Is this bootstrap argument correct?

The setup. Assume $\Omega \subset \mathbb{R}^3$ bounded and bilipschitz equivalent to the unit cube and has smooth boundary. Let $v \in W^{1,2}(\Omega,\mathbb{C})$ be a weak solution of $$\begin{cases} -\Delta v = g & \text{in } \Omega \\ v=0 & \text{on } \partial \Omega\end{cases} \tag{1}$$

Now assume we can show that $g \in L^{12/11}(\Omega)$. From standard elliptic theory (theorem 7.4 in Giaquinta's introduction to elliptic systems) we may infer that $v \in W^{2,12/11}(\Omega)$.

The problem. Now, I even know that $D g (:=\nabla g) \in L^{12/11}(\Omega)$ and want to infer that $v \in W^{3,12/11}$ and I want to use a bootstrap argument to do so.

My attempt. Since $v$ is a weak solution to (1) we know that $$\int_\Omega Dv Dw = \int_\Omega g w$$ for all $w \in H_0^1(\Omega)$.

We choose such test function $w \in C_c^\infty(\Omega)$ and set $$u:=-Dw.$$

Inserting $u$ for $w$ in the equation above, we get $$-\int_\Omega DvD^2w = -\int_\Omega gDw.$$ Integration by parts yields $$\int_\Omega D^2 v Dw = \int_\Omega Dg w.$$

This means that $\widetilde v = Dv$ is also weak solution to $$\begin{cases} -\Delta \widetilde v = Dg & \text{in } \Omega \end{cases}$$

The question. How would one then conclude that $\widetilde v = Dv \in W^{2,12/11}(\Omega)$? Note that we don't have boundary conditions anymore. Would it be easier to withdrawing to an open subset $U \subset \Omega$? How would that exact argument go?

• How did you concluded that $Dv$ is a solution of $(2)$? – Tomás Aug 20 '13 at 15:13
• I disagree with "From standard elliptic theory we may infer ..." If you want $v\in W^{2,p}$ globally in $\Omega$, you need some assumptions on the geometry of $\partial \Omega$. If you only assume $\Omega$ bounded, then the best you can get is $W^{2,p}$ on compactly contained subdomains. Is this what you are after? (I recall your other question with $\omega \Subset \Omega$) – user90090 Aug 20 '13 at 17:38
• Well, that's my question. I dont think it must satisfies the boundary condition. Anyway, as @user90090 pointed out, if you have aenough regularity in the boundary, then your assertion is true. You can see the proof in Brezis book chapter 9 in regularity part. Take a look there, it will help you. – Tomás Aug 20 '13 at 19:08
• If you need only interior regularity, then the boundary regularity is not need. The proof that can be found in Brezis book can easily be adapted to your more simple case, however, maybe there is a more elementary proof, I really don't know it. But it is good to know that in Brezis proof, both interior regularity and regularity up to the boundary are treated, in basically, the same way. – Tomás Aug 21 '13 at 19:52
• I think that the term distributional solution is better applied for $\overline{v}$. By using density argument, You can show that your equation is also true in the more Big space $H_0^1$. Then you can use the saem reasoning of Brezis for interior regularity. Anyway, wait more, maybe someone can give a better answer. – Tomás Aug 22 '13 at 12:43

As I understand, the question is: If $g\in W^{1,p}(\Omega)$ can we show that $v\in W^{3,p}(\Omega)$, where $1<p<\infty$?
To answer this let me first assume that the boundary of $\Omega$ is smooth. Nonsmooth boundaries will be discussed afterwards.
We start by considering a sequence $\{g_n\}\subset C^\infty(\bar\Omega)$ of smooth functions, converging to $g$ in $W^{1,p}$, and solve $$\Delta v_n = g_n\quad\textrm{in }\Omega, \qquad v_n=0\quad\textrm{on }\partial\Omega.$$ Because $g_n$ are smooth, by for instance the $L^2$ theory all $v_n$ are smooth. Moreover, as you mentioned, the basic $L^p$ theory gives the estimate $$\|v_n\|_{W^{2,p}} \leq c\|g_n\|_{L^p}.\qquad\qquad(*)$$ Recall also the estimate $$\|u\|_{W^{k,p}} \leq c\|\Delta u\|_{W^{k-2,p}} + c\|u\|_{W^{k,p}(\partial\Omega)} + c\|u\|_{L^p},\qquad\qquad(**)$$ that is true for any $u\in W^{k,p}(\Omega)$ with $k\geq2$ and $1<p<\infty$. By applying this estimate to $v_n$, we have $$\|v_n\|_{W^{3,p}} \leq c\|g_n\|_{W^{1,p}} + c\|v_n\|_{L^p} \leq c\|g_n\|_{W^{1,p}}.$$ So $v_n$ are uniformly bounded in $W^{3,p}$, which means that there is $w\in W^{3,p}(\Omega)$ such that $v_n\to w$ weakly in $W^{3,p}$, and by compactness, strongly in $W^{2,p}$. This implies that $\Delta w = g$ in $\Omega$, because $$\|\Delta w - g\|_{L^p} \leq \|\Delta(w-v_n)\|_{L^p} + \|g_n-g\|_{L^p} \leq \|w-v_n\|_{W^{2,p}} + \|g_n-g\|_{L^p}.$$ Trace of $w$ on $\partial\Omega$ is zero because the trace is preserved under weak convergence (or if you want, the trace map is continuous from $W^{1,p}(\Omega)$ to $L^p(\partial\Omega)$). To conclude, we have $w=v$, and thus $v\in W^{3,p}(\Omega)$.
Now let me discuss a bit about nonsmooth boundaries. A good (and perhaps primary) reference on this subject is Grisvard's book. Of course, you have the standard interior estimates regardless of the boundary regularity. Only the estimates up-to-boundary depend on how regular the boundary is. In Grisvard's book, the basic $L^p$ theory, including the estimate $(*)$, was proved assuming that the boundary $\partial\Omega$ is $C^{1,1}$, and the estimate $(**)$ was proved for $C^{\,2,1}$ boundaries. I think you get the drift: If you want $v\in W^{k,p}$ then you need a $C^{\,k-1,1}$ boundary. This cannot be improved in general: For instance in polyhedral domains, generally $v\not\in H^2(\Omega)$ even for smooth right hand side $g$. For convex domains though, assuming for instance a Lipschitz condition, at least $(*)$ is true. I expect $(**)$ to be also true. You might be able to prove it if you follow the steps in Grisvard.
• @BeniBogosel: For $\phi$ an arbitrary smooth function on the boundary, we have $\int_{\partial\Omega}v_n\phi=0$. By weak convergence, $\int_{\partial\Omega}w\phi=0$. Since $\phi$ is arbitrary, $w=0$ on $\partial\Omega$. – timur Nov 28 '14 at 20:43