<p>I will try to address your queries in the order you've asked them by providing the necessary definitions and procedures to find the answers you're looking for.</p>
<ol>
<li>The first approach is valid and it is true that a cobb-douglas function which is homogeneous of degree 1 is indeed concave and therefore quasi-concave. This follows from the following theorem:</li>
</ol>
<blockquote>
<p><span class="math-container">$f_1$</span> is non-decreasing, non-negative real valued function defined on <span class="math-container">$\mathbb{R}_+^n$</span> and <span class="math-container">$f_1$</span> is quasi-concave and homogeneous of degree 1. Then <span class="math-container">$f_1$</span> is a concave function. <span class="math-container">$^1$</span></p>
</blockquote>
<ol start="2">
<li><p>A function is indeed said to be quasi-concave if all its upper contour sets are convex sets. But your procedure doesn't show that instead, it shows that the level curves are convex <em>(There's a difference between the convexity of a function and convex sets)</em>.</p>
<p>I'll try to clear your confusion by first providing a definition of convex sets and then providing a precise definition of quasi-concavity using which you can show that <span class="math-container">$w:\mathbb{R}_+^n\to \mathbb{R}$</span> defined as <span class="math-container">$w(x,y)=x^ay^{1-a}$</span> is quasi-concave. However, it is useful to know the implication <span class="math-container">$\textbf{concavity} \implies \textbf{quasi-concavity}$</span> so you don't have to check for quasi-concavity separately if you know the function is indeed concave.</p>
</li>
</ol>
<blockquote>
<p>a set <span class="math-container">$S$</span> is said to be a <em><strong>convex set</strong></em> if for any <span class="math-container">$x', x'' \in S$</span> we have <span class="math-container">$\lambda x'+(1-\lambda)x''\in S \quad \forall \; \lambda \in [0,1]$</span></p>
</blockquote>
<blockquote>
<p>a function <span class="math-container">$f_2:A\to \mathbb{R}$</span> defined on the convex set <span class="math-container">$A\subset \mathbb{R}^n$</span>. Then <span class="math-container">$f_2$</span> is said to be <em><strong>quasi-concave</strong></em> if every upper contour set of <span class="math-container">$f_2$</span> is convex. i.e., <span class="math-container">$S(\alpha)=\{x\in S \mid f_2(x)\geq\alpha\}$</span> is a convex set <span class="math-container">$\forall \; \alpha\in\mathbb{R}$</span></p>
</blockquote>
<ol start="3">
<li><p>The function <span class="math-container">$f: \mathbb{R}_+^2\to \mathbb{R}$</span> defined as <span class="math-container">$f(x,y)=3xy$</span> is quasi-concave but not concave. First, let me use the definition provided above to prove it is quasi-concave.</p>
<p>a) <em>To Prove:</em> <span class="math-container">$f: \mathbb{R}_+^2\to \mathbb{R}$</span> defined as <span class="math-container">$f(x,y)=3xy$</span> is quasi-concave<br />
<span class="math-container">$\bullet$</span> we need to show that <span class="math-container">$S(\alpha)=\{(x,y)\in \mathbb{R}_+^2 \mid 3xy\geq \alpha \}$</span>  is a convex set for all <span class="math-container">$\alpha \in \mathbb{R}$</span></p>
<p><span class="math-container">$\bullet$</span> pick arbitrary <span class="math-container">$(x',y'), (x'',y'') \in S(\alpha)$</span> and arbitrary <span class="math-container">$\lambda \in [0,1]$</span> <span class="math-container">$\bullet
\begin{align} \\
&amp; (x',y') \in S(\alpha) \implies 3x'y'\geq \alpha \implies x'y' \geq \frac{\alpha}{3} &amp; (1)\\
&amp; (x'',y'') \in S(\alpha) \implies 3x''y'' \geq \alpha \implies x''y''\geq \frac{\alpha}{3} &amp;(2) \end{align}$</span></p>
<p><span class="math-container">$\bullet$</span> consider, <span class="math-container">$(x''',y''')=\lambda (x',y')+(1-\lambda)(x'',y'')$</span></p>
<p><span class="math-container">$\begin{align} 
x'''y'''&amp; =\lambda^2x'y'+(1-\lambda)^2x''y''+\lambda(1-\lambda)(x''y'+x'y'')\\
&amp;\geq  \frac{\lambda^2\alpha}{3}+ \frac{(1-\lambda)^2\alpha}{3}+\frac{2\lambda(1-\lambda)\alpha}{3}=\frac{\alpha}{3} &amp; [\text{from } (1) \; \&amp; \;(2)] \\
\end{align}$</span>
<span class="math-container">$\therefore \quad x'''y''' \geq \frac{\alpha}{3}\implies (x''',y''')\in S(\alpha)$</span></p>
<p><span class="math-container">$\bullet$</span> Therefore, we have shown that <span class="math-container">$S(\alpha)$</span> is convex <span class="math-container">$\forall \; \alpha \in \mathbb{R}$</span> i.e., <span class="math-container">$f$</span> is quasi-concave</p>
<p>b) To show that <span class="math-container">$f$</span> is not concave let me first define concavity</p>
<blockquote>
<p>let <span class="math-container">$f_3:A\to \mathbb{R}$</span> be a function defined on convex set <span class="math-container">$A\subset \mathbb{R}^n$</span>. We say that <span class="math-container">$f_3$</span> is concave if <span class="math-container">$f_3(\lambda x' +(1-\lambda) x'')\geq \lambda f_3(x') + (1-\lambda)f_3(x'') \quad \forall \; x',x''\in A$</span> and <span class="math-container">$\forall \;\lambda \in [0,1]$</span></p>
</blockquote>
<p>since the above definition holds for all <span class="math-container">$x', x'' \in A$</span> and for all <span class="math-container">$\lambda \in [0,1]$</span> if we can show there exists at least one such <span class="math-container">$x', x'' \in A$</span> and <span class="math-container">$\lambda \in [0,1]$</span> for which the above doesn't hold then we can say that <span class="math-container">$f_3$</span> is not concave</p>
<p>pick <span class="math-container">$(x',y')=(0,0), \;(x'',y'')=(1,1)$</span> and <span class="math-container">$\lambda =\frac{1}{2}$</span>.<br />
<span class="math-container">$\therefore \quad f(\frac{1}{2},\frac{1}{2})=\frac{3}{4}&lt;\frac{1}{2}f(0,0)+\frac{1}{2}f(1,1)=\frac{3}{2}$</span><br />
Thus, we have shown <span class="math-container">$f$</span> is quasi-concave but not concave.</p>
</li>
<li><p>If I understand you correctly, I think what you are asking is whether the following implication is true or not:</p>
<p>If a production function <span class="math-container">$F:\mathbb{R}_+^2\to \mathbb{R}$</span> is homogenous of degree less than or equal to 1 and satisfies <span class="math-container">$F(0,0)=0$</span> then <span class="math-container">$F$</span> is concave.</p>
<p>The above is not true in general. For example, take <span class="math-container">$F(x,y)=\sqrt{x^2+y^2}$</span> This function exhibits constant returns to scale (i.e., homogeneous of degree 1) and satisfies <span class="math-container">$F(0,0)=0$</span>, but you can verify that the function is not concave (in the same way as we did in 3.b)). In fact, this function is convex.</p>
</li>
</ol>
<hr />
<p><sup> Reference for definitions and theorems used in the above answer (including proof of <span class="math-container">$^1$</span>): <a href="https://www.youtube.com/playlist?list=PLUJGfL_499TLYEd-9IO1DmQKwm6Mtdt4u" rel="nofollow noreferrer">https://www.youtube.com/playlist?list=PLUJGfL_499TLYEd-9IO1DmQKwm6Mtdt4u</a> </sup></p>

CC BY-SA 4.0
2023-05-26T21:22:19.893Z
2023-05-27T08:09:12.300Z
2023-05-27T08:09:12.300Z
mynameparv
<p>I'm faced with questions that want me to show that a utility or production function is either concave, or if not then quasi-concave so that we can apply the KKT conditions.</p>
<p>For example the production function <span class="math-container">$w(x,y) = x^{1-a}y^a$</span> plus the condition  <span class="math-container">$0 &lt; a &lt; 1$</span></p>
<p>The answer demonstrates that this is concave by showing that the Hessian matrix is NSD. This requires some annoying algebra, would the following simpler approaches also be valid:</p>
<ol>
<li><strong>Known properties of Cobb-Douglas functions:</strong></li>
</ol>
<ul>
<li>Noting that <span class="math-container">$1-a = b &gt; 0$</span> and hence we have a Cobb-Douglas production function with CRS. Therefore it is concave?</li>
</ul>
<ol start="2">
<li><strong>Showing Quasi-concavity - through convex upper contour sets:</strong></li>
</ol>
<ul>
<li><span class="math-container">$y = \frac{w^{\frac{1}{a}}}{x^{\frac{1-a}{a}}}$</span>   which is decreasing in <span class="math-container">$x$</span>, and has a negative first derivative and positive second derivative and is thus convex. Hence the upper contour sets are convex?</li>
</ul>
<ol start="3">
<li>I'm particularly interested in my upper contour set argument above. Please let me know if it can be more precise or better. A similar question  stated that for the function <span class="math-container">$f(x,y) = 3xy$</span> <strong>&quot;<em>The maximand is differentiable and quasi-concave, as can be seen noting the upper contour sets are convex.</em>&quot;</strong></li>
</ol>
<ul>
<li>This was the only explanation provided. I'm assuming this is because of the argument i have made e.g.  <span class="math-container">$y = \frac{f}{3x}$</span> for any fixed <span class="math-container">$f$</span>, has a negative first derivative and a positive second derivative and is therefore convex?</li>
</ul>
<ol start="4">
<li>Can i use the following theorem to show concavity, starting with DRS or CRS.</li>
</ol>
<blockquote>
<p>If the production Function <span class="math-container">$f(K,L)$</span> has <span class="math-container">$f(0,0) = 0$</span>, and is concave then it has decreasing or constant returns to scale .</p>
</blockquote>
<p>Does this also tells us that if the production function has DRS or CRS, and has  <span class="math-container">$f(0,0) = 0$</span>, then it is concave?</p>
<ul>
<li>E.g. our first function <span class="math-container">$w(x,y) = x^{1-a}y^a$</span></li>
</ul>

<mathematical-economics><utility><production-function><optimization><lagrangian>
Arguments for Concavity or Quasi-concavity
CormJack
# Arguments for Concavity or Quasi-concavity

(Asked by: username_0 on 2023-05-26)

I'm faced with questions that want me to show that a utility or production function is either concave, or if not then quasi-concave so that we can apply the KKT conditions.

For example the production function $w(x,y) = x^{1-a}y^a$ plus the condition $0 < a < 1$

The answer demonstrates that this is concave by showing that the Hessian matrix is NSD. This requires some annoying algebra, would the following simpler approaches also be valid:

1. **Known properties of Cobb-Douglas functions:**

* Noting that $1-a = b > 0$ and hence we have a Cobb-Douglas production function with CRS. Therefore it is concave?

2. **Showing Quasi-concavity - through convex upper contour sets:**

* $y = \frac{w^{\frac{1}{a}}}{x^{\frac{1-a}{a}}}$ which is decreasing in $x$, and has a negative first derivative and positive second derivative and is thus convex. Hence the upper contour sets are convex?

3. I'm particularly interested in my upper contour set argument above. Please let me know if it can be more precise or better. A similar question stated that for the function $f(x,y) = 3xy$ **"*The maximand is differentiable and quasi-concave, as can be seen noting the upper contour sets are convex.*"**

* This was the only explanation provided. I'm assuming this is because of the argument i have made e.g. $y = \frac{f}{3x}$ for any fixed $f$, has a negative first derivative and a positive second derivative and is therefore convex?

4. Can i use the following theorem to show concavity, starting with DRS or CRS.

> If the production Function $f(K,L)$ has $f(0,0) = 0$, and is concave then it has decreasing or constant returns to scale .

Does this also tells us that if the production function has DRS or CRS, and has $f(0,0) = 0$, then it is concave?

* E.g. our first function $w(x,y) = x^{1-a}y^a$

**Tags:** mathematical-economics, utility, production-function, optimization, lagrangian


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## 1 answers

### Answer by username_1 on 2023-05-26. Score: 4 | Accepted answer

I will try to address your queries in the order you've asked them by providing the necessary definitions and procedures to find the answers you're looking for.

1. The first approach is valid and it is true that a cobb-douglas function which is homogeneous of degree 1 is indeed concave and therefore quasi-concave. This follows from the following theorem:

> $f\_1$ is non-decreasing, non-negative real valued function defined on $\mathbb{R}\_+^n$ and $f\_1$ is quasi-concave and homogeneous of degree 1. Then $f\_1$ is a concave function. $^1$

2. A function is indeed said to be quasi-concave if all its upper contour sets are convex sets. But your procedure doesn't show that instead, it shows that the level curves are convex *(There's a difference between the convexity of a function and convex sets)*.

   I'll try to clear your confusion by first providing a definition of convex sets and then providing a precise definition of quasi-concavity using which you can show that $w:\mathbb{R}\_+^n\to \mathbb{R}$ defined as $w(x,y)=x^ay^{1-a}$ is quasi-concave. However, it is useful to know the implication $\textbf{concavity} \implies \textbf{quasi-concavity}$ so you don't have to check for quasi-concavity separately if you know the function is indeed concave.

> a set $S$ is said to be a ***convex set*** if for any $x', x'' \in S$ we have $\lambda x'+(1-\lambda)x''\in S \quad \forall \; \lambda \in [0,1]$

> a function $f\_2:A\to \mathbb{R}$ defined on the convex set $A\subset \mathbb{R}^n$. Then $f\_2$ is said to be ***quasi-concave*** if every upper contour set of $f\_2$ is convex. i.e., $S(\alpha)=\{x\in S \mid f\_2(x)\geq\alpha\}$ is a convex set $\forall \; \alpha\in\mathbb{R}$

3. The function $f: \mathbb{R}\_+^2\to \mathbb{R}$ defined as $f(x,y)=3xy$ is quasi-concave but not concave. First, let me use the definition provided above to prove it is quasi-concave.

   a) *To Prove:* $f: \mathbb{R}\_+^2\to \mathbb{R}$ defined as $f(x,y)=3xy$ is quasi-concave  
   $\bullet$ we need to show that $S(\alpha)=\{(x,y)\in \mathbb{R}\_+^2 \mid 3xy\geq \alpha \}$ is a convex set for all $\alpha \in \mathbb{R}$

   $\bullet$ pick arbitrary $(x',y'), (x'',y'') \in S(\alpha)$ and arbitrary $\lambda \in [0,1]$ $\bullet
   \begin{align} \\
   & (x',y') \in S(\alpha) \implies 3x'y'\geq \alpha \implies x'y' \geq \frac{\alpha}{3} & (1)\\
   & (x'',y'') \in S(\alpha) \implies 3x''y'' \geq \alpha \implies x''y''\geq \frac{\alpha}{3} &(2) \end{align}$

   $\bullet$ consider, $(x''',y''')=\lambda (x',y')+(1-\lambda)(x'',y'')$

   $\begin{align}
   x'''y'''& =\lambda^2x'y'+(1-\lambda)^2x''y''+\lambda(1-\lambda)(x''y'+x'y'')\\
   &\geq \frac{\lambda^2\alpha}{3}+ \frac{(1-\lambda)^2\alpha}{3}+\frac{2\lambda(1-\lambda)\alpha}{3}=\frac{\alpha}{3} & [\text{from } (1) \; \& \;(2)] \\
   \end{align}$
   $\therefore \quad x'''y''' \geq \frac{\alpha}{3}\implies (x''',y''')\in S(\alpha)$

   $\bullet$ Therefore, we have shown that $S(\alpha)$ is convex $\forall \; \alpha \in \mathbb{R}$ i.e., $f$ is quasi-concave

   b) To show that $f$ is not concave let me first define concavity

   > let $f\_3:A\to \mathbb{R}$ be a function defined on convex set $A\subset \mathbb{R}^n$. We say that $f\_3$ is concave if $f\_3(\lambda x' +(1-\lambda) x'')\geq \lambda f\_3(x') + (1-\lambda)f\_3(x'') \quad \forall \; x',x''\in A$ and $\forall \;\lambda \in [0,1]$

   since the above definition holds for all $x', x'' \in A$ and for all $\lambda \in [0,1]$ if we can show there exists at least one such $x', x'' \in A$ and $\lambda \in [0,1]$ for which the above doesn't hold then we can say that $f\_3$ is not concave

   pick $(x',y')=(0,0), \;(x'',y'')=(1,1)$ and $\lambda =\frac{1}{2}$.  
   $\therefore \quad f(\frac{1}{2},\frac{1}{2})=\frac{3}{4}<\frac{1}{2}f(0,0)+\frac{1}{2}f(1,1)=\frac{3}{2}$  
   Thus, we have shown $f$ is quasi-concave but not concave.
4. If I understand you correctly, I think what you are asking is whether the following implication is true or not:

   If a production function $F:\mathbb{R}\_+^2\to \mathbb{R}$ is homogenous of degree less than or equal to 1 and satisfies $F(0,0)=0$ then $F$ is concave.

   The above is not true in general. For example, take $F(x,y)=\sqrt{x^2+y^2}$ This function exhibits constant returns to scale (i.e., homogeneous of degree 1) and satisfies $F(0,0)=0$, but you can verify that the function is not concave (in the same way as we did in 3.b)). In fact, this function is convex.

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Reference for definitions and theorems used in the above answer (including proof of $^1$): <https://www.youtube.com/playlist?list=PLUJGfL_499TLYEd-9IO1DmQKwm6Mtdt4u>


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