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I had an exam question that for a open bounded smooth(boundary) region $U\subset \mathbb R^3$, does it hold that $W^{1,p}(U)\subset L^\infty(U)$ for $p\in(1,3]$ ? In these situations I can apply no embedding theorem, neither can I find anything like $1/|x|^\lambda$ to deny the embedding.

user74489
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  • The case $p=3$ is critical, so you will need to be careful there. In the case $p<3$ you should be able to find a suitable counterexample using an appropriate power of $|x|$. – Ian Mar 02 '18 at 03:40

1 Answers1

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Without loss of generality we can assume that $U$ contains the origin. Since $U$ is bounded, there must be some $a \in (0,\infty)$ such that $U \subset B_a(0).$

Fix $\lambda > 0,$ so that the function $f(x) = |x|^{-\lambda}$ is not in $L^\infty(U)$. Since $|Df|$ is proportional to $|x|^{-\lambda - 1},$ we can use spherical coordinates to see that $$\int_U |Df|^p \lesssim \int_0^a r^{-p(\lambda+1)}4\pi r^2\,dr,$$ which converges iff $2-p(\lambda+1)>-1.$ Similarly we see that $\int_U |f|^p$ converges if $2-p\lambda > -1,$ which is strictly easier to achieve. Thus we can conclude $f \in W^{1,p}(U)$ if $p(\lambda+1)<3,$ which is achievable with positive $\lambda$ so long as $p<3.$

It remains to rule out the critical case $p=3,$ which power functions alone cannot do. The standard counterexample here is apparently $f(x)=\log \log (1+|x|^{-1});$ hopefully you can verify that this is in $W^{1,3}(U)$ but not bounded.

Anthony Carapetis
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