Closed map on Banach Space

Question

Let $X,Y$ be Banach spaces and $T \in B(X,Y)$. Show that if $T$ sends every bounded closed subsets of $X$ onto closed sets of $Y$ then $T(X)$ is closed.

It's true when the map is injective, but is it true in general?

Answer 1

Given the answer here Bounded linear operator maps norm-bounded, closed sets to closed sets. Implies closed range?, the claim is true for injective $T$.

Now let $T$ be arbitrary. We work in the quotient space $X/N(T)$. For $x\in X$ let $$ [x] = x + N(T) \in X/N(T). $$ The quotient space is a Banach space under the norm $$ \|[x]\|_{X/N(T)}:=\inf_{y\in N(T)}\|x-y\|_X. $$ Define $\hat T: X/N(T)\to Y$ by $$ \hat T([x]) = Tx. $$ Then $\hat T$ is injective, moreover $R(T)=R(\hat T)$. It remains to prove that $\hat T$ maps bounded and closed sets to closed sets.

Let now $M\subset X/N(T)$ be closed and bounded, $[z_n] \in M$ be given with $\hat T[z_n]\to y$. Since $M$ is bounded, we can choose $z_n\in [z_n]$ such that $Z:=\{z_n,\ n\in \mathbb N\}$ is bounded in $X$. By construction $Tz_n \to y$, thus $y\in cl(T(Z))$, and the assumption on $T$ implies $y\in T(cl(Z))$. Thus, there is a sequence $\tilde z_n\in Z$ with $\tilde z_n \to z\in cl(Z)$ and $Tz=u$. Now, $\tilde z_n\to z$ implies $[\tilde z_n]\to [z]$ in $X/N(T)$. Since $[\tilde z_n]\in M$ this implies $[z]\in M$ and $\hat T[z]=u$.