# Is a sequence of disjointly supported functions in $L^\infty$ complemented?

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 $$C[0,1] = \overline{\mathrm{span}(f_n)} \oplus \{f \in C[0,1]\colon f(x_n) = 0, n = 1,2,\dots \}.$$

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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

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.

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

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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
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

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).

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