This is my first post, so apologies if this is a bad post.
I'm reading "Quantum Entropy and its use' by M. Ohya and D. Petz. Theorem 5.3 states
Let $M_1$ and $M_2$ be von Neumann algebras with positive normal functionals $\phi_1, \omega_1$ and $\phi_2, \omega_2$ respectively. Let $\alpha:M_1 \to M_2$ be a [unital] Schwarz map [as in, $\alpha(x^*x) \ge \alpha(x)^* \alpha(x)$ for all $x\in M_1$] such that $\phi_2 \circ \alpha \le \phi_1$ and $\omega_2 \circ \alpha \le \omega_1$. Then $S(\omega_1, \phi_1 )\le S(\omega_2, \phi_2)$ [$S$ is the relative entropy]
brackets mine.
This seems to yield the following contradiction:
Say $M_1 = M_2$ and $\alpha = id$ the identity map. Then $\alpha$ is a unital Schwarz map and the following holds for $0<\epsilon<1$: $\phi \circ \alpha \le \epsilon^{-1} \phi$ and $\omega \circ \alpha \le \epsilon^{-1} \omega,$ for any two positive normal functionals $\phi$ and $\omega$. Then the theorem as stated would imply $S(\omega, \phi) \le \epsilon S(\omega, \phi)$, hence $S(\omega, \phi) \equiv 0$, which is a contradiction.
Is there an extra assumption that I'm missing in the statement of this theorem? There is no errata for this book (that's publicly available, at least) so any help would be greatly appreciated.