# Show $\{u_n\}$ orthonormal, A compact implies $\|Au_n\| \to 0$

I’m having a bit trouble with this homework exercise.

Let $\mathcal{H}$ be a Hilbert space and $\{u_n\}_{n=1}^\infty$ an
orthonormal sequence in $\mathcal{H}$. Let $A$ be a compact operator
on $\mathcal{H}$. Show that $\|Au_n \| \to 0$ as $n\to \infty$.

My book defines a compact operator as an operator $A$ such that whenever $f_n$ is bounded, then $Af_n$ has a convergent subsequence (equivalently, the image of $A$ is relatively compact).

It seems I must somehow combine the fact that $Au_n$ has a convergent subsequence with the fact that $\{u_n\}$ is orthonormal. This is where I get stuck. Maybe I can somehow use the fact that $\|u_n-u_m \| = \sqrt{2}\,$ for $m \neq n$.

#### Solutions Collecting From Web of "Show $\{u_n\}$ orthonormal, A compact implies $\|Au_n\| \to 0$"

I suppose it is clear for you that a compact operator is clearly bounded so continuous.
Suppose that the $(Au_n)$ has a convergent subsequence towards $v$ not zero. For the sake of simplicity, let us note also $(Au_n)$ this subsequence. $(u_n)$ is then also an orthonormal sequence in $\mathcal{H}$. Set $v_n=\frac{1}{n}\sum_{k=n}^{2n}u_k$. It is clear that the sequence $(v_n)$ converges towards $0$. But $(Av_n)$ converge towards $v$ which is not zero. QEA.