Approximation Theory, Projections and Hyperplanes

Question

Let $X$ be a Banach space and let $Y = \ker f \subset X$ be a hyperplane in $X$ for some nonzero $f \in X'$ . Prove that if $P\colon X \longrightarrow Y$ is a continous projection, than there exists $w \in X$ such that $f (w) = 1$ and $P(x) = x − f (x)w$.

Answer 1

Let $u\in X-Y$, $P(u-P(u))=P(u)-P^2(u)=0$. Remark that $f(u-P(u))=f(u)\neq 0$. We write $w={{u-P(u)}\over{f(u-P(u))}}$. We have $f(w)=1$ and $P(w)=0$.

Let $x\in X$, $f(x-f(x)w)=f(x)-f(x)f(w)=0$. This implies that $x-f(x)w\in Y$ and $P(x-f(x)w)=x-f(x)w$.

We have $x=x-f(x)w+f(x)w$, $P(x)=P(x-f(x)w)+P(f(x)w)=P(x-f(x)w))=x-f(x)w.$