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Take $p\in (1,\infty)\setminus \{2\}$. Let $X$ be a subspace of $\ell_p^n$ and let $U\colon X\to \ell_p^m$ ($m\geqslant n$) be a linear isometry. Is it possible to extend $U$ to a (non-surjective) linear isometry $\hat{U}\colon \ell_p^n\to \ell_p^m$?

For contractions this is not necessarily true however the counterexample I know is not isometric. It seems to me that it should not be true as Lamperti's theorem limits the form for isometries on $\ell_p^n$ and the subspaces cab be quite weird.

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    $\begingroup$ Check out the Plotkin-Rudin theorem for positive results for infinite dimensional $L_p$ spaces. $\endgroup$
    – Bill Johnson
    Commented Feb 2, 2018 at 14:16

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No, take $m=n=2$ and let $X$ be the $x$-axis. For every line $Y$ through the origin there is an isometry from $X$ to $Y$, and if $p\neq 2$ they obviously don't extend to isometries from $\ell^n_p$ to itself.

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