 Solution

Suppose a bounded linear isomorphism $T:{\ell }^p\to Y\subset {\ell }^2$ with bounded inverse existed. There would be constants $0<c\le C$ such that $c\Vert x{\Vert }_p\le \Vert Tx{\Vert }_2\le C\Vert x{\Vert }_p$. Put $y_j=T{e}_j$. Averaging the squared Hilbert norm over all $2^n$ independent sign choices cancels cross terms, giving the generalized parallelogram identityEvery signed sum of the original $e_j$ has $p$-norm $n^{1/p}$. The left side is therefore between $c^2{n}^{2/p}$ and $C^2{n}^{2/p}$, while the right side is between $c^{2}n$ and $C^{2}n$. Thus $n^{2/p}$ and $n$ would be bounded constant multiples of one another for all $n$, impossible unless $p=2$. Consequently no such normed-space isomorphism exists for $p\ne 2$.