Is the inclusion of Lebesgue spaces compact? [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. 
 A: 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.
A: Some further remarks. A nice aspect of this question is that it provides an example of how some general (and non-constructive) theorems of Functional Analysis often show us the way to a constructive answer to concrete problems. Let's focus on the case of the inclusion $L^\infty(\Omega)$ in $L^1(\Omega)$ (any other non-compactness follows from  $L^\infty(\Omega)\to L^p (\Omega)\to L^q(\Omega)\to L^1(\Omega)$ by composition).  By the Fréchet-Kolmogorov theorem, a bounded set $B\subset L^1(\Omega)$ (with $\Omega$ a bounded subset of $\mathbb{R}^n$) is relatively compact if and only if it is "equicontinuous-$L^1$," meaning that
$$\sup _{f\in B}\\ \omega_f(\delta) \\ =o(1),\qquad  \mathrm {as}\\ \delta\to0.$$
Here $\omega_f$ is the "modulus of continuity-$L^1$" of the function $f$, namely
$$\omega_f(\delta)= \sup _{|h|\leq \delta}\| f-f(\cdot-h)  \|_1\\ .$$
This condition is clearly not satisfied by the unit ball $B$ of $L^\infty,$ because for any $\delta$ there is in $B$ a function whose support is disjoint from a $\delta$ translate of it. In fact  $\| f-f(\cdot-h)  \|_1=|\Omega|$ holds, for instance, for the characteristic function of a suitable measurable set. So this answers the question, and also suggests an answer independent from the FK thm, by exhibiting directly a non-compact sequence of oscillating functions, like in Denis Serre's answer.
