# $\ell_{\infty}$ and the bounded approximation property

I know from several references that the Banach space of bounded sequences, namely $\ell_{\infty}$, has the bounded approximation property. But I do not have a reference where this assertion is proven.

So I would be very grateful if you can give me a reference for the result asserting that $\ell_{\infty}$ has the bounded approximation property ?

If you do not have a reference but you know how to do it, please provide me a sketch of your proof.

I just recall here the definition of the approximation property. We say that a Banach space $X$ has the approximation property (AP) if for every $\varepsilon>0$, for every compact set $K \subset X$, there exists a finite rank operator $T \in \mathcal B(X)$ such that $\|Tx-x\| \leq \varepsilon$ for every $x \in K$. Let $\lambda \geq 1$, if in the above definition T can always be chosen so that $\|T\|\leq \lambda$, then we say that $X$ has the $\lambda$-bounded approximation property ($\lambda$-(BAP)).

That $C(K)$-spaces have BAP follows from the existence of partitions of unity. See this proof on p. 256 of Köthe's Topological Vector Spaces II. Once you unwrap the proof, you will actually see that it shows that $C(K)$-spaces have 1-BAP.
Of course, $\ell_\infty=C(\beta \mathbb{N})$
Not only that I think every subspace of $L_\infty$- space also has A.P.
• Every separable Banach space is isomorphic to a subspace of $L_\infty$ ... so you claim every separable Banach space as A.P.? My references do not agree with that. – GEdgar Jul 22 '17 at 17:39
• yes. my mistake. What about $L_1$? Is every subspace of an $L_1$ space has the A.P.? – Mathbuff Jul 23 '17 at 21:03