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

^{[1]}, 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:

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

- ↑ In reality it would be sufficient the sole continuity of $g$, but in this case the proof becomes longer.