Are $\ell^p$ and $L^p [0,1]$ isometric with p distinct from 2?

Question

For values of $p$ distinct from $2$, are $\ell^p$ and $L^p [0,1]$ isometric under some conditions? In that case, what is the isometry $T:\ell^p \to L^p$?

Answer 1

This is not true, a proof can be found in Theorem 1.11 in the lecture notes.

In particular, one can show that $\ell^2$ is isometric to a subspace of $L^p([0,1])$, $1\le p < \infty$, but not isomorphic to a subspace of $\ell^p$, $p \ne 2$.

Answer 2

For $p>1$, Kchinchine's inequality implies that the closed linear span of Rademacher functions in $L_p[0,1]$ is a complemented subspace isomorphic to $\ell_2$.

However, every complemented subspace of $\ell_p$ is isomorphic to $\ell_p$.