# Quasiconcavity and homogeneity

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.

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

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.