Let $X$ be a Banach Space and $P:X \to X$ be a linear map satisfying $P^2=P$. Suppose Range$(P)$ and Ker$(P)$ is closed then prove that $P$ is continuous.
Clearly $V$=Ker$(P)$+Img$(P)$ .I think this problem is a direct consequence of closed graph theorem in functional Analysis.But I am unable to prove.Please help!
Take a sequence $\{f_n\}$ in $X$ such that $f_n\to g$ and $P(f_n)\to h$. Then we have that $f_n-P(f_n)\to g-h$. Note that $f_n-P(f_n)\in \ker P$ since $P(f_n-P(f_n))=P(f_n)-P^2(f_n)=P(f_n)-P(f_n)=0$. Since $\mathrm{Ran}(P)$ and $\ker(P)$ are closed, and $P(f_n)\in\mathrm{Ran}(P)$, we must have that $h\in\mathrm{Ran}(P)$ and $g-h\in\ker P$. But then we must have that $0=P(g-h)=P(g)-P(h)$, and so $P(g)=P(h)=h$ since $P|_{\mathrm{Ran}(P)}=Id$ (thanks to Valdinês Júnior for pointing this out). Therefore the graph of $P$ is closed, and now apply closed graph theorem.