# Grad, Ctsly differentiable function on Rn

#### tierhopeful

EDIT: Sorry, I think I posted this in the wrong section. I have moved it to "
Analysis, Topology and Differential Geometry". Please delete it!

Hi,
I have a past tier exam problem which I would like to check my solution for.

The question: Let p be real. Suppose $$\displaystyle f:\mathbb{R}^n-0\to \mathbb{R}$$ 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?

I would then have $$\displaystyle x f'(\lambda x) = p\lambda^{p-1}f(x)$$, and taking $$\displaystyle \lambda=1$$ gives the desired result.

Onto the general case, fixing $$\displaystyle x=(x_1,\ldots, x_n)\in \mathbb{R}-0$$ and defining $$\displaystyle g(\lambda) = f(\lambda x)$$ again, we have $$\displaystyle g'(\lambda) = p\lambda^{p-1}f(x)$$ and on the other hand, by the chain rule,
$$\displaystyle g'(\lambda) = \frac{d}{d\lambda}f(\lambda x_1, \ldots, \lambda x_n) = x_1 \tfrac{\partial f}{\partial x_1}(\lambda x) + \ldots + x_n\tfrac{\partial f}{\partial x_n}(\lambda x) = x\cdot \nabla f(\lambda x)$$. Taking $$\displaystyle \lambda=1$$ again gives the result.