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What is the derivatibe of 0

  1. Jul 8, 2010 #1
    I would have said it is 0, but then why is it that a twice derivable function is a function like, for example, f(x) = x. I've been studying maths for a while, but I had never asked myself this question until I came across the Heaviside function:

    [tex] H(x) = 1 \ if\ x\geq 0\ and\ H(x)=0\ if\ x<0[/tex]

    The derivative of this function (in the distributional sense) is the Dirac delta function:

    Let [tex]\varphi \in S [/tex], the Schwartz space then:

    [tex] \langle T_{H}^{(1)},\varphi \rangle = - \int_{- \infty}^{+ \infty} H(x) \varphi '(x) dx = - \int_{0}^{+ \infty} \varphi '(x) dx = -[\varphi (x)]_{0}^{\infty} = \varphi (0) = \langle \delta, \varphi \rangle = \delta [/tex]

    And not,

    [tex] \langle T_{H}^{(1)},\varphi \rangle = \int_{- \infty}^{+ \infty} H'(x) \varphi (x) dx = \int_{0}^{+ \infty} 0 \cdot \varphi (x) dx + \int_{- \infty}^{0} 0 \cdot \varphi (x) dx = 0 [/tex]

    Which implies that H(x) is not derivable in the 'normal' sense (which implies 0 doesn't have a derivative).

    EDIT: I just realized the Heaviside function is not differentiable at 0 (it is not even continuous)!
    Last edited: Jul 8, 2010
  2. jcsd
  3. Jul 8, 2010 #2


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    No, it doesn't imply that 0(x) doesn't have a derivative at all.
    It implies that the Heaviside function doesn't have a derivative in the normal sense.

    The dirac delta is the WEAK derivative of the Heaviside function, H'(x) in your second line is a meaningless expression for a function that was to be defined at ALL points.
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