Let us consider the function $f(x)=e^{-|x-a|}\chi_{\{x\leq a\}}$. I want to compute the distributional derivative. Let us consider a test function $\phi\in\mathcal{C}_0^{\infty}(\mathbb{R})$. Now \begin{align*} \int_{-\infty}^{\infty}f(x)\phi'(x)\ dx &=\int_{-\infty}^ae^{-|x-a|}\phi'(x)\ dx\\ &=\int_{-\infty}^ae^{x-a}\phi'(x)\ dx\\ &=\left[e^{x-a}\phi(x)\right]_{-\infty}^a-\int_{-\infty}^ae^{x-a}\phi(x)\ dx\\ &=\phi(a)-\int_{-\infty}^ae^{x-a}\phi(x)\ dx\\ &=-\left(\int_{-\infty}^ae^{x-a}\phi(x)\ dx-\phi(a)\right)\\ &=-\left(\int_{-\infty}^{\infty}(e^{-|x-a|}\chi_{\{x\leq a\}}-\delta(x-a)\phi(x)\ dx\right). \end{align*}Hence distributional derivative $$f'(x)=e^{x-a}-\delta(x-a).$$Is that logical or correct?


1 Answer 1


Foreword: I use " ' " as the distributional derivative.

Also $e^{-|x-a|}\chi_{x\leq a} = e^{x-a}\chi_{x\leq a}$

Consider this, shortly:

$$\left(e^{-|x-a|}\chi_{x\leq a}\right)' = e^{x-a}\chi_{x\leq a} + e^{x-a}\delta(x-a) = e^{x-a}\chi_{x\leq a} - \delta(x-a) $$

  • $\begingroup$ @JonathanZsupportsMonicaC, aha... but/and I wonder whether that might be a typo... $\endgroup$ Commented Nov 7, 2022 at 22:58
  • $\begingroup$ Yes there was typo. But why in my calculation minus sign coming before dirac delta. Any wrong? $\endgroup$
    – Jacobi
    Commented Nov 8, 2022 at 0:18
  • $\begingroup$ No, this is the above computation that is wrong, the derivative of $\chi_{x\leq a}$ is $-\delta(x-a)$ (the function is decreasing). Your computation @Jacobi is correct. $\endgroup$
    – LL 3.14
    Commented Nov 8, 2022 at 9:10

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