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I know that for any finite dimensional subspace $F$ of a banach space $X$, there is always a closed subspace $W$ such that $X=W\oplus F$, that is, any finite dimensional subspace of a banach space is topologically complemented.

However, I wonder whether we can put some condition on the complemented subspace. The problem I am working on is the following:

Let $X$ be an infinite dimensional subspace. Suppose we have \begin{equation*} X=\overline{F_1\oplus F_2\oplus F_3\oplus\cdots} \end{equation*} where all $F_j$'s are finite dimensional subspaces of $X$ with dimensions larger than 1.

Can we find a closed subspace $W$ such that $X=F_1\oplus W$ and $W\supset \overline{F_2\oplus F_3\oplus\cdots}$?

Or equivalently, can we find a vector $x\in F_1$ such that it lies outside the closed linear span of $F_j$ $(j\neq 1)$?

Thanks!

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What does your infinite direct sum actually mean? Is this an algebraic direct sum, in which case your subspace $X$ has a countably infinite Hamel basis and so cannot be a complete normed space? Or is it to be understood as some kind of limit? –  user16299 Jun 14 '12 at 3:17
    
@YemonChoi Thanks! I have corrected that. –  Hui Yu Jun 14 '12 at 4:02
    
@HuiYu I somehow oversimplified your question, when I was thinking about it. I've deleted my post since it does not answer your question. Sorry for the confusion. –  Martin Sleziak Jun 14 '12 at 21:20
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1 Answer

up vote 3 down vote accepted

Edit: I'm still not completely sure I understand your notation. For this answer, $F \oplus G$ means the algebraic sum of $F, G$, i.e $F \oplus G = \{x + y : x \in F, y \in G\}$, with the requirement that $F \cap G = \{0\}$, and $F = F_1 \oplus F_2 \oplus \dots$ means $$F = \bigcup_{n \ge 1} F_1 \oplus \dots \oplus F_n.$$

The answer to your question is: not necessarily. For instance, it may be that $F_2 \oplus F_3 \oplus \dots$ is dense in $X$.

To be explicit, take $X = C([0,1])$. I don't quite understand the point of wanting the $F_n$ to have dimension larger than 1, and it will make this example a little messier, but for $n \ge 2$ let $F_n$ be the span of $x^{2n-4}$ and $x^{2n-3}$. Then $F_2 \oplus F_3 \oplus \dots$ is precisely the space $P$ of polynomials, which by the Weierstrass approximation theorem is dense in $X$. Let $g_1, g_2$ be bump functions supported in $[0,1/4], [3/4,1]$ respectively, and let $F_1$ be the span of $g_1$ and $g_2$. Every function in $F_1$ vanishes identically on $[1/4, 3/4]$ so if it is a polynomial it is 0. Thus $F_1 \cap P = \{0\}$, so $F_1 \oplus F_2 \oplus \dots$ is still a direct sum, and it is still dense in $X$ (since it contains $P$), so your hypotheses are satisfied. But the only $W$ that contains $\overline{F_2 \oplus F_3 \oplus \dots} = X$ is $X$ itself, and so $W \cap F_1 = F_1$.

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A little bit sad that I have to look for a new approach on my problem. But thanks very much, Nate! This is a good example. –  Hui Yu Jun 15 '12 at 5:01
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A variant using a similar idea: Let $e_n = x^{n}$ for $n \geq 0$. The span of $\{e_n\}_{n \geq 0}$ is dense in $C[0,1]$ by Weierstrass. Put $f_n = e_0 + \frac{1}{n+1} e_{2n} = 1 + \frac{1}{n+1} x^{2n}$, $g_n = e_1 + \frac{1}{n+1}e_{2n+1} = x + \frac{1}{n+1} x^{2n+1}$ for $n \geq 0$ and let $F_n$ be the span of $\{f_{n},g_{n}\}$. Then $F_0 \oplus F_1 \oplus \cdots$ is dense and as $f_n \to e_0 = 1$ and $g_n \to e_1 = x$ we also have that $F_1 \oplus F_2 \oplus \cdots$ is dense in $C[0,1]$. This works verbatim with any normalized Schauder basis $\{e_n\}_{n\geq 0}$ of a Banach space $X$. –  t.b. Jun 15 '12 at 8:39
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