<p>The only thing you need to change in your proof is what you suggest yourself in the second-to-last sentence under &quot;my question&quot;. Your proof shows that <span class="math-container">$g\circ f$</span> is homogeneous of degree 1 and that <span class="math-container">$g\circ f$</span> is a monotonic transformation of (is &quot;equivalent to&quot;) <span class="math-container">$f$</span>.</p>

CC BY-SA 4.0
2023-08-31T11:03:20.613Z
2023-08-31T11:03:20.613Z
smcc
<h2><em><strong>Question</strong></em></h2>
<p>Proof that any function <span class="math-container">$f: \mathbb{R}^1 \to \mathbb{R}^1$</span> with <span class="math-container">$f' &gt; 0$</span> everywhere is equivalent to a homogeneous function of degree one.</p>
<h2><em><strong>My Attempt</strong></em></h2>
<p>Note that all strictly monotonic functions are invertible because they are guaranteed to have a one-to-one mapping from their range to their domain. Thus, if <span class="math-container">$f' &gt; 0$</span> for all <span class="math-container">$x \in \mathbb{R}^1$</span>, <span class="math-container">$f$</span> is invertible and its inverse <span class="math-container">$g$</span> satisfies <span class="math-container">$g' &gt; 0$</span> for all <span class="math-container">$x \in \mathbb{R}^1$</span>. Then <span class="math-container">$g$</span> is a monotonic transformation. Therefore, <span class="math-container">$g(f(x)) = x$</span> is equivalent to <span class="math-container">$f$</span>. Since <span class="math-container">$(g \circ f)(tx) = tx$</span>, <span class="math-container">$g$</span> is homogeneous of degree <span class="math-container">$1$</span>. Thus, <span class="math-container">$f$</span> is equivalent to a homogeneous function of degree <span class="math-container">$1$</span>.</p>
<h2><em><strong>My Question</strong></em></h2>
<p>I am not sure if my way of doing this is correct, especially for where I put &quot;since <span class="math-container">$(g \circ f)(tx) = tx$</span>, <span class="math-container">$g$</span> is homogeneous of degree <span class="math-container">$1$</span>.&quot; Basically, I am not sure how to get that <span class="math-container">$g$</span> is homogeneous of degree <span class="math-container">$1$</span>. Perhaps should I have written <span class="math-container">$(g \circ f)$</span> is homogeneous of degree <span class="math-container">$1$</span>, and that <span class="math-container">$f$</span> is equivalent to <span class="math-container">$(g \circ f)$</span>? Could someone please help me with it? Thanks a lot in advance!</p>

CC BY-SA 4.0
2023-08-31T07:38:42.210Z
I don't think this is true. Take $f(x) = x^3 + 2x$ that has $f'(x) = 3 x^2 + 2 > 0$ but $f$ is not homogeneous of degree 1.
tdm
CC BY-SA 4.0
2023-08-31T10:07:29.847Z
I found the exercise that Beerus is trying to solve. It uses the word "equivalent" to mean that two functions have the same level sets (represent the same preferences). So, the exercise is to show that for any function $f$  with $f'>0$ there exists another function $h$ that is homogeneous and is related to $f$ by a monotonic transformation.
smcc
<mathematical-economics>
Prove Any Strictly Increasing Function is Equivalent to A Homogeneous Function of Degree 1 - Mathematics for Economists (Simon & Blume) Exercise 20.14
Beerus
# Prove Any Strictly Increasing Function is Equivalent to A Homogeneous Function of Degree 1 - Mathematics for Economists (Simon & Blume) Exercise 20.14

(Asked by: username_0 on 2023-08-31)

## ***Question***

Proof that any function $f: \mathbb{R}^1 \to \mathbb{R}^1$ with $f' > 0$ everywhere is equivalent to a homogeneous function of degree one.

## ***My Attempt***

Note that all strictly monotonic functions are invertible because they are guaranteed to have a one-to-one mapping from their range to their domain. Thus, if $f' > 0$ for all $x \in \mathbb{R}^1$, $f$ is invertible and its inverse $g$ satisfies $g' > 0$ for all $x \in \mathbb{R}^1$. Then $g$ is a monotonic transformation. Therefore, $g(f(x)) = x$ is equivalent to $f$. Since $(g \circ f)(tx) = tx$, $g$ is homogeneous of degree $1$. Thus, $f$ is equivalent to a homogeneous function of degree $1$.

## ***My Question***

I am not sure if my way of doing this is correct, especially for where I put "since $(g \circ f)(tx) = tx$, $g$ is homogeneous of degree $1$." Basically, I am not sure how to get that $g$ is homogeneous of degree $1$. Perhaps should I have written $(g \circ f)$ is homogeneous of degree $1$, and that $f$ is equivalent to $(g \circ f)$? Could someone please help me with it? Thanks a lot in advance!

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: I don't think this is true. Take $f(x) = x^3 + 2x$ that has $f'(x) = 3 x^2 + 2 > 0$ but $f$ is not homogeneous of degree 1.  
> **username_2**: I found the exercise that Beerus is trying to solve. It uses the word "equivalent" to mean that two functions have the same level sets (represent the same preferences). So, the exercise is to show that for any function $f$  with $f'>0$ there exists another function $h$ that is homogeneous and is related to $f$ by a monotonic transformation.  

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## 1 answers

### Answer by username_2 on 2023-08-31. Score: 2 | Accepted answer

The only thing you need to change in your proof is what you suggest yourself in the second-to-last sentence under "my question". Your proof shows that $g\circ f$ is homogeneous of degree 1 and that $g\circ f$ is a monotonic transformation of (is "equivalent to") $f$.


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