$$ \newcommand{\R}{\mathbb{R}} \newcommand{\ra}{\rightarrow} \newcommand{\d}{\text{d}} $$
Most statements of Euler's theorem for homogeneous functions I found restrict the domain of
- the considered function, and/or
- the homogeneity degree, and/or
- the rescaling parameter.
For example
- Wikipedia states the theorem for positively homogeneous functions $f: \R^n \setminus {0} \ra \R$
- Apostol at pag. 287 considers only positive rescaling parameters.
I don't see what goes wrong with this generality:
Definition
Let $k$ be an integer. A differentiable function $f: \R^n \ra \R$ is $k$-homogeneous if $$ f(ax) = a^kf(x) $$ for all $x \in \R^n$ and all $a \in \R \setminus {0}$.
Euler's Theorem
Let $k$ be an integer and let $f: \R^n \ra \R$ be a differentiable function. Then $f$ is $k$-homogeneous if and only if $$ x \cdot \d{f}(x) = kf(x) $$ for all $x \in \R^n$.
Proof
Let $f$ be $k$-homogeneous. Then for any nonzero real $a$, $f(ax) = a^k f(x)$. Differentiate with respect to $a$: $$ \d{f}(ax) \cdot x = ka^{k-1}f(x) $$ In particular this must be true for $a$ = 1, so $$ \d{f}(x) \cdot x = kf(x) $$
Conversely, let the condition above hold; then for any real $a$, $ax \cdot \d{f}(ax) = k f (ax)$.
Define $g(a) = f(ax)$ for any fixed $x \in \R^n$. I guess here it is actually necessary to consider $g(a) = f(ax)$ for any fixed $x \in \R^n \setminus {0}$. Differentiate $g$ with respect to $a$, and assume $a \neq 0$: $$ \frac{\d{}}{\d{a}}g(a) = \d{f}(ax) \cdot x = \frac{\d{f}(ax) \cdot ax}{a} = \frac{k}{a}f(ax) = \frac{k}{a}g(a) $$ The solution of this ODE is $g(a) = g(1) a^k$, i.e. $f(ax) = a^k f(x)$ for all nonzero $a$.
To wrap up
- I'm not sure whether it is necessary to state the theorem excluding the origin of $\R^n$ from the domain of $f$;
- The only constraint on $a$ seems to be $a \neq 0$;
- I see no constraint on the homogeneity degree $k$, namely the theorem seems to hold for $k>0$, $k=0$ and $k<0$.
Is this correct?
