# Talk:Locally convex topological vector space

## Started

Just started with the basics here (it was on Wikipedia:Requested articles/mathematics). Most of the Topological vector space article is actually on LCTVSs, so I think rather than take this page much further on its own, some material be transferred. --AndrewKepert 03:30, 11 Mar 2004 (UTC)

I've merged in the locally convex page. This leaves a couple of problems.

(1) Slight conflict in definitions to resolve (one convex nhd versus a base of such.

(2) The phrasing about all examples at topological vector space being LC is obviously not a secure thing to do.

Charles Matthews 06:57, 4 May 2004 (UTC)

Hi Charles

Yes, I agree that rationalisation between the two pages is necessary, but didn't get around to it myself. However, LCTVSs require a base of convex neighbourhoods -- otherwise every TVS is locally convex (f'rinsance, the whole space is a convex nbhd of 0). Where did you pick up that only a single nbhd was necessary - the source will need fixing. I will fix the wording here, while keeping your informal/formal characterisation.

I also agree that my note that "all examples" is dangerous, especially as I then went on to add the non-locally-convex example to the TVS page ... oh well.

-- Andrew Kepert 07:16, 4 May 2004 (UTC)

PS. I had been considering moving chunks of stuff from the TVS article to here, but on second thoughts, I don't think it (the TVS article) should be gutted. It is the most general class here, so to have an overview of everything that happens within it (all the way down to the most special, such as Hilbert spaces) is sensible. It could do with a reorganise though. -- Andrew Kepert 07:34, 4 May 2004 (UTC)

The bug with the defn was from the initial version of locally convex - a contributor only around for about a week. Charles Matthews 07:38, 4 May 2004 (UTC)

Ah sorry - I misread your original note as to what you were merging with (I thought TVS), and so my "I agree" above is talking crossed purposes. Not to worry - all moving forward. Andrew Kepert 08:13, 4 May 2004 (UTC)

I am not sure where to raise this, but this talk page has seen some recent activity. The link to semi-norm appearing in the article redirects to normed vector space, however, that article does not mention semi-norms anywhere (unless I've overlooked something). What is the reason for the redirect? -- Jitse Niesen 21:08, 10 May 2004 (UTC)

OK thanks for picking that up - you are right. When I wrote the page I didn't check that all links were appropriate. I will fix. -- Andrew Kepert 04:33, 11 May 2004 (UTC)
OK, created Seminorm -- Andrew Kepert 06:27, 11 May 2004 (UTC)

## local basis from seminorms not clear

The second phrase of the "Definition" is a bit confusing, especially the "this" seems to refer to an already given local basis, while there is clearly no connection between the newly introduced family N and the existing topology of the TVS. I will try to clarify; please check if you agree to changes. MFH: Talk 18:32, 21 Apr 2005 (UTC)

I clarified the construction. Is it clearer now ? MathMartin 14:48, 24 Apr 2005 (UTC)

## too specific category

I object somehow against changing to the "more specific category" of function spaces; even if this is the main application, nothing requires /a priori/ a LCTVS to be a function space (in the usual sense --- of course, one could say that even the real numbers are also a function space, when seen as quotient of sequence spaces...) MFH: Talk 22:45, 19 September 2005 (UTC)

Yes, I guess I was wondering if this objection might be raised. I can put it back, but need your help with one thing, then. The Category:Functional analysis has 120 articles in it, and I was hoping to create a subcategory into which some of these could be moved. I was thinking along the lines of "those articles having more of a topological slant than usual" as being the subcategory, but admit this is rather vague and poorly defined. Any suggestions for a better split? linas 23:58, 19 September 2005 (UTC)
OK I'll put it back. linas 03:17, 23 September 2005 (UTC)

## Product spaces

Is the product of two LCTVS's also a LCTVS (assuming the product topology) ? Seems to be so for Frechet spaces, not clear if this is a general property. linas 03:17, 23 September 2005 (UTC)

Sure it is. Just take take the union of the two families of seminorms with pα(x,y)=pα(x) and qβ(x,y)=qβ(y). A seminorm is, after all just projection onto one component, and a product is just a space with extra components. I expect that only countable products of Fréchet spaces will be Fréchet spaces. An uncountable product might not have a countable family. I'm not sure I can prove it doesn't though, so maybe you're right, and any product of Frechet is Frechet. -lethe talk 04:06, 8 January 2006 (UTC)

## counter examples

It would be nice and instructive to have some counter-examples in addition to the omnipresent "almost all ... are ..." (e.g. LCTVS which are not semi-normable, TVS which are not LC,...). If s.o. has some (not too weird) ideas, I would appreciate. — MFH: Talk 12:53, 27 September 2005 (UTC)

Any vector space with the discrete topology is an example of a TVS which is not LC, since the "absorbent" property is not fulfilled. Or am I missing something? --Roentgenium111 (talk) 00:23, 9 April 2010 (UTC)

## large rewrite

I've just finished a pretty major rewrite of this article, and I welcome feedback and criticism. Since the diff between the versions would be hard to compare, you'll just have to read the new version, But let me enumerate points that were lost in my changes:

### definition

#### 1

part of the definition of the convex base said

I haven't replaced that in my text. I think this follows from saying you have a local base, right? I'm not sure why it was included in the first place, I'm hoping someone can clue me in if I'm missing something.

#### 2

I didn't keep "Conversely for every non-empty family of subsets ${\displaystyle {\mathcal {B}}}$ of a vector space ${\displaystyle X}$ with the above properties, there exists exactly one topology ${\displaystyle \tau }$ so that ${\displaystyle (X,\tau )}$ is a locally convex topological vector space and ${\displaystyle {\mathcal {B}}}$ a locally convex basis."

There seems to be some theorem in there, like every base induces a unique topology or something. I can't see what that's doing here either.

### Constructing a locally convex basis

This section was largely left out, because the new article puts much higher emphasis on seminorms, taking them as a definition of LCTVS. The way to construct a convex base is still mentioned, though perhaps with less length, with different notation, and different style.

### Properties

I left out these two properties, but I would very much like to put them back in:

1. A locally convex space is seminormable if and only if there exists a bounded neighbourhood for zero.
2. A locally convex space is semimetrizeable if and only if the topology can be defined by a countable family of semi-norms.

The reason I took them out is that they contradict the definitions I'm using. For example, according to my definition, every LCTVS is seminormable, not just the ones with bounded base. And every LCTVS with countable family of seminorm is metrizable, not just semimetrizable. (Actually, I don't even know what semimetrizable means).

I'd love if someone could help me figure out what definitions those statements relied on.

Nevermind about the first one. I think author meant "normable". -lethe talk 11:26, 8 January 2006 (UTC)
(semi-) normable means the topology of the topological vector space can be given using a (semi-) norm. Not every LCTVS is (semi-) normable. Of course you can pick any of the (semi-) norms which are associated with a LCTVS to create a (semi-) normed vector space but this (semi-) normed vector space does not have the same topology as the LCTVS. MathMartin 14:24, 8 January 2006 (UTC)
OK, so seminormable means topology given by a single norm. I get it. I assumed it allowed families, which explains my error. And about assuming you meant normable, I guess that only applies in the Hausdorff case, which means I've made another error. Lemme fix. -lethe talk
I gather the Hausdorffness is what's going on with semimetrisability as well. I fix that too. --lethe talk 19:43, 8 January 2006 (UTC)
No, I'm still confused. According to the article metric, a semimetric is a metric without the triangle inequality. Without the triangle inequality, the topology is not convex. I think maybe it was supposed to be pseudometric. MathMartin, can you confirm? -lethe talk 19:58, 8 January 2006 (UTC)
I went ahead and changed the article, assuming pseudometric. -lethe talk 20:36, 8 January 2006 (UTC)

And OK, other than that, I didn't remove anyone's work from the article, though I did change the style and emphasis quite a lot. So what do you think? -lethe talk 04:00, 8 January 2006 (UTC)

Most of the material on the page was from a rewrite I did a few month ago. I am busy so I do not have time to work on the article so I will just give a few short notes on your rewrite. I answered you question above and fixed your error in the introduction of the article.

The definition section became clearer but is now too long and detailed. I think the subject is specialized enough so that we can assume the reader is familiar with the basic concepts which are now explained in detail in the definition.

MathMartin 14:44, 8 January 2006 (UTC)

I very quickly skimmed the current article. I don't quite agree with the statement "we can assume the reader is familiar with the basic concepts which are now explained in detail in the definition". While I'm familiar with the concepts, its handy to have a "precise" definition on the spot. While you may recommend that I then go and read the articles on convex or seminorm to get these, I note that those articles tend to provide general definitions, sometimes so general that they can be subtly inaccurate or misleading for the topic at hand. (This is a problem for a variety of articles, not just this one). I don't think it hurts to have a precise recap here, of defn's that are directly applicable to the LCTVS, using the terminology and notation specific to this topic. Or at least, that's my gut impression; I have not examined the article carefully. linas 16:50, 8 January 2006 (UTC)
I'm usually the guy who likes to keep unnecessary definitions out of related articles, but somehow it seemed appropriate to define convex, absorbent, balanced and seminorm here in the article, but I think including the definitions here is a good idea because there are two alternative versions. A person who knows topology may not be very familiar with seminorms, while a person who knows analysis may not be familiar with local bases and balanced sets. If you think the definition is too long-winded, it can certainly be cut down. I might have erred on the side of too many definitions. -lethe talk 01:12, 9 January 2006 (UTC)

Elroch, you write "Revert. I think my edit was misread - the circle is specified as being the one *in* the 1-dimensional vector space generated by x". Right you are. Apologies. -lethe talk + 02:28, 4 May 2006 (UTC)

## Terminology

I prefer the definition of "balanced" that was previously in the article, as it is the most obvious use of the word, but a check of various sources indicates there is agreement on the modified definition I have added. The previous definition is now called "circled". Elroch 14:06, 4 May 2006 (UTC)

## absorbent sets, examples of non-locally convex spaces

1. The definition of "absorbent set" given in the article contradicts the definition given in the article on absorbent sets.

2. I was hoping to find some good examples of non-locally convex spaces. Unfortunately, the article didn't provide enough details about the two non-examples for me to see understand why they are non-examples. I know case of the Lp spaces for p<1 is discussed in some detail in Rudin's _Functional Analysis_, but I have not seen the second example elsewhere, and it looks non-trivial to work out the details based on the remarks in the article. —Preceding unsigned comment added by 128.237.251.63 (talk) 04:24, 2 October 2007 (UTC)

## Initial topology of seminorms

Currently the section "Seminorms" contains the sentence "In other words, it is the coarsest topology for which all the seminorms are continuous." I think the phrase "for which all the seminorms are continuous" should be replaced with something like "for which all the pseudometrics induced by the seminorms are continuous".

For example, if we consider the set ${\displaystyle \mathbb {R} ^{2}}$ with a one-element seminorm family consisting of the Euclidean norm then all points with equal distance from 0 would be topologically indistinguishable in the coarsest topology of ${\displaystyle \mathbb {R} ^{2}}$ for which the Euclidean norm (as a map from ${\displaystyle \mathbb {R} ^{2}}$ to ${\displaystyle \mathbb {R} }$) is continuous. But this is not the initial topology meant in this article, is it? --Jaan Vajakas (talk) 14:30, 10 February 2008 (UTC)

in this context, one is only interested in TVS topologies. so we take a basis of neighborhoods at 0 and then translate. with R^2 and the Euclidean norm, this results in the usual topology. Mct mht (talk) 09:23, 23 February 2008 (UTC)
All right, I understand. But I still think the sentence should be clarified. The next sentence ("That the vector space operations are continuous...") concludes that the topological space we get is a TVS, so before that the implicit assumption that one is only interested in TVS topologies should not be made. The referenced article Initial topology also does not mention the initial topology of a family of seminorms, so a novice (like me) might first try to apply the definition of initial topology of a family of functions literally, by regarding seminorm as a function from the topological space to ${\displaystyle \mathbb {R} }$.Jaan Vajakas (talk) 20:00, 24 February 2008 (UTC)
I totally agree with Jaan. As it stands in the article together with "In other words, it is the coarsest topology for which all the seminorms are continuous." is just wrong. It should be "In other words, it is the coarsest vector space topology for which all the seminorms are continuous." or something like that. The next sentence should then explain how to obtain this topology by the neighborhood at zero construction. Put in this way it is correct and understandable.
If someone wants to be more precise about the initial topology as such then the initial structure taken for the seminorms is taken in the category of pointed filtered spaces, i.e. tuples (X,*,U(*)) with * in X and a *-fixed Set-filter U(*) as objects and arrows the maps of pointed spaces continuous at the specific point. But I think it is not appropriate to discuss it like this in the article because it is a little bit too far off the introductory level. Filip --195.158.178.82 (talk) 13:50, 24 January 2009 (UTC)
I essentially agree with what you say, especially about not going into the category language here. But you loose a bit by requiring vector space topology; for example the next sentence in the former text (that we indeed define a TVS topology in this way) becomes meaningless. So I tried to simply say that one wants all translates of the semi-norms to be continuous. We'll see if there are some reactions. --Bdmy (talk) 15:12, 24 January 2009 (UTC)

## Nonexamples of locally convex spaces

Shouldn't the quasinorm for the ${\displaystyle L^{p}(0,1)}$ spaces be ${\displaystyle ||f||_{p}=\int _{0}^{1}|f(x)|^{p}\,dx}$ (that is, shouldn't the pth root be removed)? Otherwise it would seem that ${\displaystyle ||f||_{p}}$ is a true norm. Cpryby (talk) 06:27, 11 April 2009 (UTC)

Most authors have the quasi-norm homogenous, ${\displaystyle \|\lambda x\|=|\lambda |\,\|x\|}$ so they do have the ${\displaystyle 1/p}$ power. If you take ${\displaystyle p=1/2}$ (for example) and the ${\displaystyle \ell ^{p}}$ norm on ${\displaystyle \mathbf {R} ^{2}}$, you get for two basis vectors ${\displaystyle e_{1}=(1,0)}$ and ${\displaystyle e_{2}=(0,1)}$
${\displaystyle \|e_{1}+e_{2}\|=(1+1)^{2}=4>\|e_{1}\|+\|e_{2}\|=2,}$
so the triangle inequality fails. --Bdmy (talk) 08:00, 11 April 2009 (UTC)

## Change the definition?

Given the terminology, it seems very strange not to give the "obvious" definition: a TVS with a base of convex sets. If desired, one can remark that the balanced and absorbent conditions can also be required (but are redundant). Am I missing something? 128.178.14.170 (talk) 08:45, 2 March 2010 (UTC)

I think so. Why do you think these properties are redundant? The "absorbent" property is needed to exclude e.g. the discrete topology on a C-vector space. As for "balanced", perhaps you made the same mistake I initially made, thinking that any convex set including the origin is automaticlaly balanced? This is not so, take e.g. the interval [-1,2] in R, which is not balanced. And the topology given on R by making the sets [0, 1/n] a basis of neighbourhoods of 0 is convex, but not balanced. (Neither is it absorbent.) I'm still thinking on an example of a convex absorbent, non-balanced space... --Roentgenium111 (talk) 00:18, 9 April 2010 (UTC)
Quoting Rudin, every open neigborhood of 0 contains a balanced neighborhood of 0 and every convex neighborhood of 0 contains a balanced convex neighborhood of 0 [Thm.1.14]. Also, every open neigborhood of 0 is absorbent [Thm.1.15a], so non-balanced or non-absorbent TVS or LCTVS make no sense. Lapasotka (talk) 18:53, 29 July 2011 (UTC)

## Another improvement that should be made

Beside the general definition (see previous comment), I also find the following a bit misleading:

"Although in general such spaces are not necessarily normable, the existence of a convex local base for the zero vector is strong enough for the Hahn-Banach theorem to hold, yielding a sufficiently rich theory of continuous linear functionals."

Why not simply give it straight: local convexity is *equivalent* to the natural Hahn-Banach theorem. Normability is a distraction. —Preceding unsigned comment added by 128.178.14.170 (talk) 08:49, 2 March 2010 (UTC)

## On completeness

I think there is something wrong with the following passages. The parts I have problem with are underlined.

• A Cauchy net in a locally convex space is a net {xκ}κ such that for every ε > 0 and every seminorm pα, there exists a κ such that for every λ, μ > κ, pα(xλxμ) < ε. In other words, the net must be Cauchy in all the seminorms simultaneously. The definition of completeness is given here in terms of nets instead of the more familiar sequences because unlike Fréchet spaces which are metrisable, general spaces may be defined by an uncountable family of pseudometrics. Template:Underline A locally convex space is complete if and only if every Cauchy net converges.
• An important function space in functional analysis is the space D(U) of smooth functions with compact support in URn. A more detailed construction is needed for the topology of this space because the space C0(U) is not complete in the uniform norm. The topology on D(U) is defined as follows: for any fixed compact set K ⊂ U, the space C0(K) of functions ƒ ∈ C0(U) with supp(ƒ) ⊂ K is a Fréchet space with countable family of seminorms ||ƒ||m = supx |Dmƒ(x)| (these are actually bona fide norms, and the space C0(K) with the ||·||m norm is a Banach space Dm(K)) . Given any collection {Kλ}λ of compact sets, directed by inclusion and such that ∪λKλ = U, then the C0(Kλ) form a direct system, and D(U) is defined to be the limit of this system. Such a limit of Fréchet spaces is known as an LF space. More concretely, D(U) is the union of all the C0(Kλ) with the final topology which makes each inclusion map C0(Kλ)↪D(U) continuous. Template:Underline The space is not metrisable, and so it is not a Fréchet space. The dual space of D(Rn) is the space of distributions on Rn.

As far as I understand, D(U) is an example of a non-metrizable complete locally convex vector space. (Non-metrizability implies that the topology D(U) is not induced by a countable family of seminorms.) Furthermore, it seems that D(K) is a Frechét-Urysohn space, which means that closures of its subsets coincide with their sequential closures. Proof. Let f be in the closure of AD(U). Then fC0(K) for some compact set KU. Since C0(K)⊂D(U) is closed, the C0(K)-closure of AC0(K) coincides with its D(U)-closure. Since C0(K) is metrizable, there exists {fj}j=0C0(K)∩A such that fjf in C0(K), and hence also in D(U). This proves that D(U) is Frechét-Urysohn.□ This indicates that the topology of D(U) is quite well characterized by convergent sequences. Note that we only used the fact that a sequence {fj}j=0 converges to f in D(U) if and only if there exists a compact KU which contains the support of every fj and fjf in C0(K). So how about the Cauchy-nets? Assume that every Cauchy-sequence in a topological vector space X converges. If there exists a Cauchy-net {xκ} of X which does not converge, then there exists a neighborhood UX of 0 such that for every κ there exists λ>κ with fλX\U. Hence we can pick a Cauchy-sequence {xκ(j)}j=0⊂{xκ} which does not converge in X, a contradiction. Am I missing something? If not, these passages should certainly be modified. Lapasotka (talk) 15:30, 27 July 2011 (UTC) Corrected the issue of completeness. 18.111.56.150 (talk) 02:30, 24 February 2013 (UTC)

## Redirect from Locally convex basis

I got redirected from locally convex basis to here but we do not have a single occurence of the word basis in the whole article... Should there be a new article about convex basis? At least we should kill the redirect, shouldn't we? IXhdBAH (talk) 12:37, 12 December 2011 (UTC)

## Is the space of test functions complete?

"this means that all Cauchy sequences converge, but not necessarily more general Cauchy nets" But Treves, Topological Vector Spaces, Distributions and Kernels (Theorem 13.1) says that all (strict) LF spaces are complete, and in Example II that follows, it says that the space of test functions is LF space. 18.111.33.87 (talk) 04:51, 22 January 2013 (UTC)