Let $ K \in B(H)$ be a compact operator on a Hilbert space. We call a map $x: I \to H$ a net, if $I$ is a (not necessary countable) directed set. One can view nets as a generalization of sequences.
Let $(v_i)_{i \in I}$ be a bounded net in $H$ that converges weakly to $0$.
Prove that the net $(Kv_i)_{i \in I}$ converges to zero in norm.
Conversely, if $K \in B(H)$ has the property that $\lim_{i \in I} || Kv_i || = 0 $, whenever $(v_i)_{i \in I}$ is a bounded net converging weakly to $0$, then $K$ is compact.
I know that for every functional $\omega \in H^* $ the sequence given by $\omega(v_i)$ must converge to $0$.
Moreover I know that a set X is compact if and only if any net in X has a convergent subnet.
But I have no idea how to use this in some way to show either that the image net converges or to proof the other direction.
Best Answer
I dont see the reason to use nets since you are in a normed space and convergence is described nicely by sequences but since you wanna mess around with nets, i will write the first direction by using nets and the other one with sequences.
The direction "$\implies$" doesnt require the assumption of a Hilbert space, it works also for Banach spaces. The argument is the following:
Suppose that $K\in \mathcal{B}(H)$ is a compact operator and that there exists a bounded net, say $(u_i)_{i\in I}$ that converges weakly to $0$ but $\lim_{i}Ku_i \nrightarrow 0$. Then, there is an $\epsilon>0$ and a subnet $(u_{i_\mu})_{\mu\in M}$ of $(u_i)_{i\in I}$ such that $$\tag{1}||Ku_{i_\mu}||\geq \epsilon\ \ \text{for every}\ \mu \in M$$ Now, $(u_i)_{i\in I}$ is bounded, which means that there is a constant $C>0$ such that $||u_i||\leq C$ for every $i\in I$. This means that the net $(u_i)_{i\in I}$ belongs the open ball $B(0,2C)$ centered at $0$ of radius $2C$. Hence, the net $(Ku_{i_\mu})_{\mu \in M}$ belongs to $K\bigl(B(0,2C)\bigr)$. Now, by the compactness of $K$ we know that $\overline{K\bigl(B(0,2C)\bigr)}$ is compact. Hence, the net $(Ku_{i_\mu})_{\mu \in M}$ has a strongly convergent subnet, say $(Ku_{i_{\mu_\sigma}})_{\sigma \in \Sigma}$ to a point $y\in H$. Now, $(u_{i_\mu})_{\mu \in M}$ converges weakly to $0$, which means that $\langle u_{i_{\mu_\sigma}}, h \rangle \to 0\ $ for every $h\in H$. Using the adjoint operator $K^*$ of $K$ we see that $$\lim_{\sigma \in \Sigma}|\langle Ku_{i_{\mu_\sigma}}, h \rangle|=\lim_{\sigma \in \Sigma}|\langle u_{i_{\mu_\sigma}}, K^*h\rangle|=0$$ Which means that $(Ku_{i_{\mu_\sigma}})_{\sigma \in \Sigma}$ converges weakly to $0$. But we already know that $(Ku_{i_{\mu_\sigma}})_{\sigma \in \Sigma}$ converges strongly to $y$. Hence, $y=0$. But this means that $(Ku_{i_{\mu_\sigma}})_{\sigma \in \Sigma}$ can be arbitrary close to $0$ which contradicts $(1)$.
For the other direction "$\impliedby$" suppose that $K$ has the property that $\lim_{i\in I}||Ku_i||=0$ for every bounded net $(u_i)_{i\in I}$ which is weakly convergent to $0$. To show that $K$ is compact we only need to show that $K(B_H)$ is totally bounded (check this) where $B_H$ is the closed unit ball $\{h\in H:\, ||h||\leq 1\}$ of $H$. Supoose that $K(B_H)$ is not totally bounded. Then there exists an $\epsilon>0$ such that $K(B_H)$ cannot be covered by finitely many balls with centers from $K(B_H)$ and radius $\epsilon>0$. This means that for some $x_1 \in B_H$ the open ball $B(Kx_1,\epsilon)$ does not cover $K(B_H)$. Hence, there is some $x_2\in B_H$ such that $||Kx_1-Kx_2||\geq \epsilon$. Now, the open balls $B(Kx_1,\epsilon),\, B(Kx_2,\epsilon)$ cannot cover $K(B_H)$, hence there is some $x_3\in B_H$ such that $||Kx_1-Kx_3||,\, ||Kx_2-Kx_3||\geq \epsilon$. Continuing this way, recursively, we construct a sequence $x_n \in B_H$ such that $$\tag{2} ||Kx_m-Kx_n||\geq \epsilon\ \text{ for every }\ n\neq m$$ Now, $x_n$ belongs to the closed unit ball $B_H$. Since, $H$ is a Hilbert space, it is also reflexive, this means that $B_H$ is weakly compact. Hence, $x_n$ has a weakly convergent subsequence, say $x_{k_n}$ converges weakly to $x$. Then, $z_{n}=x_{k_n}-x$ converges weakly to $0$. From our hypothesis, it follows that $\lim_{n\to \infty}||Kz_n||=0$. But, for every $n\neq m$ $$||Kz_n- Kz_m||=||Kx_{k_n}-Kx_{k_m}||$$ which contradicts $(2)$. Hence, $K(B_H)$ is totally bounded which implies that $\overline{K(B_H)}$ is compact.