Sum of restricted strictly convex functions

Question

Suppose that $\boldsymbol{x}\in\mathbb{R}^p$. Also, assume $\Omega\subseteq \{1,2,\ldots,p\}$ is a subset of the indices, and $\Omega^c$ denotes the complement of $\Omega$. The notation $\boldsymbol{x}_\Omega$ denotes the restriction of $\boldsymbol{x}$ to the index set $\Omega$, i.e., $\boldsymbol{x}_\Omega\in\mathbb{R}^{|\Omega|}$. Suppose that $f$ and $g$ are both convex functions of $\boldsymbol{x}$. Furthermore, assume that $f$ is strictly convex along $\boldsymbol{x}_\Omega$, and $g$ is strictly convex along $\boldsymbol{x}_{\Omega^c}$, that is for $t\in[0,1]$: \begin{align*} f\left( t\begin{bmatrix} \boldsymbol{x}_{1\Omega}\\ \boldsymbol{x}_{\Omega^c} \end{bmatrix} + (1-t) \begin{bmatrix} \boldsymbol{x}_{2\Omega}\\ \boldsymbol{x}_{\Omega^c}\end{bmatrix}\right) &< tf\left( \begin{bmatrix} \boldsymbol{x}_{1\Omega}\\ \boldsymbol{x}_{\Omega^c} \end{bmatrix}\right) + (1-t)f\left( \begin{bmatrix} \boldsymbol{x}_{2\Omega}\\ \boldsymbol{x}_{\Omega^c} \end{bmatrix} \right),\\ % g\left( t\begin{bmatrix} \boldsymbol{x}_{\Omega}\\ \boldsymbol{x}_{1\Omega^c} \end{bmatrix} + (1-t) \begin{bmatrix} \boldsymbol{x}_{\Omega}\\ \boldsymbol{x}_{2\Omega^c}\end{bmatrix}\right) &< tg\left( \begin{bmatrix} \boldsymbol{x}_{\Omega}\\ \boldsymbol{x}_{1\Omega^c} \end{bmatrix}\right) + (1-t)g\left( \begin{bmatrix} \boldsymbol{x}_{\Omega}\\ \boldsymbol{x}_{2\Omega^c} \end{bmatrix} \right). \end{align*} Is there a quick prove that $f+g$ is a strictly convex function of $\boldsymbol{x}$?

Answer 1

This is not correct, even for one function alone. Indeed, let $f \colon \mathbb R^2 \to \mathbb R$ be defined via $f(x,y) = (x-y)^2$. Then, $f$ is strictly convex in $x$ and strictly convex in $y$, but not strictly convex in $(x,y)$. In order to answer the precise question you stated, we can consider $g = f$.