Talk:Normed vector space

Explicit about the underlying field

I think this article should be explicit about the underlying field of the vector space. The norm places additional constraints for the field. A norm space is a certain kind of a topological vector space, and topological vector spaces are defined over topological fields. The norm outputs a real number, and somehow the norm has to be consistent with scaling by a scalar. I'm not sure what the most general route is, anyone? I'm thinking there should be something like normed fields, of which the reals and complex numbers are examples.

--Kaba3 (talk) 21:59, 21 March 2013 (UTC)[reply]

Merging

The articles Normed vector space and Linear Algebra/Normed Vector Space need to be merged. (I wasn't aware the latter existed when I wrote the former.) The final article should be placed at Normed vector space, in accordance with the usual naming conventions (and also because it's impossible to add a /Talk to Linear Algebra/Normed Vector Space - the name is too long, apparently).
Zundark, 2001-08-13

As no-one else seemed interested, I did it myself. The other subpages of Linear Algebra should also be moved in due course.
Zundark, 2001-08-16

Complete ??

Can someone give one more word on the reason why normed vector space is not complete? Since complete normed vector space is Banach space. Thanks. Jackzhp 23:58, 28 October 2006 (UTC)[reply]

Diagram

It is requested that a mathematical diagram or diagrams be included in this article to improve its quality. Specific illustrations, plots or diagrams can be requested at the Graphic Lab.
For more information, refer to discussion on this page and/or the listing at Wikipedia:Requested images.

I'm thinking of a diagram with a bunch of vectors radiating from the origin with some depiction of a ruler measuring lengths of vectors and distances from tip to tip measuring. —Ben FrantzDale 21:04, 6 May 2007 (UTC)[reply]

I am not sure that this is a great idea. Most work on Normed vector spaces is in infinite dimensional vector spaces. Andrew Kepert 09:44, 7 May 2007 (UTC)[reply]

(replying to self...)

Actually, I think the thing that would improve the page is an explicit "Examples" section as appears on similar pages, in particular including some non-complete and non-separated examples. I propose this list should be after the "Definition" section, and be a fairly short list along the lines:

Any example of a Banach space
The space c₀₀ of all (Real or Complex) sequences of finite support with supremum norm ||a||_∞ = max_i | a(i) |. This space is not complete, since there are Cauchy sequences which do not converge. For instance the sequence a_n where a_n(i) = 1/i ; if i ≤ n and a_n(i) = 0 otherwise.
The finite rank operators on a Hilbert space, with operator norm
The L^p spaces as described below

Some of these could actually be illustrated!

The only other thought I had on a diagram to illustrate this is that feeling for how normed spaces work is via the geometric characterisations of Hahn-Banach, etc, with unit balls, hyperplanes, etc.

Andrew Kepert 10:12, 7 May 2007 (UTC)[reply]

Stupid question?

Cite:

All norms on a finite-dimensional vector space are equivalent from a topological point as they induce the same topology (although the resulting metric spaces need not be the same)

Does it mean that the metric spaces are not homeomorphic?! Or simply they are not isomorphic? The latter, I suppose. Urzyfka —Preceding signed but undated comment was added at 15:44, 23 September 2007 (UTC)[reply]

Seminorm /semi norm consistency

Both forms, "seminorm" and "semi norm" are used in the article. And "semi-normed" also appears. For consistency a single form should be chosen.

The norm article uses "seminorm", so I suggest that.

--84.9.73.5 (talk) 12:24, 1 January 2008 (UTC)[reply]

Done. --Hans Adler (talk) 15:46, 17 April 2009 (UTC)[reply]

Continuity

I studied mathematics to second-year undergraduate level only, and that was over 30 years ago and I never did any topology, but I still think there is an error in the sentence "This topology is precisely the weakest topology that makes ||·|| continuous." The weakest topology that makes ||·|| continuous would have as a basis the family consisting of all sets of the form {x: a<||x||<b}, and, unless I am very much mistaken, the intended topology would also include balls around points other than the origin. Surely it is the *distance* function that needs to be continuous (in both arguments);

Rdbenham (talk) 21:29, 9 November 2008 (UTC)[reply]

the statement is essentially correct. in the context of the article, the relevant topologies are the ones that make the underlying vector space a topological vector space. and any such topology is specified by a neighborhood basis around the origin. Mct mht (talk) 21:43, 9 November 2008 (UTC)[reply]

In my understanding, the weakest topology that makes ||·|| continuous would be the initial topology induced by ||·||. Take for instance R with the usual absolute value |·| as norm. Of course, ]-1,0[ is open with respect to the usual topology on R, but it's not open with respect to the initial topology induced by |·|.

It is not clear from the context that we also want to preserve the continuity of the vector space operations. On the contrary, the next sentence in the article states that this is the case for the natural topology. Redfrettchen (talk) 19:47, 11 August 2010 (UTC)[reply]

Introduction

In the introduction it is said that "Multiplying a vector by a positive number changes its length without changing its direction." However, the following formula states that |α||x| = |αx| for all α. If it is assumed that the scalar α is a real number, then clearly this equation is not equal to the verbal statement, since it is possible that α = 0 which is not a positive number. I don't know which statement needs fixing, so someone else will have to do it. 88.115.70.173 (talk) 21:28, 14 October 2010 (UTC)[reply]

Fixed. Boris Tsirelson (talk) 11:45, 15 October 2010 (UTC)[reply]