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How do I show that an operator with finite range is compact?

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What does closable mean? –  Sanchez 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
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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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