Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
@Leonid: The Poincaré inequality presupposes that $f\in W^{1,p}\subset L^p$. 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 '10 at 22:30

3 Answers 3

(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.

share|improve this answer

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 $p \in [1,\infty]$.

share|improve this answer

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:

Vladimir G. Maz'ja, Sobolev spaces. Translated from the Russian by T. O. Shaposhnikova. Springer Series in Soviet Mathematics. Springer-Verlag, Berlin, 1985. ISBN: 3-540-13589-8

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.