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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!

nekodesu
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  • Why do you want to apply closed graph? $A$ is bounded by assumption. – Jochen Mar 21 '18 at 09:22
  • @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

2 Answers2

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Hints only.

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 $|A\sum x_n-y|\to 0.$

Dap
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  • 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
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Built on the hints in Dap's answer. Basically mimicking the proof for the Open Mapping Theorem.

Note: here the statement "$\text{Ran}(A)$ is of the Second Category" must be understood as "$\text{Ran}(A)$, equipped with the subspace topology, is a Second Category space". It should not be understood as being a Second Category subset of $Y$.

Proof. Firstly, the closure of the range $\overline{\text{Ran}(A)}$, with its subspace topology, is a Banach space. By the Baire Category Theorem, it is of Second Category. Also, the subspace topology of $\text{Ran}(A)$, inherited from $Y$, coincides with that inherited from the closure $\overline{\text{Ran}(A)}$. If $\text{Ran}(A)$ is a Second Category space, it is then a Second Category subset in Banach space $\overline{\text{Ran}(A)}$. Hence, since $$\text{Ran}(A)=\cup_{n\in\mathbb{N}}A(nB_X(1))=\cup_{n\in\mathbb{N}}nA(B_X(1)),$$ there exists $n_0\in\mathbb{N}$ such that $\text{Int} \overline{n_0 A(B_X(1))}\neq \emptyset$ in terms of the subspace topology of $\overline{\text{Ran}(A)}$. Then there exists some $r_0>0$ and $y_0\in Y$ such that $B_Y(y_0, r_0)\cap\overline{\text{Ran}(A)} \subset \overline{n_0 A(B_X(1))}.$ It can therefore be shown that, for every $L>0$, $B_Y(\frac{r_0}{2n_0}L)\cap \overline{\text{Ran}(A)}\subset \overline{A(B_X(L))}.$

So for every $y\in \overline{\text{Ran}(A)}\setminus\{0\}$, we have $y\in \overline{B_Y(\|y\|_Y)}\cap \overline{\text{Ran}(A)}\subset \overline{A(B_X(\frac{2n_0}{r_0}\|y\|_Y))}$. That is, there exists $x\in B_X(\frac{2n_0}{r_0}\|y\|_Y)$ such that $\|y-Ax\|_Y<\frac{\|y\|_Y}{2}$. But this means $$y-Ax\in B_Y(\frac{\|y\|_Y}{2})\cap \overline{\text{Ran}(A)}\subset \overline{A(B_X(\frac{2n_0}{r_0}\frac{\|y\|_Y}{2}))}.$$ Hence, we can inductively choose a sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ satisfying $\|x_n\|_X< \frac{2n_0}{r_0}\frac{\|y\|_Y}{2^{n-1}}$ and $$\|y-A(x_1+\cdots +x_n)\|_Y<\frac{\|y\|_Y}{2^{n-1}}.$$ Hence, by the completeness of $X$, $\sum_n x_n$ converges absolutely to some $\xi\in X$. By continuity of $A$, taking the limit gives $0_Y=\lim_{n\to\infty}(y-A(x_1+\cdots +x_n))=y-A\xi,$ and so $y=A\xi\in \text{Ran}(A)$.

Hence $\text{Ran}(A)=\overline{\text{Ran}(A)}$ is closed. $\square$

user760
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