# Uniform Boundedness: Nets

I thought so far that the uniform boundedness principle applies (according to the proof I know) to any net of bounded operators. Now I read that this works for sequences only. Can you shortly explain me an example without going into details? Just to make sure we're talking about the same: $$T_\lambda\in\mathcal{B}(E,F):\quad\|T_\lambda x\|<\infty\quad(x\in E)\implies \|T_\lambda\|_{\lambda\in\Lambda}<\infty$$ where $\|T_\lambda\|_{\lambda\in\Lambda}:=\sup_{\lambda\in\Lambda}\|T_\lambda\|$.

Thank you very much!

• Here's a related post. – David Mitra Jul 22 '15 at 23:12

• Aaah right because of the subnet issue. So can it happen that $\#\{\lambda\Lambda:\lambda\ngeq\lambda_\varepsilon\}=\infty$? But shouldn't it still hold that $\|T\|<\infty$ supposed $T_\lambda x\to Tx$ for all $x\in E$?? – C-Star-W-Star Jul 23 '15 at 8:42
• I do not understand (the notation of) the first question. Concerning the second question: the proof of this fact for sequences which I know, utilizes that the convergence $T_\lambda x \to Tx$ implies boundedness of $T_\lambda x$ (w.r.t. $\lambda$) which might not hold for nets. A counterexample could be the following: take a unbounded net of functionals $f_\lambda \in X^*$ which converges to $f$. Then, $f_\lambda(x) \to f(x)$, but $f_\lambda(x)$ cannot be bounded w.r.t. $\lambda$ (for all $x$), since this would imply (by the uniform boundedness principle) the boundedness of $f_\lambda$. – gerw Jul 23 '15 at 14:03
• I meant that by convergence of the net: For fixed $x\in E$ and for any $\varepsilon>0$ there is an index $\lambda_\varepsilon\in\Lambda$ such that for all $\lambda\geq\lambda_\varepsilon$ one has $\|Tx-T_\lambda x\|<\varepsilon$. That bounds $\|T_\lambda x\|\leq\varepsilon+\|Tx\|$ for all $\lambda\geq\lambda_\varepsilon$. However there may be, in contrast to sequences, still infinitely many $\lambda\ngeq\lambda_\varepsilon$. So one cannot conclude boundedness of the whole net $\{T_\lambda x\}_{\lambda\in\Lambda}$. – C-Star-W-Star Jul 23 '15 at 14:31
• Ah I see the problem: That spoils the argument as one cannot form a sort of diagonal subnet that is simultaneously bounded for all $x\in E$. – C-Star-W-Star Jul 23 '15 at 14:40