Concrete example: closed subspace of finite codimension admits complement

functional-analysispartial differential equations

This is an example from Section 2.4 of Brezis's "functional analysis":

Suppose $E$ is a Banach space, and let $N \subset E^*$ of dimension $p$. Then $G := \{x \in E: \langle f, x \rangle = 0 \ \forall f \in N \}$ is closed and of codimension $p$. Indeed, let $f_1, \cdots, f_p$ be a basis for $N$. Then there exists $e_1, \cdots, e_p \in E$ such that $\langle f_i, e_j \rangle = \delta_{ij}$. It is easy to check that $e_i$'s are linearly independent and $\text{span}\{e_1, \cdots, e_p\}$ is a complement of $G$.

Question: I could show the existence of $e_{i}$, and they are linearly independent. But I don't see $\text{span}\{e_1, \cdots, e_p\}$ is a complement of $G$. How could I check it? Could anyone give me some hint on this?

Best Answer

Write $$H=\operatorname{span}\{e_1,\ldots,e_p\}.$$

If $x\in G\cap H$, then $x=\sum_1^px_je_j$; using that $x\in G$, you get that $0=f_j(x)=x_j$. So $x=0$. That is, $G\cap H=0$.

Given any $x\in E$, we have $y=x-\sum_jf_j(x)e_j\in G$. So $$x=y+\sum_jf_j(x)e_j\in G+H.$$

The idea is that the map $P:x\longmapsto \sum_jf_j(x)e_j$ is a projection.

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