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It is well-known (for example, see this question) that, as a consequence of the Hahn-Banach theorem, every separable Banach space can be isometrically embedded in $\ell^\infty$. In particular, for $1\leq p<\infty$, there is an isometric embedding $\phi_p:\ell^p\to\ell^\infty$. The image $\phi_p(\ell^p)$ inherits many of the properties of $\ell^p$, such as being strictly convex when $p>1$.

However, I have not been able to construct an explicit example of such a $\phi_p$. Is the existence of $\phi_p$ provable in $\sf{ZF}$ without using Hahn-Banach?

csch2
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    The answer you cited only uses Hahn-Banach to construct norm-attaining linear functionals. For $\ell^p$, such functionals can be explicitly written down. See https://en.wikipedia.org/wiki/H%C3%B6lder%2527s_inequality#Extremal_equality – David Gao Apr 30 '24 at 02:10
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    Here's an overkill proof that such an embedding can be proved to exist in ZF: let $\varphi$ be an embedding $\mathscr{l}^{p,L}\rightarrow\mathscr{l}^{\infty,L}$, where $X^L$ is the thing that the inner model $L$ (which satisfies ZFC by Godel) thinks is $X$. "Reasonably definable" objects $X$ - including both spaces here - contain their $L$-versions as "dense subobjects." The continuous extension of $\varphi$ to all of $\mathscr{l}^p$ is an embedding of the desired type. (This is really just a specific example of Mostowski absoluteness.) – Noah Schweber May 01 '24 at 03:51

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