Modified shift operator is compact.

Question

For the operator $$T(\eta_j) = \frac{\eta_{j+1}}{j}$$ on Hilbert Space $H$ where $(\eta_j)$ is a basis. Show it is compact.

Can this work?

Define $$f = (\eta_j)_{j \geq 1}$$

$$T_N(f) = \left(\frac{1}{j} \eta_{j+1}\right)_{j=1..N}$$

$$T(f) = \left(\frac{1}{j} \eta_{j+1}\right)_{j =1..\infty}$$

So $\| T_N (f) - T(f) \|^2 = \left\| \sum_{j = N+1}^{\infty} \frac{1}{j} \eta_{j+1} \right\|^2 \leq \sum_{j = n + 1}^{\infty}\frac{1}{j^2} < \epsilon/4$

since $\sum_{j} \frac{1}{j^2}$ converges.

(alternative proof in here pg5, but there is an assumption on orthonormality).