derivative of homogeneous function

Theorem 1.

Suppose $f\mathrm{:}{\mathrm{R}}^n\mathrm{\to }{\mathrm{R}}^m$ is a differentiable positively homogeneous function of degree $r$. Then $\frac{\mathrm{\partial }\mathit{}f}{\mathrm{\partial }\mathit{}{x}^i}$ is a positively homogeneous function of degree $r\mathrm{-}\mathrm{1}$.

Proof.

By considering component functions if necessary, we can assume that $m=1$. For $\lambda \in ℝ$, let $M_{\lambda }$ be the multiplication map,

$M_{\lambda }:{ℝ}^n$ $\to$ ${ℝ}^n$

$v$ $\mapsto$ $\lambda v.$

For $\lambda >0$ and $v\in {ℝ}^n$, we have

${\displaystyle \frac{\partial f}{\partial x^i}}(\lambda v)$ $=$ ${\displaystyle \frac{\partial (f\circ M_{\lambda }\circ M_{1/\lambda })}{\partial x^i}}(\lambda v)$

$=$ ${\displaystyle \sum _{l=1}^n}{\displaystyle \frac{\partial (f\circ M_{\lambda })}{\partial x^l}}(v){\displaystyle \frac{\partial {(x\mapsto x/\lambda )}^l}{\partial x^i}}(\lambda v)$

$=$ ${\displaystyle \frac{\partial (f\circ M_{\lambda })}{\partial x^i}}(v){\displaystyle \frac{1}{\lambda }}$

$=$ ${\lambda }^{r-1}{\displaystyle \frac{\partial f}{\partial x^i}}(v)$

as claimed. ∎

Title	derivative of homogeneous function
Canonical name	DerivativeOfHomogeneousFunction
Date of creation	2013-03-22 14:45:05
Last modified on	2013-03-22 14:45:05
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	9
Author	matte (1858)
Entry type	Theorem
Classification	msc 15-00