<p>A function <span class="math-container">$f:D\rightarrow \mathbb{R}$</span> is said to be quasiconcave if the following set is a convex set for every value of <span class="math-container">$a\in\mathbb{R}$</span>:
<span class="math-container">$P_a = \{x\in D: f(x) \geq a\}$</span></p>
<p>To show that <span class="math-container">$f(x,y) =\min(x, 2y)$</span> is quasiconcave, we just need to show that <span class="math-container">$P_a = \{(x,y)\in \mathbb{R}^2: \min(x, 2y) \geq a\}$</span> is a convex set. For that we consider arbitrary <span class="math-container">$(x', y')$</span> and <span class="math-container">$(x'', y'')$</span> from the set <span class="math-container">$P_a$</span> and arbitrary <span class="math-container">$\lambda\in [0,1]$</span> and show that
<span class="math-container">$\lambda (x', y')+(1-\lambda)(x'', y'')$</span> is in <span class="math-container">$P_a$</span>. Observe that <span class="math-container">$x'\geq \min(x', 2y')\geq a$</span> and <span class="math-container">$x''\geq \min(x'', 2y'')\geq a$</span>, so <span class="math-container">$\lambda x'+(1-\lambda)x''\geq a$</span>. Likewise, <span class="math-container">$\lambda 2y'+(1-\lambda)2y''\geq a$</span>. Therefore, it follows that <span class="math-container">$\min(\lambda x'+(1-\lambda)x'',2(\lambda y'+(1-\lambda)y'')) \geq a$</span> and consequently, <span class="math-container">$\lambda (x',y')+(1-\lambda)(x'', y'')$</span> is in <span class="math-container">$P_a$</span>.</p>

CC BY-SA 4.0
2021-04-05T10:26:20.723Z
2021-04-05T10:34:39.133Z
2021-04-05T10:34:39.133Z
Amit
<p>I have a question that asks:</p>
<p>&quot;Let there be two goods 1 and 2.Let <span class="math-container">$x$</span> and <span class="math-container">$y$</span> denote their respective quantities.<span class="math-container">$(x,y)$</span> represents a bundle. Suppose a consumer’s preferences over bundles in <span class="math-container">$R^2_+$</span>  can be represented by the utility function <span class="math-container">$U(x,y)=min(x,2y)$</span>. Prove that the consumer’s preferences over bundles in <span class="math-container">$R^2_+$</span> are convex by proving that the utility function is quasiconcave in <span class="math-container">$R^2_+$</span>&quot;</p>
<p>I was taught that if a utility function is quasiconcave, then <span class="math-container">$\begin{vmatrix} 0 &amp; U_x &amp; U_y \\ U_x &amp; U_{xx} &amp; U_{xy} \\ U_y &amp; U_{yx} &amp; U_{yy} \\\end{vmatrix} &gt; 0 \forall (x, y) \in R^2_+$</span></p>
<p>However, when I start doing the partial derivatives, I notice that I get:</p>
<p><span class="math-container">$\frac{\partial U}{\partial x} = \begin{cases}1, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases},$</span>
<span class="math-container">$\frac{\partial U}{\partial y} = \begin{cases}0, &amp;&amp; x \le 2y \\ 2, &amp;&amp; x &gt; 2y\end{cases},$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial x^2} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases},$</span>
<span class="math-container">$\frac{\partial U}{\partial y^2} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases},$</span></p>
<p><span class="math-container">$\frac{\partial U}{\partial xy} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases},$</span>
<span class="math-container">$\frac{\partial U}{\partial yx} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases}$</span></p>
<p>I then got <span class="math-container">$\begin{vmatrix} 0 &amp; U_x &amp; U_y \\ U_x &amp; U_{xx} &amp; U_{xy} \\ U_y &amp; U_{yx} &amp; U_{yy} \\\end{vmatrix} = \begin{cases}0, &amp;&amp; x \le 2y \\ 0, &amp;&amp; x &gt; 2y\end{cases}$</span></p>
<p>This is where I am stuck. I am unsure if:</p>
<p>a) I made a mistake somewhere</p>
<p>b) my method is wrong</p>
<p>c) Everything I've done so far is correct, but I need to do another step</p>
<p>Please let me know if I need to provide more information, thanks</p>

CC BY-SA 4.0
2021-04-04T06:41:58.057Z
min(x, 2y) is a concave function and every concave function is quasiconcave. You can watch the following playlist for the proof that min(x, 2y) is concave : https://www.youtube.com/watch?v=uzDtvgUcZj8&list=PLUJGfL_499TLYEd-9IO1DmQKwm6Mtdt4u
Amit
CC BY-SA 4.0
2021-04-04T08:56:39.333Z
Or you can just use another definition of quasiconcavity, one that does not rely on differentiability.
Giskard
CC BY-SA 4.0
2021-04-04T08:58:57.930Z
That $U$ is quasi-concave means that for any $s,t$ in the convex domain (here $\mathbb{R}_+^2$) and any $\lambda$ in $[0,1]$ (you could, equivalently, use an open interval), 
$U\big(\lambda s+(1-\lambda)t\big)\geq\min\big\{U(s),U(t)\big\}.$
It is wonderful that there exit calculus characterizations for differentiable functions, but this function is very much non-differentable, and all the action is at a kink. Start directly with the definition.
Michael Greinecker
CC BY-SA 4.0
2021-04-04T12:21:42.837Z
Thanks @MichaelGreinecker that makes sense
DoubleRainbowZ
<utility><consumer-theory><preferences>
How to prove that a utility function U(x,y)=min(x,2y) is quasiconcave?
DoubleRainbowZ
# How to prove that a utility function U(x,y)=min(x,2y) is quasiconcave?

(Asked by: username_0 on 2021-04-04)

I have a question that asks:

"Let there be two goods 1 and 2.Let $x$ and $y$ denote their respective quantities.$(x,y)$ represents a bundle. Suppose a consumer’s preferences over bundles in $R^2\_+$ can be represented by the utility function $U(x,y)=min(x,2y)$. Prove that the consumer’s preferences over bundles in $R^2\_+$ are convex by proving that the utility function is quasiconcave in $R^2\_+$"

I was taught that if a utility function is quasiconcave, then $\begin{vmatrix} 0 & U\_x & U\_y \\ U\_x & U\_{xx} & U\_{xy} \\ U\_y & U\_{yx} & U\_{yy} \\\end{vmatrix} > 0 \forall (x, y) \in R^2\_+$

However, when I start doing the partial derivatives, I notice that I get:

$\frac{\partial U}{\partial x} = \begin{cases}1, && x \le 2y \\ 0, && x > 2y\end{cases},$
$\frac{\partial U}{\partial y} = \begin{cases}0, && x \le 2y \\ 2, && x > 2y\end{cases},$

$\frac{\partial U}{\partial x^2} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$
$\frac{\partial U}{\partial y^2} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$

$\frac{\partial U}{\partial xy} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases},$
$\frac{\partial U}{\partial yx} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases}$

I then got $\begin{vmatrix} 0 & U\_x & U\_y \\ U\_x & U\_{xx} & U\_{xy} \\ U\_y & U\_{yx} & U\_{yy} \\\end{vmatrix} = \begin{cases}0, && x \le 2y \\ 0, && x > 2y\end{cases}$

This is where I am stuck. I am unsure if:

a) I made a mistake somewhere

b) my method is wrong

c) Everything I've done so far is correct, but I need to do another step

Please let me know if I need to provide more information, thanks

**Tags:** utility, consumer-theory, preferences
> **Comments:**
> **username_1**: min(x, 2y) is a concave function and every concave function is quasiconcave. You can watch the following playlist for the proof that min(x, 2y) is concave : https://www.youtube.com/watch?v=uzDtvgUcZj8&list=PLUJGfL_499TLYEd-9IO1DmQKwm6Mtdt4u  
> **username_2**: Or you can just use another definition of quasiconcavity, one that does not rely on differentiability.  
> **username_3**: That $U$ is quasi-concave means that for any $s,t$ in the convex domain (here $\mathbb{R}_+^2$) and any $\lambda$ in $[0,1]$ (you could, equivalently, use an open interval),  $U\big(\lambda s+(1-\lambda)t\big)\geq\min\big\{U(s),U(t)\big\}.$ It is wonderful that there exit calculus characterizations for differentiable functions, but this function is very much non-differentable, and all the action is at a kink. Start directly with the definition.  
> **username_0**: Thanks @MichaelGreinecker that makes sense  

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## 1 answers

### Answer by username_1 on 2021-04-05. Score: 3

A function $f:D\rightarrow \mathbb{R}$ is said to be quasiconcave if the following set is a convex set for every value of $a\in\mathbb{R}$:
$P\_a = \{x\in D: f(x) \geq a\}$

To show that $f(x,y) =\min(x, 2y)$ is quasiconcave, we just need to show that $P\_a = \{(x,y)\in \mathbb{R}^2: \min(x, 2y) \geq a\}$ is a convex set. For that we consider arbitrary $(x', y')$ and $(x'', y'')$ from the set $P\_a$ and arbitrary $\lambda\in [0,1]$ and show that
$\lambda (x', y')+(1-\lambda)(x'', y'')$ is in $P\_a$. Observe that $x'\geq \min(x', 2y')\geq a$ and $x''\geq \min(x'', 2y'')\geq a$, so $\lambda x'+(1-\lambda)x''\geq a$. Likewise, $\lambda 2y'+(1-\lambda)2y''\geq a$. Therefore, it follows that $\min(\lambda x'+(1-\lambda)x'',2(\lambda y'+(1-\lambda)y'')) \geq a$ and consequently, $\lambda (x',y')+(1-\lambda)(x'', y'')$ is in $P\_a$.


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