Is the inclusion of Lebesgue spaces compact? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:52:36Z http://mathoverflow.net/feeds/question/42615 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42615/is-the-inclusion-of-lebesgue-spaces-compact Is the inclusion of Lebesgue spaces compact? Willie Wong 2010-10-18T11:01:07Z 2010-10-18T18:18:41Z <p><em>[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.]</em></p> <p>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</p> <p>$$L^p(\Omega) \to L^q(\Omega)$$</p> <p>is continuous for $\infty \geq p \geq q \geq 1$. </p> <p><strong>Question</strong>: Is the inclusion completely continuous (i.e. compact)? If not, what is a simple counterexample? </p> <p>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 $q &lt; p$, the sequence converges strongly to 0. </p> http://mathoverflow.net/questions/42615/is-the-inclusion-of-lebesgue-spaces-compact/42616#42616 Answer by Denis Serre for Is the inclusion of Lebesgue spaces compact? Denis Serre 2010-10-18T11:07:26Z 2010-10-18T11:07:26Z <p>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.</p> http://mathoverflow.net/questions/42615/is-the-inclusion-of-lebesgue-spaces-compact/42670#42670 Answer by Pietro Majer for Is the inclusion of Lebesgue spaces compact? Pietro Majer 2010-10-18T17:11:28Z 2010-10-18T18:18:41Z <p>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 $$\sup _{f\in B}\ \omega_f(\delta) \ =o(1),\qquad \mathrm {as}\ \delta\to0.$$</p> <p>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.</p>