# Arguments for Concavity or Quasi-concavity

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?
1. 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?
1. 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?
1. 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 $$w(x,y) = x^{1-a}y^a$$

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. $$^1$$

1. 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 $$\forall \; \alpha\in\mathbb{R}$$

1. 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.

2. 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.

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