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.



Compact operators on Lebesgue spaces
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>


