*[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.