Let $H$ denote a separable Hilbert space with an orthonormal basis $\{e_k\}_k\in \mathbb N$ and consider a linear, bounded operator $A:H \to H$ such that: $Ae_k=\lambda_k e_k$. Show that $T$ is a compact operator provided that $\lambda_k \to 0$ as $k\to \infty$.

Following the argument which can be found here or in other similar posts, I want to define a sequence of operators $A_n$ which will be of finite Rank (thus compact) and then using the fact that $\lambda_k \to 0$, to deduce that $A_n$ converges to $A$ and hence $A$ is also compact.

My question is: How should I define $A_n$?

Unfortunately I can't think anything that fits here. The fact that I have $A$ mapping $e_k$ and not a general $x \in l^2$ confused me a lot.

Any help or hint is much appreciated. Thanks in advance!


1 Answer 1


Any element $x$ of $H$ has an expansion $x = \sum\limits_{k=1}^{\infty} a_i e_i$. Define $A_n(\sum\limits_{k=1}^{\infty} a_i e_i)= \sum\limits_{k=1}^{n} a_i\lambda_ie_i$. The range of $A_n$ is contained is the span of $\{e_1,e_2,...,e_n\}$. I leave it to you to verify that $\|A-A_n\| \leq \sup \{|\lambda_i|: i>n\} \to 0$.

  • $\begingroup$ $\|(A-A_n)(\sum a_ie_i)\|^{2} = \sum\limits_{k=n+1}^{\infty} |\lambda_i a_ie_i|^{2} \leq sup_{i>n} |\lambda_i|^{2} \|x\|^{2}$ if $x=\sum a_ie_i$. $\endgroup$ Jan 30, 2019 at 12:20
  • $\begingroup$ Yes sorry I didn't read that $e_n$ is orthonormal basis. But then indeed it's true what you have written there. Sorry again for the misunderstanding. (+1) $\endgroup$
    – Shashi
    Jan 30, 2019 at 12:21

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