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

  • $\begingroup$ 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 : youtube.com/… $\endgroup$ – Amit Apr 4 at 6:41
  • 1
    $\begingroup$ Or you can just use another definition of quasiconcavity, one that does not rely on differentiability. $\endgroup$ – Giskard Apr 4 at 8:56
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    $\begingroup$ 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. $\endgroup$ – Michael Greinecker Apr 4 at 8:58
  • $\begingroup$ Thanks @MichaelGreinecker that makes sense $\endgroup$ – DoubleRainbowZ Apr 4 at 12:21

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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