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It is clear that if $ T: X \rightarrow X $ is a bijective compact operator, where $ X $ is a Banach space, then $ \dim(\text{Range}(T)) = \dim(X) $, which implies that $ \dim(X) $ must be $ < \infty $.

How do I prove the converse: If $ \dim(X) < \infty $, then there exists a bijective compact operator $ T: X \rightarrow X $?

Thank you!

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There's some sort of typo in the first sentence of your question. – Christopher A. Wong Dec 21 '12 at 10:55
@ChristopherA.Wong : i think now its correct . – Theorem Dec 21 '12 at 11:00
The first statement, namely proving that the existence of a bijective compact operator $ T: X \rightarrow X $ implies that $ \dim(X) < \infty $, is actually a non-trivial result. It is a more difficult problem than the one that the OP is posing. Every operator on a finite-dimensional Banach space is compact. Hence, the identity operator on a finite-dimensional Banach space is a bijective compact operator. – Haskell Curry Dec 22 '12 at 0:41
In fact, a finite-dimensional Banach space $ X $ is nothing other than a finite-dimensional Euclidean space. This follows from the fact that all norms on a finite-dimensional vector space (over $ \mathbb{R} $ or $ \mathbb{C} $) are equivalent. Hence, every bijective operator on $ X $ can be represented as an invertible matrix with respect to a fixed finite basis of $ X $. As such, there are $ 2^{\aleph_{0}} $-many bijective compact operators on a finite-dimensional Banach space, with the identity operator being one of them. – Haskell Curry Dec 22 '12 at 0:54
up vote 5 down vote accepted

I suppose the theorem you want to prove is this one:

Let $(X,\Vert \cdot \Vert)$ be a Banach space. There exists a linear continuous operator $T \colon X \to X$ compact if and only if $\dim X <+\infty$.

One way (if) is clear: indeed, if $\dim X<+\infty$ then every operator $T \colon X \to X$ is compact (since its range is finite dimensional: this is a well-known sufficient condition for compactness). For example, take identity of $X$: it is bijective (obviously!) and compact.

Now, the other way (only if): suppose $T\colon X \to X$ is bijective and compact. There exists $T^{-1}$ and, moreover, it is continuous: so $TT^{-1}=\text{id}_X$ is compact, since $\mathcal K(X)$ is a closed ideal in $\mathcal L(X)$. In particular, the closure of the unit ball of $X$ is compact, hence the space $X$ is finite dimensional.

Hope this helps.

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But if $T$ is compact then is it required that rank of $T$ is always finite dimensional ? – Theorem Dec 21 '12 at 11:10
No. The implication you ask for should be the easy one. – flavio Dec 21 '12 at 11:17
I perfectly agree with you, jiku1797. Anyway, I've added some details. – Romeo Dec 21 '12 at 11:23
@Romeo: My comment was meant to answer Theorem's misunderstanding regarding compact operators. Your comment is completely fine! – flavio Dec 21 '12 at 11:29
@Romeo : Thanks! – Theorem Dec 21 '12 at 11:40

Theorem, it is not true that $ \dim(\text{Range}(T)) = \dim(X) $ implies that $ \dim(X) < \infty $. Another way of reasoning is as follows. Let $ T: X \rightarrow X $ be a bijective compact operator. Then by the Bounded Inverse Theorem, $ T^{-1} $ exists and is continuous. Hence, $ T $ is also a homeomorphism. Let $ B_{X} $ be the closed unit ball of $ X $. Then $ T[B_{X}] $ is closed and has compact closure, which implies that $ T[B_{X}] $ is compact (a closed subset of a compact space is also compact). However, as $ T $ is a homeomorphism, $ B_{X} $ must then be compact. Hence, as a consequence of Riesz's Lemma, $ \dim(X) < \infty $.

Of course, Romeo's answer is very slick in the sense that reasoning with the ideal $ \mathcal{K}(X) $ saves us a lot of work.

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