Talk:M. Riesz extension theorem
|WikiProject Mathematics||(Rated Start-class, Low-importance)|
In the proof, why must the sup be finite?
- You first choose x' such that x'-y belongs to K. This is possible by assumption. Now whenever y-x is in K, then so is x'-x, and so the sup is smaller than \phi(x'). — Preceding unsigned comment added by 126.96.36.199 (talk) 14:22, 13 April 2012 (UTC)
The theorem as stated in this article is wrong. A counterexample can already be found in 2-space, taking K to be the upper halfplane with the negative x-axis removed. If F is the x-axis, then the positive functional φ=X can not be extended to a positive functional on the plane.
There are also counterexamples with closed convex cones, starting in dimension three.
Thanks for pointing this out. There are probably a couple of research papers out there, which are just citing Wikipedia... —Preceding unsigned comment added by Andreas thom (talk • contribs) 18:56, 2 March 2010 (UTC)
- "If F is the closed positive x-axis" -> "If F is the x-axis"
- correct as stated in the article. x is not a positive functional on the whole axis, this is exactly the point of the example.
- "the additional assumption that for every there exist such that and " seems redundant, and equivalent to
- I am not sure I understand your comment (K is not a linear space). I guess one side is sufficient, but is this a reason for an expert tag?
- It is fullfilled for F=the x-axis in and , but for , it is not true that "Every point in is a positive linear multiple of either or for some "
- At first sight (very quick and superficial), I was not able to locate this theorem in "M.Riesz, Sur le problème des moments". Can you please give a page number ?
- Anne Bauval (talk) 09:59, 17 September 2011 (UTC)
- Try the third paper, page 2.
- Sasha (talk) 16:24, 17 September 2011 (UTC)
- "If F is the closed positive x-axis", to me (and probably other readers), means , which is not a linear subspace
- My rewriting of "the additional assumption" is :
- , and (using only that E and F are linear spaces, of course not K)
- I really think that E=K+F is a simpler formulation than "for every there exist such that and ", but I admit it is a matter of taste.
- As indicated in the comment of my edit here, the main reason for the expert tag was the flaw in the proof (see counterexample above).
- Anne Bauval (talk) 20:45, 17 September 2011 (UTC)