7
$\begingroup$

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!

$\endgroup$
0

2 Answers 2

12
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Thanks a lot. You answered all possible questions on my mind about this subject. $\endgroup$ Nov 6, 2017 at 18:46
  • $\begingroup$ Ok, I thought I understood those but it seems like I didn't. I understand your intuition and what you are trying to do but I did not understand how you derived this inequalities. These two in particular θmin[f(x0),g(x0)]+(1−θ)min[f(x1),g(x1)]≤θf(x0)+(1−θ)f(x1)≤f(xθ) θmin[f(x0),g(x0)]+(1−θ)min[f(x1),g(x1)]≤θg(x0)+(1−θ)g(x1)≤g(xθ) $\endgroup$ Nov 7, 2017 at 17:46
  • $\begingroup$ 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. $\endgroup$
    – Zhuoran He
    Nov 7, 2017 at 17:49
  • $\begingroup$ Oh I see it now. I was confused left side but min[f(x),g(x)]≤f(x) always hold so left side holds too. Thank you very much again! $\endgroup$ Nov 7, 2017 at 18:33
7
$\begingroup$

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.

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .