Why doesn't $c_0$ admit a complement in $l^\infty$?

Question

The projection theorem shows that every closed linear subspace $ M $ of hilbert space $H$ has at least one complementry closed linear subspace namely $M^\perp$.

But in some Banach spaces a closed subspace may fail to have complementry closed linear subspace; for instance, the closed subspace $c_0$ of banach space $l^\infty$ is not complemented in $l^\infty$.

I wonder why $c_0$ have not a complemented linear subspace in $l^\infty$.

Answer 1

That $c_0$ is not complemented in $\ell^\infty$ was originally proved by Phillips (1940) but a shorter proof was published by Robert Whitley:

Projecting $m$ onto $c_0$, Amer. Math. Monthly, 73, No. 3 (Mar., 1966), 285-286.

I sketch this one, following the presentation in Conway's A course in Functional Analysis.

Step 1: There is an uncountable family of infinite subsets of $\mathbb N$ with mutually finite intersections. (Proof: switch from $\mathbb{N}$ to $\mathbb{Q}$, pick convergent sequences to irrationals. More details in this question.)

Step 2: Let $T: \ell^\infty \to \ell^\infty$ be an operator such that $\ker T\supseteq c_0$. Then there is an infinite subset $A$ of $\mathbb{N}$ such that $T(x) = 0$ for all $x$ supported on $A$. (Proof below.)

Step 3: If $c_0$ were complemented in $\ell^\infty$, there would be a continuous projection $P: \ell^\infty \to \ell^\infty$ with range $c_0$. Applying Step 2 with $T = I-P$ we would find an infinite set $A$ such that $(I-P)(x) = 0$ for all sequences supported in $A$. But then $P(\chi_A) = \chi_A\notin c_0$, a contradiction. $\Box $

To prove Step 2, consider a family $\{A_i\}_{i \in I}$ as in Step 1. Suppose to the contrary that for each $i \in I$ we can find $x_i \in \ell^\infty$ supported on $A_i$ such that $T(x_i) \neq 0$, in particular $x_i \notin c_0$. Normalizing $x_i$, we may assume that $\|x_i\|=1$ for all $i \in I$.

Since $I$ is uncountable there must be $n \in \mathbb{N}$ such that $I_n := \{i \in I\,:\,(Tx_i)(n) \neq 0\}$ is uncountable. Since $I_n$ is uncountable, there must be $k$ such that $I_{n,k} := \{i \in I\,:\,|(Tx_i)(n)| \geq 1/k\}$ is uncountable.

Now fix $n$ and $k$ such that $I_{n,k}$ is uncountable. Let $J \subset I_{n,k}$ be finite and consider $y = \sum_{j \in J} \operatorname{sign}{[(Tx_j)(n)]} \cdot x_j$. Note that $$ (Ty)(n) = \sum_{j \in J} \operatorname{sign}{[(Tx_j)(n)]}\cdot (Tx_j)(n) \geq \frac{\# J}{k} $$ by our choice of $y$. Since $A_i \cap A_j$ is finite for $i \neq j$, we can write $y = f + z$ where $f$ has finite support and $\|z\| \leq 1$. Thus $T(y) = T(f) + T(z) = T(z)$ by hypothesis on $T$ and therefore $\|T(y)\| \leq \|T\| \|z\| \leq \|T\|$. This yields $\# J \leq \|T\| k$ contradicting that $I_{n,k}$ is infinite.

Answer 2

A proof can be found in the book

R. S. Phillips: On linear transformations, Trans. Amer. Math. Soc. 48 (1940), 516-541 .

Answer 3

There is one more rather elementary proof of it (that works for a broader class of subspace of $\ell_\infty$) that you may find here:

A letter concerning Leonetti's paper ‘Continuous Projections onto Ideal Convergent Sequences’, Results in Mathematics 74, 12 (2019), 4 pp. arXiv:1810.09383.

The idea is that if it were complemented, you would have found a non-separable weakly compact of $\ell_\infty$ (as there is one in $\ell_\infty/c_0$ and this space would be a direct summand of $\ell_\infty$), which is impossible.