Talk:Nuclear operators between Banach spaces

Projective tensor product?
The article uses the term "projective tensor product" which is not defined either here or elsewhere on Wikipedia. Presumably, this is a completed tensor product so that should be clarified as well. kapil (talk) 03:06, 25 October 2017 (UTC)

is it a dual basis?
The article uses a basis {fn}n for space B and a basis {fn*}n for A*. Now, the fact that the symbol f is used for both bases makes me want to think that one is the dual basis of the other. However, given that they're in different spaces, this doesn't seem possible. If they are not meant to be dual bases, I would really prefer it if different symbols were used. I'd like to change it, but I'm not sure. -lethe talk [ +] 17:54, 7 March 2006 (UTC)
 * You are right, these should be called g or some such. I'm not sure what I was thinking. Maybe I wanted to avoid confusion with the g earlier in the article. Maybe I changed notation half way through, and got sloppy. linas 02:54, 8 March 2006 (UTC)
 * Actually, it was User:R.e.b. that introduced this confusion, circa October 2005, as a by-product of other changes. I fixed it, I believe. linas 16:51, 12 March 2006 (UTC)
 * Thank you, linas. -lethe talk [ +] 23:49, 12 March 2006 (UTC)

In the formula $$\mathcal{L} = \sum_{n=1}^N \rho_n \langle f_n, \cdot \rangle g_n$$

I'm supposing $$\langle f_n, \cdot \rangle$$ means the adjoint of $$f_n$$?

Is there a clearer way of writing this?

198.228.228.164 (talk) 15:55, 6 February 2014 (UTC) Collin237
 * The use of dot for a missing argument is respectable notation, in my view. I wouldn't use a star for the canonical isometry of a Hilbert space with its dual, even though it is one of the two definitions of adjoint when viewed correctly (and is conjugate linear). 84.226.185.221 (talk) 15:05, 17 September 2015 (UTC)

preamble incoherent with content
The preamble says "this article concentrates on the general case, for the [ Hilbert case see ] trace class operator", however the article body considers the "general"(?) (i.e. non Hilbert) case only in the last subsection. IMHO this is either content duplication and incoherent with the specification at the beginning of the article, or, maybe more reasonably, the last phrase of the intro should be deleted and this page more seen as an "overview" page with two see the main article ... crossrefs. &mdash; MFH:Talk 21:54, 4 September 2007 (UTC)

Nuclear operator of order q between Banach spaces
The formula $$\inf \left\{ p\le 1 : \sum_n |\rho_n|^p < \infty \right\} = q$$ seems to be incorrect because the article claims that an operator is Hilbert-Schmidt iff $$q=2$$. Perhaps it should be $$\inf \left\{ p\ge 1 : \sum_n |\rho_n|^p < \infty \right\} \le q$$? --Jaan Vajakas (talk) 17:34, 17 October 2012 (UTC)

Although my specialty isn't analysis, the formula is obviously wrong. I changed it to $$ p\ge 1$$. The order of a nuclear operator is greater than or equal to 1. — Preceding unsigned comment added by 2001:DA8:D800:107:E1BB:43D5:78EA:8E71 (talk) 07:34, 1 January 2013 (UTC)
 * It can't be right to use the inf. For then an operator that is nuclear of order q is NOT nuclear for any order r > q, so they don't form a vector space. Also, Hilbert-Schmidt operators would not actually require that the sum of the squares converge, only that the sum of the p_th powers converge for each p> 2. Similar problem for trace class (i.e. nuclear) -- the definition would not coincide with nuclear of order 1. 84.226.185.221 (talk) 14:54, 17 September 2015 (UTC)

Very, very, very sloppy article
Nowhere in this article is there anything presented as the definition of nuclear operators.

Plus, the typography is sloppy as well.

This article needs a lot of work. It is a very bad idea to have a math article about "nuclear operators" without a section that defines "nuclear operators".50.205.142.50 (talk) 22:39, 22 May 2020 (UTC)
 * + Why is is called "..between Banach spaces" when the article is mostly about Hilbert spaces!--Tensorproduct (talk) 21:10, 28 February 2023 (UTC)