<p>Note first that the function <span class="math-container">$r\mapsto A(r)^{\mu/p}$</span> is strictly increasing on <span class="math-container">$\mathbb{R}_+$</span>. When you look at the minimum in the definition of quasi-concavity, you can therefore ignore this part and it suffices to show that the function
<span class="math-container">$$(x_1,x_2,\ldots,x_n)\mapsto a_1x_1^p+a_2x_2^p+...+a_nx_n^p$$</span>
is quasi-concave. Actually, the function is even concave. The sum of concave functions is concave, so it suffices to show that the function
<span class="math-container">$$(x_1,x_2,\ldots,x_n)\mapsto a_ix_i^p$$</span>
is concacave for each <span class="math-container">$i=1,\ldots,n$</span>.</p>

CC BY-SA 4.0
2021-05-08T16:55:47.140Z
Correct me if I'm wrong but your argument seems to be valid only for $p \in (0,1]$? For example if $p < 0$ then $A(r)^{\mu/p}$ is not increasing. Also if $p > 1$ then $x^p$ is convex?
tdm
CC BY-SA 4.0
2021-05-09T17:19:47.660Z
Yes, but that assumption is made in the question.
Michael Greinecker
CC BY-SA 4.0
2021-05-11T06:05:01.530Z
Ah ok, I saw the $p \ne 0$ qualification but looked over the $p \in (0,1]$ condition. Sorry.
tdm
CC BY-SA 4.0
2021-05-08T13:48:55.657Z
2021-05-08T13:48:55.657Z
Michael Greinecker
<p>I have a question I have been asked to solve:</p>
<p>Given that <span class="math-container">$(a_1, a_2,...,a_n)\in R_{++}^n$</span> and <span class="math-container">$(x_1, x_2,...,x_n)\in R_{++}^n$</span>, and <span class="math-container">$A&gt;0, \mu &gt;0, p \neq 0$</span>, if there is a function <span class="math-container">$f(x)=A(a_1x_1^p+a_2x_2^p+...+a_nx_n^p)^\frac{\mu}{p} \space \space \forall x \in R_{++}^n$</span>, show that function <span class="math-container">$f$</span> is a quasiconcave function over <span class="math-container">$R_{++}^n$</span> when <span class="math-container">$p \in (0,1]$</span>.</p>
<p>I understand that to show a function is quasiconcave, it needs to satisfy this definition:</p>
<p><span class="math-container">$f:_+^→$</span> is quasiconcave on <span class="math-container">$_+^$</span> if and only if <span class="math-container">$∀,∈_+^$</span> and for all <span class="math-container">$\in(0,1)$</span></p>
<p><span class="math-container">$f(+(1−))≥\{f(),f()\}$</span></p>
<p>I am wondering how I would use this definition to show that a generalised function <span class="math-container">$f$</span> is quasiconcave, or if I should be approaching this from another concept/definition.</p>

CC BY-SA 4.0
2021-05-08T13:33:05.410Z
What do you mean by "generalized" function? This is an ordinary function.
Michael Greinecker
CC BY-SA 4.0
2021-05-08T13:39:02.723Z
Sorry, I omitted CES when typing it out. It should read generalised CES function $f$
DoubleRainbowZ
<microeconomics><ces-function>
How to prove a generalised function is quasiconcave?
DoubleRainbowZ
# How to prove a generalised function is quasiconcave?

(Asked by: username_0 on 2021-05-08)

I have a question I have been asked to solve:

Given that $(a\_1, a\_2,...,a\_n)\in R\_{++}^n$ and $(x\_1, x\_2,...,x\_n)\in R\_{++}^n$, and $A>0, \mu >0, p \neq 0$, if there is a function $f(x)=A(a\_1x\_1^p+a\_2x\_2^p+...+a\_nx\_n^p)^\frac{\mu}{p} \space \space \forall x \in R\_{++}^n$, show that function $f$ is a quasiconcave function over $R\_{++}^n$ when $p \in (0,1]$.

I understand that to show a function is quasiconcave, it needs to satisfy this definition:

$f:\_+^→$ is quasiconcave on $\_+^$ if and only if $∀,∈\_+^$ and for all $\in(0,1)$

$f(+(1−))≥\{f(),f()\}$

I am wondering how I would use this definition to show that a generalised function $f$ is quasiconcave, or if I should be approaching this from another concept/definition.

**Tags:** microeconomics, ces-function

> **Comments:**
> **username_1**: What do you mean by "generalized" function? This is an ordinary function.  
> **username_0**: Sorry, I omitted CES when typing it out. It should read generalised CES function $f$  

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## 1 answers

### Answer by username_1 on 2021-05-08. Score: 5 | Accepted answer

Note first that the function $r\mapsto A(r)^{\mu/p}$ is strictly increasing on $\mathbb{R}\_+$. When you look at the minimum in the definition of quasi-concavity, you can therefore ignore this part and it suffices to show that the function
$$(x\_1,x\_2,\ldots,x\_n)\mapsto a\_1x\_1^p+a\_2x\_2^p+...+a\_nx\_n^p$$
is quasi-concave. Actually, the function is even concave. The sum of concave functions is concave, so it suffices to show that the function
$$(x\_1,x\_2,\ldots,x\_n)\mapsto a\_ix\_i^p$$
is concacave for each $i=1,\ldots,n$.

> **Comments:**
> **username_2**: Correct me if I'm wrong but your argument seems to be valid only for $p \in (0,1]$? For example if $p < 0$ then $A(r)^{\mu/p}$ is not increasing. Also if $p > 1$ then $x^p$ is convex?  
> **username_1**: Yes, but that assumption is made in the question.  
> **username_2**: Ah ok, I saw the $p \ne 0$ qualification but looked over the $p \in (0,1]$ condition. Sorry.  

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