Let $X$ be a banach space, and let $U$ be a finite dimensional subsapce, then there is a closed subspace $V$ s.t $X=U\bigoplus V$

MY attempt:

Let $U=Span\{v_1,...,v_n\}$ and consider the following bounded operators $F:U \to R^n$, $F_k$ being linear functionals,

$F(x)=F(\alpha_1 v_1+...+\alpha_n v_n)=(F_1(x),...,F_n(x))=(\alpha_1,..,\alpha_n)$. Now by Banach extension theorem we can extend each $F_k$ and so $F$ to the entire $X$. Note $F$ still remains bounded.

Now let $V=\ker(F)$. I claim this choice of $V$ works. Let $x\in X$ then $F(x)=(\alpha_1,..,\alpha_n)$ and so $x=\alpha_1v_1...+(x-\alpha_1v_1...)$. Uniquness is easy to see.

Does this work?

  • 1
    $\begingroup$ It is correct but Hahn Banach Thereom has to be appllied to each component since the theorem applies only to linear functionals. $\endgroup$ Jul 8, 2020 at 8:12

1 Answer 1


Your idea is almost correct but as mentioned in the comments you can only use Hahn-Banach to linear functionals, so Consider $\{e_1,...e_n\}$ to be a basis for $U$ by the Hahn-Banach Theorem you can get linear functionals $\{f_1,...,f_n\}$ such that $f_i(e_j)=\delta_{ij}$ and then consider $V=\cap_{i=1}^{n}kerf_i$.


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