I am studying homothetic functions using Mathematics for Economists by Simon and Blume. I am reading their proof of the following theorem:

Theorem$\quad$ Let $u: \mathbb{R}_+^\mathbf{n} \to \mathbb{R}$ be a strictly monotonic function. Then, $u$ is homothetic if and only if for all $\mathbf{x}$ and $\mathbf{y}$ in $\mathbb{R}_+^\mathbf{n}$, \begin{align} u(\mathbf{x}) \geq u(\mathbf{y}) \iff u(\alpha \mathbf{x}) \geq u(\alpha \mathbb{y}),\quad \textit{for all}\quad \alpha > 0.\tag1 \end{align}

I feel that there is a lot of typo and misleading notation in their proof, and that the proof in the textbook is not complete. So I rewrite the proof. I would like to know if my proof is rigorous and complete. I would really appreciate it if someone could help me check!

Here is my attempt:

Proof$\quad$ We first show that if $u$ satisfies (1), it is homothetic. Let $\mathbf{e}$ denote the vector $(1, 1, \dots, 1)$, that spans the diagonal $\Delta$ in $\mathbb{R}^{\mathbf{n}}$. Define function $f: \mathbb{R}_+ \to \mathbb{R}$ by \begin{align*} f(t) = u(t\mathbf{e}). \end{align*} Since $u$ is strictly increasing, so is $f$; and therefore, $f$ has a strictly increasing inverse $f^{-1}$. Then \begin{align*} f \circ (f^{-1} \circ u) = (f \circ f^{-1}) \circ u = u. \end{align*} To prove that $u = f \circ (f^{-1} \circ u)$ is homothetic, we need only show that $(f^{-1} \circ u)$ is homogeneous.

$\quad$ For any scalar $a$, the function $a \mapsto f^{-1}(a)$ tells how far up the diagonal $\Delta$ the level set $u^{-1}(a)$ meets $\Delta$. Consequently, $f^{-1}(u(\mathbf{x}))$ tells how far up $\Delta$ the $u$-level set through $\mathbf{x}$ crosses $\Delta$. Analytically, $t = f^{-1}(u(\mathbf{x}))$ is the solution of \begin{align} u(\mathbf{x}) = u(t \mathbf{e}).\tag2 \end{align} Let $\alpha > 0$ be a scalar. By (1), \begin{align} u(\mathbf{x}) = u(t \mathbf{e}) \implies u(\alpha \mathbf{x}) = u(\alpha t \mathbf{e}).\tag3 \end{align} But (3) indicates that $s = \alpha t = \alpha f^{-1}(u(\mathbf{x}))$ is the solution of \begin{align} u(\mathbf{\alpha x}) = u(s \mathbf{e}).\tag4 \end{align} Since (2) indicates that $s = f^{-1}(u(\alpha \mathbf{x}))$ is also the solution of (4), we have that \begin{align*} f^{-1}(u(\alpha \mathbf{x})) = \alpha f^{-1}(u(\mathbf{x})); \end{align*} thus, $(f^{-1} \circ u)$ is homogeneous of degree one. Since $(f^{-1} \circ u)$ is homogeneous and $f$ is increasing, $u = f \circ (f^{-1} \circ u)$ is homothetic.

$\quad$ To prove the converse, suppose first that $u$ is linear homogeneous, that is homogeneous of degree one, and that $\alpha > 0$. These two properties yield \begin{align*} u(\mathbf{x}) \geq u(\mathbf{y}) & \iff \alpha u(\mathbf{x}) \geq \alpha u(\mathbf{y})\\ & \iff u(\alpha \mathbf{x}) \geq u(\alpha \mathbf{y}) \end{align*} so, property (1) holds.

$\quad$ More generally, suppose that $u$ is homothetic, so that $u = g_1 \circ v$, with $g_1$ increasing and $v$ homogeneous of degree $k$. Write $v$ as $g_2 \circ h$, where $g_2(z) = z^k$ and $h(\mathbf{x}) = v(\mathbf{x})^\frac{1}{k}$. We check that $h$ is homogeneous of degree one: \begin{align*} h(\alpha \mathbf{x}) & = v(\alpha \mathbf{x})^\frac{1}{k}\\ & = \left(\alpha^k v(\mathbf{x})\right)^\frac{1}{k}\\ & = \alpha (v(\mathbf{x}))^\frac{1}{k}\\ & = \alpha h(\mathbf{x}). \end{align*} We also have that $g_2$ is increasing. Thus, we can write $u$ as $u = p \circ h$ with $p \equiv g_1 \circ g_2$ increasing and $h$ linear homogeneous.

$\quad$ Once again, suppose $\alpha > 0$. Since $p$ is strictly increasing, it has a strictly increasing inverse $p^{-1}$. Therefore, \begin{align*} u(\mathbf{x}) \geq u(\mathbf{y}) & \iff p^{-1}(u(\mathbf{x})) \geq p^{-1}(u(\mathbf{y}))\\ & \iff h(\mathbf{x}) \geq h(\mathbf{y})\\ & \iff \alpha h(\mathbf{x}) \geq \alpha h(\mathbf{y})\\ & \iff h(\alpha \mathbf{x}) \geq h(\alpha \mathbf{y})\\ & \iff p(h(\alpha \mathbf{x})) \geq p(h(\alpha \mathbf{y}))\\ & \iff u(\alpha \mathbf{x}) \geq u(\alpha \mathbf{y}) \end{align*} and so $u$ satisfies property (1).

Thanks a lot for any help!


1 Answer 1


Looks rigorous and complete to me. And it is indeed clearer than the corresponding proof in Simon and Blume.


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