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

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?

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

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 $$V=\cap_{i=1}^{n}kerf_i$$.