Compactness of $(x_1,x_2,...)\mapsto(0,x_1,x_2/2...)$

Question

I read that the linear operator in the Hilbert space $\ell_2$ defined by $(x_1,x_2,...,x_n,...)\mapsto(0,x_1,x_2/2,...,x_n/n,...)$ is compact.

I wanted to prove it by proving that the image of the closed unit sphere is totally limited, but I am not sure that is the right path and I am getting nothing...

Moreover, I would like to learn to find its spectrum, but my book does not give examples about how to find it...

$\infty$ thank you for any help!

Answer 1

Hint:

Let $\{x(0,i)\}_{i=1}^\infty = (x_1(0,i),x_2(0,i),\dots)$ be an arbitrary sequence in the unit ball.

For all $n \geq 0$, let $\{x(n+1,i)\}_{i=1}^\infty = (x_1(n+1,i),x_2(n+1,i),\dots)$, be a subsequence of $\{x(n,i)\}_{i=1}^\infty$ such that $\{x_{n+1}(n+1,i)\}_{i=1}^\infty$ converges.

Now, consider the subsequence $\{x(i,i)\}_{i=1}^\infty$ of our original sequence. Note that this sequence converges in each coordinate.

Define $$ T(x_1,x_2,...,x_n,...) = (0,x_1,x_2/2,...,x_n/n,...) $$ and show (with an appropriate $\epsilon$-$\delta$ argument) that $\{Tx(i,i)\}_{i=1}^\infty$ is Cauchy (or that it converges) with respect to the $\ell^2$ norm.