Compact operators on Hilbert Space I m working on the following problem:

Let $K:H\rightarrow H$ be a compact operator on a Hilbert space. Show that if there exists a sequence $(u_n)_n\in H$ such that $K(u_n)$ is orthonormal, then $|u_n|\rightarrow \infty$. 

Here is my argument: It is suffice to show for all $u\in H$, there exists $M&gt;0$ such that $|u_n-u|\geq M$, that is $(u_n)$ has no limit in $H$. Suppose $u_n\rightarrow u$ (all the convergence are strong in my argument), then $Ku_n\rightarrow Ku$. I claim that we can extract a subsequence $(Ku_{n_k})_k$ that is divergent. Denote $E_k=\overline{span\{Ku_{n_k}\}}$,  and we may choose $Ku_{n_k}$ such that $||Ku_{n_k}||=1$ (trivial) and $dist(Ku_{n_{k+1}}, E_k)\geq 1/2$. So $|Ku_{n_k}-Ku_{n_{k+1}}|\geq 1/2$ for all $k$, which violates the convergence of the sequence $Ku_n$. 

Since I did not use the fact that $K$ is a compact operator, there must be something wrong with the proof. Can someone let me know which part is problematic? Any hints for the right approach?
I'd proceed as follows:


An orthonormal set has no limit point.  
Therefore, any set containing an infinite orthonormal set is not compact.
If $|u_n|$ is bounded, then $K$ maps a bounded set to a set whose closure is not compact.


Your proof got confused in the first step
Remember that a compact operator is continuous, You use continuity when said 
$Ku_{n}\rightarrow Ku$. Thus the proof is fine.