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


1. $X'$ is closed, $X/X' \cong \ell\_1 \Rightarrow X'$ is complementary;
2. $X' \cong \ell\_\infty \Rightarrow X'$ 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.



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


