# Homogeneous functions of one or more variables

Let us begin with the definition of homogeneous function.

Theorem (Homogeneous function)

A function $f(x)$ is said to be homogeneous if:

$f(\lambda x)=g(\lambda )f(r)\quad \quad \forall \lambda \in \mathbb {R}$ where $g$ is, for now, an unspecified function (we will shortly see that it has a precise form).

An example of homogeneous function is $f(x)=Ax^{2}$ ; in fact:

$f(\lambda x)=A\lambda ^{2}x^{2}=\lambda ^{2}f(x)$ and so in this case $g(\lambda )=\lambda ^{2}$ .

A very interesting property of homogeneous functions is that once its value in a point $x_{0}$ and the function $g(\lambda )$ are known, the entire $f(x)$ can be reconstructed; in fact, any $x$ can be written in the form $\lambda x_{0}$ (of course with $\lambda =x/x_{0}$ ), so that:

$f(x)=f(\lambda x_{0})=g(\lambda )f(x_{0})$ We now want to show that $g(\lambda )$ has a precise form.

Theorem

The function $g(\lambda )$ as in the definition of homogeneous function is:

$g(\lambda )=\lambda ^{p}$ Proof

From the definition of homogeneous function, for $\lambda ,\mu \in \mathbb {R}$ we have on one hand that:

$f(\lambda \mu x)=f(\lambda (\mu x))=g(\lambda )f(\mu x)=g(\lambda )g(\mu )f(x)$ but also:
$f((\lambda \mu )x)=g(\lambda \mu )f(x)$ and so:
$g(\lambda \mu )=g(\lambda )g(\mu )$ If we now suppose $g$ to be differentiable, then differentiating with respect to $\mu$ this last equation we get:
$\lambda g'(\lambda \mu )=g(\lambda )g'(\mu )$ where by $g'$ we mean the derivative of $g$ with respect to its argument. Setting $\mu =1$ and defining $p:=g'(1)$ we have:
$\lambda g'(\lambda )=g(\lambda )p\quad \Rightarrow \quad {\frac {g'(\lambda )}{g(\lambda )}}={\frac {p}{\lambda }}$ which yields:
${\frac {d}{d\lambda }}\ln g(\lambda )={\frac {p}{\lambda }}\quad \Rightarrow \quad \ln g(\lambda )=p\ln \lambda +c\quad \Rightarrow \quad g(\lambda )=e^{c}\lambda ^{p}$ Now, $g'(\lambda )=pe^{c}\lambda ^{p-1}$ , so since $g'(1)=p$ by definition we have $p=pe^{c}$ and thus $c=0$ . Therefore:
$g(\lambda )=\lambda ^{p}$ A homogeneous function such that $g(\lambda )=\lambda ^{p}$ is said to be homogeneous of degree $p$ .

In case $f$ is a function of more than one variable $x_{1},\dots ,x_{n}$ , the definition of homogeneous function changes to:

$f(\lambda x_{1},\dots ,\lambda x_{n})=g(\lambda )f(x_{1},\dots ,x_{n})$ 1. In reality it would be sufficient the sole continuity of $g$ , but in this case the proof becomes longer.