let $e(j)$ denote the canonical Hilbert base of $l^2(\mathbb{C})$, then define a linear operator by $T(e(j))=\frac{1}{1+j} e(2j)$. I know that this operator is compact. My question occurs within the proof of it:
Define the linear operator of finite rank $T_k(e(j))=T(e(j))$ if $j<k$ and $0$ is $j\ge k$. With this operator one can show that the operator norm $$\|T-T_k\|=\frac{1}{k+1}$$ and hence is zero for $k\to \infty$. This shows the compactness of $T$.
My question: It is already obvious that $\lim_{k\to\infty}$ $(T_k-T)=0$ and hence the norm $$\|\lim_{k\to\infty}T-T_k\|=\lim_{k\to\infty}\|T-T_k\|=0.$$ Is this a coincidence that I can interchange limit and norm here?
I assume not, because I can make the $T_k$-construction for any linear operator and hence any operator is compact, which is not the case. But why does it seem to work here?
Thanks!
No, it's not a coincidence. The definition of "$\lim (T-T_k)=0$" is "$\lim||T-T_k||=0$".
No, you can't do the same for any operator. You can't show that $||T-T_k||\to0$. (Let $Tx=x$. What is $||T-T_k||$?)