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.



Extending linear isometries from subspaces of $\ell_p^n$
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.


