# $(\Delta u_n)_n$ bounded in $L^\infty(\Omega) \Rightarrow (\nabla u_n)_n$ locally bounded in some space?

Consider a regular bounded open set $\Omega\subset\mathbb{R}^3$, and a set of regular scalar functions $(u_n)_n\in\mathscr{C}^\infty(\Omega)$ such as $\|\Delta u_n\|_{L^\infty(\Omega)} \leq C$. Is it possible to show that $\|\nabla u_n\|_{L^p_{\text{loc}}(\Omega)}$ is also bounded for some $p\in ]1,\infty[$ ? Or the values on the boundary are too important ?

-
How is the norm in $L^p_{\operatorname{loc}}$ defined? –  Davide Giraudo Jun 29 '12 at 12:11
You're right, Davide, of course I meant $\nabla u_n$ bounded in the mentionned space (which is not normed, but is Fréchet). Anyway... this seems to be false, even in in dimension 1 (see below) –  xounamoun Jun 29 '12 at 13:13
We can work with harmonic function (which obviously satisfy the bounded Laplacian condition but with unbounded gradient). For example $u_n(x)=nx_1$ is such that $\Delta u_n=0$ but $\int_K|u_n|^p=n^p\int_K|x_1|^p$ so for a compact set with non-empty interior, the sequence $\{\lVert u_n\rVert_{p,K}\}$ is not bounded for any $p$.
Sorry for the silly question. Obviously wrong, even in dimension 1 : the sequence $f_n:x\mapsto x^2+nx$ have bounded second derivative though $(f'_n)_n$ is not bounded in $L^1_{\text{loc}}(\mathbb{R})$...
May be it can be mended by demanding additionally that $L_\infty$ or $L^p_{\operatorname{loc}}$ norm of $u$ is bounded. –  Andrew Jun 29 '12 at 17:14