Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a compact Hausdorff topological set, and $Y$ be its closed subset. Is the ideal of functions vanishing on $Y$ $$ I=\{f\in C(X):\ \forall y\in Y\ f(y)=0\} $$ complementable (as a closed subspace) in $C(X)$ (as a Banach space)?

This is true in the case when $X\subseteq {\mathbb R}^n$ (this follows from: Stein. Singular integrals... VI 2.2), but what about general case?

share|improve this question
3  
Counterexamples are already presented below, but on the positive side, it is true whenever $Y$ is compact metrizable. In such a case, there is a unital positive linear operator $T\colon C(Y)\to C(X)$ such that $T(f)|_Y=f$ for all $f\in C(Y)$. –  Narutaka OZAWA Aug 12 at 23:11
    
@NarutakaOZAWA, could you, please, give the reference? –  Sergei Akbarov Aug 13 at 18:02
2  
The result Taka mentioned is due to K. Borsuk, Bull. Internat. Acad. Polon. Sci. Sér. A No. 113 (1933), 1–10. –  Bill Johnson Aug 13 at 22:04
2  
Borsuk assumed that $X$ is separable; Dugundji proved that it is enough to assume that only $Y$ is separable: J. Dugundji, An extension of Tietze's theorem. Pacific Journal of Mathematics 1 (1951), no. 3, 353--367. There are two nice papers, one due to Pełczyński and the second one due to Haydon, about spaces which satisfy the above-mentioned theorem. Google for the term: "Dugundji space". –  Tomek Kania Aug 14 at 13:42

2 Answers 2

up vote 6 down vote accepted

Not in general.

It's well-known in Banach space theory that the ideal $c_0$ in $\ell^\infty$ is not complemented (see e.g. Albiac & Kalton).

By the Gelfand representation, $\ell^\infty \simeq C(\beta \mathbb{N})$ as a $C^\ast$-algebra. This maps $c_0$ to the ideal of functions on $\beta \mathbb{N}$ that vanish on $\beta \mathbb{N} \setminus \mathbb{N}$.

share|improve this answer
    
But $\beta{\mathbb N}\setminus{\mathbb N}$ is not closed in $\beta{\mathbb N}$. –  Sergei Akbarov Aug 12 at 18:39
2  
@SergeiAkbarov: Actually, it is. $\mathbb{N} \subset \beta \mathbb{N}$ consists of isolated points, so it's open. And in any case, your ideal $I$ only depends on the closure of the set. –  Alexander Shamov Aug 12 at 18:47
    
Yes, excuse me and thank you! –  Sergei Akbarov Aug 13 at 6:04

Alexander Shamov answered your question with a classical example, but you might be interested in a modern example constructed by Piotr Koszmider. There is an infinite connected compact Hausdorff space $K$ s.t. $C(K)$ has no complemented subspaces that are both infinite dimensional and infinite codimensional. In particular, if $L$ is a closed subset of $K$ s.t. the ideal of functions vanishing on $L$ is complemented, then $L$ is finite.

Koszmider, Piotr(BR-SPL) Banach spaces of continuous functions with few operators. (English summary) Math. Ann. 330 (2004), no. 1, 151–183.

share|improve this answer
    
Yes, that's unexpected. Thank you, Bill. –  Sergei Akbarov Aug 13 at 18:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.