Sobolev inequality with a constant independent of the support of the function

Question

I am revising our PDE module and I came across the following version of the Sobolev inequality: $$ \exists C>0 \forall u\in W^{1,p}_0(\mathbb{R}^n):\, \|u\|_{L^{p*}}\le \|Du\|_{L^p}\, , $$ where $p*$ is the usual coefficient. Now what stunned be that this constant shall not depend on the support of $u$. In Evan's book I found only proofs for bounded domains, for which I can totally agree, or for continuously differentiable $u$'s. But in general, couldn't I take $u$ to be constant to one on the set $[-M,M]$ and then going linearily down with slope one and remaining zero after hitting zero? In that case $M$ to infinity would produce a contradiction, wouldn't it?

Answer 1

This is only true for the critical exponent $p^*=np/(n-p)$ and $p<n$. The one-dimensional counterexample you sketched doesn't really apply. (Although, for $n=1$ it's okay to use $p=1$ and $p^*=\infty$; then the statement is true with $C=1/2$.)

At the critical exponent, both sides of the inequality scale the same way: replacing $u$ with $u(\lambda x)$ yields the same power of $\lambda$ on both sides. Hence, whatever constant works for the unit ball also works for all bounded domains, hence for all compactly supported functions. And then it works for $W^{1,p}_0(\mathbb{R}^n)$ by the density of compactly supported functions.