If $T$ is compact, then $\|T e_n \| \to 0$ for any othonormal sequence $e_n$.

Question

The following question is a part of question 2.4.6 of Conway's functional analysis:

Let $\mathcal{H}$ and $\mathcal{K}$ be Hilbert spaces, and let $T: \mathcal{H} \rightarrow \mathcal{K}$ be a bounded linear operator. Show that if $T$ is a compact operator, then $$ \lim _{n \rightarrow \infty}\left\|T\left(e_{n}\right)\right\| =0. $$

This question been extensively answered on the stack exchange (for example here) : A Question on Compact Operators.

One thing that is unappealing about the answers that I have seen is that they use what seem like "big-gun" results about weakly convergent sequences and compact operators, which a reader of Conway's book up to this point would not have seen yet! I've devised an alternate proof that uses only basic principles below. Is it correct? If so, could any light be shed as to how it relates to the weak-convergence argument?

Proof: $T$ compact implies that $\sup _{\|x\| = 1}\left\|T{x}\right\| \leq M<\infty$ for some $M$, and that $Te_n$ has a convergent subsequence $ T e_{n_k} $. Suppose $\left\|T e_{n_k}-h\right\| \rightarrow 0$, but $h \neq 0$. By restricting to a further subsequence if necessary, we may assume $\left\|T_{ {e_{n_k} }}-h\right\| \leq 2^{-k}$. Define $$ x_{m}=\sum_{j=1}^{m} \frac{c}{j} e_{n_{j}} \text {, where } \frac{1}{c}=\left(\sum_{j=1}^{\infty} \frac{1}{j^{2}}\right)^{1/2} \text {. } $$ clearly $\left\|x_{m}\right\| \leq 1$ For all $m$. But $$ \begin{aligned} M & \geqslant \| T_{x_m}\|=\| \sum_{j=1}^{m} \frac{c}{j} (T_{e n j}-h+h) \| \\ &=\left\|\sum_{j=1}^{m} \frac{c}{j}\left(T_{e_{n_j}}-h\right)+h \sum_{j=1}^{m} \frac{c}{j}\right\| =:\left\|A_{m}+B_{m}\right\| \end{aligned} $$

since $\|h\| \neq 0, \quad\left\|B_{m}\right\| \rightarrow \infty$ as $m \rightarrow \infty$. By the triangle inequality $\sup_m\left\|A_{m}\right\| \leqslant \sum_{j=1}^{\infty} \frac{c}{j} 2^{-j}<\infty$. This implies $\left\|A_{m}+B_{m}\right\| \rightarrow \infty$ as $m \rightarrow \infty$, a contradiction. Hence $h=0$, and every subsequence of $Te_n$ must have a further subsequence converging to 0, giving $$ \| Te_n \| \rightarrow 0 \quad \text { as } n \rightarrow \infty \text {. } $$

Answer 1

I didn't check the details of your answer so I cannot say if it is correct or not, but it seems overly complicated to me. The approach that feels natural to me (and doesn't use any big guns) is the following:

Assume that $Te_n\not\to0$ in norm. Thus there exists a subsequence $(e_{n_k})$ of $(e_n)$ and some $\varepsilon>0$ such that $\|Te_{n_k}\|\ge\varepsilon$ for all $k$. Since $(e_{n_k})$ is bounded and $T$ is compact, find a convergent subsequence of $(Te_{n_k})$. In order to avoid confusion, we can assume without loss of generality that $(Te_{n_k})$ converges itself to a limit, say $x\in H$, so $\|x\|\ge\varepsilon$.

But here is the problem: since for any $w\in H$ we have $\|w\|^2=\sum_{n=1}^\infty|\langle w,e_n\rangle|^2$, we have $\langle w,e_n\rangle\to0$. Now if $z\in H$, we have $$\langle x,z\rangle=\lim_{k\to\infty}\langle Te_{n_k},z\rangle=\lim_{n\to\infty}\langle e_{n_k},T^*z\rangle=0$$ and thus for $z=x$ we get $\|x\|=0$, a contradiction.

Don't be intimidated by the words weak convergence, all we have used is Parseval's identity and the fact that bounded operators on Hilbert spaces have adjoints! these are elementary facts in my opinion.

Answer 2

Hint: By Bessel's theorem, for each $x\in X$, $\langle e_n,x\rangle\xrightarrow{n\rightarrow\infty}0$ . Every subsequence of $\{Te_n:n\in\mathbb{N}\}$ has a subsequence $Te_{n'}$ that converges. Say, $Te_{n'}$ converges to $y$. Then $\langle Te_{n'},y\rangle=\langle e_{n'},T^*y\rangle\xrightarrow{n'\rightarrow\infty}0$ This means that $\langle y,y\rangle=0$ and so $y=\mathbf{0}$. This implies that $Te_n\xrightarrow{n\rightarrow\infty}\mathbf{0}$.