Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider an uncountable set $I$ and let $A=\mbox{Fin}(I)$ be the family of finite subsets of $I$ ordered by inclusion. Let $E$ be a normed space and $F$ be a Banach space. Suppose moreover we have a net $(T_\alpha)_{\alpha\in A}$ of bounded operators between $E$ and $F$. I want to show that $(T_\alpha)_{\alpha\in A}$ is convergent to a certain operator $T$. Is there any version of Banach-Steinhaus theorem valid in this case? That is, what I can show is the fact that $(T_\alpha x)_{\alpha\in A}$ is convergent in $Y$ to $(Tx)_{\alpha\in A}$ for each $x\in X$. Can I conclude that $(T_\alpha)_{\alpha\in A}\to T$ ?

share|improve this question
add comment

1 Answer

You cannot conclude this at all, even when you have a sequence. See the corollary and the note here.

The pointwise limit defines a bounded operator, but the limit of the sequence of operators does not necessarily converge (in the norm topology) to the operator defined by the pointwise limit.

share|improve this answer
1  
Banach-Steinhaus requires $\{T_\alpha x:\alpha\in A\}$ to be bounded for each $x$, and convergent nets need not be bounded in general, so there potentially is a difference. –  Jonas Meyer Nov 19 '11 at 21:29
    
@JonasMeyer Nice catch. I don't immediately see how to fix this potential problem, so I'll edit the answer accordingly. –  Jonathan Gleason Nov 19 '11 at 21:52
    
Yes, I have $\sup_{\alpha \in A, \|x\|\leq 1}\|T_\alpha x\|<\infty$. So, is $T$ a limit in this case? –  bellotojimenez Nov 20 '11 at 7:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.