1
$\begingroup$

How to prove that if $f$ is strictly quasi-concave and homogeneous of degree 1, then $f$ is concave? It was left as an exercise by Silberberg & Suen (2001), p.140.

I simply could not elaborate any sketches to leave here as a starting point.

$\endgroup$
  • $\begingroup$ Isn't this a mathematics question rather than an economics one? $\endgroup$ – Mozibur Ullah Aug 14 '18 at 23:56
1
$\begingroup$

Note that strictness of quasi-concavity is not required, unless you want to obtain strict concavity as a result.

Take any $x,y\in\mathbb R^n$. Observe that homogeneity of degree 1 (HD1) implies that \begin{equation} f(x/f(x))=f(x)/f(x)=1=f(y/f(y)). \end{equation} For any $\alpha\in(0,1)$, let \begin{equation} \theta=\frac{\alpha f(x)}{\alpha f(x)+(1-\alpha)f(y)}.\tag{1} \end{equation} Note that $\theta$ also lives in the (open) unit interval. Thus, by quasi-concavity, we have for every $\theta\in(0,1)$, \begin{equation} f\left(\theta\frac{x}{f(x)}+(1-\theta)\frac{y}{f(y)}\right)\ge\min\left\{\frac{x}{f(x)},\frac{y}{f(y)}\right\}=1. \end{equation} Expanding the LHS using $(1)$, we get \begin{equation} f\left(\frac{\alpha x+(1-\alpha)y}{\alpha f(x)+(1-\alpha)f(y)}\right)\ge1. \end{equation} Invoking HD1 again, we have \begin{equation} \frac{f(\alpha x+(1-\alpha)y)}{\alpha f(x)+(1-\alpha)f(y)}\ge1 \quad\Leftrightarrow\quad f(\alpha x+(1-\alpha)y)\ge \alpha f(x)+(1-\alpha)f(y), \end{equation} which means $f$ is concave.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.