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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?

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  • $\begingroup$ What's your argument for injective $T$? $\endgroup$ – tomasz Nov 28 '14 at 6:25
  • $\begingroup$ @tomasz would you mind going through the proof that $T(B)$ is closed for any closed, bounded $B$? The fact that $T(\ell_2)$ is not closed seems straightforward. $\endgroup$ – Jason Nov 28 '14 at 7:59
  • $\begingroup$ @tomasz This $T$ does not satisfy the assumption: Take the unit sphere (it is closed and bounded), $T(e_n)\to 0$, but $0$ is not in the image of the unit sphere under $T$. $\endgroup$ – daw Nov 28 '14 at 8:43
  • $\begingroup$ Here is the proof for injective $T$: math.stackexchange.com/questions/183146/… $\endgroup$ – daw Nov 28 '14 at 8:50
  • $\begingroup$ $S:=\{x\in l^2: \|x\|=1\}$, $0\in \overline{(T(S))}$, $0\not\in T(S)$, $S$ is closed and bounded, but $T(S)$ is not closed. $\endgroup$ – daw Nov 28 '14 at 8:55
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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 $\hat T[z]=u$.

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