Kernel of $T$ is closed iff $T$ is continuous

I know that for a Banach space $X$ and a linear functional $T:X\rightarrow\mathbb{R}$ in its dual $X'$ the following holds: \begin{align}T \text{ is continuous } \iff \text{Ker }T \text{ is closed}\end{align} which probably holds for general operators $T:X\rightarrow Y$ with finite-dimensional Banach space $Y$. I think the argument doesn't work for infinite-dimensional Banach spaces $Y$. Is the statement still correct? I.e. continuity of course still implies the closedness of the kernel for general Banach spaces $X,Y$ but is the converse still true?

Solution 1:

The result is false if $Y$ is infinite dimensional. Consider $X=\ell^2$ and $Y=\ell^1$ they are not isomorphic as Banach spaces (the dual of $\ell^1$ is not separable). However they both have a Hamel basis of size continuum therefore they are isomorphic as vector spaces. The kernel of the vector space isomorphism is closed (since is the zero vector) but it can not be continuous.

Solution 2:

No, this is not true. Here is a counterexample for every infinite-dimensional Banach space $Y$.

Take a discontinuous functional $\phi\in Y^*$ (this is always possible if $Y$ is infinite-dimensional, by the axiom of choice). Let $A:=\ker \phi$, and $B$ any algebraic supplement. Note that $A$ is non-closed, and $B$ is closed in view of $\dim B=\text{codim} A=1$.

Consider the projection $p:Y\to Y$ onto $A$, w.r.t. the algebraic direct sum decomposition $Y=A\oplus B$. Now $\ker p=B$ is closed, but $\ker(1-p)=A$ is not, so $p$ is discontinuous.