Let $T$ be a bounded linear operator between Banach space $ X \to Y$ such that $\dim \ker T<\infty$ and the range $R(T)$ is closed . Prove the complement $C$ (such that $X = \ker T \oplus C$ ) is closed.
For Hilbert space, this is automatic since the complement can be chosen to be $\ker T ^\perp$.For general Banach space the closed range assumption is needed. I can prove it as follows since $C $ toplogically isomorphic to $Y/R(T)$ since $Y$ is Banach space and $R(T)$ is closed we know $C$ is in fact a Banach space hence as subset of Banach space $X$ must be closed.
Is my proof correct.Is there some alternative proof for example using the duality theory?Or something else?Is there example that when $R(T)$ is not closed or $\ker(T)$ is not finite dimensional such that $C$ is not closed ?
Best Answer
Let $v_1,v_2,\ldots, v_n$ be a basis in $\ker T.$ By the Hahn-Banach theorem there exist continuous linear functionals $\varphi_1,\varphi_2,\ldots, \varphi_n$ such that $\varphi_i(v_j)=\delta_{ij}.$ Then $C=\bigcap_{j=1}^n \ker \varphi_j .$ Indeed for every $x\in X$ we have $$x=\sum_{j=1}^n \varphi_j(x)v_j+\left [x-\sum_{j=1}^n \varphi_j(x)v_j\right ].$$ The second term belongs to $C,$ while the first one to $\ker T.$ The intersection of $C$ and $\ker T$ is trivial, as for $x\in C$ the first term is equal $0.$
In general every finite dimensional subspace $Y$ of $X$ has a closed complement subspace.