Does there exist $q>p$ such that the canonical inclusion $L^q(0,1) \to L^p(0,1)$ is compact?
My answer is no.
Since we know that $L^\infty (0,1) \to L^p(0,1)$ is not compact, take $\{\sin(nx)\}$ as a counter example; this counter example will also work for $L^q$.
Is my argument correct? And what if I change "compact" to "weakly compact"? Because then my counter example would fail.
Another question, is there any connection between the weak topology on $L^p$ and the strong topology on $L^q$? My guess is no, since the weak topology is not metrizable.
If $1\le p<q<\infty$, then the embedding $$L^q(0, 1)\subset L^p(0, 1)$$ trivially is weakly compact. Indeed, an $L^q$-bounded sequence is $L^p$-bounded too, and by Banach-Alaoglu's theorem (which applies since $q\ne 1, \infty$) it has weakly convergent subsequences.