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Math Help - weak derivative

  1. #1
    Junior Member
    Joined
    May 2008
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    50

    weak derivative

    I have the function f(x)= x+1 for x greater or equal to 0 and -x-1 for x less than 0 ,
    I have to prove that f has no weak derivative.

    I computed <Df,g>= 2g(0)+\int_0^\infty g(x)dx - \int_{-\infty}^0 g(x)dx , where g is a test function. But how do I prove that this is of the form \int_{-infty}^\infty h(x)g(x)dx and that Df is not in L^1_{loc}(R)?

    Thanks.
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  2. #2
    Super Member
    Joined
    Apr 2009
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    México
    Posts
    721

    Re: weak derivative

    Notice that f has classical derivatives on (-\infty,0) and on (0,\infty), so any weak derivative of f must coincide, on these intervals, with the strong derivative, so the only possible choice of derivative is g(x)=1 if x>0 and g(x)=-1 for x<0. Now take h \in C_c^{\infty}(\mathbb{R}), we calculate

    \int_{\mathbb{R}} fh'= 2\left( xh(x)|_{0}^{\infty}-\int_0^{\infty} h(x)dx\right) - 2h(0)
    \int_{\mathbb{R}} gh= 2\int_0^{\infty} h(x)dx

    By the definition of weak derivative, the sum of these two quantities must be zero, so we arrive at h(0)=0. Since h was arbitrary we get our contradiction. Notice that the contradiction works only because we allow the support of h to be outside the sets where we know the function is differentiable.
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