$\ell_1$ and $\ell_\infty$ as complementary subspaces of Banach space

Question

Let $X$ be a Banach space, and let $X'\subset X$ - its subspace. Then the following propositions are true:

$X'$ is closed, $X/X' \cong \ell_1 \Rightarrow X'$ is complementary;
$X' \cong \ell_\infty \Rightarrow X'$ is complementary.

For the second one there is an option to try using Hahn-Banach theorem, as we do the same for proof of finite-dimensional subspaces.

More precisely, we can extend an identity operator $\ell_\infty \rightarrow \ell_\infty$ to the norm-one operator $X \rightarrow \ell_\infty$, but by the definition of complement subspace we should find closed subspace (such subspace is a kernel of extension, as I know, but why does it form a closed subspace?) and moreover show that $\ell_\infty$ is closed.

How $\ell^1$ is a complement of $X'$? You do not have a priori $\ell^1$ in $X$. — Fedor Petrov
– Fedor Petrov, Commented Nov 7, 2017 at 13:01
@FedorPetrov I meant that $X'$ complement should be $X/X'$, but $X/X'\cong \ell_1$, that is — Evgeny
– Evgeny, Commented Nov 7, 2017 at 14:08
The first statement is true, due to the lifting property of $\ell_1$, but it does not hold for other $\ell_p$. — Mateusz Wasilewski
– Mateusz Wasilewski, Commented Nov 7, 2017 at 15:34
I give both as exercises when I teach the second semester of real analysis. — Bill Johnson
– Bill Johnson, Commented Nov 7, 2017 at 16:12

Pietro Majer · Accepted Answer · 2017-11-07 20:12:47Z

For (1), use (0): for any Banach space $X$, $L(\ell_1,X)$ is isometrically isomorphic to the space $\ell_\infty(X)$ of bounded sequences in $X$. You may easily define a concrete isometry and its inverse. As a consequence: any surjective bounded linear operator $S:X\to\ell_1$ is a left inverse: use (0) to define a right inverse $R:\ell_1\to X.$ The proof of (2) is on the same lines: you need to prove that any injective bounded linear operator $R:\ell_\infty\to X$ is a right inverse.

(Also recall that: for bounded operators on Banach spaces, $S:X\to Y$ is a left inverse if and only if it is surjective and its kernel splits; $R:Y\to X$ is a right inverse if and only if it is injective and its range splits. Moreover, if $SR=I_X$, the splitting is $X=\operatorname{ker S}\oplus\operatorname{ran R}$ with projectors$[S,R]$ and $RS$).

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