Joint Convexity Proof

Question

Let $x$ be $n \times 1$ vector and $Y$ be $n \times n$ matrix. Prove that $f(x,Y) = x'Y^{-1}x$ is jointly convex in $x$ and $Y$ when $Y \succ 0$.

Answer 1

Are you leaving out some assumptions? What if $n=1$ and $Y$ is negative?

EDIT: With the new assumptions, we may consider a line segment through $(x,Y)$ space: $x = x_0 + t x_1$, $Y = Y_0 + t Y_1$, where $Y_0$ is positive definite and $Y_1$ is symmetric. It is enough to show that $\left.\dfrac{d^2}{dt^2} (x' Y^{-1} x)\right|_{t=0} \ge 0$. Now if $Z = Y_0^{-1/2} Y_1 Y_0^{-1/2}$, $w_0 = Y_0^{-1/2} x_0$ and $w_1 = Y_0^{-1/2} x_1$, $$Y^{-1} = (Y_0^{1/2} (I + t Z) Y_0^{1/2})^{-1} = Y_0^{-1/2} (I - t Z + t^2 Z^2 + \ldots) Y_0^{-1/2}$$
so $$ x' Y^{-1} x = (w_0 + t w_1)' (I - t Z + t^2 Z^2 + \ldots) (w_0 + t w_1)$$ and $$\eqalign{\left.\dfrac{1}{2} \dfrac{d^2}{dt^2} (x' Y^{-1} x)\right|_{t=0} &= w_0' Z^2 w_0 - w_1' Z w_0 - w_0' Z w_1 + w_1' Z^2 w_1 \cr &= (Z w_0 - w_1)'(Z w_0 - w_1) \ge 0}$$ Thus it is indeed jointly convex.

Answer 2

The simplest way to prove this is to take advantage of the fact that a function is convex if and only if its epigraph is convex. In this case, the epigraph is $$\left\{(x,Y,z)\in\mathbb{R}^n\times\mathbb{R}^{n\times n}\times\mathbb{R}\,|\,Y\succ 0,~x^TY^{-1}x\leq z\right\}$$ But consider this linear matrix inequality: $$\begin{bmatrix} Y & x \\ x^T & z \end{bmatrix} \succeq 0$$ Combine this with our knowledge that $Y\succeq 0$, and the Schur complement rule for semidefinite matrices gives us this: $$Y\succ 0, ~ \begin{bmatrix} Y & x \\ x^T & z \end{bmatrix} \succeq 0 \quad\Longleftrightarrow\quad Y\succ 0, ~ z - x^TY^{-1}x \geq 0 \quad\Longleftrightarrow\quad Y\succ 0, ~x^TY^{-1}x \leq z$$ Therefore, the epigraph is the intersection of one strict LMI and one non-strict LMI, both describing convex sets. As the intersection of two convex sets, the epigraph is convex, and so the function is convex.