# An equivalent definition of a compact operator

How can I show, that the following definitions of a compact operator are equivalent?

Definition 1:

An operator $A$ on a Hilbert space $H$ is called compact if for every bounded sequence $(x_n)$ in $H$, the sequence $(Ax_n)$ contains a convergent subsequence.

Definition 2:

An operator $A$ on a Hilbert space $H$ is called compact if $x_n \rightharpoonup x$ and $y_n \rightharpoonup y$ implies $(Ax_n, y_n) \to (Ax, y)$.

I managed to prove, that Definition 1. is equivalent to ($x_n \rightharpoonup x$ implies $Ax_n \to Ax$). Can I apply this result somehow?

• I suppose by $x_n \rightharpoonup x$ you mean that $x_n$ converges weakly to $x$? – tomasz Jan 18 '17 at 20:26

Suppose $x_n \rightharpoonup x$ and $y_n \rightharpoonup y$, and let $\varepsilon>0$ be given. For any $n\in\mathbb{N}$ we have \begin{align} |\langle Ax_n,y_n\rangle-\langle Ax,y\rangle| &\leq |\langle Ax_n,y_n\rangle-\langle Ax,y_n\rangle|+|\langle Ax,y_n\rangle-\langle Ax,y\rangle| \\ &\leq\|Ax_n-Ax\|\|y_n\|+|\langle Ax,y_n\rangle-\langle Ax,y\rangle| \end{align} Since $y_n \rightharpoonup y$, the sequence $\{y_n\}$ is norm-bounded, so there is some $M>0$ with $\|y_n\|<M$ for all $n$. If $A$ satisfies definition 1., then there is some $N_1\in\mathbb N$ such that $\|Ax_n-Ax\|<\varepsilon/2M$ for $n\geq N_1$. Since $y_n \rightharpoonup y$, there is some $N_2\in\mathbb N$ such that $|\langle Ax,y_n\rangle-\langle Ax,y\rangle|<\varepsilon/2$ for all $n\geq N_2$. Then for $n\geq\max\{N_1,N_2\}$ we have \begin{align} |\langle Ax_n,y_n\rangle-\langle Ax,y\rangle| &\leq\|Ax_n-Ax\|\|y_n\|+|\langle Ax,y_n\rangle-\langle Ax,y\rangle| \\ &<M\varepsilon/2M+\varepsilon/2=\varepsilon, \end{align} and thus $(Ax_n, y_n) \to (Ax, y)$.
• I think Banach-Alaoglu is quite enough. Any bounded sequence has a weakly convergent sequence (by B-A and reflexivity), which is mapped to a convergent sequence by $A$ (just take $y_n$ constant). – tomasz Jan 18 '17 at 21:49