# Bounded linear operator (between Banach spaces) with second category range has closed range

I'm attempting a problem about closed range of a bounded linear operator.

Assume $X, Y$ are Banach spaces and $A$ is a bounded linear operator. If $\operatorname{Ran}(A)$ is of the second category, then show that $\operatorname{Ran}(A)$ is closed.

I want to use the Closed Graph Theorem and assume that $Ax_i \to y$ in $Y$, if we can show that $x_i \to x$ in $X$, then the result follows by continuity of $A$. But I'm having a hard time showing that $\{x_i\}$ is a convergent sequence in $X$, and don't know how to use the fact that $\operatorname{Ran}(A)$ is of the second category.

Any help would be appeciated!

• Why do you want to apply closed graph? $A$ is bounded by assumption. – Jochen Mar 21 '18 at 9:22
• @Jochen. The Closed Graph is a iff" statement. $A$ being continuous implies that if you have $x_i \to x$ and $Ax_i \to y$ then $Ax = y$. It remains to show that for arbitrary $Ax_i \to y$, my $x_i$ actually converges. – nekodesu Mar 21 '18 at 13:16
• But that's not true: It may very well be that $Ax_i=0$ for all $i$ but $x_i$ do not converge. Your question is about the open mapping theorem. – Jochen Mar 21 '18 at 13:27
• @Jochen. Indeed, it might not be true in general, but I'm curious how the condition that $\text{Ran}(A)$ is of the second category jumps in. – nekodesu Mar 21 '18 at 15:21
• This is classical form of the open mapping theorem, see, e.g., Rudin's Functional Analysis, Theorem 2.11. – Jochen Mar 21 '18 at 15:32

The second category assumption lets you prove that $A(B_X(1))$ is dense in some ball $B_Y(\epsilon)\cap\mathrm{Ran}(A).$ Use this to show that for $y\in\overline{\mathrm{Ran}(A)},$ you can inductively pick a sequence $x_n$ with $\sum x_n$ converging absolutely and $|A\sum x_n-y|\to 0.$
• I'm a bit confused since $y$ should already be in the closure of range of $A$. We need it to be in the range of $A$ which is stronger. – nekodesu Mar 21 '18 at 15:32
• @octoberbear: I'm saying that for every $y$ in the closure of range of $A,$ you can try to pick a series in $X$ whose sum gets mapped to $y.$ As mentioned in the comments, there's a complete proof in Rudin 2.11 – Dap Mar 21 '18 at 16:34