Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(f_n)_{n \geq 1}$ be disjointly supported sequence of functions in $L^\infty(0,1)$. Is the space $\overline{\mathrm{span}(f_n)}$ (the closure of linear span) complemented in $L^\infty(0,1)$? By complemented we mean that $L^\infty(0,1) = \overline{\mathrm{span}(f_n)} \oplus X$, where $X$ is a subspace of $L^\infty$ and $\oplus$ is direct sum.

Equivalently, we can ask if there exists a projection $P\colon L^\infty(0,1) \to \overline{\mathrm{span}(f_n)}$?

It is quite easy to prove this in $C[0,1]$. Indeed, let $(f_n)$ be disjointly supported sequence in $C[0,1]$ and fix $x_n \in \mathrm{supp}(f_n)$, $n \in \mathbb{N}$. Then the space $C[0,1]$ can be written as $$ C[0,1] = \overline{\mathrm{span}(f_n)} \oplus \{f \in C[0,1]\colon f(x_n) = 0, n = 1,2,\dots \}. $$

share|improve this question
    
You might want to reconsider your accepted answer. –  Jonas Meyer Jun 24 '11 at 9:10
    
@Jonas: Thanks for pointing this out. –  xen Jun 29 '11 at 9:04
add comment

3 Answers 3

up vote 8 down vote accepted

The answer is no. The closed linear span of such a sequence is separable, and so if $\overline{span(f_n)}$ was complemented in $L^\infty (0, 1)$ then every Banach space isomorphic to $L^\infty (0, 1)$ would contain an infinite dimensional, separable complemented subspace. In particular, since $L^\infty (0, 1)$ is isomorphic to $\ell^\infty$ (this is an old result due to Pelczynski, but a proof is given as Theorem 4.3.10 of Albiac and Kalton's text Topics in Banach space theory), if $\overline{span(f_n)}$ was complemented in $L^\infty (0, 1)$ then $\ell^\infty$ would contain an infinite dimensional, separable complemented subspace... however, this is not true since every infinite dimensional complemented subspace of $\ell^\infty$ is isomorphic to $\ell^\infty$ by a result of J. Lindenstrauss (see, e.g., Lindenstrauss and Tzafriri's book Classical Banach Spaces I, Theorem 2.a.7), hence nonseparable.

share|improve this answer
add comment

$X$ can be taken to be $\left\{ f \in L^{\infty} | \forall n,\ \int_{[0,1]} f f_n =0 \right\}$

share|improve this answer
    
Thanks! Just to be sure, if $f \in L^\infty(0,1)$ then $f = (f - g) + g$, where $g = \sum_{k \geq 1} \frac{\int f f_k}{\int f_k^2} f_k$. Then $g \in \overline{\mathrm{span}(f_n)}$ and $f-g \in X$, right? –  xen Mar 17 '11 at 20:32
    
@xen, @Plop: This does not work. Let $f_n = 1_{[\frac{1}{n+1}, \frac{1}{n})}$. If $g \in \operatorname{span}(f_n)$ then $g = 0$ a.e. on a neighborhood of $0$. So if $g \in \overline{\operatorname{span}(f_n)}$, then by uniform convergence, $\lim_{x \to 0} g(x) = 0$ (after modification on a null set). If we can write $1 = f+g$, then $f$ must have $\lim_{x \to 0} f(0) = 1$, and in particular $\int f_n f > 0$ for sufficiently large $n$. So $f$ is not in your space $X$. –  Nate Eldredge May 30 '11 at 17:09
1  
Thanks for adding that comment Nate; I was planning on coming back myself after a little sleep to write something similar. When I wrote my answer it was after 3am local time, but I had to do my duty: xkcd.com/386 –  Philip Brooker May 31 '11 at 5:21
    
Indeed $X$ supplements $\{ \sum_n \lambda_n f_n\ |\ (\lambda_n)_n\ \mathrm{bounded} \} \neq \overline{\mathrm{span}(f_n)}$. I can't delete the answer while it is accepted. –  Plop May 31 '11 at 12:16
add comment

If all but finitely many of the functions are the zero function, then the answer is yes, because any finite-dimensional subspace is complemented. But this is the trivial case.

In the nontrivial case, just note that if you normalize the nonzero functions in the sequence, they form a basic sequence $1$-equivalent to the unit vector basis of $c_0$. But we know $\overline{\text{span}(f_n)}=c_0$ is not complemented in $L_\infty=\ell_\infty$ (see, for example, Albiac and Kalton's Topics in Banach Space Theory).

share|improve this answer
add comment

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.