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Is the wiki page for the Sobolev inequality correct?

Let $p$, so that $1 \leq p < \infty$ and $\Omega$ a subset with at least one bound. There then exists a constant $C$, depending only on $\Omega$ and $p$, so that, for every function $u$ of $W_0^{1,p}$, $$ \|u\|_{L^{p} (\Omega)} \leq C \| \nabla u \|_{L^{p} (\Omega)}.$$

One can easily make the counterexample of $u \equiv 1$ on the interval $(0,1)$, which has gradient identically 0.

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  • $\begingroup$ I think it says $u$ in $W_0^{1,p}$ right? $\endgroup$ – SquirtleSquad Aug 4 '16 at 19:22
  • $\begingroup$ The zero in the subscript means vanishing at the boundary. $\endgroup$ – SquirtleSquad Aug 4 '16 at 19:24
  • $\begingroup$ Also, the derivatives are understood in a weak sense. So, I'm not sure $1$ even has a weak derivative (looking at the definition of weak derivative). $\endgroup$ – SquirtleSquad Aug 4 '16 at 19:38
  • $\begingroup$ Shouldn't weak derivatives match the classical derivative if the latter exists? $\endgroup$ – user369210 Aug 5 '16 at 0:52
  • $\begingroup$ en.wikipedia.org/wiki/Weak_derivative Judging from the formula, if this were the case, then the R.H.S. would have to be 0, which seems wrong. $\endgroup$ – SquirtleSquad Aug 5 '16 at 1:04
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Your function $u$ satisfies $u \in W^{1,p}(0,1)$, but not $u \in W_0^{1,p}(0,1)$. Indeed, $u$ is continuous and has boundary values $1$. If it would be in $W_0^{1,p}(0,1) \cap C([0,1])$, it would satisfy $u(-1) = u(1) = 0$.

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