# Is the max-min a convex combination of the minima?

Let $$f_1,\ldots,f_n$$ be convex continuous functions defined in a convex compact domain $$C\subseteq \mathbb{R}^d$$, and let $$x_i := \arg\min_{x\in C} f_i(x).$$

Let $$g$$ be a convex function defined by $$g(x) := \max_{i}f_i(x)$$, and let $$y := \arg\min_{x\in C} g(x).$$

Is $$y$$ a convex combination of $$x_1,\ldots,x_n$$?

Here is an illustration of the question for $$n=2$$ functions in dimension $$d=1$$:

EDIT: We can assume that $$C$$ is compact and the functions $$f_i$$ are continuous.

• Do we have any other asusmptions on $C$ or the $f_i$? e.g. is $C$ closed or compact? are $f_i$ lower semicontinuous or coercive?
– Zim
Jun 6, 2022 at 8:50
• @Zim we can assume that $C$ is compact and that the $f_i$ are continuous. Edited. Jun 7, 2022 at 5:07

The minimiser of the maximum $$y$$ does not have to be a convex combination of the $$x_i$$. Take $$f_1(u,v)=(u-v)^2+\frac{v^2}{100}$$, $$f_2(u,v)=(u+v-1)^2+\frac{v^2}{100}$$. That is, $$f_1$$ and $$f_2$$ are elongated valleys in different directions, where the floor of each valley has a slight depression on the $$u$$-axis. The minimum of $$\max\{f_1,f_2\}$$ lies in the intersection of the 'valleys', off the $$u$$-axis.