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Euler’s Homogeneous Function Theorem and Its Gradient Identity

Article Quant Q&A · Author: SRKX
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Summary

The document explains Euler’s theorem for differentiable homogeneous functions. If scaling every input by a positive factor scales the function by that factor raised to degree k, then the dot product of the input vector with the function’s gradient equals k times the function value. For degree one, this yields the decomposition into inputs multiplied by their corresponding partial derivatives.

The proof differentiates the scaling relation with respect to the scale parameter and evaluates at one. It also sketches the converse by solving the resulting differential equation along each ray from the origin. A Euclidean norm is given as an example of a nonlinear degree-one homogeneous function. The result requires continuous differentiability on the stated domain; the note is mathematical background and does not develop a specific quantitative finance application.

Key ideas

  • A homogeneous function scales predictably when all its inputs are multiplied by a common positive factor.
  • For a differentiable degree-k function, the input-gradient dot product equals k times the function value.
  • The identity follows by differentiating the scaling relation with respect to its scale parameter.
  • The theorem also works in reverse under the stated regularity conditions.
  • Degree-one homogeneity does not imply that a function is linear.

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# What is exactly Euler's decomposition?


# 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 by olaker (score 15, accepted)

https://quant.stackexchange.com/a/8912

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.