# $min\{f(x,y),g(x,y)\}$ is also quasiconcave for $f(x,y)$ and $g(x,y)$ quasiconcave functions

Assume that $$f(x,y)$$ and $$g(x,y)$$ are quasiconcave functions, then $$min\{f(x,y),g(x,y)\}$$ is also quasiconcave.

What I did

Assume that $$\{(u,v), (z,t)\}$$ are domain of f and g for $$0<\theta<1$$

$$f$$ is quasiconcave then $$f(\theta (u,v)+(1-\theta)(z,t))=f(\theta u+(1-\theta)z, \theta v+(1-\theta)t)>\min( f(u,v), f(z,t) )=f(z,t)$$

$$g$$ is quasiconcave then $$g(\theta (u,v)+(1-\theta)(z,t))=g(\theta u+(1-\theta)z, \theta v+(1-\theta)t)>\min( g(u,v), g(z,t) )=g(z,t)$$

I assume that $$m(x,y)=min(f(x,y), g(x,y))$$

I want to show that $$m(\theta (u,v)+(1-\theta)(z,t))=m(\theta u+(1-\theta)z, \theta v+(1-\theta)t)>\min( m(u,v), m(z,t) )$$

$$m(\theta (u,v)+(1-\theta)(z,t))=m(\theta u+(1-\theta)z, \theta v+(1-\theta)t)=min(f(\theta u+(1-\theta)z, \theta v+(1-\theta)t), g(\theta u+(1-\theta)z, \theta v+(1-\theta)t))=$$

And then, how can I proceed this proof?

Any helps will be appreciated. Thank you.

*duplicated question

## 2 Answers

To show that $$\min(f(x,y), g(x,y))$$, consider the upper level set $$P^a = \left\{(x,y) \in \mathbb{R}^2 | \min(f(x,y), g(x,y)) \geq a\right\}$$

We'll show that this is a convex set for every $$a\in\mathbb{R}$$.

$$\begin{eqnarray*} P^a & = & \left\{(x,y) \in \mathbb{R}^2 | \min(f(x,y), g(x,y)) \geq a\right\} \\ & = & \left\{(x,y) \in \mathbb{R}^2 | f(x,y) \geq a\right\} \cap \left\{(x,y) \in \mathbb{R}^2 | g(x,y) \geq a\right\}\end{eqnarray*}$$ which is the intersection of two convex sets because $$f$$ and $$g$$ are both quasi-concave, and hence it is also convex. Therefore, $$\min(f(x,y), g(x,y))$$ is quasi-concave.

• Thank you so much professor! I see! Commented Oct 15, 2022 at 16:15

You don't need to prove anything!

$$H(x,y) = \min\{g(x,y), f(x,y)\}$$is either equal to $$g(x,y)$$ or to $$f(x,y)$$, which are both quasi-concave by definition!