$C[0,1]$ doesn't contain a complemented subspace isomorphic to $l^1$

Solution 1:

If $\ell_1$ were complemented in $C[0,1]$, $\ell_\infty$ would be isomorphic to a (complemented) subspace of $C[0,1]^*$. The dual space $C[0,1]^*$ is however weakly sequentially complete being isomorphic to an uncountable $\ell_1$-sum of $L_1$'s. Certainly $\ell_\infty$ is not weakly sequentially complete.

Solution 2:

Note that $C[0,1]$ is an $\mathscr{L}_\infty$ space, then so does all its complemented subspaces. On the other hand $\ell_1$ is an $\mathscr{L}_1$-space. It is remains to recall that no infinite dimensional Banach space can be $\mathscr{L}_1$ and $\mathscr{L}_\infty$-space at the same time.

To learn more on $\mathscr{L}_p$-spaces see New classes of $\mathscr{L}_p$-spaces. J. Bourgain.