Given the following operator: $T:\ell^2\to\ell^2$, which acts in the following way on the standard basis vectors: $$Te_{2k-1}=\frac{1}{k}(e_{2k-1}-ie_{2k})$$ $$Te_{2k}=\frac{1}{k}(ie_{2k-1}+e_{2k})$$ I need to prove that this operator is compact by showing that it is the limit of finite rank operators. My attempt:
Define $(T_nx)_j=(Tx)_j$, for $j\le 2n$, and for $j\gt 2n: (T_nx)_j = 0$. Intuitively this should converge to $T$ In the operator norm and all these operators $(T_n)_{n\ge1}$ are of finite rank thus $T$ is compact!
Let us try and show that $\|T-T_n\|\to0$, for $n\to \infty$. $$\|T-T_n\|=\sup_{\|x\|=1}\|(T-T_n)x\|$$, but from here I am lost I don’t see how this expression goes to $0$. Can anyone show me how it works?
So you found a good possible approximation by finite rank operators. Now just notice that $T-T_n$ is zero on the $2n$ first coordinates. But for the other coordinates ($k>2n$), because of the $1/k$ in the definition of the operator, you always have $$ \sum_{k=1}^\infty\left|((T-T_n)x)_{k}\right|^2 = \sum_{k=2n+1}^\infty\left|((Tx)_{2k}\right|^2 ≤ \sum_{k=2n}^\infty\frac{2\,|x_k|^2}{(k-2)^2} ≤ \frac{4\,\|x\|^2}{(n-2)^2} $$ and similarly for the odd coordinates. Therefore $\|T-T_n\| ≤ \frac{2}{n-2}$