# Show $T$ is compact

$H$ 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

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Try to find finite rank operators $T_n \colon H \to K$ such that $T_n \to T$ in $L(H,K)$. – martini Nov 26 '12 at 14:13
$T_n x = \sum_{k=1}^{n} <x,u_k>v_k$ has rank at most n and converges to T as n goes to infinity, right? – hello123 Nov 26 '12 at 14:22
That would be the obvious choice. You of course have to check it does work. – martini Nov 26 '12 at 14:25
Check that $T_n\rightarrow T$ as n goes to $\infty$ ? – hello123 Nov 26 '12 at 14:32
@hello123 You can answer your own question. – Davide Giraudo Nov 26 '12 at 20:56

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.