Solving Max-Min problem by first doing min and then max

Question

Assume that we have the following max-min optimization prblem \begin{align} \max_i \min_x f_i(x),\quad i\in\{1,2,\cdots,I\} \end{align} where the optimization problem $\min_x f_i(x)$ is solvable and its optimal solution for each $i$ is $x_i^\ast$. Do we find the optimal value $x^\ast$ of max-min based on $x_i^\ast$.

Answer 1

For each $i$, solve $v_i^* = \min_x f_i(x)$ with $x_i^* = \text{argmin}_x \text{ } f_i(x)$.

Then take $i^* = \text{argmax}_i\text{ } v_i^*$, and the corresponding $x_{i^*}^*$ is the optimal $x$ for $i^*$.