|
1
4
|
Let $\Omega$ be an open set in $R^n$, and $f \in L^1_{loc}(\Omega)$, such that foreach multiindex $\alpha\in N^n$, $|\alpha| = l$ f has weak derivative $D^\alpha f$ in $L^p(\Omega)$, with $1\leq p\leq \infty$. In general it is not true that $f\in L^p(\Omega)$, but it have to be true that $f\in L^p_{loc}(\Omega)$. How can be shown that? |
|||
|
1
|
(Original answer edited to make it shorter) It suffices to show this for $l = 1$. It also suffices to show that $f$ is locally in $L_q$ for some $q \ge p$. But this follows immediately by the Sobolev inequality. |
|||
|
|
|
0
|
Nicolo, I'm not entirely sure if I understand your question but I will attempt an answer. What you're saying seems to be trivial actually. Just take any constant function $f=C$. Then $f \in L_{loc}^1(\Omega)$ and $f$ has a weak derivative lying in every $L^p(\Omega)$. Moreover $f \notin L^p(\Omega)$ but $f \in L_{loc}^p(\Omega)$ for all $p \in [1,\infty]$. |
||
|
|
|
0
|
Actually more is true. It suffices to assume that $f$ is a distribution on $\Omega\subset\mathbb{R}^n$ such that all its distributional derivatives of order $l$ are in $L^p(\Omega)$. Then $f\in L^p_\mathrm{loc}(\Omega)$. A proof based on convolution and the fundamental solution of the polyharmonic operator can be found in the Theorem of Section 1.1.2 in the following book:
|
||
|
|
$f\in W^{1,p}\subset L^p$. Here we don't know that, hence it seems a bit harder. I suspect, though, that an approximation argument using convolution with a mollifier and using Poincaré for each approximand will resolve it. – Harald Hanche-Olsen Apr 4 2010 at 22:30