# Convergence on locally convex spaces

I'm new on the locally convex spaces. I know that if $$X$$ is a vector space and $$S$$ an irreducible set of seminorms defined in $$X$$, $$(X,S)$$ is a locally convex vector space. The first question is, how the convergence is defined in $$(X,S)$$ and how is a Cauchy sequence defined? And second, yet is not clear for me which are the elements that are in the topology induced by the seminorms in $$S$$? If you could recommend me a book about this I'll be grateful.

• seminorms induce a uniformity (and thus a notion of Cauchy sequences and filters/nets) and a uniformity induces a topology. Both are standard constructions. – Henno Brandsma Feb 28 at 5:14

First you need to define the topology $$\mathscr{T}$$ on $$X$$. For each seminorm $$\rho\in S$$, $$\epsilon>0$$, and point $$x\in X$$, define the open ball $$B(x,\rho,\epsilon)=\{y\in X\ |\ \rho(y-x)<\epsilon\}\ .$$ Take the collection of all such balls. That is a subbasis for the topology $$\mathscr{T}$$. Namely, by taking finite intersections and then arbitrary unions of these balls you get all the open subsets of $$X$$.

Convergence is defined in the usual way. A sequence $$(x_n)$$ in $$X$$ converges to some $$x\in X$$, iff $$\forall V\ {\rm open\ neighborhood\ of\ } x, \exists N, \forall n\ge N,\ x_n\in V$$ Similarly unsurprising is the definition of $$(x_n)$$ being Cauchy: $$\forall V\ {\rm open\ neighborhood\ of\ } 0, \exists N, \forall m,n\ge N,\ x_m-x_n\in V$$

That being said, a few remarks are in order. In this general setting, sequences are not the best thing to work with, especially if $$S$$ is uncountable. You should use nets instead, with similar definitions for being Cauchy and convergence. A good introductory reference is the book by Osborne on "Locally Convex Spaces". To go a bit further, "Introduction to Functional Analysis" by Meise and Vogt is also good.

• Thank you for your answer, and the continuity of $f \rightarrow \mathbb{R}$ in $x_0$ as follows right? $\forall \epsilon > 0 \hspace{.1cm} \exists q \in S: \hspace{.1cm} q(x-x_0) \leq 1 \Rightarrow |f(x) - f(x_0)| \leq \epsilon$ – The Student Mar 3 at 2:13
• You mean $f:X\rightarrow\mathbb{R}$, right? Your definition of continuity is correct except for the quantifier $\exists q\in S$. It should be $\exists q\in \bar{S}$ where $\bar{S}$ is the set of all seminorms on $X$ that continuous with respect to the topology $\mathscr{T}$ defined by $S$. BTW, you used the word "irreducible" to describe $S$. What do you mean by that? – Abdelmalek Abdesselam Mar 3 at 16:51
• A set S of seminorm is irreducible if it verifies 3 properties: 1.- $\forall t \geq 0$ and $q \in S: tq \in S$. 2.- $If q \in S$ and $p$ is a seminorm in $X$ such that $p \leq q$: $q \in S$. 3.- $sup(q,p) \in S \hspace{.1cm} \forall q,p \in S$. For 1, can be proved that every seminorm in $X$ is continuos and moreover is uniformely continuos. – The Student Mar 3 at 18:18
• OK then with this definition of "irreducible" then $\bar{S}=S$ and we agree. Where did you see this terminology? It sounds weird to me because irreducible usually means minimal, whereas your $S$ is maximal as far as having the property of defining the topology. – Abdelmalek Abdesselam Mar 3 at 19:36
• The term "set irreducible" of seminorms is in the book "Locally Convex Spaces and Linear Partial Differential Equations" by Francois Treves. – The Student Mar 5 at 4:00

A basis for the topology on $$(X,S)$$ consists of the sets of the form $$\{x \in X : \|x-x_0\|_1 < \varepsilon, \ldots, \|x-x_0\|_n < \varepsilon\}$$ for some $$x_0 \in X$$, $$\varepsilon > 0$$ and a finite collection of seminorms $$\|\cdot\|_1, \ldots, \|\cdot\|_n \in S$$.

Using this you can show that a net $$(x_\lambda)_{\lambda\in \Lambda}$$ converges to $$x \in X$$ if and only if $$\|x- x_\lambda\| \to 0$$ for every seminorm $$\|\cdot\| \in S$$.

Similarly, a net $$(x_\lambda)_{\lambda\in \Lambda}$$ is Cauchy if and only if for every $$\varepsilon > 0$$ and seminorm $$\|\cdot\| \in S$$ there exists $$\lambda_0 \in \Lambda$$ such that $$\lambda, \mu \ge \lambda_0 \implies \|x_\lambda-x_\mu\| < \varepsilon$$.