# Homogeneous functions of one or more variables

Let us begin with the definition of homogeneous function.

Theorem (Homogeneous function)

A function ${\displaystyle f(x)}$ is said to be homogeneous if:

${\displaystyle f(\lambda x)=g(\lambda )f(r)\quad \quad \forall \lambda \in \mathbb {R} }$

where ${\displaystyle g}$ is, for now, an unspecified function (we will shortly see that it has a precise form).

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

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

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

${\displaystyle f(x)=f(\lambda x_{0})=g(\lambda )f(x_{0})}$

We now want to show that ${\displaystyle g(\lambda )}$ has a precise form.

Theorem

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

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

Proof

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

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

A homogeneous function such that ${\displaystyle g(\lambda )=\lambda ^{p}}$ is said to be homogeneous of degree ${\displaystyle p}$.

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

${\displaystyle f(\lambda x_{1},\dots ,\lambda x_{n})=g(\lambda )f(x_{1},\dots ,x_{n})}$

1. In reality it would be sufficient the sole continuity of ${\displaystyle g}$, but in this case the proof becomes longer.