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Math Help - 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 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 0,\infty)\to \mathbb{R}" alt="g0,\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.

    Is the above jump to Rn-0 ok?

    Thanks so much for your help!
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  2. #2
    MHF Contributor

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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 f:\mathbb{R}^n-0\to \mathbb{R}
    Do you mean \mathbb{R}^n- \{0\}, the set of all real numbers except 0?
    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 0,\infty)\to \mathbb{R}" alt="g0,\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!

    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.

    Is the above jump to Rn-0 ok?

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