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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – tdm
    Commented Aug 31, 2023 at 7:38
  • 2
    $\begingroup$ 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. $\endgroup$
    – smcc
    Commented Aug 31, 2023 at 10:07

1 Answer 1


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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