Euler’s theorem on homogeneous functions

Theorem 1 (Euler).

Let $f\mathit{}\mathrm{(}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_k\mathrm{)}$ be a smooth homogeneous function of degree $n$. That is,

$$f(t{x}_1,\mathrm{\dots },t{x}_k)=t^{n}f(x_1,\mathrm{\dots },x_k).$$(*)

Then the following identity holds

$$x_1\frac{\partial f}{\partial x_1}+\mathrm{\cdots }+x_k\frac{\partial f}{\partial x_k}=nf.$$

Proof.

By homogeneity, the relation ((*) ‣ 1) holds for all $t$. Taking the t-derivative of both sides, we establish that the following identity holds for all $t$:

$$x_1\frac{\partial f}{\partial x_1}(t{x}_1,\mathrm{\dots },t{x}_k)+\mathrm{\cdots }+x_k\frac{\partial f}{\partial x_k}(t{x}_1,\mathrm{\dots },t{x}_k)=n{t}^{n-1}f(x_1,\mathrm{\dots },x_k).$$

To obtain the result of the theorem, it suffices to set $t=1$ in the previous formula.∎

Sometimes the differential operator $x_1{\displaystyle \frac{\partial }{\partial x_1}}+\mathrm{\cdots }+x_k{\displaystyle \frac{\partial }{\partial x_k}}$ is called the Euler operator. An equivalent way to state the theorem is to say that homogeneous functions are eigenfunctions of the Euler operator, with the degree of homogeneity as the eigenvalue.