<p>Consider the upper level set
<span class="math-container">\begin{eqnarray*}S_a&amp;=&amp;\left\{(x_1,x_2)\in\mathbb{R}^2_+|2x_1x_2+x_1+2x_2\geq a\right\}\\ &amp;=&amp; \left\{(x_1,x_2)\in\mathbb{R}^2_+|2(x_1+1)\left(x_2+\frac{1}{2}\right)\geq a+1\right\}\\ &amp;=&amp; \left\{(x_1,x_2)\in\mathbb{R}^2_+|(x_1+1)\left(x_2+\frac{1}{2}\right)\geq \dfrac{a+1}{2}\right\}\\ &amp;=&amp; \left\{(u,v)-\left(1,\frac{1}{2}\right)\in\mathbb{R}^2_+|uv\geq \dfrac{a+1}{2}\right\} \\ &amp;=&amp; \left\{(u,v)\in[1,\infty)\times[\frac{1}{2},\infty)|uv\geq \dfrac{a+1}{2}\right\} -\left(1,\frac{1}{2}\right) \ \ \ [\text{Element-wise subtraction}]\end{eqnarray*}</span>
which is a convex set if and only if
<span class="math-container">$$\left\{(u,v)\in[1,\infty)\times[\frac{1}{2},\infty)|uv\geq \dfrac{a+1}{2}\right\}=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq \dfrac{a+1}{2}\right\}\bigcap [1,\infty)\times[\frac{1}{2},\infty)$$</span> is a convex set.
So, it is sufficient to show</p>
<p><span class="math-container">$T_k=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq k\right\}$</span></p>
<p>is a convex set for all <span class="math-container">$k\in\mathbb{R}$</span>.</p>
<p>We'll now show that.</p>
<p>For <span class="math-container">$k\leq 0$</span>, <span class="math-container">$T_k=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq k\right\}=\mathbb{R}^2_+$</span> is a convex set.</p>
<p>Now consider for any <span class="math-container">$k&gt;0$</span>, the set
<span class="math-container">$T_k=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq k\right\}$</span>. To show that <span class="math-container">$T_k$</span> is convex, consider any <span class="math-container">$(u',v')\in T_k$</span> and <span class="math-container">$(u'',v'')\in T_k$</span>, and any <span class="math-container">$\lambda\in (0,1)$</span>, we'll show that <span class="math-container">$(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T_k$</span></p>
<p>To show this, we'll need the following inequality:
Without loss of generality letting <span class="math-container">$k\leq u'v'\leq u''v''$</span>, the following holds:
<span class="math-container">$u'v''+u''v'=u'v'(\frac{v''}{v'}+\frac{u''}{u'})\geq u'v'(\frac{u'}{u''}+\frac{u''}{u'})\geq 2k$</span>
[Please note that <span class="math-container">$(\frac{u'}{u''}+\frac{u''}{u'})\geq 2$</span> because <span class="math-container">$\min_{a&gt;0} a + \frac{1}{a} = 2$</span>]</p>
<p>Therefore, we also have <span class="math-container">$u'v''+u''v'\geq 2k$</span> along with <span class="math-container">$u'v'\geq k$</span>, <span class="math-container">$u''v''\geq k$</span>.</p>
<p>Now, we'll show that <span class="math-container">$(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T_k$</span>. Here is the argument:
<span class="math-container">\begin{eqnarray*}
&amp;&amp; (\lambda u'+(1-\lambda)u'')(\lambda v'+(1-\lambda)v'') \\ &amp;=&amp;
\lambda^2 u'v'+ \lambda(1-\lambda)(u'v''+u''v')+ (1-\lambda)^2u''v''\\ &amp;\geq &amp; \lambda^2 k+ \lambda(1-\lambda)(2k)+ (1-\lambda)^2k \\ &amp;=&amp; (\lambda^2 + 2\lambda(1-\lambda)+ (1-\lambda)^2)k \\ &amp;=&amp; k
\end{eqnarray*}</span>
Therefore, <span class="math-container">$(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T_k$</span>.
So, <span class="math-container">$T_k=\left\{(u,v)\in\mathbb{R}^2_+|uv\geq k\right\}$</span>
is a convex set for all <span class="math-container">$k\in\mathbb{R}$</span>.</p>

CC BY-SA 4.0
2024-05-07T05:53:26.410Z
2024-05-07T06:04:42.000Z
2024-05-07T06:04:42.000Z
Amit
<p>Alternatively, you could check the bordered hessian:
<span class="math-container">$$
\begin{bmatrix} 0 &amp; \frac{\partial f}{\partial x_1} &amp; \frac{\partial f}{\partial x_2}\\
\frac{\partial f}{\partial x_1} &amp; \frac{\partial^2 f}{\partial x_1 \partial x_1} &amp; \frac{\partial^2 f}{\partial x_1 \partial x_2}\\ \frac{\partial f}{\partial x_2} &amp; \frac{\partial f}{\partial x_1 \partial x_2} &amp; \frac{\partial^2 f}{\partial x_2 \partial x_2}\end{bmatrix}
$$</span>
The determinant of this matrix is strictly positive for <span class="math-container">$x_1, x_2 &gt; 0$</span>, which shows the function is quasi-concave on <span class="math-container">$\mathbb{R}^2_{++}$</span>.</p>

CC BY-SA 4.0
2024-05-08T07:21:06.970Z
2024-05-08T07:21:06.970Z
tdm
<p>I have a utility function, and I want to prove that it is a quasi-concave function:
<span class="math-container">$$ u(x_1,x_2)= 2x_1x_2+x_1+2x_2 $$</span>
I do this by showing that the set of points where the utility is larger than or equal to any given level forms a convex set, and denote this level of utility by <span class="math-container">$U^{x}$</span> The set of points where the utility function is larger than or equal to can be represented as:
<span class="math-container">$$ S ={ (x_1,x_2)|2x_1x_2+x_1+2x_2 ≥ u^{x}} $$</span>
So I have to prove that the set S is convex for all possible values <span class="math-container">$u^{x}$</span>
I consider two arbitrary points (<span class="math-container">$x_1^{a},x_2^{a}$</span>) and (<span class="math-container">$x_1^{b},x_2^{b}$</span>) in the set S that satisfy:
<span class="math-container">$$[2x_1^{a}x_2^{a}+x_1^{a}+2x_2^{a} ≥ u^{x} ]$$</span>
<span class="math-container">$$[2x_1^{b}x_2^{b}+x_1^{b}+2x_2^{b} ≥ u^{x} ]$$</span>
Now, if I take a convex combination of those points:
<span class="math-container">$$(x_1,x_2)= \lambda(x_1^{a},x_2^{a})+(1-\lambda)(x_1^{b},x_2^{b}$$</span> where <span class="math-container">$(0≤\lambda≤1)$</span>. This lies within the line segment connecting (<span class="math-container">$x_1^{a},x_2^{a}$</span>) and (<span class="math-container">$x_1^{b},x_2^{b}$</span>)</p>
<p>Now I evaluate the utility function at the point <span class="math-container">$(x_1,x_2)$</span>:
<span class="math-container">$$u(x_1,x_2)= 2x_1x_2+x_1+2x_2$$</span></p>
<p>I now substitute the expression for <span class="math-container">$(x_1,x_2)$</span>:</p>
<p><span class="math-container">$$u(x_1,x_2)= 2 (\lambda x_1^{a}+(1-\lambda)(x_1^{b})(\lambda x_2^{a}+(1-\lambda)(x_2^{b}))) + \lambda x_1^{a}+(1-\lambda)(x_1^{b} + 2 \lambda x_2^{a}+(1-\lambda)(x_2^{b})) $$</span></p>
<p>I know that I now have to show that u(x_1,x_2) is larger than or equal to <span class="math-container">$u^{x}$</span>, but I am not sure how to proceed further. I would greatly appreciate any help.</p>
<p>EDIT: Solved!</p>

CC BY-SA 4.0
2024-05-06T14:33:54.953Z
For quasi-concavity, you have to show that the set of points where the utility is larger or equal (!) to a certain level is convex.
Michael Greinecker
CC BY-SA 4.0
2024-05-06T14:40:20.957Z
Thank you, I have updated my answer.
Noah
CC BY-SA 4.0
2024-05-06T15:44:47.773Z
Something to consider is that you have *three* (!) more left parentheses in your final equation than right parentheses. Now I am a man of left wing leanings, but this seems too radical even to me.
Giskard
CC BY-SA 4.0
2024-05-06T20:06:38.833Z
If you solved it, you can post an answer to your own question. Other readers might be interested.
VARulle
<microeconomics><mathematical-economics><optimization>
Proving quasi-concavity for a utility function
Noah
# Proving quasi-concavity for a utility function

(Asked by: username_0 on 2024-05-06)

I have a utility function, and I want to prove that it is a quasi-concave function:
$$ u(x\_1,x\_2)= 2x\_1x\_2+x\_1+2x\_2 $$
I do this by showing that the set of points where the utility is larger than or equal to any given level forms a convex set, and denote this level of utility by $U^{x}$ The set of points where the utility function is larger than or equal to can be represented as:
$$ S ={ (x\_1,x\_2)|2x\_1x\_2+x\_1+2x\_2 ≥ u^{x}} $$
So I have to prove that the set S is convex for all possible values $u^{x}$
I consider two arbitrary points ($x\_1^{a},x\_2^{a}$) and ($x\_1^{b},x\_2^{b}$) in the set S that satisfy:
$$[2x\_1^{a}x\_2^{a}+x\_1^{a}+2x\_2^{a} ≥ u^{x} ]$$
$$[2x\_1^{b}x\_2^{b}+x\_1^{b}+2x\_2^{b} ≥ u^{x} ]$$
Now, if I take a convex combination of those points:
$$(x\_1,x\_2)= \lambda(x\_1^{a},x\_2^{a})+(1-\lambda)(x\_1^{b},x\_2^{b}$$ where $(0≤\lambda≤1)$. This lies within the line segment connecting ($x\_1^{a},x\_2^{a}$) and ($x\_1^{b},x\_2^{b}$)

Now I evaluate the utility function at the point $(x\_1,x\_2)$:
$$u(x\_1,x\_2)= 2x\_1x\_2+x\_1+2x\_2$$

I now substitute the expression for $(x\_1,x\_2)$:

$$u(x\_1,x\_2)= 2 (\lambda x\_1^{a}+(1-\lambda)(x\_1^{b})(\lambda x\_2^{a}+(1-\lambda)(x\_2^{b}))) + \lambda x\_1^{a}+(1-\lambda)(x\_1^{b} + 2 \lambda x\_2^{a}+(1-\lambda)(x\_2^{b})) $$

I know that I now have to show that u(x\_1,x\_2) is larger than or equal to $u^{x}$, but I am not sure how to proceed further. I would greatly appreciate any help.

EDIT: Solved!

**Tags:** microeconomics, mathematical-economics, optimization
> **Comments:**
> **username_1**: For quasi-concavity, you have to show that the set of points where the utility is larger or equal (!) to a certain level is convex.  
> **username_0**: Thank you, I have updated my answer.  
> **username_2**: Something to consider is that you have *three* (!) more left parentheses in your final equation than right parentheses. Now I am a man of left wing leanings, but this seems too radical even to me.  
> **username_3**: If you solved it, you can post an answer to your own question. Other readers might be interested.  

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## 2 answers

### Answer by username_4 on 2024-05-07. Score: 1

Consider the upper level set
\begin{eqnarray\*}S\_a&=&\left\{(x\_1,x\_2)\in\mathbb{R}^2\_+|2x\_1x\_2+x\_1+2x\_2\geq a\right\}\\ &=& \left\{(x\_1,x\_2)\in\mathbb{R}^2\_+|2(x\_1+1)\left(x\_2+\frac{1}{2}\right)\geq a+1\right\}\\ &=& \left\{(x\_1,x\_2)\in\mathbb{R}^2\_+|(x\_1+1)\left(x\_2+\frac{1}{2}\right)\geq \dfrac{a+1}{2}\right\}\\ &=& \left\{(u,v)-\left(1,\frac{1}{2}\right)\in\mathbb{R}^2\_+|uv\geq \dfrac{a+1}{2}\right\} \\ &=& \left\{(u,v)\in[1,\infty)\times[\frac{1}{2},\infty)|uv\geq \dfrac{a+1}{2}\right\} -\left(1,\frac{1}{2}\right) \ \ \ [\text{Element-wise subtraction}]\end{eqnarray\*}
which is a convex set if and only if
$$\left\{(u,v)\in[1,\infty)\times[\frac{1}{2},\infty)|uv\geq \dfrac{a+1}{2}\right\}=\left\{(u,v)\in\mathbb{R}^2\_+|uv\geq \dfrac{a+1}{2}\right\}\bigcap [1,\infty)\times[\frac{1}{2},\infty)$$ is a convex set.
So, it is sufficient to show

$T\_k=\left\{(u,v)\in\mathbb{R}^2\_+|uv\geq k\right\}$

is a convex set for all $k\in\mathbb{R}$.

We'll now show that.

For $k\leq 0$, $T\_k=\left\{(u,v)\in\mathbb{R}^2\_+|uv\geq k\right\}=\mathbb{R}^2\_+$ is a convex set.

Now consider for any $k>0$, the set
$T\_k=\left\{(u,v)\in\mathbb{R}^2\_+|uv\geq k\right\}$. To show that $T\_k$ is convex, consider any $(u',v')\in T\_k$ and $(u'',v'')\in T\_k$, and any $\lambda\in (0,1)$, we'll show that $(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T\_k$

To show this, we'll need the following inequality:
Without loss of generality letting $k\leq u'v'\leq u''v''$, the following holds:
$u'v''+u''v'=u'v'(\frac{v''}{v'}+\frac{u''}{u'})\geq u'v'(\frac{u'}{u''}+\frac{u''}{u'})\geq 2k$
[Please note that $(\frac{u'}{u''}+\frac{u''}{u'})\geq 2$ because $\min\_{a>0} a + \frac{1}{a} = 2$]

Therefore, we also have $u'v''+u''v'\geq 2k$ along with $u'v'\geq k$, $u''v''\geq k$.

Now, we'll show that $(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T\_k$. Here is the argument:
\begin{eqnarray\*}
&& (\lambda u'+(1-\lambda)u'')(\lambda v'+(1-\lambda)v'') \\ &=&
\lambda^2 u'v'+ \lambda(1-\lambda)(u'v''+u''v')+ (1-\lambda)^2u''v''\\ &\geq & \lambda^2 k+ \lambda(1-\lambda)(2k)+ (1-\lambda)^2k \\ &=& (\lambda^2 + 2\lambda(1-\lambda)+ (1-\lambda)^2)k \\ &=& k
\end{eqnarray\*}
Therefore, $(\lambda u'+(1-\lambda)u'', \lambda v'+(1-\lambda)v'')\in T\_k$.
So, $T\_k=\left\{(u,v)\in\mathbb{R}^2\_+|uv\geq k\right\}$
is a convex set for all $k\in\mathbb{R}$.


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### Answer by username_5 on 2024-05-08. Score: 0

Alternatively, you could check the bordered hessian:
$$
\begin{bmatrix} 0 & \frac{\partial f}{\partial x\_1} & \frac{\partial f}{\partial x\_2}\\
\frac{\partial f}{\partial x\_1} & \frac{\partial^2 f}{\partial x\_1 \partial x\_1} & \frac{\partial^2 f}{\partial x\_1 \partial x\_2}\\ \frac{\partial f}{\partial x\_2} & \frac{\partial f}{\partial x\_1 \partial x\_2} & \frac{\partial^2 f}{\partial x\_2 \partial x\_2}\end{bmatrix}
$$
The determinant of this matrix is strictly positive for $x\_1, x\_2 > 0$, which shows the function is quasi-concave on $\mathbb{R}^2\_{++}$.


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