Question about convergence of nets in $X^*$

Question

Let $X$ be a Banach space with dual $X^*.$ Let ${a_i}$ and ${b_i}$ be two nets in $X^*$ such that $a_i$ is norm-bounded in $X^*$ and $a_i \to a$ and $a_i + b_i \to 0$, where in both the convergence is understood in terms of weak-star Topology.

My question is that from above can we conclude that the $(b_i)$ is norm-bounded or at least admits a bounded subnet ? What if $X$ be reflexive ?

You can slightly modify my example from https://math.stackexchange.com/a/158933/822. — Nate Eldredge, Jun 14 '19 at 21:30

score 5 · Accepted Answer · answered Jun 14 '19 at 21:30

5

No. Not even if $a_i=0$ for all $i$.

If $b_i$ converges weak* to $0$ it does not follow that $b_i$ is bounded. It does not follow that $b_i$ has a bounded subnet. The thing to observe: every weak* neighborhood of $0$ is an unbounded set. Using that we can build a net $b_i$ that converges weak* to $0$ but $\|b_i\|$ converges to $\infty$.

Try it yourself!

answered Jun 14 '19 at 21:30

GEdgar

117,296

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HOWEVER. A sequence $b_i$ that converges weak* is, indeed, bounded. – GEdgar Jun 14 '19 at 21:34
In the sequence case if put $b_n = n e_n \in \ell^2 $ I think is counter example again! Am I missing something ? – Red shoes Jun 18 '19 at 17:53
Yes, $n e_n$ does not converge weak*. Indeed, $(1,2^{-3/4},3^{-3/4},\cdots) \in l^2$, right? – GEdgar Jun 18 '19 at 18:05
Indeed. Thanks, is there any well-known result that confirms that? – Red shoes Jun 18 '19 at 18:12
Uniform Boundedness Principle + "a convergent sequence of scalars is bounded". – GEdgar Jun 18 '19 at 18:16

Question about convergence of nets in $X^*$

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