Differentiable function on Rn

May 2010
3
0
(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 \(\displaystyle f:\mathbb{R}^n-0\to \mathbb{R}\) 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.

Is the above jump to Rn-0 ok?

Thanks so much for your help!
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
(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 \(\displaystyle f:\mathbb{R}^n-0\to \mathbb{R}\)
Do you mean \(\displaystyle \mathbb{R}^n- \{0\}\), the set of all real numbers except 0?
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?

One simply stays in the 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.

Is the above jump to Rn-0 ok?

Thanks so much for your help!
 
May 2010
3
0
Yes I meant \(\displaystyle \mathbb{R}^n- \{0\}\); please forgive my poor notation. I'm glad my work was ok, thanks HallsofIvy!