Consider two Banach spaces $E,F$ and a net $T_\alpha : E \to F$ of continuous operators.

I know that for each $x \in E$ the net $T_\alpha (x)$ is convergent in $F$ and it is easy to show that the limit $L(x)$ is a linear function.

Can one deduce that $L$ is bounded?

I thought originally that this is an obvious consequence of the Uniform Bounded Principle, but unfortunatelly it is not.

P.S. this problem comes from some question on convergence in a non-metrisable LCTVS, which I reduced to the simpler looking question above. Because of this I cannot work with sequences.

  • $\begingroup$ As pointed out below, this is false, even for one-dimensional range space. Indeed, the continuous dual (even of a lcs) is always dense in the algebraic dual for the topology of pointwise convergence. I added this comment to suggest a caveat. The negative claim depends on the fact that these two duals are distinct. There are, however, models of set theory in which this is false. One further remark—there are many important classes of non-metrisable lcs‘s where continuity follows from sequential continuity (especially relevant in the theory of distributions and measure theory). $\endgroup$ – user131781 Jun 18 '19 at 13:45

No, this works for sequences but not general nets. For a counterexample, let $T: E \to \mathbb{R}$ (or $\mathbb{C}$) be an unbounded linear functional. For each finite dimensional subspace $E_0$ of $E$, the restriction of $T$ to $E_0$ is bounded so by the Hahn-Banach theorem it can be extended to a bounded linear functional on $E$. Ordering by inclusion of finite dimensional subspaces then yields a net of bounded linear functionals which converges pointwise to $T$.

This is why the uniform boundedness principle for sequences is so amazing!

  • $\begingroup$ BTW, if you have a uniform bound on the norms of the $T_\alpha$ then the desired conclusion does hold, of course. In case that might help in your situation. $\endgroup$ – Nik Weaver Jun 18 '19 at 14:02
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    $\begingroup$ The uniform boundedness principle says that every pointwise bounded set of operators is uniformly bounded. Contraty to sequences pointwise concergent nets need not be pointwise bounded! $\endgroup$ – Jochen Wengenroth Jun 18 '19 at 21:16
  • $\begingroup$ @JochenWengenroth oops, you are right, I was thinking of Banach-Steinhaus. $\endgroup$ – Nik Weaver Jun 19 '19 at 2:48
  • $\begingroup$ Ty... The issue is exactly the uniform bound on the norms, if I have that then the proof to what I seek is pretty simple. I thought that UBP gives me the uniform bound, which is true if I work with sequences, but I have to work with nets... Your example really clarifies that my approach cannot work. $\endgroup$ – Nick S Jun 19 '19 at 8:46
  • $\begingroup$ You are welcome! $\endgroup$ – Nik Weaver Jun 19 '19 at 10:37

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