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In Brezis Proposition 9.3 they are saying that if $u\in W^{1,p}(\Omega)$ with $1<p\le \infty$, then for all $\omega\subset\subset\Omega$, and for all $h\in \mathbb R^n$ with $|h|<\text{dist}(\omega,\partial\Omega)$ we have

$$\Vert\tau_hu-u\Vert_{L^p(\omega)}\le \| \nabla u\|_{L^p(\omega)} |h|,$$

where $\tau_hu(x)=u(x+h)$.

I had understood the proof for $p<\infty$. However, for the case $p=\infty$ they are saying that the result follows from the $p<\infty$ case by taking $p\rightarrow\infty$. Can someone explain how we can pass to the limit and get the result?

Sahab
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1 Answers1

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Just use this nice result. Assume $u \in L^\infty$. Then it is in $L^p$ for all $p$ and

$$\Vert\tau_hu-u\Vert_{L^p(\omega)}\le \|\nabla u \|_{L^p(\omega)}|h|\le |\operatorname{Vol}(\omega)|^{1/p} \| \nabla u\|_{L^\infty(\omega)} |h|$$

for all $p$. Take $p\to \infty$ give

$$ \Vert\tau_hu-u\Vert_{L^\infty(\omega)}\le \|\nabla u \|_{L^\infty(\omega)}|h|$$