The famous Pitt's theorem asserts that if $p>q$ then each bounded operator $T\colon \ell^p\to\ell^q$ is compact. Since $\ell^p$ and $\ell^q$ are incomparable ($p\neq q$, $p,q\geq 1$), each operator $T\colon \ell^p\to\ell^q$ is strictly singular anyway.
Now I want to ask about $L^p(\mu)$ (put any assumptions on $\mu$ as you wish). Under what conditions on $p$ and $q$ each weakly compact operator $T\colon L^p(\mu)\to L^q(\mu)$ is compact?
or $1\leq q < \infty$ then every linear operator $L^p\to L^q$ is weakly compact. This is clear when either $p$ or $q$ lies in $(1,\infty)$, because then either the domain or range is reflexive; and the only other case is $p=\infty$, $q=1$ when my claim follows from Grothendieck's theorem (one can also prove it without G's thm, but I don't know the argument without looking it up).
– 2011-11-02