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Let $F$ be a finite dimensional subspace of an infinite dimensional Banach space $X$, we know that $F$ is always topologically complemented in $X$, that is, there is always a closed subspace $W$ such that $X=F\oplus W$.

I am thinking about the converse. Suppose $W$ is a subspace of $X$ such that $X=F\oplus W$ for some finite dimensional subspace $F$. Is $W$ necessarily closed?

I guess the answer should be negative but I cannot find such an example. Can somebody give a hint?

Thanks!

Martin Sleziak
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Hui Yu
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1 Answers1

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If $f:X\to\mathbb C$ is a discontinuous linear functional, then $\ker f$ is not closed. If $v$ is in $X\setminus \ker f$, then $F=\mathbb C v$ and $W=\ker f$ gives a counterexample. ($X$ is the internal direct sum of $F$ and $W$ as vector spaces, but it is not a topological direct sum.)

Jonas Meyer
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  • Thanks! But actually I am thinking about a slightly different problem. Do you mind looking at this earlier post of mine? http://math.stackexchange.com/questions/158077/finding-the-topological-complement-of-a-finite-dimensional-subspace – Hui Yu Jun 14 '12 at 23:40
  • @HuiYu: I just looked there, and it looks like Nate gave it a good answer (after you posted your comment). – Jonas Meyer Jun 15 '12 at 04:56
  • @JonasMeyer If $\dim F= \dim (\mathbb C v) = 1$ then how are you sure that it is complemented with $\ker f$ ? if $\dim (X/\ker f) \ge 2$ then it is not true that $X = F \oplus \ker f$, or did I misunderstood something ? – Physor Nov 07 '21 at 10:22