equality in noncommutative Hölder inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:25:40Z http://mathoverflow.net/feeds/question/66716 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66716/equality-in-noncommutative-holder-inequality equality in noncommutative Hölder inequality BigBill 2011-06-02T07:32:21Z 2012-05-29T15:19:30Z <p>Let $1\leq p,q,r\leq \infty$ such that $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$. Let $S_p$ denote the Schatten space. For any $x\in S_p$ and any $y\in S_q$ we have $$||xy||_{S_r} \leq ||x||_{S_p}||y||_{S_q}$$ (noncommutative Hölder's inequality).</p> <p>Does it exists necessary and sufficient conditions on $x,y$ in order to have an equality in this inequality?</p> <p>More generally, I ask the same question replacing $S_p$ by the noncommutative $L_p$-space $L_p(M)$ associated with a semifinite von Neumann algebra $M$ equipped with a normal semifinite faithful trace $\tau$.</p> http://mathoverflow.net/questions/66716/equality-in-noncommutative-holder-inequality/66762#66762 Answer by Martin Argerami for equality in noncommutative Hölder inequality Martin Argerami 2011-06-02T18:30:55Z 2011-06-02T20:53:34Z <p>The necessary and sufficient condition is that $|x|^p$ and $|y^*|^q$ are proportional. This can be deduced from <a href="http://archive.numdam.org/article/BSMF_1953__81__9_0.pdf" rel="nofollow">Dixmier's paper</a> (although it is not clearly stated that way there; it is based on Proposition 8). It probably also appears in more modern treatments (Nelson, Terp, Haagerup, Hiai, Kosaki, etc.) but I don't have the sources here to check that. </p>