# About Proof of Theorem 20.8 in Mathematics for Economists by Simon and Blume

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!