Projection of $\ell^\infty$ to $c_0$

Question

We have the following conjecture that we need to find a counterexample for:

"Let $X$ and $Y$ be normed spaces, and let $W \subset X$ be a linear subspace. Suppose that $T_W \in B(W,Y)$. Then $T_W$ can be extended to an element $T\in B(X,Y)$."

Our assignment gave us the hint to consider $Y = W$ and $T_W$ the identity map on $W$. Now it is easy to see that if such an extension for the identity exists, then it is a projection, and since projections are nice on finite dimensional spaces, this rules out all vector spaces $X$ with $\dim(X)<\infty$.

Looking around on MSE, it seems the following counterexample would work: take $T_W$ the identity map on $W = c_0$, considered as a linear subspace of $X=\ell^\infty$. The problem then reduces to showing that there is no projection from $\ell^\infty$ on $c_0$, and it seems that the way to prove this is by using something called "Phillip's Lemma", however searching for this lemma on the internet yields almost no results except for other MSE posts, and the proof and statement of this lemma seem quite involved. Is there any other way of showing there is no projection from $\ell^\infty$ into $c_0$?

Other counterexamples that work are also welcome!

There is a relatively short and very neat proof in section $2.5$ of the book "Topics in Banach Space Theory" by Albaic and Kalton. — Aweygan, Dec 16 '17 at 18:03
Unfortunately it happens to be Theorem 2.5.5 (Phillips-Sobczyk, 1940-1), which builds on Theorem 2.5.4 and Lemma 2.5.3, whose proofs are exactly the humongous proof of this post (Phillip's Lemma), which he is not asking for: https://math.stackexchange.com/questions/132520/complement-of-c-0-in-ell-infty — The Phenotype, Dec 17 '17 at 16:09

score 2 · Answer 1 · answered Dec 17 '17 at 15:19

2

Consider $V=\prod_{i\in \mathbb{Z}} \mathbb{R} e_i$ where $\{e_i\}$ is the standard basis of $V$, i.e., $e_i$ is the vector in $V$ with all entries equal to $0$ except the $i$-th entry eqauls to $1$. Using the usual Euclidean norm on $V$ as an $\mathbb{R}$-vector space. Then consider $W=\bigoplus_{i\in \mathbb{Z}}\mathbb{R} e_i$ as an $V$-subspace, and denote $Id_W$ as the identity map on $W$. Then there does not exist any extension from $V$ to $W$. If there exists such extension, then every $e_i\mapsto e_i$ (because $e_i\in W$). Then this extension should be identity. However, $W\subsetneq V$, so contradiction.

answered Dec 17 '17 at 15:19

Brilliant! Thanks mate. – The Brainlet Exterminator Dec 17 '17 at 15:52
1

$\mathbb{Z}$ should be $\mathbb{N}$, right? – Dec 17 '17 at 16:30
1

@Insertnamehere right. – Dec 17 '17 at 16:32
4

$V$ has an uncountable basis, so ${e_i}$ don't form a basis for $V$; they do for $W$. A basis is a linearly independent subset $B\subset V$ s.t. any $v\in V$ may be written as a finite linear combination of basis vectors.. – Dec 17 '17 at 16:58
2

Alright! Interesting discussion boys, how about we meet up in real life soon? – The Brainlet Exterminator Dec 17 '17 at 17:06
@Insertnamehere right! – Dec 17 '17 at 17:39
1

What norm on $V$ are you talking about, exactly? If it is true that the extension would have to be the identity that depends on boundedness - there certainly are unbounded extensions, since as noted the $e_i$ are not a basis for $V$. – David C. Ullrich Dec 17 '17 at 19:32

David C. Ullrich · Accepted Answer · 2017-12-19T15:25:41.417

Edit: Please note that my comments on the other answer are left over from a point when the other answer was the "accepted" one - at that time it seemed important to point out it was wrong, lest readers be misled by the checkmark.

The [other] answer isn't right. Or at the very least it needs substantial clarification: there's really no such thing as the standard Euclidean norm on the given $V$.

What norm we're talking about is crucial, because the non-existence of an extension depends on the fact that we're talking about bounded operators; in straight linear algebra with no topology any linear map defined on a subspace extends linearly to the containing space. [In particular, to make the argument in the other answer correct we need to know that the span of the $e_n$ is dense in $V$; this depends on what the norm is. For example, the non-density of the span of the $e_n$ in $\ell^\infty$ is why it's not trivial to show there's no projection from $\ell^\infty$ onto $c_0$.]

But the idea in the [other] answer works: Say $V$ is a Banach space and $W\ne V$ is a dense subspace; then it's clear that the identity map in $W$ cannot be extended to a bounded map from $V$ to $W$. [Hmm, regardless of whether that's actually the idea behind the other answer, I realized it was so after reading the other answer...]

More interesting would be an example where $V$ and $W$ are both complete.

Projection of $\ell^\infty$ to $c_0$

2 Answers2