3
$\begingroup$

I have an elementary question about finite-dimensional subspaces and finite-codimensional subspaces. This question may be known.

Question. Let $U$ be a finite-dimensional subspace of an infinite-dimensional Banach space $X$. Let $\epsilon>0$. Is there a finite-codimensional subspace $V$ of $X$ such that $U\cap V=\{0\}$ and $$\|u\|\leq (1+\epsilon)\|u+v\|$$ for every $u\in U, v\in V$ ?

Improve this question
$\endgroup$
3
  • $\begingroup$ (Of course the extra condition on $U \cap V$ is unnecessary, since, if $u \in U \cap V$, then your inequality gives $\|u\| \le (1 + \epsilon)\|u - u\| = 0$. Also there's no need to specify infinite dimensionality, since, if $X$ is finite dimensional, then you may take $V = 0$.) $\endgroup$
    – LSpice
    May 18, 2020 at 8:35
  • $\begingroup$ The condition $U\cap V=\{0\}$ is necessary, LSpice. $\endgroup$
    – Dongyang Chen
    May 18, 2020 at 8:44
  • $\begingroup$ My inequality implies that $U\cap V=\{0\}$. You are right, LSpice. But how to prove the result? $\endgroup$
    – Dongyang Chen
    May 18, 2020 at 8:47

1 Answer 1

Reset to default
1
$\begingroup$

Let $\epsilon>0$ and let $(u_{i})_{i=1}^{n}$ be a $\delta$-net for $S_{U}(\delta>0$ will be specified later). For each $i$, choose $x^{*}_{i}\in S_{X^{*}}$ such that $\langle x^{*}_{i},x_{i}\rangle=1$. Let $V=\cap_{i=1}^{n}Ker(x^{*}_{i})$. Then $V$ is the required finite-codimensional subspace.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.