Talk:Sequence space

Question
Is a sequence space a space composed of sequences or is it a space composed of the functions that form the sequences from N --> C? It's not really clear. And as I'm just learning this topic myself I'm not ready to jump in and change anything. futurebird (talk) 13:36, 28 November 2007 (UTC)
 * When you get down to it, the usual definition of sequence is a function from the natural numbers. In any case, it makes no difference whether you think of the space as a collection of countably infinite ordered sets or as a collection of functions from N. nadav (talk) 14:02, 28 November 2007 (UTC)
 * Okay, that makes sense, can you look at my recent edit to check that I didn't make matters worse? futurebird (talk) 14:04, 28 November 2007 (UTC)
 * I think it made it more confusing. The article already says directly that a sequence space is some subspace of the space of functions from N to C. nadav (talk) 14:09, 28 November 2007 (UTC)

Okay, thanks. I still think the wording could use some work, though. It seems to come to the main definition in as round about way. I don't know maybe you can look it since you seem to know a bit more about the topic. futurebird (talk) 14:24, 28 November 2007 (UTC)

Scalars -> elements of K
"Denote by K^N the set of all sequences of scalars...": IMO this is imprecise as a notion of "scalars" only exists in the context of a vector space which the definition introduces only after saying what a sequence is. It would be more precise IMO to say "Denote by K^N the set of all sequences of elements of K..." — Preceding unsigned comment added by 189.70.56.183 (talk) 13:49, 26 August 2012 (UTC)

Error in proof of isometry
I believe there's an error here:

In fact, taking y to be the element of ℓp with
 * $$y_n = \begin{cases}0&\rm{if}\ x_n=0\\

x_n^{-1}|x_n|^q &\rm{if}\ x_n\not=0 \end{cases}$$ gives Lx(y) = ||x||q

because the result is actually the q-norm of x raised to the q. (But the norm of y is that as well so dividing by that would work out). Excuse my lazy correction if it is a correction; I am inexperienced both in the math and in wikipedia.

65.36.74.84 (talk) 01:03, 27 September 2012 (UTC)

Other sequence spaces
I doubt there is an isometry between bounded series and bounded sequences : consider the constant sequence 1, 1, 1 ... the serie is obviously divergent

What the author may have wanted to say is to define a sequence as the partial sums of a serie, but then one should not write everything with x_n — Preceding unsigned comment added by Noix07 (talk • contribs) 10:59, 10 May 2014 (UTC)


 * The existence of an isometry does not mean that they are the same space. These spaces are isometric: see Dunford and Schwartz, IV.13.13.   Sławomir Biały  (talk) 11:21, 10 May 2014 (UTC)

I don't have access to that book at the moment, but it is written "isometric isomorphism". I will probably change it in a few days if no protest, into something like $$ y_n = \sum_{i=1}^n x_i $$ — Preceding unsigned comment added by 134.100.125.222 (talk) 19:47, 12 May 2014 (UTC)


 * You seem to be confused on the notion of isometric isomorphism. The statement is not that the identity map is an isomorphism of one space to the other.  They are different vector spaces.  The statement is that there exists an isometric isomorphism from one space to the other.  In fact, that isomorphism is precisely the one that is already described in the article, and it is the same as the one you just wrote down.   Sławomir Biały  (talk) 20:26, 12 May 2014 (UTC)

I think I got it. The problem was not isometry, it was the very definition of bs and its norm. — Preceding unsigned comment added by 85.183.206.130 (talk) 08:33, 13 May 2014 (UTC)

"ℓ∞" listed at Redirects for discussion
The redirect [//en.wikipedia.org/w/index.php?title=%E2%84%93%E2%88%9E&redirect=no ℓ∞] has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at  until a consensus is reached. 1234qwer1234qwer4 19:09, 12 April 2023 (UTC)