MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

[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.

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
You beat me by seconds :) – Piero D'Ancona Oct 18 '10 at 11:09
Oh dash it, I forgot to test in frequency space. Thanks. =) – Willie Wong Oct 18 '10 at 11:22

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.