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

<p>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. </p>

CC BY-SA 4.0
2018-08-15T18:12:19.563Z
2018-08-15T18:12:19.563Z
Herr K.
<p>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 &amp; Suen (2001), p.140.</p>

<p>I simply could not elaborate any sketches to leave here as a starting point.</p>

CC BY-SA 4.0
2018-08-14T23:56:22.203Z
Isn't this a mathematics question rather than an economics one?
Mozibur Ullah
<production-function>
Quasiconcavity and homogeneity
Muradim
# Quasiconcavity and homogeneity

(Asked by: username_0 on 2018-08-14)

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.

**Tags:** production-function
> **Comments:**
> **username_1**: Isn't this a mathematics question rather than an economics one?  

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## 1 answers

### Answer by username_2 on 2018-08-15. Score: 3 | Accepted answer

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.


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