Let $X,Y$ be NLS , $T:X \to Y$ be a linear map such that $T^{-1}(\{y\})$ is closed for every $y \in Y$ and $T$ has closed graph , then is it true that $T$ is continuous ?
I know that the statement is true if $Y$ is finite dimensional because in that case $\ker T$ closed implies continuity of $T$ , and I also know that if $X,Y$ are Banach spaces then it is true by closed graph theorem . But I don't know what happens in general . Since $T$ has closed graph , I know that $T$ maps compact sets to closed sets ; for general metric spaces I know that a function having closed graph and carrying compact sets to "compact "sets must be continuous , so it seems we are very close . Though for $\mathbb R$ and only maps ( not linear ) the statement is false as can be seen from $f : \mathbb R \to \mathbb R$ as $f(x)=1/x , \forall x\ne 0 ; f(0)=0$
Let $S$ be any injective bounded linear operator of Banach spaces $Y \to Z$ such that $S(Y)$ is not closed, take $X = S(Y)$ (with the norm of $Z$) and $T = S^{-1}$.