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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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    @Jonas: Thanks for pointing this out.2011-06-29

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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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    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.2011-05-31
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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).