mathematics - What is exactly Euler's decomposition?

I have often seen the following statement in different paper:

As $\sigma$ is homogeneous and of degree 1, we use Euler decomposition and write $\sigma(x)=\sum_{i=1}^n x_i \frac{\partial \sigma(x)}{\partial x_i}$

The thing is, I am trying to find on wikipedia where this comes from and I can't quite find it. Could someone point me to a reference where this Euler decomposition is explained or give me a brief explanation here?

Answer

A function $f : \mathbb R^n\backslash\{0\} →\mathbb R$ is called (positive) homogeneous of degree $k$ if

$$f(\lambda \mathbf x) = \lambda^k f(\mathbf x) \,$$ for all $\lambda > 0$. Here $k$ can be any complex number. The homogeneous functions are characterized by

Euler's Homogeneous Function Theorem. Suppose that the function $f : \mathbb R^n \backslash\{0\} →\mathbb R$ is continuously differentiable. Then $f$ is homogeneous of degree $k$ if and only if

$$\mathbf{x} \cdot \nabla f(\mathbf{x})\equiv \sum_{i=1}^nx_i\frac{\partial f}{\partial x_i} = kf(\mathbf{x}).$$

The result follows at once by differentiating both sides of the equation $f(\lambda \mathbf x) = \lambda^k f(\mathbf x)$ with respect to $\lambda$, applying the chain rule, and choosing $\lambda=1$.

The converse holds by integrating. Specifically, let $\phi(\lambda) = f(\lambda \mathbf{x})$. Since $\lambda \mathbf{x} \cdot \nabla f(\lambda\mathbf{x})= k f(\lambda \mathbf{x})$, we have that

$$\phi'(\lambda) = \mathbf{x} \cdot \nabla f(\lambda \mathbf{x}) = \frac{k}{\lambda} f(\lambda \mathbf{x}) = \frac{k}{\lambda} \phi(\lambda).$$ Thus, $\phi'(\lambda) - \frac{k}{\lambda} \phi(\lambda) = 0.$ This implies $\phi(\lambda) = \phi(1) \lambda^k$. Therefore, $$f(\lambda \mathbf{x}) = \phi(\lambda) = \lambda^k \phi(1) = \lambda^k f(\mathbf{x}),$$ so $f$ is homogeneous of degree $k$.

Edit. Contrary to popular belief, there exist nonlinear homogeneous functions of degree $1$, e.g.

$$f(\mathbf x)=\sqrt{\sum_{i=1}^n x_i^2}.$$