# Differentiable function on Rn

• May 13th 2010, 05:29 PM
tierhopeful
Differentiable function on Rn
(I moved this from the calculus section)

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

The question: Let p be real. Suppose $f:\mathbb{R}^n-0\to \mathbb{R}$ is continuously differentiable, and satisfies
$f(\lambda x) = \lambda^pf(x)$ for all $x\neq 0$ and for all $\lambda>0$.
Let $\nabla f(x)$ denote the gradient of f at x and $\cdot$ the dot product. Prove that
$x \cdot \nabla f(x) = pf(x)$ for all $x\neq 0$.

I first considered the n=1 case: Fixing $x\neq 0$, define $g:(0,\infty)\to \mathbb{R}$ by $g(\lambda) = f(\lambda x)$. We have $g(\lambda) = \lambda^p f(x)$ for all $\lambda>0$. Differentiating with respect to $\lambda$ gives $g'(\lambda) = p\lambda^{p-1}f(x)$.

On the other hand, $g'(\lambda) = \tfrac{d}{d\lambda} f(\lambda x) = x f'(\lambda x)$. 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 $x f'(\lambda x) = p\lambda^{p-1}f(x)$, and taking $\lambda=1$ gives the desired result.

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

Thanks so much for your help!
• May 14th 2010, 01:04 AM
HallsofIvy
Quote:

Originally Posted by tierhopeful
(I moved this from the calculus section)

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

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

Do you mean $\mathbb{R}^n- \{0\}$, the set of all real numbers except 0?
Quote:

is continuously differentiable, and satisfies
$f(\lambda x) = \lambda^pf(x)$ for all $x\neq 0$ and for all $\lambda>0$.
Let $\nabla f(x)$ denote the gradient of f at x and $\cdot$ the dot product. Prove that
$x \cdot \nabla f(x) = pf(x)$ for all $x\neq 0$.

I first considered the n=1 case: Fixing $x\neq 0$, define $g:(0,\infty)\to \mathbb{R}$ by $g(\lambda) = f(\lambda x)$. We have $g(\lambda) = \lambda^p f(x)$ for all $\lambda>0$. Differentiating with respect to $\lambda$ gives $g'(\lambda) = p\lambda^{p-1}f(x)$.

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

One simply stays in the restricted domain!

Quote:

I would then have
Quote:

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

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

Yes I meant $\mathbb{R}^n- \{0\}$; please forgive my poor notation. I'm glad my work was ok, thanks HallsofIvy!