Weighted shift operator is Hilbert-Schmidt

Question

If $W : \ell^2 \to \ell^2$ is the weighted shift operator defined by $$W(x_1,x_2,x_3,\ldots)=(0,x_1,\frac 12x_2,\frac 13x_3,\ldots),$$ how can I show that $W$ is Hilbert-Schmidt?

If I have $\{e_n\}_{n=1}^\infty$ as the orthonormal basis of $\ell^2$, then I need to show that $\sum_{j=1}^\infty \|We_j\|^2 < \infty$.

So far, I have: $$\sum_{j=1}^\infty \|We_n\|^2=\sum_{j=1}^\infty \left( \frac 1j e_{jn} \right)^2.$$ If this is correct so far, what can I do next?

Answer 1

First note that $$W(e_n)=W(\underbrace{0,\ldots,0}_{n-1 \text{ terms}},1,0,\ldots)=(\underbrace{0,\ldots,0}_{n \text{ terms}},\frac 1n,0,\ldots)$$ so that $$\sum_{n=1}^\infty \|We_n\|^2=\sum_{n=1}^\infty \frac 1{n^2} < \infty.$$

Answer 2

You are done, just need to clarify your notation.

You have $\|W\|_{HS}^2=\sum_{n=1}^{\infty}\|We_n\|^2= \sum_{j=1}^{\infty}\frac{1}{j^2} < \infty$.