ to $L^2(\partial \Omega)$, and furthermore $$\begin{pmatrix} \Delta \atop > \gamma\end{pmatrix} $$ is an isomorphism from $D^\frac{1}{2}(\Omega)$ to $\Xi^{-\frac{3}{2}}(\Omega)\times > L^2(\partial \Omega)$.

(Here $\Xi^{-\frac{3}{2}}(\Omega)$ is a rather ugly interpolation space, which however is nice enough to contain $L^2(\Omega)$).

In particular, it follows that

Corollary. $u\in H^\frac{1}{2}(\Omega)$ whenever $u$ satisfies $\Delta u=f$ for some $f\in > L^2(\Omega)$.

Many results of Lions-Magenes' have been extended to the case of $C^{1,1}$-domains, or even to general convex bounded domains, most notably in Grisvard's "Elliptic problems in nonsmooth domains", but I was not able to find an extension of the above Corollary. What I am interested in is simply the case of the hypercube $\Omega:=(0,1)^N$, that is I am asking the following

Question. Let $u$ solve $\Delta u=f$ for some $f\in L^2\left((0,1)^N\right)$. Is it true that $u\in H^\frac{1}{2}\left((0,1)^N\right)$?

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Do you want to add a boundary condition, or is this result independent of that? –  Daniel Spector Feb 20 '13 at 18:37
    
it is independent of the boundary condition, that is exactly the point. –  Delio M. Feb 20 '13 at 18:46
    
I think it should be worded differently, because if $u \in H^\frac{1}{2}$ then $u \in L^2$, and by the PDE this implies $u \in H^1$ (or better). Do you mean the fractional semi-norm of $u$ is finite, in this case? –  Daniel Spector Feb 21 '13 at 8:01
    
I don't see how this implies $u\in H^1$ (or better) by the general PDE theory. All I am aware of is that you do have $H^1$ (indeed, usually even $H^2$) if you can apply Gauß-Green so that you can set up a variational formulation and apply Lax-Milgram. But in this case this is not possible, since we have no idea of what happens at the boundary. –  Delio M. Feb 21 '13 at 8:20
    
I guess then the first question is what do Lions-Magenes mean by $\Delta u=f$ for $f \in L^2(\Omega)$? Likely these quantities are defined almost everywhere and so it is in some integral sense. Maybe the $C^\infty$ assumption is to do the extension and use Fourier transforms or some other technique like this on the whole space? I will look for a copy of it in the library and get back to you. –  Daniel Spector Feb 21 '13 at 9:14
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2 Answers

Ok. So I thought a little more carefully about it, though I still have not seen the Lions-Magenes book, but I think $C^\infty$ boundary is related to the above result holding for more general spaces (for example, $D^{n+\frac{1}{2}}$ for $n \in \mathbb{N}$. If this is the case, the smoothness is probably needed for an extension as I mentioned above.

Is your corollary straight out of the book? Since the logical conclusion of the theorem you mentioned is

1) If $u \in H^\frac{1}{2}$ and $\Delta u$ is in a funny space, then $u|_{\partial \Omega} \in L^2(\partial \Omega)$.

and

2) If $\Delta u$ is in a funny space and $u|_{\partial \Omega} \in L^2(\partial \Omega)$ then $u \in H^\frac{1}{2}$.

That is, from my reading of the theorem (and thinking about Craig's comment) is that you do need the boundary data. However, $C^\infty$ should not be needed for the boundary, only smooth enough to extend an $H^\frac{1}{2}$ function to all of $\mathbb{R}^N$. Hence, the cube should be sufficient.

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I tried to add a comment but it wouldn't let me, so I am adding a comment in the answer box.

Something to me seems a bit funny (at least to me) in the first corollary stated above. Are you assuming any apriori conditions on $u$ or are you just assuming $ \Delta u = f$ in $ \Omega$ with $ f \in L^2(\Omega)$. If you are not assuming any conditions is it true? For instance take $ u(x)=|x|^{2-N}$ with $ \Omega$ some domain with $ 0 \in \partial \Omega$. In dimensions $N \ge 4$ we have $ u $ not even in $L^2\Omega)$. And if you don't like taking the singularity on the boundary take it at point outside the set and approaching the boundary.

Or maybe i am completely missing the point.

Craig

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Then again, for the function you mention, $\Delta u$ is not in $L^2(\Omega)$ either. The singularity at $0$ will mean the Laplacian exists in the sense of distributions and is a measure, which is not in the space of square integrable functions (i.e. $f=\delta_0$). –  Daniel Spector Mar 5 '13 at 14:24
    
true. But the test functions won't see it (assuming we are using compactly supported test functions). Or instead take $ x_m \rightarrow 0 $ with $ x_m \notin \Omega$ and set $ u_m(x)= |x-x_m|^{2-N}$. One will not be able to get an $L^2$ estimate on $u_m$ and yet $ \Delta u_m = 0 $ in $ \Omega$. –  Craig Mar 5 '13 at 18:28
    
I guess I am saying that we do not have $\Delta u =0$ or $\Delta u_m=0$. In fact, we have $\Delta u = \delta_0$ and $\Delta u_m = \delta_{x_m}$, so the equation satisfied by $u,u_m$ is with a right hand side in $(C_0(\Omega))^\prime$, and therefore the standard estimates can not be used, and something else is needed. –  Daniel Spector Mar 13 '13 at 8:36
    
How do you define a continuous linear function on $C_0$? Any reasonable definition is going to have $ \Delta u_m=0$ in the dual of C_0. I think the issue is with the interpretation of the original result. –  Craig Mar 13 '13 at 20:14
    
Ok. Sorry for the delay. I read what you wrtoe more closely and apologize for missing the $0 \in \partial \Omega$. However, the question you mention is not unique, since in general, the solution $u=0$ is a regular solution. So maybe the question is not correct in asserting solvability in the right space versus the operator is bounded from one space to the other (and in fact, as you show, could be bounded on some functions not in this space). –  Daniel Spector Mar 16 '13 at 16:38
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