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Thread: Differentiable function on Rn

  1. #1
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    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 $\displaystyle f:\mathbb{R}^n-0\to \mathbb{R}$ 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 $\displaystyle g0,\infty)\to \mathbb{R}$ 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!
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  2. #2
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    Quote Originally Posted by tierhopeful View Post
    (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 $\displaystyle g0,\infty)\to \mathbb{R}$ 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!
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  3. #3
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    Yes I meant $\displaystyle \mathbb{R}^n- \{0\}$; please forgive my poor notation. I'm glad my work was ok, thanks HallsofIvy!
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