Suppose $f$ is a matrix convex function over symmetric, positive semidefinite matrices with spectra in some interval $I$ [1]. That is, for $A,B\succeq 0$ with spectra in $I$, and any $\theta\in[0,1]$,

$$ \theta f(A)+(1-\theta)f(B)\preceq f(\theta A + (1-\theta) B)\,\,. $$

$f(X)=X^{-1/2}$ is one such example.

If the directional gradient at $X$ towards $\Delta$, $\partial f(X ; \Delta )$, exists such that $X+\Delta$ is in the domain of $f$ for psd $\Delta$, does a subgradient-like inequality such as

$$ f(B)+\partial f(B; A-B)\preceq f(A) $$

follow? This is proven in [2] in the case of $\mathrm{rank}\,(A-B)=1$ and $f\in\mathcal{C}^1$ (equation 3.4), but I don't understand why the rank constraint is essential.

For the example, $f(X)=X^{-1/2}$ over $(0,\infty)$, we'd have $\partial f(X;\Delta)=-X^{-1/2}\left[(X\oplus X)^{-1}\Delta\right]X^{-1/2}$ where $(X\oplus X)^{-1}\Delta$ is the solution $Y$ to the continuous Lyapunov equation $XY+YX=\Delta$, which would give rise to an explicit perturbative upper bound on $f(X)$.

[1]: Concavity of Certain Maps on Positive Definite Matrices and Applications to Hadamard Products, Ando 1979, https://www.sciencedirect.com/science/article/pii/0024379579901794

[2]: Trace-Inequalities and Matrix-Convex Functions, Ando 2010, https://fixedpointtheoryandapplications.springeropen.com/articles/10.1155/2010/241908