Example of when the minimum of two convex functions is not convex

Question

My understanding is that taking the minimum of two (or more) functions is like creating a union of the functions which in some cases would result in a non-convex function. I can draw it out with some simple 2d functions but I can't figure out how to show it with equations. Thanks in advance for the help!

Answer 1

Minimum of $|x|$ and $|1-x|$ is not convex. (In fact it is strictly concave on $(0,1)$). Write $\frac 1 2$ as $\frac 1 2 (0)+\frac 1 2 (1)$. Convexity would mean $\frac 1 2 \leq \frac 1 2 (0)+\frac 1 2 (0)$!

Answer 2

The minimum $h:\Bbb R \to \Bbb R$ of $f(x)=(x-1)^2$ and $g(x)=(x+1)^2$ is not convex at $[-1,1]$.

Proof: $h(0)>h(-1)$ and $h(0)>h(1)$.

Answer 3

In $[0,1]$, $$f:=x-x^3,\\g:=(1-x)-(1-x)^3,\\h:=\min(f,g)$$

Now the average of $h\left(\dfrac1{\sqrt 3}\right)$ and $h\left(1-\dfrac1{\sqrt 3}\right)$ is $\dfrac{2}{3\sqrt3}$, but $h\left(\dfrac12\right)=\dfrac38$, which is smaller.