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I'm stuck on kind of a silly functional analysis problem. Suppose that $T:D(T) \subset X \to Y$ is a closable linear operator, where $X$ and $Y$ are Banach spaces and $R(T)$ is finite dimensional. I need to show that $T$ is continuous, but for some reason we are not allowed to use the fact that a linear operator with finite dimensional range is compact.

My approach so far has been to pick an arbitrary sequence $\{x_n\} \subset D(T)$, with $x_n \to 0$. Then I want to show that $\{Tx_n\} \subset R(T)$ is Cauchy, so that the limit will exist (since $R(T)$ is a finite dimensional linear subspace of the Banach space Y, and thus complete). Then since $T$ is closable, I will have $\lim_{n \to \infty}Tx_n=0$. Since the sequence $\{x_n\}$ was arbitrary, then $T$ will be continuous at $0$ and hence continuous. However, I can't seem to show that $\{Tx_n\}$ converges.

Am I taking the wrong approach, or just missing something?

EDIT: For the person asking what closable means: on page 3 there is a useful criteria.

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What does closable mean? – user27126 Dec 6 '12 at 4:50
Given that discontinuous linear functionals exist, it's good that you're not allowed to use the "fact" that a linear operator with finite dimension range is compact: it isn't true. – Nate Eldredge Dec 6 '12 at 5:16

You won't be able to show that $\{T x_n\}$ is Cauchy without using the closability. For a discontinuous (hence non-closable) operator, we can have, for instance, $x_n \to 0$ but $\|T x_n\| \to \infty$.

But here's a thought. Suppose to the contrary that $T$ is discontinuous. Then by appropriate scaling you can find a sequence $x_n \to 0$ but $\|T x_n\| = 1$ for all $n$. Since $R(T)$ is finite dimensional and hence locally compact, you can pass to a subsequence so that $T x_n$ converges to some $y$, which must have $\|y\| = 1$ and in particular $y \ne 0$. But this contradicts closability.

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