I've studied Functional Analysis and Sobolev Spaces from different references and what I noticed is that sometimes people consider $L^1_{loc}$ to be defined by integrating on bounded domain and sometimes on compact one. The difference is crucial since, for instance, for the function $\frac{1}{x}$ we get that for the "bounded" definition $$ \frac{1}{x} \not\in L^1_{loc}((0,1)) $$ but for the compact definition $$ \frac{1}{x} \in L^1_{loc}((0,1)) $$
$L^1_{loc}$ bounded or compact?
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functional-analysis
pde
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0Maybe the think of a a bounded domain s.t. $\bar{U} \subset \subset \Omega$? – 2017-02-15
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0Usually: compact. $1/x$ is perfectly fine as an element of $L^1_{\mathrm{loc}}(0,1)$, its basically the same as $x$ on $(0,\infty)$ which is $L^1_{\mathrm{loc}}$. – 2017-02-15