# weak star and strong convergence of net in Banach spaces

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.

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).