I was wondering if the following statement is true:
Let $\psi,\phi$ two positive normal functionals on a von Neumann algebra such that $\psi(A)\leq \phi(A)$ for every possitive element $A$. Let $H$ be a operator $0\leq H\leq 1$ such that $\phi(HAH)\leq\psi(A)$ for every positive element $A$. Then, if there exists a positive $B$ such that $\phi(HBH)<\psi(B)$, there exists a subprojection $P$ of a spectral projection $E^H_{(\lambda,\lambda+\epsilon)}$ of H such that $\lambda^2\phi(P)<\psi(P)$.
My first attemp to prove it was using the operator $f(H)=\sum_{i=0}^N x_i E^H_{x_{i+1}-x_i}$, $\{x_i\}_i$ a partition of $[0,1]$ with $x_0=0$ and $x_N=1$, or the spectral decomposition of the operator $H$, to construct a contradicition, but the crossed terms in the expansion of $\phi\left(f(H)Bf(H)\right)$ create problems.
A second attemp was using a known result that there exists $C>0$ such that $\phi(HAH)<\psi(A)$ for every $A\leq C$.
I would appreciate if someone give me a tip or a counter example.