Proof that smooth positive degree $m$ homogeneous function is polynomial of degree $m$ and $m$ is a positive integer

Question

So I know that by Euler's homogeneous function theorem $m$ is a positive number, but why is it an integer? And how to prove that $f$ is polynomial of degree $m$?

Answer 1

so in order to solve this I used Euler's homogeneous function theorem

to prove that a the derivative of positive homogeneous function of degree $m$ is also homogeneous function of degree $m-1$ (basically one needs to take the derivative of $\partial x_{i}$ on both sides and get this result)

after that we realize that we can continue this process but what happens when the degree goes below zero? the function is not defined ... (see what happens for $\lim a \to 0$)

thank you very much for the help :)

Answer 2

First note that by Euler's homogeneous function theorem we have that all partial derivatives of $f$ with order $k<m$ are homogeneous of order $m-k$, thus all these vanish at $0$. Now note that the partial derivatives of order $m$ are homogeneous of order $0$, hence the differential $d^mf$ satisfies $d^mf(tx)=d^mf(x)$ for all $x$ and $t$, and making $t\to0$, by continuity we have $d^mf(x)=d^mf(0)$ for all $x$.

Then we can apply the Taylor's formula with Lagrange's rest arround $0$ to obtain

$$f(x)=f(0)+df(0).x+\dfrac{1}{2}d^2f(0).x^2+...+\dfrac{1}{m!}d^mf(tx).x^m,\text{ with }t\in]0,1[$$

and we can concluded that $f(x)=\dfrac{1}{m!}d^mf(0).x^m$ for all $x$, where $d^kf(0)$ is a $k$-linear form such that $d^kf(0).x^k$ is a polynomial of degree $m$ in the coordinates of $x$.

Answer 3

Here is a variation of the above answer by Ludwik indicating also the sufficient differentiability degree.

Lemma. Let $f:\mathbb{R}^{n}\to \mathbb{R}$ be a $C^{k}$ homogeneous function of integer degree $k\geq0$. Then $f$ is a homogeneous polynomial of degree $k$.

Proof. If $k=0$, then the assumption that $f$ is $C^{0}$ (i.e. continuous) and homogeneous of degree $0$, means that $f(tx)=t^0f(x)=f(x)$ for all $t\geq0$ and $x\in \mathbb{R}^{n}$. In particular, for $t=0$ we get that $f(x)=f(0)$ for all $x\in \mathbb{R}^{n}$, so $f$ is constant.

For $k>0$, applying $\tfrac{\partial}{\partial x_i}$ to both sides of the identity $f(tx)=t^kf(x)$, we get $tf'_{x_i}(tx) = t^kf'_{x_i}(tx)$, whence $f'_{x_i}(tx) = t^{k-1}f'_{x_i}(tx)$, i.e. every partial derivative of $f'_{x_i}$ is a homogeneous $C^{k-1}$ function. Then, by induction on $k$, $f'_{x_i}$ for each $i=1,\ldots,n$, must be a homogeneous polynomial of degree $k-1$. Hence, by Euler's identity, $f(x) = \tfrac{1}{n}\sum\limits_{i=1}^{n} x_i f'_{x_i}(x)$ is a homogeneous polynomial of degree $k$.