I have been trying to prove the following fact in elementary operator theory without success:
For any two positive operators $0 \leq R \leq S$, we have $\Vert R \Vert_2 \leq \Vert S \Vert_2$, where $\Vert S \Vert_2^2 = Tr(S^* S) = \sum_{i=1}^\infty\Vert S e_i \Vert^2$ for some orthonormal basis of a separable Hilbert Space (if the sum converges). Even more generally, could this be shown for Schattern $p-$norms?
It seems naive, but I couldn't find the right tool to deal with this, any hints are appreciated.
Best Answer
Inside the trace, things commute (cyclically, so in general one needs to be careful, but with only two operators there's no issue).
So (commuting things inside the trace in the first equality in the second line) \begin{align} \|S\|_2^2-\|R\|_2^2 &=\operatorname{Tr}(S^2)-\operatorname{Tr}(R^2) =\operatorname{Tr}(S^2-R^2)\\[0.3cm] &=\operatorname{Tr}((S-R)(S+R)) =\operatorname{Tr}((S-R)^{1/2}(S+R)(S-R)^{1/2})\geq0. \end{align}