Borderline Sobolev embedding in dimension 2

Question

In this paper¹ on p22, the author claims (between equations (40) and (41)) that:

for every $r > 2$, there is a number $S_r >0$ such that for every $u \in W^{1,2}( \Omega)$, where $\Omega$ is the open interval $(0,3) \times (0,1)$ inside $\mathbb{R}^2$, one has $u \in L^r( \Omega)$ with

$$ \| u \|_{L^r(\Omega)} \le S_r \| u \|_{W^{1,2}(\Omega)}.$$

This is important published paper so I assume it correct. But I don't understand. Since $\Omega$ is 2-dimensional, this is borderline case of Sobolev embedding theorem, and I don't see how claimed inequality follows.

¹Alberto Abbondandolo, Matthias Schwarz: On the Floer homology of cotangent bundles, https://arxiv.org/abs/math/0408280

In dimension two, $W^{1,2}$ is embedded in every $L^r$ space. — Siminore, Oct 03 '15 at 14:11
A proof in Haim Brezis' book of functional analysis, for instance. — Siminore, Oct 03 '15 at 16:19

score 5 · Answer 1 · answered Oct 03 '15 at 19:12

On a bounded domain, Sobolev spaces are nested in the same way and for the same reason that Lebesgue spaces are nested: $W^{k,p}\subset W^{k,q}$ if $p\ge q$. (More precisely, the identity map is a bounded operator from $W^{k,p}$ to $W^{k,q}$.)

So, you can obtain the claimed inequality by considering the composition of inclusion $W^{1,2} \to W^{1,q}$ with the familiar Sobolev embedding $W^{1,q}\to L^{2q/(2-q)}$, where $q<2$ is chosen so that $\frac{2q}{2-q}=r$. Since $\frac{2q}{2-q}$ can be arbitrarily large, this shows $W^{1,2}$ embeds into $L^r$ for every finite $r$.

More can be said about the borderline case: see Trudinger's theorem.

Borderline Sobolev embedding in dimension 2

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