# derivative of homogeneous function

###### Theorem 1.

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

###### Proof.

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

 $\displaystyle M_{\lambda}\colon\mathbbmss{R}^{n}$ $\displaystyle\to$ $\displaystyle\mathbbmss{R}^{n}$ $\displaystyle v$ $\displaystyle\mapsto$ $\displaystyle\lambda v.$

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

 $\displaystyle\frac{\partial f}{\partial x^{i}}(\lambda v)$ $\displaystyle=$ $\displaystyle\frac{\partial(f\circ M_{\lambda}\circ M_{1/\lambda})}{\partial x% ^{i}}(\lambda v)$ $\displaystyle=$ $\displaystyle\sum_{l=1}^{n}\frac{\partial(f\circ M_{\lambda})}{\partial x^{l}}% (v)\,\frac{\partial(x\mapsto x/\lambda)^{l}}{\partial x^{i}}(\lambda v)$ $\displaystyle=$ $\displaystyle\frac{\partial(f\circ M_{\lambda})}{\partial x^{i}}(v)\,\frac{1}{\lambda}$ $\displaystyle=$ $\displaystyle\lambda^{r-1}\frac{\partial f}{\partial x^{i}}(v)$

as claimed. ∎

Title derivative of homogeneous function DerivativeOfHomogeneousFunction 2013-03-22 14:45:05 2013-03-22 14:45:05 matte (1858) matte (1858) 9 matte (1858) Theorem msc 15-00