I'm studying Sobolev Space and I have a question about the definition:

Def.: The Sobolev Space $W^{k,p}(U)$ consists of all locally summable functions $u:U\to \mathbb{R}$ such that for each multiindex $\alpha$ with $|\alpha|\geq k,$ $D^\alpha u$ exist in the weak sense and belongs to $L^p(U).$

Observation: If $k=0$ and $p=2$, then $W^{0,2}(U)=L^2(U)$.

We have that $ u\in W^{k,p}(U)$ since $u\in L^1_{loc}(U),$ so every $u\in L^2(U)$ belong to $L^1_{loc}(U)?$

How can I show that? Thanks.

  • $\begingroup$ Every $u\in L^2(U)$ is in $L^1_{loc}(U)$, which has nothing to do with Sobolev space. $\endgroup$ – user99914 Jun 2 '15 at 1:35
  • $\begingroup$ It's a question about the definition, just it. $\endgroup$ – Irddo Jun 2 '15 at 1:37
  • $\begingroup$ So are you asking why an $L^2$ function is in $L^1_{loc}(U)$? $\endgroup$ – user99914 Jun 2 '15 at 1:48
  • $\begingroup$ Yes, is it. Cause, in the book $W^{0,2}(U)$ is identified with $L^2(U)$, and I don't see why $(\int_U |f(x)|^2dx)^{1/2}$ finite implie $\int_V |f(x)|dx$, for every $V\subset\subset U$. Sorry if I was not clear. $\endgroup$ – Irddo Jun 2 '15 at 1:54

Indeed this is just Holder's inequality: Pick $V \subset U$ with $|V|<\infty$, then

$$||u||_{L^1(V)}=\int_V |u| dx\le \sqrt{\int_V 1^2 dx}\sqrt{\int_V |u|^2 dx} = \sqrt{|V|}\ ||u||_{L^2(U)}$$

then $u \in L^1_{loc}(U)$.


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