Let $\rho,\sigma$ are positive semidefinite operator, let $f: (0;+\infty) \rightarrow \mathbb{R}$ is operator convex function, in A. Lesniewski and M. B. Ruskai. Monotone Riemannian metrics and relative entropy on noncommutative probability spaces. J. Math. Phys., 40:5702–5724, 1999. it can be written as
$$f(x) = f(1) + f'(1)(x-1) + c(x-1)^2 + \int_{[0;+\infty]} \dfrac{(x-1)^2}{x+s} d \lambda(s), \quad x \in (0;+\infty), c \ge 0$$
and a positive measure $\lambda$ on $[0;+\infty)$ satisfying $\int_{[0;+\infty)} (1+s)^{-1} d \lambda(s) < +\infty$.
I attempt to proof $f(\sigma^{-1/2} \rho \sigma^{-1/2}) = f(\sigma^{-1} \rho)$ using trace function. $\text{Tr} (\sigma^{-1/2} \rho \sigma^{-1/2}) = \text{Tr} (\sigma^{-1} \rho)$ is true. So, when we work with the operator convex function $f$, does $\text{Tr} f(\sigma^{-1/2} \rho \sigma^{-1/2}) = \text{Tr} f(\sigma^{-1} \rho)$ still true?