Talk:Banach space

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[edit] 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

[edit] 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 \|A\| = \sup_{x \in B} \|Ax\|, where B is the unit ball in V, so B is the set of all x ∈ V with ||x|| ≤ 1. -- Jitse Niesen 15:10, 28 Dec 2004 (UTC)

[edit] 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 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 Ap(D) ? linas 14:05, 19 September 2005 (UTC)

[edit] 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

[a,b] in the real line, using the Dirac measure

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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.