derivative of homogeneous function

Theorem 1.

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

Proof.

By considering component functions if necessary, we canassume that $m=1$.For $\lambda \in ℝ$, let $M_{\lambda }$ be themultiplication 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.∎