Assume $\Omega$ is some open, bounded domain with smooth boundary – say $\Omega = B(0,1) \subset \mathbb{R}^3$. Let $v$ be a solution of the Laplace equation \begin{equation} \begin{cases} \Delta v =0 & \mbox{on } \Omega \\ v=f|_{\partial\Omega} & \mbox{on } \partial \Omega \end{cases} \end{equation} and assume furthermore $f \in L^1_{loc} (\overline{\Omega}) \cap L^4(\overline{\Omega})$ and $Df \in L^2(\overline{\Omega})$.
Can one prove that $v \in L^4$?
What I found so far: this works for $f \in C(\overline{\Omega})$ by the Poisson Integral Formula, but my $f$ is not as smooth (as I can't assert that $f \in H^4$).
Any help is much appreciated!
Best Answer
First note that $f\in L^4(\Omega)$ implies $f\in L^1(\Omega)$, hence the hypothesis $f\in L^4(\Omega)\cap L^1_{loc}(\Omega)$ is unnecessary. Now, consider the problem
$$\tag{P} \left\{ \begin{array}{cc} \Delta v=0 &\mbox{ in $\Omega$} \\ v=f &\mbox{ in $\partial\Omega$} \end{array} \right. $$
It is well know that if $f\in H^{1/2}(\partial\Omega)$, then problem $(P)$ has a unique solution $v\in H^1(\Omega)$ satisfying $(P)$ in the weak sense. But $H^1(\Omega)$ is contained in $L^{2^\star}$ (Sobolev Embedding), where in our case $$\tag{1}2^\star=\frac{2N}{N-2}=6$$
From $(1)$ we conclude that $v\in L^4(\Omega)$.
Remark 1: $f\in L^4(\Omega)$ with $Df\in L^2(\Omega)$ implies that $f\in H^1(\Omega)$, which implies that $\operatorname{trace}(f)\in H^{1/2}(\partial\Omega)$
Remark 2: To solve problem $(P)$ we procced as follows:
Claim: The solution $v\in H^1$ is characterized by $$\tag{2}\int_\Omega |\nabla v|^2=\min\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$$
Denote $\mathcal{K}=\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$
First note that $\mathcal{K}$ non empty, because $\operatorname{trace}:H^1(\Omega)\to H^{1/2}(\Omega)$ is surjective, hence, we can take a minimizing sequence in $\mathcal{K}$, i.e. a sequence $u_n\in\mathcal{K}$ satisfying $$\int_\Omega |\nabla u_n|^2\to \inf\mathcal{K}$$
Try to prove that $\|\nabla u_n-\nabla u_m\|_2\to 0$ as $n,m\to\infty$. Note that $u_m-u_n\in H_0^1(\Omega)$, hence, by Poincare inequality we can conclude that $$\|u_n-u_m\|_{1,2}\to 0$$
This implies the existence of some $v\in H^1(\Omega)$ such that $u_n\to v$ in $H^1$. Now you can conclude.
Remark 3: Let me propose you another way to solve this problem.
Let $K=\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$ and define $F:K\to\mathbb{R}$ by $$\tag{3}F(u)=\int_\Omega|\nabla u|^2$$
First note that $K$ is closed and convex. Now, try and show:
I - $F$ is coercive, i.e. if $\|u\|_{1,2}\to\infty$ in $K$, then $F(u)\to\infty$,
II - $F$ is weakly lower semicontinuous, i.e. if $u_n\in K$ weakly converges to $u\in K$, then $F(u)\leq\liminf F(u_n)$,
III - $F$ is convex.
I, II and III implies that $F$ is minimized by some $v\in K$ which implies that $F'(v)=0$.