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!
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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.
Zhuoran He
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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.
Kim Jong Un
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- 1
- 24
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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)
– Selçuk Öztürk Nov 07 '17 at 17:46