TITLE: Show $T\colon H\rightarrow K$ is defined by $Tx=\sum_{n=1}^{\infty} \langle x,u_n\rangle v_n$ is compact
QUESTION [6 upvotes]: $H$ and $K$ are Hilbert Spaces, $(u_n)$ and $(v_n)$ are sequences in $H$ and $K$ respectively. $\sum_{n=1}^{n=\infty} \|u_n\|\|v_n\| $ converges.
$T\colon H\rightarrow K$ is defined by $Tx=\sum_{n=1}^{\infty} \langle x,u_n\rangle v_n$.
I need to show that $T$ is compact, and I am frankly clueless.
All I can think to say is that the first sum converging means each series is bounded, but I don't know if that is even relevant.
And hints/help would be appreciated.
Thanks

REPLY [2 votes]: For $u\in H$, $v\in K$ denote by $u\bigcirc v$ the rank one operator defined by
$$
u \bigcirc v:H\to K:x\mapsto \langle x, u\rangle v
$$
Obviously $\Vert u \bigcirc v\Vert\leq \Vert u\Vert\Vert v\Vert$. Then for each $N\in\mathbb{N}$ consider operator
$$
T_N=\sum\limits_{n=1}^N u_n\bigcirc v_n
$$
It is of finite rank as the finite sum of rank one operators. Since the series $\sum\limits_{n=1}^\infty\Vert u_n\Vert\Vert v_n\Vert$ converges then
$$
\lim\limits_{N\to\infty}\sum\limits_{n=N+1}^\infty\Vert u_n\Vert\Vert v_n\Vert=0\tag{1}
$$
On the other hand
$$
\Vert T-T_N\Vert=
\left\Vert \sum\limits_{n=N+1}^\infty u_n\bigcirc v_n\right\Vert\leq 
\sum\limits_{n=N+1}^\infty\Vert u_n\bigcirc v_n\Vert\leq 
\sum\limits_{n=N+1}^\infty\Vert u_n\Vert \Vert v_n\Vert\tag{2}
$$
From $(1)$ and $(2)$ it follows that $\lim\limits_{n\to\infty}\Vert T -T_N\Vert=0$, i.e. $T$ is a limit of finite rank operators in the topology of $\mathcal{B}(H,K)$. Hence $T$ is compact.