I can’t remember anymore, but something compelled me to consult Lokenath’s and Mikusinski’s text *Introduction to Hilbert Spaces with Applications *and I stumbled upon a cute property of compact operators, which I had forgotten. In case you forget stuff like I do, remember that an operator on a normed space is compact if, for every bounded sequence , the sequence has a convergent subsequence. The property I had forgotten is that if is a separable, infinite-dimensional Hilbert space and is a compact linear operator, then is necessarily unbounded.

When I saw this result stated, I must admit that it took me a while to figure out the proof because I had also forgotten compact operators map weakly convergent sequences to strongly convergent sequences. So, I would like to take this post towards presenting (in an organized fashion), the a proof of the aforementioned property.

We begin by proving the main auxiliary result, which is the weak-to-strong mapping property of compact operators.

Theorem 1.An operator on a Hilbert space is compact if and only if maps weakly convergent sequences into strongly convergent sequences:

*Proof.* Suppose is compact and . If , then there exists some and a subsequence such that for all . Since weakly convergent sequences are bounded by the Banach-Steinhaus theorem, has a strongly convergent subsequence . By hypothesis of weak convergence, for every ,

which implies that and therefore . Since weak limits are unique and strong convergence implies weak convergence, . But since is a subsequence of , this is a contradiction.

Now suppose that implies that . For any bounded sequence , we need to show that the sequence has a convergent subsequence. By hypothesis that is separable, it has a complete orthonormal system . Let be constant such that , for all . By Cauchy-Schwarz,

so there exists a subsequence such that the sequence converges. Suppose we have chosen nested subsequences such that the sequences converges as . Since , there exists a convergent subsequence . By induction, we obtained a countable collection of nested subsequences such that the Fourier coefficients converge as .

Define a new sequence by taking the diagonal of the subsequences. It is immediate that exists for every . By Bessel’s inequality, for any ,

Letting , we obtain

An application of Parseval’s indetity yields that the series is convergent in , and we denote its limit by . I claim that , for every , as . Indeeed, this follows immediately from observing that

as . Since is dense in , it follows that . Hence, . Recalling that is a subsequence of completes the proof.

Cor 2.Let be as above. If is compact any is an orthonormal sequence, then .

*Proof.* We only need to show that an orthonormal sequence . By Bessel’s inequality, , which implies that

An operator defined on a Hilbert space is said to be unbounded, if the range is not a bounded subset. Since is bounded if and only if is continuous at the origin , an operator is unbounded if and only if there exists a sequence but , as .

The differential operator on is a standard example of an unbounded operator; we will prove this result in a later post. More generally, any compact, invertible operator on an infinite-dimensional (separable) Hilbert space has an unbounded inverse . Indeed, necessarily has an orthonormal sequence . For each , there exists a unique such that . Since and is compact, . But , since for all .

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