A non-commutative Radon-Nikodym derivative. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:44:26Z http://mathoverflow.net/feeds/question/23154 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23154/a-non-commutative-radon-nikodym-derivative A non-commutative Radon-Nikodym derivative. Andre 2010-05-01T01:57:37Z 2010-05-01T08:20:02Z <p>In <a href="http://www.ams.org/journals/bull/1965-71-01/S0002-9904-1965-11265-4/S0002-9904-1965-11265-4.pdf" rel="nofollow">this</a> classic paper, Sakai proves the following Radon-Nikodym theorem:</p> <blockquote> <p>Let $M$ be a von Neumann algebra, and let $\phi$ and $\psi$ be two normal positive linear functionals on $M$. If $\psi \leq \phi$, then there is a positive operator $t_0\in M$ such that $0 \leq t_0 \leq 1$, and $\psi(x) = \phi(t_0 x t_0)$ for all $x \in M$.</p> </blockquote> <p>The paper provides no uniqueness result. One would naively expect that any two such operators $t_0$ and $t_1$ would satisfy $\phi((t_1-t_0)^2)=0$. I can find no such statement in the literature. Is this true?</p> <p>Please note that $\phi$ is not assumed to be faithful.</p> http://mathoverflow.net/questions/23154/a-non-commutative-radon-nikodym-derivative/23158#23158 Answer by Jon for A non-commutative Radon-Nikodym derivative. Jon 2010-05-01T04:12:33Z 2010-05-01T04:12:33Z <p>Does this paper have any relevance? </p> <p><a href="http://arxiv.org/PS_cache/math-ph/pdf/0303/0303056v3.pdf" rel="nofollow">http://arxiv.org/PS_cache/math-ph/pdf/0303/0303056v3.pdf</a></p> http://mathoverflow.net/questions/23154/a-non-commutative-radon-nikodym-derivative/23161#23161 Answer by Dmitri Pavlov for A non-commutative Radon-Nikodym derivative. Dmitri Pavlov 2010-05-01T08:20:02Z 2010-05-01T08:20:02Z <p>Such t_0 is unique if its support is at most p, where p is the support of ϕ. Note that we can replace t_0 by pt_0p and the support of pt_0p is at most p.</p> <p>Without this additional condition t_0 is highly non-unique, because we can replace t_0 by t_0 + q, where q is an arbitrary self-adjoint element with support at most 1-p such that t_0 + q ≥ 0. Using simple algebraic manipulations one can show that all solutions can be obtained in this way.</p> <p>See Lemma 15.4 (page 104) in Takesaki's book “Tomita's theory of modular Hilbert algebras and its applications”. Electronic version: <a href="http://gen.lib.rus.ec/get?md5=ACC2A399A5C65C5CB2CCEE7CBEB3FAC3" rel="nofollow">http://gen.lib.rus.ec/get?md5=ACC2A399A5C65C5CB2CCEE7CBEB3FAC3</a> [Note that Takesaki implicitly assumes that φ_0 is faithful, hence you need to introduce an additional condition on the support of h.]</p> <blockquote> <p>One would naively expect that any two such operators t_0 and t_1 would satisfy ϕ((t_1−t_0)^2)=0. </p> </blockquote> <p>This is a trivial corollary of the above statement characterizing all possible solutions.</p>