weak star and strong convergence of net in Banach spaces

Question

In Banach spaces, the following result is well-known:

(1) Let $X$ be a Banach space. Let $\{x_n\}\subset X$ and $\{x^*_n\}\subset X^*$ be such that $x_n \rightarrow x$ (convergence with respect to strong topology on $X$) and $x^*_n\overset{\ast}{\rightharpoonup} x^*$ (convergence with respect to weak star topology on $X^*$). Then we have, $\langle x^*_n, x_n\rangle\rightarrow \langle x^*,x\rangle$.

The proof for the above result is based on the fact that if $x^*_n\overset{\ast}{\rightharpoonup} x^*$ then $\{x^*_n\}$ is bounded. We have known that this fact does not hold if we replace the sequence by net (based on the following counterexample: Must a weakly or weak-* convergent net be eventually bounded?)

My questions are:

1) Whether or not the result (1) still holds if the sequence is replaced by the net (see the following for definition: https://en.wikipedia.org/wiki/Net_(mathematics))?

2) In the case (1) is false for the net, could we construct a counterexample? And what more assumptions are added such that (1) is true for the net.

Thank you for all discussions on this topic.

Answer 1

The couterpart of result (1) can fail if the sequence is replaced by the net. Our counterexample is based on Nate Eldredge’s counterexample. Direct a set $I=I’\times\Bbb N$ by the preorder $\preceq’$ defined by

$$(U’,n’) \preceq’ (V’, m’) \mbox{ iff } U’ \preceq V’ \mbox{ and } m’\ge n’.$$

For each $U\in\mathcal U$ pick $x_U\in X$ such that $\|x_U\|=1$ and $\langle f_U, xU\rangle\ne 0$. Define nets indexed by $I’$ putting $x^*_{(U,n,n’)}=f_{U,n}=nf_U$ and $x_{(U,n,n’)}=\frac 1{n’}x_U$ for for each $(U,n,n’)\in I$. Clearly, the net $\{ x_{(U,n,n’)}: (U,n,n’)\in I’\}$ converges to the zero. Since the net $\{f_{U,n}:(U,n)\in I\}$ converges to the zero, the net $\{ x^*_{(U,n,n’)}: (U,n,n’)\in I’\}$ converges to the zero too. On the other hand, for each $(U,n,n’)\in I’$ and each natural $m$ we have $(U,n,n’)\preceq’ (U,m,n’)$ and $\langle x^*_{(U,m,n’)}, x_{(U,m,n’)}\rangle=\langle mf_U, \frac 1{n’}x_U \rangle= \frac {m}{n’} \langle f_U, x_U \rangle$, which has an absolute value bigger than $1$ for a sufficiently big $m$.

The couterpart of result (1) holds when the directed set $(I,\le)$ of the net has countable cofinalty, that is there exists a countable set $D$ of $I$ such that for each $n\in I$ there exists $d\in D$ with $d\ge n$. Indeed, suppose to the contrary that $\langle x^*_n, x_n\rangle\not\rightarrow \langle x^*,x\rangle$. Then there exists $\varepsilon>0$ such that for each $n\in I$ there exists $n’\ge n$ such that $|\langle x^*_n, x_n\rangle - \langle x^*,x\rangle|\ge\varepsilon$. Let $\{d(k):k\in\Bbb N\}$ be any enumeration of the set $D$. Then by indution we can build a sequence $\{n(k):k\in\Bbb N\}$ of elements of $I$ such that for each $k$ we $n(k)\ge d(k)$ and $|\langle x^*_{n(k)}, x_{n(k)}\rangle - \langle x^*,x\rangle|\ge\varepsilon$. But a sequence $\{x_{n(k)}\}$ converges to $x$ and a sequence $\{x^*_{n(k)}\}$ converges to $x^*$, a contradiction with result (1).