if $T\in B(H)$ has closed range then show that $T ^*$ also have closed range.

Question

if $T\in B(H)$ has closed range then show that $T ^*$ also have closed range.

we are given $$R(T)=(Ker(T^*))^{\perp}$$ and we want to show $$R(T^*)=(Ker(T))^{\perp}$$

its easy to see $$R(T^*)\subset (Ker(T))^{\perp}$$. but how to show other way around.any suggestion.

or there is some other better way to do this .thanks in advanced.

Answer 1

The statement you want to prove is (a special case of) the closed range theorem. Let me sketch the proof.

The map $\hat T\colon (\ker T)^{\perp}\longrightarrow \operatorname{ran} T,\,x\mapsto Tx$ has a bounded inverse by the open mapping theorem. For $y$ in the closure of $\operatorname{ran} T^\ast$ define $$ \phi\colon \operatorname{ran} T\longrightarrow \mathbb{K},\,Tz\mapsto \langle \hat T^{-1}Tz,y\rangle. $$ The map $\phi$ is linear and bounded, thus there exists $x\in \operatorname{ran} T$ such that $\phi(Tz)=\langle Tz,x\rangle$ for all $z\in H$. It follows that $$ \langle Tz,x\rangle=\phi(Tz)=\langle \hat T^{-1}Tz,y\rangle=\langle z,y\rangle, $$ where we used that $y\in \operatorname{ran}T^\ast\subset (\ker T)^{\perp}$. Therefore, $y=T^\ast x\in \operatorname{ran} T^\ast$.