I'm currently studying for a functional analysis exam and I noticed that often when it is asked to prove that an operator is compact I run into issues:

Let H be a real Hilbert space. Use the theorem of Banach-Alaoglu to show the following operator is compact:

$ Tx:=\sum_{k=1}^{\infty}<x,x_n>y_n$ where $ x_n$ and $ y_n$ are sequences in H with $\sum_{k=1}^{\infty}||y_n||*||x_n||<\infty$.

I know that for a bounded sequence $(x_n)_{n \in \mathbb{N}}$ I need to show that that $(Tx_n)_{n \in \mathbb{N}}$ has a strongly convergent subsequence. B-A provides me a weakly convergent subsequence of $(x_n)_{n \in \mathbb{N}}$ and a weakly convergent subsequence of $(Tx_n)_{n \in \mathbb{N}}$ because via Cauchy-Schwarz it follows that T is bounded as well. How do I continue from here? While I'm foremost interested in knowing how to solve this via B-A I'd be interested in hearing other approaches that might come in handy for this type of exercise as well.


1 Answer 1


A different approach: Let $T_N(x)= \sum\limits_{n=1}^{N} \langle x, x_n \rangle y_n$. Then,$ T_n$ has finite rank and all finite rank operators are compact. Also, $\|T-T_N\|\leq \sum\limits_{n=N+1}^{\infty} ||x_n\|\|y_n\|\to 0$ as $ N \to\infty$. This implies that $T$ is compact.

I am using two basic facts about compact operators:

  1. Any finite rank operator is compact.

  2. The limit in operator norm of a sequence of compact operators is compact.

  • 2
    $\begingroup$ It should perhaps be mentioned that the set of compact operators is closed in the norm topology. $\endgroup$ Commented Feb 26, 2023 at 5:00

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