Convexity of an inner product value function

Question

Let the function $g : \mathbb{R}^m \to \mathbb{R}^n$ be such that each of its elements $g_i$ is strictly convex, and let $f : \mathbb{R}_{\geq 0}^m \times \mathbb{R}_{\geq 0}^n \to \mathbb{R}$ be defined by

$$ f (x,y) := y^\top g(x),$$

Consider the value function

$$ v(y) = \min_{x\in C(y)} f(x,y),$$

where $C(y)$ is compact, and $C$ is convex in the sense that for any $0\leq\lambda\leq 1$, and $y,y'\in\mathbb{R}_{\geq 0}^n$, we have that $x\in C(y)$ and $x'\in C(y')$ implies $$ \lambda x+(1-\lambda)x' \in C(\lambda y+(1-\lambda)y'). $$ I would like to show that $v$ is convex.

I have shown the result in a special case where $g(x)=x^\top Gx$ with $G$ positive definite and for a specific $C(y)$ but I'm hoping that it holds more generally.

EDIT: I have updated the notion of convexity for $C$. See Hans Engler's response below.

Answer 1

Assume $f(x,y)$ is jointly convex in $x,y$ and continuous then, By jointly convex, we mean, $$f(\lambda x_1 + (1-\lambda) x_2,\lambda y_1 + (1-\lambda) y_2) \leq \lambda f(x_1,y_1) + (1-\lambda) f(x_2,y_2)$$

Let $$v(y_1) = \min_{x \in C(y_1)} f(x,y_1) = f(x_1,y_1)$$ $$v(y_2) = \min_{x \in C(y_2)} f(x,y_2) = f(x_2,y_2)$$ $$v(\lambda y_1 + (1-\lambda) y_2) = \min_{x \in C(\lambda y_1 + (1-\lambda) y_2)} f(x,\lambda y_1 + (1-\lambda) y_2) \leq f(\lambda x_1 + (1-\lambda) x_2,\lambda y_1 + (1-\lambda) y_2) \leq \lambda f(x_1,y_1) + (1-\lambda) f(x_2,y_2) = \lambda v(y_1) + (1-\lambda) v(y_2)$$

Hence $v(.)$ is convex if $f(x,y)$ is jointly convex and continuous.

Joint convexity is also necessary, since we can assign $C(y_1)=\{x_1\}$ and $C(y_2)=\{x_2\}$ and this assignment forces convexity property of $v(y)$ to imply $f(x,y)$ to be jointly convex w.r.t $(x_1,y_1)$ and $(x_2,y_2)$ and this assignment is completely arbitrary.

Answer 2

This cannot be correct in any sense unless you make some assumptions on how $C(y)$ depends on $y$. For a counterexample, consider the 1-dim case with $$ g(x) = x^2, \; f(x,y) = x^2y\; \text{for}\; x, y \ge 0, \quad C(y) = \begin{cases} [1,2] \quad (y \in \mathbb{Q}) \\ [3,4] \quad (y \notin \mathbb{Q}) \end{cases} $$ Then $v(y) = y$ if $y \in \mathbb{Q}$ and $v(y) = 9y$ if $y \notin \mathbb{Q}$, which is certainly not a convex function of $y$.