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Let $E_1$ be a (nontrivial) vector space, $P$ be a family of seminorms on $E_1$, $\tau_1$ denote the topology generated by $P$, $(E_2,\tau_2)$ be a topological space, $f:E_1\to E_2$ at $x\in E_1$.

Can we show that $f$ is $(\tau_1,\tau_2)$-continuous at $x$ if and only if $f$ is $(p,\tau_2)$-continuous for all $p\in P$?

EDIT: If not, is the proof in the reference below wrong or is there anything special in that situation?

The following instance of this setting (found in Linde's Probability in Banach Spaces) is confusing me:

enter image description here enter image description here

In his proof, the author has found a single $K$ for which $\hat\mu$ is (uniformly) $p_K$-continuous.

Translated to the general setting in this question, this seems to indicate that it's sufficient to have $(p,\tau_2)$-continuity for a single $p\in P$. Is this really true?

On the other hand, if I got it right, he has even shown uniform continuity. Is this the crucial ingredient?

0xbadf00d
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  • Good question. It's been ages since I last saw this stuff but I would totally bet that uniformity is key here. – Giuseppe Negro Dec 18 '20 at 15:42
  • I tried something, but I can't conclude. For that I should review my functional analysis and I would love so, but I haven't time right now. Let me write down what I have been thinking. I would like to prove that $\hat\mu$ is continuous at $0$, then it would presumably be easy to conclude that it is continuous everywhere. So we need to prove that, for all $\epsilon>0$, the set $${a'\in E', :, \lvert \hat\mu(a')-1\rvert<\epsilon }$$ is an open neighborhood of the origin of $(E', \tau_c(E', E))$. [...] – Giuseppe Negro Dec 18 '20 at 16:01
  • A basis of neighborhoods of the origin is given by $$N_{a_1, \ldots, a_n;\delta}={ a'\in E'\ :\ \max_{a_1,\ldots , a_n} \lvert \langle a', a_j\rangle \rvert <\delta}.$$Now I really think that the condition they find in the excerpt you posted suffices to prove that the set of my previous comment contains a set of this kind. – Giuseppe Negro Dec 18 '20 at 17:09
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    @GiuseppeNegro Thank you for sharing your thoughts. At the moment, I don't see that it is easier to show continuity at $0$. Please take a look at the question I've asked for that particular problem: https://math.stackexchange.com/q/3879201/47771. – 0xbadf00d Dec 19 '20 at 18:17
  • @GiuseppeNegro I think I was able to prove the claim; hopefully I did no mistake: https://math.stackexchange.com/a/3955938/47771. – 0xbadf00d Dec 20 '20 at 10:22

1 Answers1

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Only one direction is true.

If $f$ is $(p,\tau_2)$-continuous for any $p \in P$, then $f$ is $(\tau_1,\tau_2)$-continuous. This follows simply from the fact that $\tau_1$ is finer (larger) than the topology of $p$.

The other direction is not true in general. Let $p,q$ be two nonequivalent norms on a vector space $E$. W.l.o.g. assume that there is no constant $M>0$ such that $p \le Mq$. Let $\tau_1$ be the topology generated by $P=\{p,q\}$ and $\tau_2$ the standard topology on $\Bbb{R}$. Then $p : E \to \Bbb{R}$ is $(\tau_1,\tau_2)$-continuous but not $(q,\tau_2)$ continuous.

mechanodroid
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  • Thank you for your answer. Actually, that is what I expected, but what am I missing then in the cited proof of the reference? Is the proof wrong or is there anything special in the situation? – 0xbadf00d Dec 19 '20 at 18:05
  • Please take a look at the answer I've provided to the question whether or not the argument in the cited excerpt is correct: https://math.stackexchange.com/a/3955938/47771. Would be great if you could share what you think about it. – 0xbadf00d Dec 20 '20 at 10:24
  • @0xbadf00d I may be missing something, but doesn't the proof use the direction which does hold? It concludes that $\hat{\mu}$ is $(p_K,\tau_2)$-continuous so it is $(\tau_1,\tau_2)$-continuous. – mechanodroid Dec 20 '20 at 14:56
  • Are you referring to my proof in the other question? And did you intend to write "the direction wich does not hold"? In any case, please take note of the latest edit of my answer and please let me know if you still have doubts. – 0xbadf00d Dec 20 '20 at 18:42
  • @0xbadf00d No, I was referring to the proof in the screenshot in this question. No, I meant the direction which does hold. – mechanodroid Dec 20 '20 at 18:48
  • You have shown that if $f$ is $(p,\tau_2)$-continuous for all $p\in P$, then $f$ is $(\tau_1,\tau_2)$-continuous. Can we conversely show that if $f$ is $(\tau_1,\tau_2)$-continuous, then $f$ is $(p,\tau_2)$-continuous for some $p\in P$? – 0xbadf00d Dec 27 '20 at 15:43
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    @0xbadf00d No, consider my example with $P = {p,q}$ where $p,q$ are two non-comparable norms. Then $f : E \to \Bbb{R}$ given by $f = p+q$ is $(\tau_1,\tau_2)$-continuous but not $(p,\tau_2)$ nor $(q,\tau_2)$-continuous. – mechanodroid Dec 27 '20 at 16:44
  • You're right, but do you agree that the claim is true when $P$ is the family of seminorms generating the topology of compact convergence on $C(X,Y)$ for a topological space $X$ and a normed space $Y$? It should be true: If $(Z,d)$ is a metric space, $f\in C(X,Y)$ and $F:C(X,Y)\to Z$ is $(P,d)$-continuous at $f$, then for each fixed $\varepsilon>0$, there is a $n\in\mathbb N$, compact $K_1,\ldots,K_n\subseteq X$ and $\delta>0$ with $$d(F(f),F(g))<\varepsilon;;;\text{for all }g\in f+\delta\bigcap_{i=1}^nU_{p_{K_i}}\tag1,$$ – 0xbadf00d Jan 05 '21 at 11:25
  • where $U_{p_{K_i}}$ is the closed unit ball with respect to the seminorm $$p_{K_i}(g):=\sup_{x\in K_i}\left|f(x)\right|Y;;;\text{for }g\in C(X,Y).$$ Now, $$K:=\bigcup{i=1}^nK_i$$ is compact as well and it's easy to see that $$U_{p_k}=\bigcap_{i=1}^nU_{p_{K_i}}\tag2.$$ Thus, $F$ is $(p_K,d)$-continuous at $f$. Or am I missing something? – 0xbadf00d Jan 05 '21 at 11:25
  • @0xbadf00d I don't think it holds. Your compact set $K$ depends on $\varepsilon$, you should pick a compact set $K \subseteq X$ such that for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $g \in C(X,Y)$ holds $$p_K(f-g) < \delta \implies d(F(f),F(g)) < \varepsilon.$$ – mechanodroid Jan 08 '21 at 22:33
  • @0xbadf00d I believe I have a counterexample. Take something like $F : C_b(\Bbb{R},\Bbb{R}) \to \Bbb{R}$ given by $$F(f) = \sum_{n=1}^\infty \frac1{2^n}f(n)$$ where $C_b(\Bbb{R},\Bbb{R})$ is the space of all bounded functions $\Bbb{R} \to \Bbb{R}$ endowed with the topology of uniform convergence on compact sets (induced by seminorms $P$). Then $F$ can be seen to be $P$-continuous by the Lebesgue dominated convergence theorem but for any fixed compact set $K \subseteq \Bbb{R}$ it isn't $p_K$-continuous as you cannot control what happens outside $K$ and $K$ is bounded. – mechanodroid Jan 08 '21 at 22:53
  • Now $F$ is a continuous linear functional on a subspace $C_b(\Bbb{R},\Bbb{R})$ of a locally convex space $C(\Bbb{R},\Bbb{R})$ so by Hahn-Banach it extends to a continuous functional $F' : C(\Bbb{R},\Bbb{R}) \to \Bbb{R}$ which again isn't $p_K$-continuous for any fixed $K$. – mechanodroid Jan 08 '21 at 22:55
  • I know that $K$ depends on $\varepsilon$. What I've shown is that if $f\in C(X,Y)$ and $F$ is $(P,d)$-continuous at $f$, then $$\forall\varepsilon>0:\exists K\subseteq X\text{ compact}:\exists\delta>0:\forall g\in f+\delta U_{p_K}:d(F(f),F(g))<\varepsilon\tag3.$$ And that does precisely imply what I previously claimed: If $F$ is $(P,d)$-continuous at $f$, then $F$ is $(p_K,d)$-continuous for some compact $K\subseteq X$. Am I missing something? – 0xbadf00d Jan 09 '21 at 11:44
  • @0xbadf00d No, this doesn't imply that $F$ is $(p_K,d)$-continuous. You need $$\exists K \subseteq X \text{ compact} : \forall \varepsilon > 0 : \exists \delta > 0 : \forall g \in f + \delta U_{p_K} : d(F(f),F(g)) < \varepsilon.$$ – mechanodroid Jan 09 '21 at 14:45
  • You're right. However, what I've shown should yield that $F$ being $(P,d)$-continuous at $f$ is equivalent to $(3)$. Do you agree? – 0xbadf00d Jan 13 '21 at 10:41