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This might be a silly question, but what motivates the name "locally convex" for locally convex spaces?

The definition in terms of semi-norms seems to have nothing to do with convexity or with the other definition involving neighborhood bases -- and the neighborhood basis definition makes little sense to me either, because it refers to sets which are "absorbent", "balanced", and convex.

Why the restrictions? And then why aren't they called "locally absorbent, balanced, and convex spaces"? And why do we never here about the terms absorbent or balanced in any other context?

Also, I know that Banach spaces are locally convex, but this just confuses me further -- what do Banach spaces have to do with convexity? And why are locally convex spaces a natural generalization of Banach spaces?

I have some vague ideas -- the Hahn-Banach theorem (and hyperplanes) are used a lot in convex programming, and the "p-norm" is only a norm for $p \ge 1$, the same values for which $x^p$ is a convex function -- are norms somehow "convex", does this follow from the triangle inequality? Then why aren't arbitrary complete metric spaces locally convex?

Any insights would be greatly appreciated.

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    $\begingroup$ Here is a link between convex sets and semi norms : A closed convex set is the intersection of all the half spaces that contains it. Half spaces can be viewed like $p^{-1}([0;\infty[)$ where $p$ is a linear form. Now the typical example of semi norm is the absolute value of a lineare form. So the equivalence between the two definitions is not so surprising. I suppose the restrictions "absorbent" and "balanced" are here to make things work and avoid pathological counterexamples. Now locally absorbant, balancer and convex space" would be a bit too long... $\endgroup$
    – Renart
    May 23, 2016 at 14:13
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    $\begingroup$ Having a local base that is both absorbent and balanced is a compulsory requirement for a topological vector space. Hence we call a space that admits a local base with an additional property of being locally convex a "locally convex space". $\endgroup$
    – BigbearZzz
    May 23, 2016 at 14:30
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    $\begingroup$ Generally, a space is called "locally foo" if every point has a neighbourhood basis consisting of neighbourhoods that are "foo". Locally connected, for example. Locally convex thus means every point has a neighbourhood basis consisting of convex sets. $\endgroup$
    – Daniel Fischer
    May 23, 2016 at 14:31
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    $\begingroup$ For more information on topological vector spaces, you may which to consult Rudin's Functional Analysis. The book covers the properties "absorbent" and "balanced" very early. $\endgroup$
    – BigbearZzz
    May 23, 2016 at 14:34

2 Answers 2

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The only reason the question seems silly is that you include the answer! A locally convex TVS is one that has a basis at the origin consisting of balanced absorbing convex sets. The reason for the emphasis on "convex" is that that's what distinguishes locally convex TVSs from other TVSs: every TVS has a local base consisting of balanced absorbing sets.

Regarding "Also, I know that Banach spaces are locally convex, but this just confuses me further -- what do Banach spaces have to do with convexity? And why are locally convex spaces a natural generalization of Banach spaces?", surely this is clear. Balls in a normed space are convex (as well as balanced and absorbing).

Why aren't arbitrary complete metric spaces locally convex? Huh? The notion of convexity makes no sense in a general metric space.

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  • $\begingroup$ This clears up most of my confusion. Why does the notion of convexity make no sense in a general metric space? Or do you mean that convexity as a notion only makes sense for linear spaces (vector spaces, not necessarily topological), because otherwise "the line between two points" is not defined? $\endgroup$
    – Chill2Macht
    May 23, 2016 at 14:54
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    $\begingroup$ Again I'm puzzled because you've answered your own question. $C$ is convex if for every $x,y\in C$ and $t\in[0,1]$ we have $tx+(1-t)y\in C$, and in a general metric space there is no such thing as $tx++(1-t)y$. $\endgroup$
    – David C. Ullrich
    May 23, 2016 at 15:00
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    $\begingroup$ I can think of a large number of things the question "Only for vector spaces, not necessarily topological? " might mean. If what you're asking is something that hasn't already been answered then ask again, a little more explicitly... $\endgroup$
    – David C. Ullrich
    May 23, 2016 at 15:48
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    $\begingroup$ Yes, as you say, convexity makes sense in any vector space. Otoh "locally convex" only makes sense for topological vector spaces. $\endgroup$
    – David C. Ullrich
    May 23, 2016 at 15:56
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    $\begingroup$ @Chill2Macht You can endowed a vector space with a metric d so that d(tx, ty) = |t| * d(x, y). Then the metric vector space is locally convex. This might be nearest to what you want. $\endgroup$
    – Michael
    Dec 1, 2022 at 13:29
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There is two different yet equivalent definition of Locally convex spaces : one in which the topology endowed by a family of semi-norms, and one in term of absorbent balanced and convex basis. The equivalence between the two definition is rather long to prove but you can find it in Rudin's Functional Analysis.

This might be of interest to you : https://en.wikipedia.org/wiki/Minkowski_functional

Why are locally convex spaces a natural generalisation of Banach spaces ? The topology of a Banach space is induced by the norm. And a norm is a family of semi-norm, namely the family containing one semi-norm which happens to be a norm.

What do Banach spaces hae to do with convexity ? Every $x$ in a Banach space has a localy base of convex sets, namely the balls centered a $x$ and of radius $r > 0$. Every open neighbourhood of $x$ contains one such ball centered at $x$.

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