Continuous logic and embeddings of Lebesgue spaces

Abstract

We use the compactness theorem of continuous logic to give a new proof that $L^r([0,1]; {\mathbb {R}})$ isometrically embeds into $L^p([0,1]; {\mathbb {R}})$ whenever $1 \le p \le r \le 2$. We will also give a proof for the complex case. This will involve a new characterization of complex $L^p$ spaces based on Banach lattices.

Appendix

The following appears to be a matter of folklore in the theory of Banach lattices.

Theorem 5.1

The complexification of $L^p(\Omega ; {\mathbb {R}})$ is $L^p(\Omega ; {\mathbb {C}})$.

We give a proof which was suggested by Glück [7] whose thesis contains a number of results on complex Banach lattices.

A Banach lattice is Dedekind complete if each nonempty set of vectors that is bounded above has a supremum. A Banach lattice is super Dedekind complete if it has the property that whenever X is a nonempty set of vectors that is bounded above, there is a countable $D \subseteq X$ so that $\sup X = \sup D$. It is well-known that $L^p$ spaces are Dedekind complete and that $\sigma$-finite $L^p$ spaces are super Dedekind complete.

Now, let $f \in L^p(\Omega ; {\mathbb {C}})$, and let $g(t) = |f(t)|$. It suffices to show that $g = \sup _\theta {\text {Re}}(e^{i \theta } f)$. Let $h = \sup _\theta {\text {Re}}(e^{i \theta } f)$. We first note that for each complex number z, $z = \sup _\theta {\text {Re}}(e^{i \theta } z)$. It follows that $g \ge h$. Suppose $g_0 \ge h$. Since the support of f is a $\sigma$-finite set, we can assume $\Omega$ is sigma-finite. Thus, $L^p(\Omega ; {\mathbb {R}})$ is super Dedekind complete, and so there is a countable set of reals D so that $h = \sup _{\theta \in D} {\text {Re}}(e^{i \theta } f)$; we can additionally assume that D is dense. Now, for each $\theta \in D$, $g_0(t) \ge {\text {Re}}(e^{i \theta }f(t))$ a.e.. Since D is countable, it follows that $g_0(t) \ge \sup _{\theta \in D} {\text {Re}}(e^{i \theta } f(t))$ a.e.. Thus, since D is dense, $g_0(t) \ge g(t)$ a.e.. We conclude that $g = h$.