In fact, we don’t need to restrict ourselves to injective Hilbert-space operators to show that if is a compact linear operator between Banach spaces, then is closed if and only if is finite-dimensional. We know that any finite-dimensional normed space is closed. To prove the reverse direction, we need a few more tools. We begin by proving an elementary yet extremely useful lemma, often called Riesz’s lemma.

Lemma 1. (F. Riesz)Let be a proper, closed subspace of a Banach space . Then for any , there exists an element such that and .

*Proof.* Let , and set . By definition of infimum, for any , there exists such that . Set , so that . Then

Varying our choice of depending on the given completes the proof.

The following lemma is a corollary of Riesz’s lemma which gives a useful criterion for a Banach space to be finite-dimensional.

Lemma 2.Let be a Banach space. If every bounded sequence has a convergent subsequence , then is finite-dimensional.

*Proof. Suppose is infinite-dimensional. *By Riesz’s lemma and induction, we can construct a sequence in such that and

By hypothesis, has a convergent subsequence . But . Hence, is not Cauchy, which is a contradiction.

For Banach spaces , denote their topological duals by and , respectively.

Theorem 3.A bounded linear operator is compact if and only if is compact.

*Proof.* Suppose that is compact, and let be a bounded sequence in . Set . Denote the closed unit ball in by . Since is a compact operator, is a compact subset of and in particular, a Banach space. Define a sequence of functions by

I claim that is an equicontinuous, uniformly bounded sequence of functions. Indeed,

and for all ,

By the Arzela-Ascoli theorem, there exists a subsequence that converges uniformly on . I claim that the sequence is Cauchy in . Indeed,

Since is complete, we see that is a convergent subsequence.

Now suppose that is a compact operator, and let be a bounded sequence in . By our first result, is compct. We know the canonical embeddings of into and into are closed subspaces and therefore Banach spaces. Let be a bounded sequence in . Since is a compact operator, has a convergent subsequence . Then for any continuos functional and ,

Hence, is the embedding of into . Thus, the limit of , which implies that exists in .

Lemma 4.Suppose is a surjective linear map between Banach spaces. Then is bounded from below.

*Proof. *The Open Mapping Theorem for Banach-space operators tells us that there exists such that

By duality, for ,

We now have the preliminary tools and are ready to prove the main result.

If is a compact operator with closed range , then is finite-dimensional.

*Proof.* First, suppose that is bounded from below: there exists a positive constant such that for all . By Lemma 2, it suffices to show that every bounded sequence has a convergent subsequence . Since is bounded from below,

which implies that is a bounded sequence in . Since is a compact operator, has a convergent subsequence .

Now consder the general case where is not necessarily bounded from below. Since is a closed subspace of , it is a Banach space. Define a compact linear operator by . By Theorem 1, the dual operator is a compact operator. Since is surjective, is bounded from below by some . Applying our first result shows that is finite-dimensional. Since is an isomorphism (boundedness from below implies injectivity), we see that is finite-dimensional. Since a finite-dimensional space is isomorphic to its dual and , we conclude that is finite-dimensional.