Showing that if $\lim_{i\to\infty}\lambda_i=0$ and $Ae_i=\lambda_ie_i$ for $A\in\mathcal{B}(H)$, then $A$ is a compact operator

Question

Let $\{e_i\}$ be a Hilbert basis for a Hilbert space $H$ over $\mathbb{C}$ and $\{\lambda_i\}$ be a sequence of complex numbers tending to zero. Suppose that $A\in\mathcal{B}(H)$ such that $Ae_i=\lambda_ie_i$. I am trying to show that $A$ is a compact operator. I know that alternatively I could try to show that if $A$ is not a compact operator, then the sequence $\{\lambda_i\}$ doesn't converge to zero, but that seems a bit too hard since there are quite a few cases for $A$ and quite a possibilities how a sequence doesn't converge to a limit.

In any case, we'd like to show that any bounded sequence is mapped to a sequence with a convergent subsequence. And this already is a problem for me, since the only results I now that guarantee the existence of a converging subsequence are those of the like of Heine-Borel theorem. But, to be constructive, I'm going to write everything that I think that we know about our situation.

Let $\{v_i\}$ is a bounded sequence in $H$. Then $v_i = \sum_{j}c_{ij}e_j$ for complex scalars $c_{ij}$ and $||v_i|| \leq M$ for all $i$ for some finite number $M$. By continuity and linearity of $A$,

$$A(v_i) = \sum_{j}c_{ij}A(e_j) = \sum_{j}c_{ij}\lambda_je_j$$

For all $j$ large enough, $|\lambda_j| < \epsilon$ for a given $\epsilon > 0$ and $\{\lambda_j\}$s are bounded by some $C\geq 0$. Therefore,

$$||A(v_i)||^2 = \sum_{j}|c_{ij}\lambda_j|^2 < C^2\sum_{j=1}^{J}|c_{ij}| + \epsilon\sum_{j=J+1}^\infty |c_{ij}|^2$$

for a suitable large index $J$.

We also know that

$$\sum_{j}|c_{ij}|^2 < \infty$$

for all $j$, meaning that tails of the sum $\sum_{j}|c_{ij}|^2$ converge to zero.

But how do we go about showing that $\{A(v_i)\}$s converge along some subsequence?

Answer 1

Hint: In the general Banach space context: Compact operators are a (operator norm) closed subspace of the bounded linear operators.

Hint 2:

Show that $A$ is the limit of operators with finite dimensional range (which are compact, why?).

Hint 3:

Show that $Ax = \sum_{i \in \mathbb{N}} \lambda_i \langle x,e_i \rangle e_i$ for any $x \in H$. $\sum_{i=1}^N \lambda_i \langle -,e_i \rangle e_i$ is a operator with finite dimensional range for any $N \in \mathbb{N}$. Show that $(\sum_{i=1}^N \lambda_i \langle -,e_i \rangle e_i)_{N \in \mathbb{N}}$ converges to $A$ in operator norm.

If you are not aware that compact operators are a (operator norm) closed subspace of the bounded linear operators, i sketch the proof below.

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Sketch of proof: Let $X,Y$ be Banach spaces and $(T_n)_{n \in \mathbb{N}}$ a sequence of compact operators with $T_n \to T$. Let $x:\mathbb{N} \to X$ be a bounded sequence. Then $x$ has a subsequence $x_1$ so that $T_1 x_1$ converges, because $T_1$ is compact. Now $x_1$ has a subsequence $x_2$ so that $T_2x_2$ converges, because $T_2$ is compact.

By following the above procedure at each stage we receive a sequence of sequences $(x_n)_{n \in \mathbb{N}}$.

Define a sequence $y : \mathbb{N} \to X$ by $y(n) = x_n(n)$ (the diagonal sequence). Then $y$ is a subsequence of $x$ and $T_ny$ converges for any $n \in \mathbb{N}$, because $y$ is a subsequence of $x_n$ after the $n$-th index.

The operator norm convergence of $(T_n)_{n \in \mathbb{N}}$ can be used together with the above to show that $Ty$ converges, which concludes the proof.

Hint:

We want to show that $Ty$ is a Cauchy sequence. For any $m ,n \in \mathbb{N}$ it is true that: $\| T_n y(m) - T y(m) \| \leq \|T_n - T\| \sup_{l \in \mathbb{N}} \| y(l) \|$. Now $\sup_{l \in \mathbb{N}} \| y(l) \|< \infty$ and $T_n \to T$. Together with the fact that for $n$ fixed, $T_n y$ is Cauchy we can conclude using a "smuggling trick".

Smuggling Hint:

For any $n,m,l \in \mathbb{N}$ it is true that $\| T y(m) - T y(l) \| \leq \| (T- T_n) y(m)\| + \| (T-T_n) y(l) \| + \| T_n y(m) - T_n y(l) \| $. Now apply the previous hint.