To Show $f(A,B)=B^{T}A^{-1}B$ is K-convex on $\mathbb{S}_{++}^n \times \mathbb{R}^{n \times n}$

Question

Here, $K = \mathbb{S}_+^n$, a set collecting all PSD matrix. And $\mathbb{S}_{++}^n$ is a set colleting all PD matrix.

I've tried to start from $f(X) = X^{-1}$ is also K-convex on $\mathbb{S}_{++}^n$. And applying the property of norm formed by inner product. However, I can't derive the desired K-convexity inequality, i.e. $z^Tf(\theta(A,B)+(1-\theta)(X,Y))z \leq z^T(\theta f(A,B)+(1-\theta)f(X,Y))z$.

And I've tried second-order condition on $g(t) = f((A,B)+t(X,Y))$. There comes out too much term; as a result, I can't show $g''(t)\geq 0$

Now I'm tring to prove this question from $\textbf{epi}_K f$ and show $\textbf{epi}_K f$ is intersection of closed halfspaces.

Answer 1

Schur complement: Consider a mtrxi $X \in \mathbb{S}^n_+$ partitioned as $\begin{bmatrix} A & B\\ B^T & C \end{bmatrix}$ where $A \in \mathbb{S}^k_+$. If $\det(A)\neq 0$, the matrix $S = C-B^TA^{-1}B$ is called the ''Schur complement'' of $A$ in $X$.

And we have Theorem: If $A$ is PD, then $X$ is PSD iff $S$ is PSD.

Consider the ${\bf{epi}}_K\;f = \{(A,B,t):\; (A,B) \in \mbox{dom}, f,\; t \in \mathbb{R^{n\times n}}, \;z^T(t - B^T A^{-1}B)z \geq 0 \} = \{(A,B,t):\; \begin{bmatrix} A & B \\ B^T & t \end{bmatrix} \mbox{is PSD}, A\;\mbox{is PD}\}$

by applying the Theorem of Schur complement.

As a result, ${\bf epi}_K f $ is convex, which implies f is K-convex.

Answer 2

By choosing $B,Y$, the inequality wlog becomes: $(\theta u + (1-\theta)v)^T (\theta A+ (1-\theta)X)^{-1} (\theta u + (1-\theta)v) \leq \theta u^TA^{-1} u + (1-\theta) v^TX^{-1} v$
Special case: $AX=XA$:
If $AX=XA$ then $A,B$ are simultaneously diagonalizable and the problem reduces to $A,X$ being diagonal matrices. The inequality becomes: $(\theta u_i + (1-\theta)v_i)^2 \frac{1}{\theta a_{ii} + (1-\theta)x_{ii}} \leq \theta u_i^2 \frac{1}{a_{ii}} + (1-\theta) v_i^2 \frac{1}{x_{ii}}.$
This follows from convexity of $f(x,y) = \frac{x^2}{y}$ on $\{(x,y): y>0\}$.