Talk:M. Riesz extension theorem

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Sup
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 78.53.96.139 (talk) 14:22, 13 April 2012 (UTC)

Theorem
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 $$ y \in E $$ there exist $$ x,x' \in F $$ such that $$x-y \in K$$ and $$y-x' \in K.$$" seems redundant, and equivalent to $$E=K+F$$
 * 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 $$\R^2$$ and $$K=\{(u,v);|v|\le u\}$$, but for $$y=(0,1)$$, it is not true that "Every point in $$E\setminus F$$ is a positive linear multiple of either $$x-y$$ or $$y-x'$$ for some $$x,x' \in K$$"
 * 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 $$F=\R^+\times\{0\}$$, which is not a linear subspace
 * My rewriting of "the additional assumption" is :
 * $$(\forall y\in E,\exists x\in F, x-y\in K)\Leftrightarrow E\subset F-K$$ and
 * $$(\forall y\in E,\exists x'\in F,y-x'\in K)\Leftrightarrow E\subset F+K$$, and (using only that E and F are linear spaces, of course not K)
 * $$E\subset F-K\Leftrightarrow -E\subset -F+K\Leftrightarrow E\subset F+K$$
 * I really think that E=K+F is a simpler formulation than "for every $$ y \in E $$ there exist $$ x,x' \in F $$ such that $$x-y \in K$$ and $$y-x' \in K$$", 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)
 * I am sorry -- I answered in a hurry, you are right indeed. I will try to clean it up a bit. Thanks! Sasha (talk) 21:51, 17 September 2011 (UTC)