# Concavity of max of two concave functions

I would like to check whether max(f(x),g(x)) is concave when f(x) and g(x) are concave on R to R. I can think it as verbally but couldn't find a mathematical solution. Any help is appreciated. Cheers!

Maximum preserves convexity and minimum preserves concavity. So the maximum of two concave functions may be neither concave nor convex. It may become double peaked. For example,

$$f(x)=\max[-|x+1|,-|x-1|]$$

has an "M"-shaped graph. The minimum of two concave functions is always concave. This is not difficult to prove. Use the definition. For concave $f(x),g(x)$, we have

$$\theta f(x_0)+(1-\theta)f(x_1)\leq f(x_\theta),$$ $$\theta g(x_0)+(1-\theta)g(x_1)\leq g(x_\theta),$$

where $x_\theta=\theta x_0+(1-\theta)x_1$ and $\theta\in[0,1]$. Therefore

$$\theta\min[f(x_0),g(x_0)]+(1-\theta)\min[f(x_1),g(x_1)]\leq\theta f(x_0)+(1-\theta)f(x_1)\leq f(x_\theta),$$

and similarly, $$\theta\min[f(x_0),g(x_0)]+(1-\theta)\min[f(x_1),g(x_1)]\leq\theta g(x_0)+(1-\theta)g(x_1)\leq g(x_\theta).$$

Therefore,

$$\theta\min[f(x_0),g(x_0)]+(1-\theta)\min[f(x_1),g(x_1)]\leq \min[f(x_\theta),g(x_\theta)],$$

which proves that $\min[f(x),g(x)]$ is concave.

• It's because $\theta\in[0,1]$. Therefore $\theta\geq 0$ and $1-\theta\geq 0$. Then using $\min[f(x),g(x)]\leq f(x)$ and $\min[f(x),g(x)]\leq g(x)$, you get the left part. The right part is just the definition of concavity. Nov 7, 2017 at 17:49
Consider the concave functions $f(x)=-x^2+3x$ and $g(x)=-x$. But $h(x)\equiv\max\{f(x),g(x)\}$ is not concave.