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Dec 7, 2011 at 15:47 vote accept Laurent Berger
Dec 6, 2011 at 23:57 comment added Bill Johnson @ Laurent. Actually there is a deep explanation for that. If you think about $\ell_2$ as being the closed span $R$ of the Rademacher functions in $L_1$ (or a sequence of IID gaussians) and $T$ is an isomorphism from $R$ into $C[0,1/2]$ (considered as a subspace of $C[0,1]$ in the obvious way), then there is an extension of $T$ to an isomorphism from $L_1$ into $C[0,1]$.
Dec 6, 2011 at 17:06 comment added Laurent Berger For some reason, it's much easier to realize $\ell^1(R)$ as a subspace of $C^0([0;1],R)$.
Dec 6, 2011 at 17:05 comment added Laurent Berger Thank you! I was hoping for an easier answer, but of course now that I think about it, a sequence of functions which form a Hilbert basis really amounts to the coordinates of a continuous surjective map to the weak unit ball, so the answer is bound to be complicated.
Dec 5, 2011 at 20:40 history answered Robert Israel CC BY-SA 3.0