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I am currently working with a class of functions, where every function looks like

$f(x)=v^T(xA+(1-x)B)^{-1}v,$

where $v\in\mathbb{R}^n$ is an arbitrary vector, $A,B \in \mathbb{R}^{n\times n}$ are positive definite matrices and $x\in(0,1)$. Is there a straight forward way to prove, that $f(x)$ is convex on $(0,1)$?

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Proving that $f$ is a convex function of $x$ is the same as proving that $g(A) := v^T A^{-1}v$ is a convex function when defined on the domain of positive definite matrices. Fix positive definite matrices $A$ and $B$ and $\lambda\in [0,1]$. For any positive definite $C$ the Schur complement condition shows that \[ \begin{bmatrix} s & v^T \\ v & C\end{bmatrix} \] is positive semidefinite if and only if $s \geq v^TC^{-1}v$. The set of positive semidefinite matrices is convex, so the matrix \[ \lambda\begin{bmatrix}v^TA^{-1}v & v^T \\ v & A\end{bmatrix}+(1-\lambda)\begin{bmatrix}v^TB^{-1}v & v^T \\ v & B\end{bmatrix} = \begin{bmatrix}\lambda v^TA^{-1}v+(1-\lambda)v^TB^{-1}v & v^T \\ v & \lambda A + (1-\lambda)B\end{bmatrix} \] is positive semidefinite. The definition of $g$ and the Schur complement condition applied to the matrix on the right combine to give: \[ \lambda g(A) + (1-\lambda)g(B) = \lambda v^TA^{-1}v+(1-\lambda)v^TB^{-1}v \geq v^T(\lambda A + (1-\lambda)B)^{-1}v = g(\lambda A + (1-\lambda)B), \] so the function $g$ is convex.

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Fact If $X, Y$ are positive definite then $tX^{-1}+(1-t)Y^{-1}-(tX+(1-t)Y)^{-1}$ is positive semidefinite for $t\in (0,1)$. $X^{-1}$ is an operator convex function over the cone of positive definite matrices.

Now for any $\lambda\in(0,1)$, $f(\lambda x+(1-\lambda )y)=v^t\{[\lambda x+(1-\lambda )y]A+[1-\lambda x-(1-\lambda )y]B\}^{-1}v=v^t\{\lambda [xA+(1-x)B]+(1-\lambda)[yA+(1-y)B]\}^{-1}v\le \lambda f(x)+(1-\lambda)f(y)$.

Done.

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