# Talk:Banach space

## Applications

As a non expert reader, I have no idea what the use of the concept of a Banach space is. Does it help in understanding elementary particles, cryptography or computer simulation of physical problems? Is the concept of a Banach space useful to prove other critical mathematical theorems?

The utility may not be the reason that mathematicians are inspired to work on Banach spaces. Any useful purposes are still helpful to most readers though to help them understand why Banach spaces are important.

I've included a first attempt at saying something about this to generate some discussion. — Preceding unsigned comment added by Chogg (talkcontribs) 11:37, 6 September 2012 (UTC)

My contribution was a bit mickey mouse, so I got rid of it.

I suppose that it is not good if every article on a detailed mathematical article has a generic statement of applications which rely on the mathematics. However, the Banach Space article's "parent" article on functional analysis could contain some accessible discussion of how it effects the man or woman on the street.

Banach spaces play a vital role in the mathematical analysis of everything from relativity to materials science. Probably part of the reason there is no "applications" section is that Banach spaces are almost too ubiquitous. I'm not saying that there shouldn't be one, just that the task of doing so seems to require an almost encyclopedic knowledge of modern analysis. Sławomir Biały (talk) 12:47, 6 September 2012 (UTC)

## joke

a clever pun, to be sure, but i'm not sure it belongs in this article.

Q: What's yellow and normed, linear, and complete?

A: A Bananach Space.

And why it's yellow? -217.118.178.238 17:57, 26 Dec 2004 (UTC)
Bananach = banana + Banach. Bananas are typically yellow, Banach spaces are normed, linear, and complete. -lethe talk

## norm in L(V,W)

dear Jitse Nielsen! please explain me, how we can divide norm of Ax (i.e., ||Ax||) to vector (not a number!) x? how we will get a number in this case? ||Ax||/x will not be a number, but ||Ax||/||x|| will. -217.118.178.238 22:14, 27 Dec 2004 (UTC)

I don't understand what you mean. You are right that one cannot divide a norm by a vector. Perhaps the notation
||A|| = sup { ||Ax|| : x in V with ||x|| ≤ 1 }
is not clear? The symbol : does not mean division, see set-builder notation. The definition for ||Ax|| can also be written as ${\displaystyle \|A\|=\sup _{x\in B}\|Ax\|}$, where B is the unit ball in V, so B is the set of all xV with ||x|| ≤ 1. -- Jitse Niesen 15:10, 28 Dec 2004 (UTC)

## Common names and notation for examples?

Do any of the examples have commonly used names, such as the "blah" space or the "blah" Banach space? For instance, I have in front of me the "Banach space of all analytic functions on a domain D with sup norm", and am fishing for a shorter term for this. Also: symbology: the text in front of me calls this space ${\displaystyle A_{\infty }(D)}$ ... Is this a common notation for this? Why infty? I presume there is a "Banach space of polynomials of degree p on domain D with sup norm" called ${\displaystyle A_{p}(D)}$ ? (Dohh its the sup norm, silly me.) linas 14:05, 19 September 2005 (UTC)

## Infinite dimensional?

In the opening paragraph, should there be a comment that Banach spaces may be finite dimensional as well as infinite dimensional? ~ Moocowpong1 00:20, 7 November 2005 (UTC)

I like the way things are now. It says "typically infinite-dimensional" which does not imply "only infinite-dimensional". And indeed, the truly useful case of Banach spaces is in the infinite-dimensional case. So I would leave it the way it is. Oleg Alexandrov (talk) 02:26, 7 November 2005 (UTC)
• Can you find a partition of a Banach space being infinite dimensional ?? --85.85.100.144 09:13, 1 May 2007 (UTC) in a similar form you can find a partition for

${\displaystyle [a,b]}$ in the real line, using the Dirac measure

## linas' recent revert

linas, it seems that you reverted some anon who removed this stuff a section saying that the trace yields an isomorphism between a space and its dual. I can't understand how this section, as it currently stands, the way you've reverted it, can be correct, so perhaps you can explain it to me?

So we have a map F: V*xV → K. This map can be called the trace map (it is the trace map in a finite dim space). By currying, you propose to give us a map V → V*. But it seems to me that currying here can only give you an element of V**. And this is borne out by the formula, because F(-,x) seems to be an element of V**: it takes an element of V* like f, and returns a scalar F(f,x)=f(x). In short, I think what you've got here is a longwinded notation to describe the well known injection V → V** given by f ↦ f(x). Have I made a mistake? -lethe talk 18:00, 22 November 2005 (UTC)

Nope, sorry, I made the mistake. The original version so breif, I thought it was trying to curry. I tried to make that clear, and as you point out, I didn't do it right. The current version looks good. linas 20:13, 22 November 2005 (UTC)
Actually, now that I think about it, I completely bloopered it, didn't I? I sat there trying to get te wording right, and it kept coming out wrong. That alone should have been a warning sign ... linas 20:15, 22 November 2005 (UTC)

## changing V' to V*

I see that recently, Oleg, that you changed V' to V* and V'' to V** in order to "make it agree with Dual space". If you read that article, you will see that both V' and V* are used, but to mean different things; the first is the continuous dual, and the second is the algebraic dual. I'm not sure whether this is standard. Seems to me that the dual space should always be a construction within the category, and so the continuous dual is the important one. Anyway, whatever notation we use we should make it clear what we're talking about. I didn't revert your edit, however, because I think the injection V → V** is true for both continuous duals and algebraic duals, and the latter subsume the former, being more general, and so your notation agrees with the most general statement. I just want to make sure that you were aware of the implications of your edit. If you were, forgive my presumption. -lethe talk 18:24, 22 November 2005 (UTC)

You are right. I was aware of the distinction between dual space and continuous dual space, but not about which notation is used for which one. There is no problem with Banach space itself, as there the * is used consistently to mean the continous dual. But I will now change it back to prime to be indeed as in dual space. Oleg Alexandrov (talk) 01:53, 23 November 2005 (UTC)
The best thing to do when there is an ambiguity is to just explain it. I've added my attempt at explanation. -lethe talk 11:55, 19 December 2005 (UTC)
Thanks! Oleg Alexandrov (talk) 16:13, 19 December 2005 (UTC)

## Nice article

This is a very fine, easy to read, well-written article. Most of the credit goes to AxelBolt I would say. Great job. Oleg Alexandrov (talk) 00:26, 13 December 2005 (UTC)

## Possible correction

I quote the article: "Since the norm induces a topology on the vector space, a Banach space provides an example of a topological vector space..." I don't understand why the norm induces a topology... I would say that a norm induces a metric... It's just an opinion from a guy that doesn't understands much of this, so I leave it to someone else. --201.253.45.251

The norm induces a metric, which induces a topology. So it's OK. --Zundark 19:06, 8 October 2006 (UTC)

## Possible correction (take two)

Small complaint about the section on Hilbert spaces, in the sketch of the proof that parallelogram identity implies inner-product norm. The article says: "the completeness of the norm extends the linearity to the whole real line" from the rationals. However, it is actually the continuity of the norm that is being used here. Suppose ax is in the space. Take a sequence of rationals q_i converging to a, and regardless of completeness q_ix -> ax. The continuity of (. , .) finishes the proof. 84.0.252.215 10:43, 28 February 2007 (UTC)

## Hamel basis generally uncountable?

I hope this isn't impolite, but I deleted the part in the definition when it says that the Baire category theorem implies that the hamel basis of an infinite dimensional banach space is uncountable. Is a separable hilbert space not an example of an infinite dimensional banach space (with the norm induced by the inner product) endowed with a countable hamel basis? Maybe it could say that the basis is not generally countable... Also, I think the baire category theorem only proves that the space itself is uncountable, which is trivially true by the fact that there is a non-degenerate homomorphism into the underlying field. I'm a little bit drunk, so please revert my change and forgive me if I'm completely off base (no pun intended). —Preceding unsigned comment added by 74.136.214.173 (talk) 06:54, 9 December 2007 (UTC)

I've reverted you. An infinite-dimensional separable Hilbert space has uncountable Hamel basis. Please don't edit mathematical articles when you're drunk. --Zundark (talk) 08:56, 9 December 2007 (UTC)
Oops, sorry! Thanks for reverting my mistake. In my inebriated state, I obviously conflated the notion of hamel basis with that of the schauder basis. I only hope that this will serve as a warning to future generations against the dangers of drunk deriving.

## Banach manifolds

Should there be a little bit about banach manifolds in this article? I would think yes but perhaps that deserves its own separate content.

Topology Expert (talk) 04:33, 10 October 2008 (UTC)

## Sequence space

I've a question about sequence spaces which most books (and this article) seem to shy away from, and its confusing me, yet it seems central to Banach spaces. The space of all real-valued sequences with sup norm, i.e. the space of all ${\displaystyle f:\mathbb {N} \to \mathbb {R} }$ with sup norm, is not a Banach space (at least not normally, not that I'm aware of), right? But why? I know that the "classical" answer would be "because N is not compact", or that by simply limiting one-self to bounded sequences, one would get a Banach space. That's the classical approach.

But the modern definition seems to be "complete normed vector space". Now, perhaps I'm being really dumb here, but the space of all real-valued sequences meets this definition: its a vector space, its normed, and its complete. Now, this space is "sick", in the sense that many of its elements have infinite norm (e.g. the sequence ${\displaystyle a_{n}=n}$ has infinite norm), but the "modern" definition doesn't rule out infinite norm, right? The word "bounded" wasn't in the definition. The space is also "sick" in that all sequences of infinite norm appear to be topologically indistinguishable -- so e.g. ${\displaystyle a_{n}}$ is not distinguishable from ${\displaystyle b_{n}=n^{2}}$ by the norm ... but so what? The "modern" definition does not require the Banach space to be separable, Hausdorff or T_1 or T_0 or anything. So what gives? What am I missing? linas (talk) 00:30, 9 November 2008 (UTC)

Maybe you should ask this at the Wikipedia:Reference desk/Mathematics for a wider response, but as far as I can see, you are missing the following:

a) A Banach space is a complete (means that there has to be a metric inducing the topology, for a start, or completenss would not make sense) normed vector space. And since any metric space is perfectly normal, it must be Hausdorff, T1 and T0 immediately. So as you say, the space given cannot be a Banach space.

b) A normed vector space is by definition, a vector space V, with a function (let A be the set of all non-negative real numbers) f: V -> A that satisfies certain axioms. Since the range of this function is all nonnegative real numbers, 'infinity' is not included in the range. Therefore, no point can have infinite norm in a Banach space (notice that this is important for otherwise, the 'metric' induced by the norm would not actually be a metric).

So, the modern definition does actually imply all the properties you mentioned. As you say, the main reason why the space you have given is not a Banach space is because some points have infinite norm (so that is why compactness of the domain is important).

Topology Expert (talk) 01:23, 9 November 2008 (UTC)

Of course, thank you, very good answer. I suppose I should have thought about it more, before asking. Perhaps I'll need to crack open a book on general topology to see if its possible to compactify a metric space by adding a point at infinity; I suppose its not, but its not immediately obvious why not. linas (talk) 16:04, 9 November 2008 (UTC)

## Pronunciation

'Banak' does not make any sense, see Stefan Banach. —Preceding unsigned comment added by 212.87.13.78 (talk) 20:21, 7 March 2010 (UTC)

## Little proveits

I kinda want to add a little "(Why?)" to the statement, in examples, that a continuous function (R->R) defined on a closed interval is bounded. I thought the mental check of the statement to be a good review: closed interval of R <=> compact <=> continuous maps compact to compact <=> target space is R, compact <=HB=> closed and bounded. (Did I get it right?? :) :) ) 128.252.164.248 (talk) 14:23, 29 January 2011 (UTC)

I realize the above comment is over a year old, but I wanted to mention a perhaps more intuitive reason. If the function is unbounded, pick a sequence of points ${\displaystyle x_{i}}$ for which the limit of the values ${\displaystyle f(x_{i})}$ is infinity. The sequence has a convergent subsequence, so we may suppose ${\displaystyle x_{i}\rightarrow x}$ by replacing the sequence with this subsequence. But then since f is continuous ${\displaystyle \lim _{i\rightarrow \infty }f(x_{i})=f(x)=\infty }$, a contradiction. The same reasoning, slightly modified, also shows that f's image is closed. While your reasoning is correct and might be shorter to write, I find it a little less intuitive.
†This is basically the Bolzano-Weierstrass theorem, though it's not hard to see why this is true "by hand". One can pick a subsequence so the points are monotonically increasing (decreasing), and this subsequence has a supremum (infimum) in the interval since the interval is closed and bounded. 208.107.152.253 (talk) 04:32, 10 March 2012 (UTC)

## "Small factual correction"

The plus sign in the polarization identity is right provided that ⟨x,y⟩ is linear in its first argument. "In some conventions, inner products are linear in their second arguments instead" (quote from Hilbert space), in which case the minus sign in the polarization identity is right. Boris Tsirelson (talk) 18:59, 31 August 2011 (UTC)

ah, in this case you are right. I added a small info above this formula. I am working with linearity in second argument and it didn't cross my mind that it could be otherwise here, so I think that this information should be close to this formula - but please feel free to rephrase it. Janek Kozicki (talk) 07:05, 1 September 2011 (UTC)

## "less technical intro paragraph" added

"A piece of paper is a complete vector space, unless you omit the points exactly on the edge." — No; it is rather a complete metric space. A bounded part of a vector space cannot be a vector space (unless it is the origin only). Boris Tsirelson (talk) 04:52, 17 June 2012 (UTC)

## Recent big revision

It doesn't seem to contain anything incorrect but has made an article mostly just a collection of theorems (which are covered in detail elsewhere). This is not a step forward. I will revert it for now. Mct mht (talk) 19:20, 9 September 2012 (UTC)

Looking at that list of theorems again, some were trivial, some were fundamental, all are intermingled on the same list. No context. I suggest any such drastic change to discussed here first (e.g. one can certainly make the case that the article should mention tensor products of Banach spaces). Mct mht (talk) 19:31, 9 September 2012 (UTC)
The introduction is far too long so I make a extra "Definition" section. If someone enter the page you need a short information.
The Examples are also in the list, so there is nothing lost, but clearly structured.
The Linear operators section contain a mistake, it is B(X,Y) for "bounded" not L(X,Y)

I splitted the dual space section to dual space and reflexivity and added information. A Section Tensor product was completely missing so Non-Archimedean banach spaces also a diagram of the hierarchy.

And just a collection of theorems, that is mathematics...I thought? I try to optimize the page. Much information and if your looking for something you can find it quick.
Your right all of them are in one list, but I didnt know how else I can organize them. The fundamental are linked and Theorem,Corollary,Proposition so you have a priority and if your looking for such information, where else can you find them? StefanEckert 20:17, 9 September 2012 (UTC)
How else to organize the theorems? Grouping them into reasonable sections (or subsections), which of course needs more effort than just the list... and also more effort than just reverting... Boris Tsirelson (talk) 21:44, 9 September 2012 (UTC)
Ok, StefanEckert's version is back. I don't want to get into an edit conflict and also because this is a good-faith attempt, I will leave it alone. I will just note that, IMHO, as an expository article, it falls painfully short of the mark and is very objectionable. I get the impression this is a student's lecture notes. A few examples of questionable features: (too many to completely list)
• Completeness being a metric and not a topological property sounds like something one might say during a lecture. This trivial fact is given its own "explanation" section.
• A bag of random theorems is simply thrown haphazardly into a section. (That Banach implies Frechet is called a "theorem", Banach-Alaoglu is listed before dual space is introduced,...)
• The diagram of structures is idiosyncratic and of debatable relevance. You ask two mathematicians for such a diagram, you will possibly get two completely different pictures. (For me personally, I might put Banach manifolds in there.)
An approximation of of a good article would be in the style of, say, the "What is..." columns in the Notices. This is not it. Mct mht (talk) 02:05, 16 September 2012 (UTC)

Someone needs to do something with this article. It's quite bad now. Sławomir Biały (talk) 00:24, 19 November 2012 (UTC)

Now better, thanks to User:Bdmy. Boris Tsirelson (talk) 17:47, 28 April 2013 (UTC)
I'd second that. Huge improvement. Mct mht (talk) 04:16, 2 May 2013 (UTC)
Well, thanks, but there is still a long way to go! Bdmy (talk) 18:08, 28 April 2013 (UTC)

Wow. The article has improved hand over fist since my last visit, thanks to Bdmy. I agree that there is still potential for substantial improvement, but it is now a very respectable encyclopedia article on the topic. For what it's worth, some of the deficiencies at this point as I see it are the following.
(1) history of Banach spaces. I've found Dunford and Schwartz generally to be a good source for history of functional analysis. It may also be worth looking at Bourbaki.
(2) some introduction for the non-mathematician. Although we should probably suppose a certain amount of mathematical maturity of anyone reading the article, I don't think we should necessarily suppose that someone wanting to know what a Banach space is knows what a metric space or Cauchy sequence is. The lead in particular should be written with a general audience in mind, so that a non-mathematician will at least have a sense of what a Banach space is and what they're good for if he or she only were to read the lead.
(3) Applications. One of the chief reasons that Banach spaces are so widely studied is their ubiquitous role in applications. For instance, basically the entire edifice of differential equations rests on Banach spaces of various kinds. Sławomir Biały (talk) 12:04, 17 June 2013 (UTC)

## Duplicated list of classical Banach spaces

Thanks goes to editor Bdmy for a initiating a nice reorganization and cleanup of this article. Regarding cleanup, I noticed that the same (or nearly the same) large table of classical Banach spaces is present in both the Examples section of this article and in the List of Banach spaces article. We probably don't need this duplication. Would it be better to delete it from this article, or perhaps keep just a few examples? Or would it be better to fold back in the content from List of Banach spaces? --Mark viking (talk) 20:06, 28 April 2013 (UTC)

Thanks for your thanks! I believe that one table is enough, and then, it is probably better and more logical to have it in List of Banach spaces, with just a link from here to there, and only some examples in the text of this article. Bdmy (talk) 20:39, 28 April 2013 (UTC)
I think it should be kept for now, but ultimately it would be better to give a few examples in greater detail for illustration purposes with a link to the main list. Sławomir Biały (talk) 12:04, 17 June 2013 (UTC)

## about my recent "General cleaning"

I do not usually totally erase contributions that other made the effort to write, so I think I have to explain two suppressions I made :

• The definition of Banach space had an inline citation attached to it, although it is just the usual definition, as it can be found in the Wikipedia article Complete metric space. Also, the choice of the book seemed a bit strange to me.
• Math Reviews says it has registered over 600.000 author names, but "Willkomm, Anton" is unknown there. I don't think that "Willkomm, Anton (1976), Über die Darstellungstheorie topologischer Gruppen in nicht-archimedischen Banach-Räumen, Dissertation, Rheinisch-Wetfälische Technische Hochschule Aachen." is a notable reference for Wikipedia, and I removed it from "References".

I feel that I finished (most of) what I wanted to do in order to meet the remarks in this Talk pages by Mct mht and Sławomir Biały. Since English is not my mother tongue, it could be useful that someone with English expertise read and correct my awkwardness.

I may made further additions later, for example to (shortly) review some of Banach space theory since around 1980. This should take more time than just reorganize.

Bdmy (talk) 21:28, 1 May 2013 (UTC) Bdmy (talk) 21:37, 1 May 2013 (UTC) Bdmy (talk) 08:07, 2 May 2013 (UTC)

Approximation problem redirects to compact operators. Approximation property is probably a better target (although that page doesn't say what the problem is either, presumably about whether CAP the identity map can be approximated by finite rank operators?). Mct mht (talk) 20:23, 2 May 2013 (UTC)
Oops, should've read the intro. Approximation property still seems more appropriate, though. Mct mht (talk) 20:39, 2 May 2013 (UTC)
I changed the link. Also, I'll have to improve my vague statement about tensor products, Grothendieck and the approximation property. Bdmy (talk) 09:50, 3 May 2013 (UTC)

## Placement in the hierarchy of mathematical structures

I propose to remove the figure at Banach space#Placement in the hierarchy of mathematical structures. In my opinion, it is not very informative. But more than this, there is no particular reason to have it here rather than at "Hilbert space", "metric space", "topological space",... If nobody objects, I'll remove it in about a week. Bdmy (talk) 18:17, 4 June 2013 (UTC)

I like the illustration, but agree that doesn't really belong here. Perhaps someday an infobox of mathematical spaces will be created that accomplishes the same purpose of placing Banach spaces in a greater context. In the meantime, I have copied the illustration and caption over to Space (mathematics)#Types of spaces, which does have a nice comparative overview of mathematical spaces and could use the illustration. I have no objections to removal. --Mark viking (talk) 21:04, 4 June 2013 (UTC)

## Next cleaning

I plan to get rid of the "Further results" section, where I have moved far down, some time ago, a bunch of randomly picked results with no context. These results were part of the "long bulleted list" that appeared in this article around September 2012. If noone objects, I'll remove it in about a week.

I did it. Bdmy (talk) 12:30, 13 July 2013 (UTC)

Besides, I'd like to find some help for the writing of a friendly introduction to this article that is becoming awfully technical (partly my fault). I certainly agree with the remarks of User:Sławomir Biały at the end of the above section "Recent big revision", but I don't feel like doing all the work alone. Bdmy (talk) 09:08, 25 June 2013 (UTC)

Help didn't come! (yet) Bdmy (talk) 12:30, 13 July 2013 (UTC)

## What does weak completeness mean in this table?

Excuse me, in which sense for example ℓ1 is weakly complete? Eozhik (talk) 09:19, 26 July 2013 (UTC)

The table has been here for a long time, and also appears at List of Banach spaces. I think it comes from Dunford and Schwartz. Clearly the meaning is weakly sequentially complete (see in the text, at Banach space#Weak convergences of sequences). There are also a few errors in the table, where duals isomorphic to ℓ are marked as being ℓ1. I had the plan to clean this some day, but it is not my highest priority (I don't appreciate this table very much. I suggested to remove it, since it appears elsewhere, but some opposed and I respected their reasons). Bdmy (talk) 09:32, 26 July 2013 (UTC)
Ah, I see... However, somebody must put "sequentially" into this place of the table (and correct the links), because otherwise this will be a source of misunderstandings. Eozhik (talk) 12:25, 26 July 2013 (UTC)