Is there any simple example or proof to show that $H^1$ fails to be embedded in $L^{\infty}$?, where, $H^1=W^{1,2}$ is a Sobolev space.
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On what domain? ā user251257 Jul 11 '20 at 11:29
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1I would suggest you consider $u(x) := |x|^{-\alpha}$ on $B(0,1) \subset \mathbb{R}^n$ for suitable $\alpha$ and $n$. ā mathdoge Jul 11 '20 at 14:16
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So, in dimension $1$, the embedding $H^1(\mathbb{R}) ā L^\infty(\mathbb{R})$ is true. In dimension $dā„ 3$ you can take $\varphi\in C^\infty_c(\mathbb{R}^d)$ and consider for example $f(x) = \varphi(x) \,|x|^{-1/4}$ (or replace $1/4$ by any $\varepsilon < d/2-1$).
But I think you are mainly interested in the case $d=2$. In this case, you can use the counterexamples from Function in $H^1$, but not continuous or Discontinuous Sobolev Function
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