Talk:Nuclear operators between Banach spaces

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is it a dual basis?

The article uses a basis {f_n}_n for space B and a basis {f_n*}_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)[reply]

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)[reply]

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)[reply]

Thank you, linas. -lethe ^{talk +} 23:49, 12 March 2006 (UTC)[reply]

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. — MFH:Talk 21:54, 4 September 2007 (UTC)[reply]

Nuclear operator of order q between Banach spaces

The formula $\inf \left\{p\leq 1:\sum _{n}|\rho _{n}|^{p}<\infty \right\}=q$ seems to be incorrect because the article claims that an operator is Hilbert-Schmidt iff $q=2$ . Perhaps it should be $\inf \left\{p\geq 1:\sum _{n}|\rho _{n}|^{p}<\infty \right\}\leq q$ ? --Jaan Vajakas (talk) 17:34, 17 October 2012 (UTC)[reply]

Although my specialty isn't analysis, the formula is obviously wrong. I changed it to $p\geq 1$ . 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)[reply]