Let $X$ be a Banach space and let $T:X \rightarrow X$ be a linear map such that $T^2=T$ and both $\text{Im}(T)$ and $\text{Ker}(T)$ are closed. Then $T$ is bounded?
I got stuck in this problem while I was studying functional analysis. Can anyone help me?
This is an easy consequence of Closed Graph Theorem. Suppose $x_n \to x$ and $Tx_n \to y$. Since the range is closed we can write $y$ as $Tz$ for some $z$. Now $Tx_n-x_n$ belongs to the kernel (because $T^{2}=T$) and hence its limit $y-x$ also belongs to the kernel. Thus $Ty=Tx$. But $Ty=T(T(z))=Tz=y$ so we get $y=Tx$ as required.
