# Proving a linear operator is compact: understanding the statement “norm limit of a sequence of finite rank operators”.

I am having serious trouble understanding the proof that an operator is compact.

This is the original question I asked and the proof is included very helpfully in the answer.

Show if $\lim_{n \to \infty} \lambda_n=0$ then $Tu=\sum^\infty_{n=1} \lambda_n \langle u,e_n \rangle e_n$ defines a compact operator.

When showing the operator $T$ is compact the main criteria mentioned is "show $T$ is the norm limit of a sequence $T_n$ of finite rank operators".

What does this statement mean?

Do we have a finite rank operator?

Apologies if I am not asking the right question because I am not sure where to begin.

In the proof, why do we have to show $$||T-T_k || \to 0$$

Is it because this shows that $T$ is the limit of a sequence of finie rank operators? So $T_k$ are the finite rank operators and $T$ is the limit?

• Finite rank operators are compact and the subspace of compact operators is closed with respect to the operator norm. – Jochen May 3 '16 at 10:06
• In my case what is the operator norm? – Al jabra May 3 '16 at 10:07
• $\|T\|_{\text{op}}=\sup\lbrace \|T(x)\|_Y: x\in X,\, \|x\|_X\le 1\rbrace$ where $T:X\to Y$. – Jochen May 3 '16 at 10:09
• I thought this whole business of finite rank operators was to do with normed spaces or Banach spaces? – Al jabra May 3 '16 at 10:10
• This is the definition. Of course "finite rank" is an algebraic condition but operator means continuous linear map between Banach spaces. – Jochen May 3 '16 at 10:12