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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!

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  • $\begingroup$ Here's a related post. $\endgroup$ – David Mitra Jul 22 '15 at 23:12
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The uniform bounded principle does work for any collection of operators, see wiki or any book on functional analysis.

Note that convergent nets may not be bounded (in contrary to convergent sequences), see the post to which David Mitra is referring.

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  • $\begingroup$ 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 $x\in E$?? $\endgroup$ – C-Star-W-Star Jul 23 '15 at 8:42
  • $\begingroup$ 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 $f_\lambda$. $\endgroup$ – gerw Jul 23 '15 at 14:03
  • $\begingroup$ 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 $\{T_\lambda x\}_{\lambda\in\Lambda}$. $\endgroup$ – C-Star-W-Star Jul 23 '15 at 14:31
  • $\begingroup$ Ah I see the problem: That spoils the argument as one cannot form a sort of diagonal subnet that is simultaneously bounded for all $x\in E$. $\endgroup$ – C-Star-W-Star Jul 23 '15 at 14:40

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