1

Recently I have often seen product estimates in the integral like the following. E.g.: they want to show that

$\displaystyle\int |f|^p \phi dx$

is bounded where $\phi$ is a test function with compact support and $f$ is a $L^p-$function. But why is this bounded? I can't use the Hölder inequality because we are already on the highest possible integration exponent on $f$. Is it because $\phi$ is bounded? But that would be weird as one then could always immediately get rid of any terms involing $\phi$ and its derivatives in weak formulations of pde's as they would all be bounded...

  • The important facts are that $\phi$ has compact support and that $f\in L^p$. You can disregard everything outisde of the support of $\phi$ for the integral. I assume you also want $\phi$ to be continuous, which would imply boundedness of $\phi$. – Keen-ameteur Dec 17 '22 at 12:35
  • Yep thanks, indeed. But since we can always "bound away" the test function terms, why do we e.g. need u to be in $L^p$ if a u is a weak solution to the p-Laplacian and the integral should be bounded, i.e. $\displaystyle\int |\nabla u|^{p-2}\nabla u \cdot \nabla \phi ,dx =0$? We can use Cauchy-Schwarz and bound $|\nabla \phi|$ away to get $\displaystyle\int |\nabla u|^{p-1}$...which would also be bounded if u only was in $L^{p-1}$...I probably might misunderstand something – HelloEveryone Dec 17 '22 at 14:58
  • Well, if you want the weak derivative to be in $L^p$, I don't think you can avoid this assumption. I'm pretty sure that if $f$ would not be in $L^p$ and the weak derivatives exist, it would imply some of those are not in $L^p$. It seems to me that you would want it to be in the appropriate Sobolev space. – Keen-ameteur Dec 17 '22 at 20:20
  • Sorry in my above post i meant of course that $\nabla u$ would only have to be in $L^{p-1}$ instead of $u$...Hm thank you! I still don't understand though why we need $\nabla u\in L^p$ when it would suffice to be in $L^{p-1}$ according to the last estimate... – HelloEveryone Dec 17 '22 at 20:28

0 Answers0