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.
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The author was supported in part by Simons Foundation Grant # 317870. The author thanks the anonymous referee for the many suggestions for improvement.
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Appendix
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 \(g = h\).
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McNicholl, T.H. Continuous logic and embeddings of Lebesgue spaces. Arch. Math. Logic 60, 105–119 (2021). https://doi.org/10.1007/s00153-020-00734-7
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DOI: https://doi.org/10.1007/s00153-020-00734-7