Let $\{f_i : i \in I\}$ be a set (finite or infinite) of functions from a convex set $\mathcal{D}\subseteq \mathbb{R}^n$ to $\mathbb{R}$ which are convex and bounded on $\mathcal{D}$. Let $g(x) : = \inf_{i \in I} f_i(x)$. Is $g$ convex? Why or Why not?
I think that $g$ is not necessarily convex because $\inf_{i \in I} f_i(z) \le \inf_{i \in I}[\lambda f_i(x) + (1-\lambda) f_i(y)] \not\le \lambda \inf_{i \in I} f_i(x) + (1-\lambda) \inf_{i \in I}f_i(y)$, where $z = \lambda x + (1-\lambda) y$ for $\lambda \in (0,1)$. To make sure that my guess is correct, I want to show the counter example, but I am struggling to come up with. I first came up with the obvious example $f_i(x) = x^i$ and $I = \mathbb{N}$ and $\mathcal{D} = (0,1)$. In this case $\inf_{i \in I} f_i(x) = 0$, but $g(x) = 0$ is still convex.
If my initial guess is correct, could you provide one counter example? or if it is wrong, how can I prove that $g(x)$ is convex?
