# 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.

## 1 Answer

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).