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How can I show that $c_{0}$ cannot be complemented in $\ell^{\infty}$? Complement in the following sense

$$c_{0}+V = \ell^{\infty}$$

And the projections are continuous.

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    $\begingroup$ This is called Phillips's lemma. A nice proof can be found in Robert Whitley, Projecting $m$ onto $c_0$, The American Mathematical Monthly Vol. 73 (3) (Mar., 1966), pp. 285-286. But most texts on basic functional analysis contain a proof, e.g. Conway. $\endgroup$
    – t.b.
    Apr 16 '12 at 14:28
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    $\begingroup$ The original reference is Phillips, On linear transformations, Trans. Amer. Math. Soc. 48 (1940), 516-541, see 7.5 on page 539. $\endgroup$
    – t.b.
    Apr 16 '12 at 14:35
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This is usually called Phillips's lemma and was proved in 7.5 on page 539 of Phillips, On linear transformations, Trans. Amer. Math. Soc. 48 (1940), 516–541 (it was also discovered by Sobczyk around the same time).

A somewhat different proof was given by Whitley in Projecting $m$ onto $c_0$, The American Mathematical Monthly Vol. 73 (3) (Mar., 1966), pp. 285–286. The proof given there is as follows (I'm following the exposition of Albiac–Kalton, Topics in Banach space theory, pp 45f with some minor modifications):

  1. Let $S$ be a countable infinite set. Then there is an uncountable “almost disjoint family” of infinite subsets of $S$: more precisely, there is a family $\{A_i\}_{i \in I}$ of subsets of $S$ such that $\# I = \# \mathbb{R}$, that $\# A_i = \# \mathbb{N}$ for all $i$ and the intersection $A_i \cap A_j$ is finite whenever $i \neq j$.

    Identify $S$ with the rational numbers in $[0,1]$ and let $I = [0,1] \smallsetminus S$. For each $i \in I$ choose a sequence $a^{(i)}_n \xrightarrow{n\to\infty} i$ from $S$ and set $A_i = \{a^{(i)}_{n}\,:\,n \in \mathbb{N}\}$ — an infinite set because $i$ is irrational. Then $\# I = \# \mathbb{R}$ and $A_i \cap A_j$ is finite whenever $i \neq j$, as desired.

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

    To see this, consider a family $\{A_i\}_{i \in I}$ of almost disjoint subsets of $\mathbb{N}$ as in 1. Assume that for each $i \in I$ we can find $x_i \in \ell^\infty$ supported on $A_i$ such that $P(x_i) \neq 0$, 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 $I_n = \{i \in I\,:\,(Px_i)(n) \neq 0\}$ is uncountable. Since $I_n$ is uncountable, there must be $k$ such that $I_{n,k} = \{i \in I\,:\,|(Px_i)(n)| \geq 1/k\}$ is uncountable.

    Now fix $n$ and $k$ such that $I_{n,k}$ is uncountable. Let $J \subset I_{n,k}$ be finite and consider $y = \sum_{j \in J} \operatorname{sign}{[(Px_j)(n)]} \cdot x_j$. Note that $$ (Py)(n) = \sum_{j \in J} \operatorname{sign}{[(Px_j)(n)]}\cdot (Px_j)(n) \geq \frac{\# J}{k} $$ 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 $P(y) = P(f) + P(z) = P(z)$ by hypothesis on $P$ and therefore $\|P(y)\| \leq \|P\| \|z\| \leq \|P\|$. This yields $$\# J \leq \|P\| k$$ whence the absurdity that $I_{n,k}$ must be finite.

    It follows that there must be a set $A = A_i$ such that $P(x) = 0$ for all sequences supported in $A$. Since all sets $A_i$ are infinite, $A$ is infinite and we are done.

  3. The subspace $c_0$ of $\ell^\infty$ is not complemented.

    If $c_0$ were complemented in $\ell^\infty$, there would be a continuous projection $Q: \ell^\infty \to \ell^\infty$ with range $c_0$. Applying 2. with $P = 1-Q$ we would find an infinite set $A$ such that $(1-Q)(x) = 0$ for all sequences supported in $A$. But this means that $Q(x) = x$ for all sequences supported in $A$. This is absurd since $A$ is infinite; not all bounded sequences supported in $A$ belong to $c_0$.

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    $\begingroup$ Small terminological glitch: independent family has another meaning: all finite Boolean combinations of sets are non-empty. What you have is an almost disjoint family. $\endgroup$ Apr 16 '12 at 17:34
  • $\begingroup$ Oh, thanks, silly me... I'll fix that in the next edit. $\endgroup$
    – t.b.
    Apr 16 '12 at 17:36
  • $\begingroup$ But is there a continuous map from $\ell^\infty \to c_0$ such that it restricts to identity on $c_0$? $\endgroup$
    – Mambo
    Feb 10 '18 at 19:51
  • $\begingroup$ I don't understand the step where you concluded that $y=f+z$ based on the fact that $A_i\cap A_j$ is finite for $i\neq j$. $\endgroup$
    – trek26
    Mar 21 '21 at 17:49
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Here is my proof. The weak topology in the space $l^\infty$ is realcompact. The weak topology in the space $l^\infty/c_0$ is not realcompact. Therefore $l^\infty/c_0$ is not isomorphic to a closed subspace of $l^\infty$.

Reference for the "weak realcompact" material:
Corson, H. H. The weak topology of a Banach space. Trans. Amer. Math. Soc. 101 1961 1–15.
Edgar, G. A. Measurability in a Banach space. II. Indiana Univ. Math. J. 28 (1979), no. 4, 559–579.

I showed this to W. B. Johnson once, he thought for a minute or so and came up with the more conventional proof.

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    $\begingroup$ Whitley phrases his proof in the following way: the dual of $\ell^\infty$ contains a countable total subset, while the dual of $\ell^\infty/c_0$ does not. The property that the dual contains a countable total subset passes to closed subspaces, hence $\ell^\infty/c_0$ can't be isomorphic to a closed subspace of $\ell^\infty$. $\endgroup$
    – t.b.
    Apr 17 '12 at 12:02
  • $\begingroup$ But is there a continuous map from $\ell^\infty \to c_0$ such that it restricts to identity on $c_0$? $\endgroup$
    – Mambo
    Feb 10 '18 at 19:52

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