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Math Help - Grad, Ctsly differentiable function on Rn

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    Grad, Ctsly differentiable function on Rn

    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 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!
    Last edited by tierhopeful; May 13th 2010 at 06:30 PM. Reason: Wrong forum topic!
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