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Let $K:{\rm L}^p({\bf R}^d)\to {\rm L}^p({\bf R}^d)$ be a bounded linear operator for every $p\in(1,\infty)$. Assume that for some $r\in(2, \infty)$ it holds that $K$ is compact on ${\rm L}^q({\bf R}^d)$ for every $q\in[2,r)$.

My question is: does it hold that $K$ is compact on ${\rm L}^{r}({\bf R}^d)$? I was thinking about the counterexample, but unsuccessfully.

Remark: I would like to apply such result (or not to apply it) on the K that is a commutator of Fourier integral operator and simple multiplication operator.

Thank You in advance.

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  • $\begingroup$ Thank you for the answer, but if $u_j\in{\rm L}^r$, then we do not know that $u_j\in{\rm L}^{2r-2}$ at all (because $r>2$). If $u_j\in{\rm L}^{2r-2}$ for $r>2$, then we could jump down to ${\rm L}^r$ case easily. I am not sure that the interpolation will do the trick here, since it will not give information at the right boundary case (the one where $q$ = r), only in the $\langle 2, r\rangle$ as you have mentioned in your comment. $\endgroup$
    – user66632
    Commented Feb 2, 2015 at 13:11
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    $\begingroup$ More is true: if $K$ is compact on $L^p$ for some $p$, then it is compact for all $1<p<\infty$. This is a results of Krasnoselʹskiĭ. See ams.org/mathscinet-getitem?mr=119086 Interestingly it is an open question whether this holds in complete generality for complex interpolation, see arxiv.org/abs/1410.4527 $\endgroup$
    – Mikael de la Salle
    Commented Feb 2, 2015 at 13:49
  • $\begingroup$ Wow, thank you, Mikael. This is more than I hoped for =) $\endgroup$
    – Semmel
    Commented Feb 2, 2015 at 18:04
  • $\begingroup$ @MikaeldelaSalle I suggest that you make this an answer, just so the question can be marked as answered :) $\endgroup$
    – Yemon Choi
    Commented Feb 5, 2015 at 23:22
  • $\begingroup$ @YemonChoi I agree; done. $\endgroup$
    – Mikael de la Salle
    Commented Feb 6, 2015 at 8:16

2 Answers 2

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More is true: if K is compact on $L^{p_1}$ for some $1\leq p_1<\infty$ and bounded on $L^{p_2}$ for some other $1 \leq p_2 \leq \infty$, then it is compact for all $p$ in the interval $[p_1,p_2)$ or $(p_2,p_1]$. This is a results of Krasnoselʹskiĭ. See http://www.ams.org/mathscinet-getitem?mr=119086

Interestingly it is an open question whether this holds in complete generality for complex interpolation, see http://arxiv.org/abs/1410.4527

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If $K$ is compact from $L^2$ to $L^2$, and continuous from $L^{2r-2}$ to $L^{2r-2}$, then it's also compact $L^r$ to $L^r$ just by interpolation of $L^2$ and $L^{2r-2}$ norms $$\|Ku\|_r\le \|Ku\|_2 ^{1/r}\|Ku\|_{2r-2}^{1-1/r}. $$ (So if $u_j\to 0$ weakly in $L^r$, then also weakly in $L^2$, the RHS of the inequality is $o(1)O(1)=o(1)$ )

edit: this for finite measure spaces…

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  • $\begingroup$ More generally: continuous on $L^s$ for $s\in[p,q]$ and compact for $s=p$ implies compact for $s\in[p,q)$, for the same reason. $\endgroup$
    – Pietro Majer
    Commented Feb 2, 2015 at 12:56

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