"Classical" Sobolev inequality

Question

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.

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). — SquirtleSquad, Aug 04 '16 at 19:38
Shouldn't weak derivatives match the classical derivative if the latter exists? — user369210, Aug 05 '16 at 00:52
https://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. — SquirtleSquad, Aug 05 '16 at 01:04
Nevermind, below it says that if it has a derivative in the classical sense then that is its weak derivative. Seems strange to me. — SquirtleSquad, Aug 05 '16 at 01:08
@SquirtleSquad shouldn't seem strange since the weak definition can be derived from the classical one. Then uniqueness gives the desired result. — user369210, Aug 05 '16 at 14:10

score 3 · Accepted Answer · answered Aug 05 '16 at 06:52

3

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 $u(-1) = u(1) = 0$.

answered Aug 05 '16 at 06:52

gerw

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"Classical" Sobolev inequality

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