6
$\begingroup$

Suppose $f \in H^1(\mathbb R^2)$, where $H^1$ is the Sobolev space, then how to use this information to bound $\Vert f \Vert_{L^q}$, where $q>2$? It seems like Sobolev embedding, but it's not.

$\endgroup$
  • $\begingroup$ @Zixiao_Liu "Also $W^{1,2}$ norm will provides a Hardy Inequality, which can bound weighted 𝐿𝑞 norm with singular weights." Are you able to provide some reference for this statement? I'm not saying it's wrong, just don't know why. $\endgroup$ – Bourne Feb 1 at 4:52
  • 1
    $\begingroup$ On Wikipedia, you can easily find: Hardy Inequality origins from 1-dimension integral by parts. Due to Szego Inequality and Riesz Rearrangement inequality, we only need to concern for radial symmetric functions that blows up at the origin. So there would be a suitable weight such that u is weighted integrable on entire $R^2$. More detail I refer to the book"Weighted inequalities of Hardy Type". Maybe I'm wrong about this idea, sorry I didn't recheck it again. $\endgroup$ – Zixiao_Liu Feb 1 at 7:03
2
+50
$\begingroup$

This is immediate from the Gagliardo-Nirenberg inequality, or from the Sobolev inequality with fractional exponents, which says in particular that $$ \|f\|_{L^q} \leq C\|f\|_{H^s} \leq C'\|f\|_{H^1} , \qquad f\in C^1_0(\mathbb{R}^2), $$ for $2\leq q<\infty$ and $s=1-\frac2q$.

$\endgroup$
  • $\begingroup$ but the question seems to be about $H^1$ i.e. $s=1$, or $q=\infty$ $\endgroup$ – Calvin Khor Feb 23 at 12:16
  • $\begingroup$ @CalvinKhor: $H^1$ is trivially embedded into $H^s$ with $s<1$. The question only asked about $q>2$, which I read as $2<q<\infty$. $\endgroup$ – timur Feb 23 at 14:45
  • $\begingroup$ Oh, I see what you're getting at now, thanks. I expected "worse" results than $H^1\subset BMO$ from using less regularity and BMO functions are only locally in $L^q$... $\endgroup$ – Calvin Khor Feb 23 at 14:55
  • $\begingroup$ @timur Does $C'$ depend on q? $\endgroup$ – Bourne Feb 26 at 5:46
  • 1
    $\begingroup$ Yes it will blow up in the limit $q\to\infty$, because $H^1$ is not embedded into $L^\infty$. $\endgroup$ – timur Feb 26 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.