<p>To show that <span class="math-container">$\min(f(x,y), g(x,y))$</span>, consider the upper level set <span class="math-container">$$P^a = \left\{(x,y) \in \mathbb{R}^2 | \min(f(x,y), g(x,y)) \geq a\right\}$$</span></p>
<p>We'll show that this is a convex set for every <span class="math-container">$a\in\mathbb{R}$</span>.</p>
<p><span class="math-container">\begin{eqnarray*} P^a &amp; = &amp; \left\{(x,y) \in \mathbb{R}^2 | \min(f(x,y), g(x,y)) \geq a\right\} \\ &amp; = &amp; \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*}</span>
which is the intersection of two convex sets because <span class="math-container">$f$</span> and <span class="math-container">$g$</span> are both quasi-concave, and hence it is also convex. Therefore, <span class="math-container">$\min(f(x,y), g(x,y))$</span> is quasi-concave.</p>

CC BY-SA 4.0
2022-10-15T16:15:02.540Z
Thank you so much professor! I see!
studentp
CC BY-SA 4.0
2022-10-15T05:36:38.450Z
2022-10-15T05:36:38.450Z
Amit
<p>You don't need to prove anything!</p>
<p><span class="math-container">$H(x,y) = \min\{g(x,y), f(x,y)\}$</span>is either equal to <span class="math-container">$g(x,y)$</span> or to <span class="math-container">$f(x,y)$</span>, which are both quasi-concave by definition!</p>

CC BY-SA 4.0
2022-10-14T21:32:37.810Z
2022-10-14T21:32:37.810Z
Matteo Bulgarelli
<p>Assume that <span class="math-container">$f(x,y)$</span> and <span class="math-container">$g(x,y)$</span> are quasiconcave functions, then <span class="math-container">$min\{f(x,y),g(x,y)\}$</span> is also quasiconcave.</p>
<p>What I did</p>
<p>Assume that <span class="math-container">$\{(u,v), (z,t)\}$</span> are domain of f and g for <span class="math-container">$0&lt;\theta&lt;1$</span></p>
<p><span class="math-container">$f$</span> is quasiconcave then <span class="math-container">$f(\theta (u,v)+(1-\theta)(z,t))=f(\theta u+(1-\theta)z, \theta v+(1-\theta)t)&gt;\min( f(u,v), f(z,t) )=f(z,t)$</span></p>
<p><span class="math-container">$g$</span> is quasiconcave then <span class="math-container">$g(\theta (u,v)+(1-\theta)(z,t))=g(\theta u+(1-\theta)z, \theta v+(1-\theta)t)&gt;\min( g(u,v), g(z,t) )=g(z,t)$</span></p>
<p>I assume that <span class="math-container">$m(x,y)=min(f(x,y), g(x,y))$</span></p>
<p>I want to show that <span class="math-container">$m(\theta (u,v)+(1-\theta)(z,t))=m(\theta u+(1-\theta)z, \theta v+(1-\theta)t)&gt;\min( m(u,v), m(z,t) )$</span></p>
<p><span class="math-container">$$m(\theta (u,v)+(1-\theta)(z,t))=m(\theta u+(1-\theta)z, \theta v+(1-\theta)t)=min(f(\theta u+(1-\theta)z, \theta v+(1-\theta)t), g(\theta u+(1-\theta)z, \theta v+(1-\theta)t))=$$</span></p>
<p>And then, how can I proceed this proof?</p>
<p>Any helps will be appreciated. Thank you.</p>
<p>*duplicated question</p>

<microeconomics><mathematical-economics><self-study>
$min\{f(x,y),g(x,y)\}$ is also quasiconcave for $f(x,y)$ and $g(x,y)$ quasiconcave functions
studentp
# $min\{f(x,y),g(x,y)\}$ is also quasiconcave for $f(x,y)$ and $g(x,y)$ quasiconcave functions

(Asked by: username_0 on 2022-10-14)

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

**Tags:** microeconomics, mathematical-economics, self-study

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## 2 answers

### Answer by username_1 on 2022-10-15. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: Thank you so much professor! I see!  

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### Answer by username_2 on 2022-10-14. Score: 1

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!


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