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!
Best Answer
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 $W \cap F_1 = F_1$.