I realized this morning that I need to clarify what I mean by “invertible” in yesterday’s post. Suppose is a separable Hilbert space and is a compact operator as before. I don’t mean that there exists an operator such that . The existence of such an operator is impossible by the Bounded Inverse Theorem. Instead, I mean that is injective, and therefore has an inverse latex defined on the range of .

The range of will be be a proper subspace of , but it will not be closed unless is of finite-rank. We know that any finite-dimensional subspace of a Banach space is closed. For the reverse direction, we need few lemmas dealing with closed operators. Recall that if is a linear operator defined in a Banach and with range in a Banach space , then is closed if

Note that the definition of closed operator is different from that of continuous in that we impose the additional hypothesis that the sequence converges and don’t *a priori *require that .

**Lemma 1. **Suppose is a closed linear operator between Banach spaces. If the inverse exists in , then is closed.

*Proof. *Suppose is a sequence in such that and . Since is closed, .

**Lemma 2. **Suppose are Banach spaces and that is an injective, closed linear operator. Then is bounded if and only if is closed.

*Proof. *First, suppose that . Let be a sequence in such that . Then is a Cauchy seqence in . For ,

which implies that is a Cauchy sequence in . By completeness, . Since is closed, .

Now suppose that is closed. Then is a Banach space and is closed by Lemma 1. We can apply the Closed Graph Theorem, which is equivalent to the Bounded Inverse Theorem, to obtain that is continuous or equivalently, bounded.

We have enough tools to show that that is necessarily finite-dimensional if is a compact, linear operator such that is a closed subspace of . A compact operator is *a fortiori *closed, and if is closed, then is a Banach space and is a closed linear operator between Banach spaces by Lemma 1. By Lemma 2, , but this contradicts the result of yesterday’s post.

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