Talk:Nuclear space

From Wikipedia, the free encyclopedia

[edit] Hilbert space

The article states:

There are no Banach spaces that are nuclear, except for the finite dimensional ones.

I'm naive, but isn't every Hilbert space going to be nuclear? Err, well, at least trace class? Or am I confused? And since Hilbert spaces are Banach spaces, the above statement sounds false to me. I'm confused ... linas 23:07, 24 October 2005 (UTC)

No, infinite dimensional Hilbert spaces are not nuclear. The identity operator of such a space is not trace class (or even compact). R.e.b. 01:09, 25 October 2005 (UTC)

Err, OK, thanks. I guess I was confused; I'll have to study a bit. linas 15:00, 25 October 2005 (UTC)


Another confusing point that I ran into (not an expert in the field, btw):

"If p is a seminorm on V, we write Vp for the Banach space given by completing V using the seminorm p."

Hence I conclude that the completion of V with respect to a norm or seminorm is a Banach space.

"There are no Banach spaces that are nuclear."

So the completion of V is not nuclear... so far so good.

"The completion of a nuclear space is nuclear (and in fact a space is nuclear if and only if its completion is nuclear)."

... right. Can anyone help me out here? Are there two meanings of completion? Moocowpong1 00:31, 17 September 2007 (UTC)

I may have an answer to my own question... is the completion in the third quote referring to this meaning, while the first two are the completion with respect to a seminorm, not the topology of the space? Moocowpong1 06:49, 27 September 2007 (UTC)

[edit] Banach spaces with seminorms?

The article states: If p is a seminorm on V, we write Vp for the Banach space given by completing V using the norm p. Now hold up, p is a seminorm, not a norm, so how on Earth is it going to make a Banach space? -lethe talk + 19:17, 7 March 2006 (UTC)

The process of completing a space automatically kills off the norm 0 vectors, so a seminorm on a space is (or more precisely induces) a norm on the competion. (Anyone confused by this is allowed to kill the norm 0 vectors before taking a completion.) R.e.b. 19:54, 2 April 2006 (UTC)

[edit] Link from "nuclear" ?

It would be useful to have a link from the disambiguation page of "nuclear" to this page (and to the page on nuclear operators). Sorry, but I myself don't know how to do that...