I came across the following question:

Given an orthonormal basis $\{e_n\}$ in a Hilbert space $H$ and a sequence $\{\lambda_n\}$ such that $\lambda_n\rightarrow 0$, we define the operator $$T:H \to H$$ as follows: $$Tu=\sum^\infty_{n=1} \lambda_n \langle u,e_n \rangle e_n$$

Show that $T$ is compact.

My method is to show that $T$ is the limit of a sequence of compact operators. Define the following sequence of finite rank operators:

$$T_Nu=\sum^N_{n=1} \lambda_n \langle u,e_n \rangle e_n$$

We thus get:

$$(T_N-T)u=\sum^\infty_{n=N} \lambda_n \langle u,e_n \rangle e_n$$ $$||(T_N-T)u||^2=\sum^\infty_{n=N} |\lambda_n|^2 |\langle u,e_n \rangle|^2$$

Since $\lambda_n\rightarrow 0$, for a large enough value of $N$, $|\lambda_n|\leq 1, \forall n\geq N$ and so $$||(T_N-T)u||^2\leq\sum^\infty_{n=N} |\langle u,e_n \rangle|^2$$

The sum above is the tail of a convergent series and thus approaches zero, and so by the comparison test for positive series, $||T_N-T||\rightarrow 0, \forall u\in H$, which means that $T_N\rightarrow T$.

Could you tell me if my proof is correct? I have already seen a different proof, but I wanted to make sure that I proved the convergence of the operator sequence correctly (mainly - wanted to make sure the my argument using the tail of a convergent series is correct).

Thanks in advance.

  • $\begingroup$ What is exactly the 'comparison test' and how is it applied? It seems your proof could be applied even for only bounded sequences as well. We need $T_n\to T$ in norm, which is not necessarily implied by ($T_nu\Tu$ for all $u$). $\endgroup$
    – Berci
    Jul 28, 2019 at 21:00
  • $\begingroup$ The comparison test for positive series - if the terms of a series are smaller than the terms of another series (from a certain point on) and the larger series converges, then the smaller series converges. Could you explain why my proof doesn't show that the sequence converges in norm? If it's true for all vectors, shouldn't the supremum of the norm of Tv where v is a unit vector also approach zero? $\endgroup$
    – GSofer
    Jul 28, 2019 at 21:09

1 Answer 1


Your proof is wrong. $\|T_n-T\|$ is the supremum of $\|(T_n-T)u\|$ over all $u$ in the unit ball. You have only shown that $(T_n-T)u \to 0$ for each fixed $u$. Compactness is not preserved under this type of convergence (which is called strong convergence). For a correct proof use the fact that $|\lambda_n| <\epsilon$ for $n$ sufficiently large and $ \sum\limits_{k=N}^{\infty} |\langle u, e_n \rangle|^{2} \leq \|u\|^{2}$ which gives $\|T_n-T\| \leq \epsilon$ for $n$ sufficiently large.

To see why your argument fails consider the case where $\lambda_n=1$ for all $n$. In this case $T=I$ which is not comapct. But if your argument works then $T_n \to I$ in norm which should give compactness of $I$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.