*[Disclaimer: this may be a very trivial question; it certainly looks like it ought to have been studied and understood. I started thinking about it this morning when writing some notes for Rellich-Kondrachov, but cannot find a simple counterexample.]*


For the time being, let us just work on $\mathbb{R}^d$ with the Lebesgue measure. It is well-known that for an open, bounded domain (hence with finite measure) $\Omega$, the inclusion


$$ L^p(\Omega) \to L^q(\Omega) $$


is continuous for $\infty \geq p \geq q \geq 1$. 


**Question**: Is the inclusion completely continuous (i.e. compact)? If not, what is a simple counterexample? 


I dug around a bit and cannot find any references to a proof (or even the statement). The usual non-compactness mechanisms, of course, do not work. Let $f\_i$ be a sequence of functions with $\|f\_i\|\_{L^p(\Omega)} \leq 1$, so by continuous inclusion it is a bounded sequence in $L^q(\Omega)$. Because $\Omega$ is compact, we cannot have the problem fixed-scale translations: if $f\_i(x) = f(x + y\_i)$ for a sequence of points $y\_i$, since $y\_i$ must have a converging subsequence, then so must $f\_i$, even in $L^p$. The other usual non-compactness mechanism is dilations. WLOG assume $0\in supp(f) \subset\subset \Omega$ and that the $supp(f)$ is convex. Then the sequence $f\_j = 2^{jd/p} f(2^j x)$ is a bounded but non-compact sequence in $L^p(\Omega)$. But in $L^q(\Omega)$ for $q < p$, the sequence converges strongly to 0. 



Is the inclusion of Lebesgue spaces compact?
Take $f\_j(x)=\sin(jx\_1)$. This is a bounded sequence in every $L^p(\Omega)$ when the domain $\Omega$ is bounded. Yet, it is non-compact in every $L^q(\Omega)$. It happens to converge weakly-star in these spaces towards $f=0$, but not strongly.


