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I'm reading through some notes one locally convex spaces ("lcs" from now on) analysis and there the following version of the Banach-Steinhaus theorem is given

Theorem (Banach-Steinhaus) $\quad$ The pointwise limit of a sequence of continuous, linear mappings from a barrelled lcs $U$ to a lcs $V$ is again a continuous, linear mapping.

followed by the remark

If we replace "sequence" with "net" this needn't be the case: For a discontinuous functional $f:U\rightarrow \mathbb{K}$ we can construct for each subspace $W\subseteq U$ a continuous, linear functional $F_W$ such that $f\big|_W=F_W\big|_W$.

Can someone explain, or give me a hint, how to make this construction from the last sentence above explicit ?

I'm also not sure how to use this to obtain a counterexample to the theorem above ? I somehow can't think of a way to make use of a point of discontinuity in $x_0\in U$ of $f$ to show that the net $(F_W)_W$ doesn't converge at all at $x_0$ (at least I intuitively think that this is the case - opposed to that the net indeed converges everywhere, but not to a continuous, linear functional).

  • I'm also not sure how to use this to obtain a counterexample to the theorem above ? You have a net of continuous linear functionals that converges pointwise to a discontinuous linear functional. That is the (counter)example. If a sequence of continuous linear functionals converges pointwise to a linear functional on a barreled space, the limit functional must be continuous. The crucial point is that convergent sequences are bounded, but convergent nets need not be bounded. – Daniel Fischer May 07 '14 at 18:40

1 Answers1

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Consider the net $FIN(U)$ of finite dimensional subspaces of $U$. For each finite dimnsional subspace $W$ the functional $f|_W$ is continuous. Extend it using Hahn-Banach theorem to get continuous functional $F_W$ on $U$ such that $F_W|_W=f|_W$. By construction $(F_W)_{W\in FIN(U)}$ is a net of continuous functionals, and because $F_W|_W=f|_W$ this net pointwise converges to the functional $f$ which is discontinuous.

Now we need to show existence of at least one discontinuous functional on infinite dimensional lcs $U$. Since $U$ is infinite dimensional it has infinite Hamel basis, say $(e_i)_{i\in I}$. Fix any neighbourhood of $0$ which we denote $B$. Since $B$ is a neighbourhood of $0$ for each $i\in I$ we can find $\lambda_i\in\mathbb{K}\setminus\{0\}$ such that $\lambda_i e_i \in B$. Denote $e_i'=\lambda_i e_i$, then $(e_i')_{i\in I}$ is a Hamel basis of $U$. Now we define discontinuous linear functional $f:U\to\mathbb{K}$. Clearly it is enough to specify values of $f$ only at basis vectors. Consider arbirary countable subset $S=\{i_1,\ldots,i_n,\ldots\}$ of $I$. Define $f(e_{i_n}')=4^n$ for each $i_n\in S$ and $f(e_i')=0$ for $i\in I\setminus S$. We claim that $f$ is discontinuous. Since $\{e_i':i\in I\}\subset B$, then $\lim\limits_{n\to\infty} 2^{-n} e_{i_n}'=0$, but $\lim\limits_{n\to\infty} f(2^n e_{i_n}')=\lim\limits_{n\to\infty} 2^{-n} f( e_{i_n}')=\lim\limits_{n\to\infty} 2^{-n} 4^n=+\infty$. This proves discontinuity of $f$.

Norbert
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  • But how do we find the discontinuous $f$ in the first place and why does $f\big|_W$ has to be continuous ? –  May 07 '14 at 18:24
  • As for the construction of discontinuous linear functional see this answer (you can easily generalize it for lcs). Any linear map on finite dimensional space is continuous, that is why $f|_W$ is continuos. – Norbert May 07 '14 at 18:27
  • Could you please explain, how to prove, that the functional from the link you provided, is discontinuous ? –  May 07 '14 at 19:28
  • ok, I'll do it${}$ – Norbert May 07 '14 at 19:29
  • Thanks for your detailed response. I have one last, quick question, before I award you your well-deserved bounty: Why does is follow from ${e_i':i\in I}\subset B$ that $\lim\limits_{n\to\infty} 2^{-n} e_{i_n}'=0$ (how can I show that the $2^{-n}e'_{i_n}$ "crawl" for sufficiently large $n$ into every neighborhood $\tilde{B}$ of $0$ ) ? –  May 07 '14 at 20:01
  • Answer to the first question. Let $x$ be represnted as linear combination of elements of the basis $(e_i){i\in I}$ as $x=\sum{i\in I}\alpha_i e_i$. In other words $(\alpha_i){i\in I}$ are coordinates of $x$ in the basis $(e_i){i\in I}$. Then $x=\sum_{i\in I}\alpha_i \lambda_i^{-1} e_i'$, i. e. $\alpha_i'=\lambda_i^{-1} \alpha_i$ for $i\in I$ are coordinates of $x$ int the basis $(e_i')_{i\in I}$. – Norbert May 07 '14 at 20:07
  • Answer to the second question. It is enough to show that $(2^{-n}B){n\in\mathbb{N}}$ is a local base of neighbourhoods of $0$. Let $B'\subset U$ be arbitrary neighbourhood of $0$. By construction $B$ is a neighbourhood of $0$, hence bounded, hence we have $s>0$ such that $t>s$ imples $B \subset tB'$. Consider $N$ such that $2^{-N}s<1$, then $B\subset 2^n B'$. Hence $B'$ contains all sets of the form $2^{-n}B$ for $n>N$. Since $B'$ is arbitrary, then $(2^{-n}B){n\in\mathbb{N}}$ is a local base of $U$. – Norbert May 07 '14 at 20:16
  • Perfect! Thanks a lot! You have your bounty! (Though in your last comment, I wonder why $B$ had to be a priori bounded; the whole set $E$ is also a neighborhood of $0$ but not necessarily bounded; but of course $B$ could be chosen in such a way as to be bounded). –  May 07 '14 at 20:29
  • I just see, that I have to wait another 21 hours, before I can award the bounty :( You shall have it tomorrow then. –  May 07 '14 at 20:30
  • Ok, no problem! – Norbert May 07 '14 at 20:41