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.
Best Answer
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 $I$.