# The vanishing property of a compact operator acting on an orthonormal basis of a Hilbert space

The question is:

Suppose $$H$$ is a Hilbert space, $$A \in K(H)$$, and $$\{e_n\}$$ is an orthonormal basis of $$H$$. Prove $$\lim\limits_{n \to \infty} \langle Ae_n, e_n \rangle = 0$$.

$$\langle \cdot, \cdot \rangle$$ is the inner product defined on $$H$$.

$$A \in K(H)$$ means $$A$$ is a compact operator (thus linear and continuous) mapping $$H$$ to itself.

There are numerous ways to define a compact operator. Suppose $$X, Y$$ are Banach spaces and $$T:X\to Y$$ is a linear operator. The followings are equivalent:

1. $$T$$ is compact
2. the closure of the image of any bounded set (or just consider the unit ball in $$X$$) is compact (i.e. let $$E$$ be bounded in $$X$$, then $$\overline{TE}$$ is compact in $$Y$$)
3. the image of any bounded set (or just consider the unit ball in $$X$$) is totally bounded (i.e. $$TE$$ is totally bounded in $$Y$$ if $$E$$ is bounded in $$X$$)
4. the image of any bounded sequence in $$X$$ has a convergent subsequence in $$Y$$ (i.e. if $$\{x_n\}$$ is bounded in $$X$$, then $$\{Tx_n\}$$ has a convergent subsequence

I totally have no idea how to prove this amazing effect.

## 1 Answer

Take $$\varepsilon>0$$. As $$A$$ is compact, there is a finite rank operator $$T$$ such that $$\|A-T\|\leq\varepsilon$$. Then, $$|\langle Ae_n,e_n\rangle|\leq|\langle Te_n,e_n\rangle|+\varepsilon$$

But as $$T$$ has finite rank, you know that $$\mathrm{Im} T\subset \mathrm{Vect}(e_1,\cdots,e_N)$$ for some $$N\geq 0$$. This will imply that $$\langle Te_n,e_n\rangle=0$$ as soon as $$n> N$$.

This shows what we wanted.

• Thanks! This proof uses the fact that finite operators are dense in the space of compact operators. But our teacher didn't cover this in class. Commented May 14, 2019 at 13:23
• I think this is a good exercise to prove this by your own, the idea is to use the precompacity criterion in Banach spaces (a subset is precompact iff it can be approached by a finite dimensional vector space as precisely as you want). This criterion itself is important and it is an important exercise to show it Commented May 14, 2019 at 13:26
• I'll try it. Thanks! Commented May 14, 2019 at 13:28
• The hypothesis is $\|A-T\|\leq\varepsilon$. So, $|\langle Ae_n,e_n\rangle|\leq |\langle (A-T)e_n,e_n\rangle|+|\langle Te_n,e_n\rangle|\leq \|A-T\|\|e_n\|\|e_n\|+|\langle Te_n,e_n\rangle|\leq\varepsilon+|\langle Te_n,e_n\rangle|$, where I use the triangular inequality, Cauchy Schwarz and the definition of the subordinate norm Commented May 14, 2019 at 15:08