Do subgradient inequalities hold for matrix convex functions?

Question

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

Answer 1

Some further research has pointed out the answer as affirmative (though I will leave the question unanswered for now, as I'm still perplexed by Ando's rank restriction, which now seems unnecessary).

Theorem V.3.3 of [1]. Suppose the matrix map $f$ is induced by a scalar function on $I\subset \mathbb{R}$ applied to its input's eigenvalues. Then $$ \partial f(X; \Delta)=f^{[1]}(X)\circ \Delta\,\,, $$ where $\circ$ is the Schur product in the eigenbasis of $X$ and $f^{[1]}$ is the first divided difference map with $f^{[1]}(X)=Uf^{[1]}(\Lambda)U^\dagger$ given a diagonalization of $X=U\Lambda U^\dagger$ and the definition of $f^{[1]}$ for diagonal matrices,

$$ f^{[1]}(\Lambda)_{ij}=f^{[1]}(\lambda_i,\lambda_j)=\begin{cases} \frac{f(\lambda_i)-f(\lambda_j)}{\lambda_i-\lambda_j}&\lambda_i\neq \lambda_j\\ f'(\lambda)&\lambda_i=\lambda_j \end{cases}\,\,. $$

Exercise V.3.15. If $f\in\mathcal{C}^1(I)$ then it is matrix convex on $I$ iff for all $A,B\succeq0$ with spectra in $I$ $$ f(A)-f(B)\succeq f^{[1]}(B)\circ (A-B)\,\,. $$

This confirms that we have an equivalent characterization of smooth matrix convex functions as in the scalar case.

[1] Matrix Analysis, Bhatia 1997, https://link.springer.com/book/10.1007/978-1-4612-0653-8