is continuously differentiable, and satisfies

$\displaystyle f(\lambda x) = \lambda^pf(x)$ for all $\displaystyle x\neq 0$ and for all $\displaystyle \lambda>0$.

Let $\displaystyle \nabla f(x)$ denote the gradient of f at x and $\displaystyle \cdot$ the dot product. Prove that

$\displaystyle x \cdot \nabla f(x) = pf(x)$ for all $\displaystyle x\neq 0$.

I first considered the n=1 case: Fixing $\displaystyle x\neq 0$, define $\displaystyle g

0,\infty)\to \mathbb{R}$ by $\displaystyle g(\lambda) = f(\lambda x)$. We have $\displaystyle g(\lambda) = \lambda^p f(x)$ for all $\displaystyle \lambda>0$. Differentiating with respect to $\displaystyle \lambda$ gives $\displaystyle g'(\lambda) = p\lambda^{p-1}f(x)$.

On the other hand, $\displaystyle g'(\lambda) = \tfrac{d}{d\lambda} f(\lambda x) = x f'(\lambda x)$.

*I do not know how to justify this step however: f is defined on R-0, so how does one go about differentiating with respect to a variable that has a restricted domain?*