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

CC BY-SA 4.0
2023-09-20T10:42:35.857Z
2023-09-20T10:42:35.857Z
VARulle
<p>I am studying homothetic functions using <em>Mathematics for Economists</em> by Simon and Blume. I am reading their proof of the following theorem:</p>
<blockquote>
<p><strong>Theorem</strong><span class="math-container">$\quad$</span> Let <span class="math-container">$u: \mathbb{R}_+^\mathbf{n} \to \mathbb{R}$</span> be a strictly monotonic function. Then, <span class="math-container">$u$</span> is homothetic if and only if for all <span class="math-container">$\mathbf{x}$</span> and <span class="math-container">$\mathbf{y}$</span> in <span class="math-container">$\mathbb{R}_+^\mathbf{n}$</span>,
<span class="math-container">\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 &gt; 0.\tag1
\end{align}</span></p>
</blockquote>
<p>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!</p>
<p>Here is my attempt:</p>
<blockquote>
<p><strong>Proof</strong><span class="math-container">$\quad$</span> We first show that if <span class="math-container">$u$</span> satisfies (1), it is homothetic. Let <span class="math-container">$\mathbf{e}$</span> denote the vector <span class="math-container">$(1, 1, \dots, 1)$</span>, that spans the diagonal <span class="math-container">$\Delta$</span> in <span class="math-container">$\mathbb{R}^{\mathbf{n}}$</span>. Define function <span class="math-container">$f: \mathbb{R}_+ \to \mathbb{R}$</span> by
<span class="math-container">\begin{align*}
    f(t) = u(t\mathbf{e}).
\end{align*}</span>
Since <span class="math-container">$u$</span> is strictly increasing, so is <span class="math-container">$f$</span>; and therefore, <span class="math-container">$f$</span> has a strictly increasing inverse <span class="math-container">$f^{-1}$</span>. Then
<span class="math-container">\begin{align*}
    f \circ (f^{-1} \circ u) = (f \circ f^{-1}) \circ u = u.
\end{align*}</span>
To prove that <span class="math-container">$u = f \circ (f^{-1} \circ u)$</span> is homothetic, we need only show that <span class="math-container">$(f^{-1} \circ u)$</span> is homogeneous.</p>
<p><span class="math-container">$\quad$</span> For any scalar <span class="math-container">$a$</span>, the function <span class="math-container">$a \mapsto f^{-1}(a)$</span> tells how far up the diagonal <span class="math-container">$\Delta$</span> the level set <span class="math-container">$u^{-1}(a)$</span> meets <span class="math-container">$\Delta$</span>. Consequently, <span class="math-container">$f^{-1}(u(\mathbf{x}))$</span> tells how far up <span class="math-container">$\Delta$</span> the <span class="math-container">$u$</span>-level set through <span class="math-container">$\mathbf{x}$</span> crosses <span class="math-container">$\Delta$</span>. Analytically, <span class="math-container">$t = f^{-1}(u(\mathbf{x}))$</span> is the solution of
<span class="math-container">\begin{align}
    u(\mathbf{x}) = u(t \mathbf{e}).\tag2
\end{align}</span>
Let <span class="math-container">$\alpha &gt; 0$</span> be a scalar. By (1),
<span class="math-container">\begin{align}
    u(\mathbf{x}) = u(t \mathbf{e}) \implies u(\alpha \mathbf{x}) = u(\alpha t \mathbf{e}).\tag3
\end{align}</span>
But (3) indicates that <span class="math-container">$s = \alpha t = \alpha f^{-1}(u(\mathbf{x}))$</span> is the solution of
<span class="math-container">\begin{align}
    u(\mathbf{\alpha x}) = u(s \mathbf{e}).\tag4
\end{align}</span>
Since (2) indicates that <span class="math-container">$s = f^{-1}(u(\alpha \mathbf{x}))$</span> is also the solution of (4), we have that
<span class="math-container">\begin{align*}
    f^{-1}(u(\alpha \mathbf{x})) = \alpha f^{-1}(u(\mathbf{x}));
\end{align*}</span>
thus, <span class="math-container">$(f^{-1} \circ u)$</span> is homogeneous of degree one. Since <span class="math-container">$(f^{-1} \circ u)$</span> is homogeneous and <span class="math-container">$f$</span> is increasing, <span class="math-container">$u = f \circ (f^{-1} \circ u)$</span> is homothetic.</p>
<p><span class="math-container">$\quad$</span> To prove the converse, suppose first that <span class="math-container">$u$</span> is <em>linear homogeneous</em>, that is homogeneous of degree one, and that <span class="math-container">$\alpha &gt; 0$</span>. These two properties yield
<span class="math-container">\begin{align*}
    u(\mathbf{x}) \geq u(\mathbf{y}) &amp; \iff \alpha u(\mathbf{x}) \geq \alpha u(\mathbf{y})\\
    &amp; \iff u(\alpha \mathbf{x}) \geq u(\alpha \mathbf{y})
\end{align*}</span>
so, property (1) holds.</p>
<p><span class="math-container">$\quad$</span> More generally, suppose that <span class="math-container">$u$</span> is homothetic, so that <span class="math-container">$u = g_1 \circ v$</span>, with <span class="math-container">$g_1$</span> increasing and <span class="math-container">$v$</span> homogeneous of degree <span class="math-container">$k$</span>. Write <span class="math-container">$v$</span> as <span class="math-container">$g_2 \circ h$</span>, where <span class="math-container">$g_2(z) = z^k$</span> and <span class="math-container">$h(\mathbf{x}) = v(\mathbf{x})^\frac{1}{k}$</span>. We check that <span class="math-container">$h$</span> is homogeneous of degree one:
<span class="math-container">\begin{align*}
    h(\alpha \mathbf{x}) &amp; = v(\alpha \mathbf{x})^\frac{1}{k}\\
    &amp; = \left(\alpha^k v(\mathbf{x})\right)^\frac{1}{k}\\
    &amp; = \alpha (v(\mathbf{x}))^\frac{1}{k}\\
    &amp; = \alpha h(\mathbf{x}).
\end{align*}</span>
We also have that <span class="math-container">$g_2$</span> is increasing. Thus, we can write <span class="math-container">$u$</span> as <span class="math-container">$u = p \circ h$</span> with <span class="math-container">$p \equiv g_1 \circ g_2$</span> increasing and <span class="math-container">$h$</span> linear homogeneous.</p>
<p><span class="math-container">$\quad$</span> Once again, suppose <span class="math-container">$\alpha &gt; 0$</span>. Since <span class="math-container">$p$</span> is strictly increasing, it has a strictly increasing inverse <span class="math-container">$p^{-1}$</span>. Therefore,
<span class="math-container">\begin{align*}
    u(\mathbf{x}) \geq u(\mathbf{y}) &amp; \iff p^{-1}(u(\mathbf{x})) \geq p^{-1}(u(\mathbf{y}))\\
    &amp; \iff h(\mathbf{x}) \geq h(\mathbf{y})\\
    &amp; \iff \alpha h(\mathbf{x}) \geq \alpha h(\mathbf{y})\\
    &amp; \iff h(\alpha \mathbf{x}) \geq h(\alpha \mathbf{y})\\
    &amp; \iff p(h(\alpha \mathbf{x})) \geq p(h(\alpha \mathbf{y}))\\
    &amp; \iff u(\alpha \mathbf{x}) \geq u(\alpha \mathbf{y})
\end{align*}</span>
and so <span class="math-container">$u$</span> satisfies property (1).</p>
</blockquote>
<p>Thanks a lot for any help!</p>

<microeconomics><mathematical-economics>
About Proof of Theorem 20.8 in Mathematics for Economists by Simon and Blume
Beerus
# About Proof of Theorem 20.8 in Mathematics for Economists by Simon and Blume

(Asked by: username_0 on 2023-09-19)

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!

**Tags:** microeconomics, mathematical-economics

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## 1 answers

### Answer by username_1 on 2023-09-20. Score: 2 | Accepted answer

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


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