# Sequence of compact operators convergence - proof verification

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).

• 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 $u$). Jul 28, 2019 at 21:00
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 $$I$$.