Since we know that a jump discontinuity will produce $\delta$ measure with the jump size. I was wondering is there a way to take the distributional derivative of $$g(x) = \left\{ \begin{array}{l l} \frac{1}{\sqrt{x}} & \quad \text{if } x\in (0,1]\\ 0 & \quad \text{if } x\in [-1,0)\\ \end{array} \right. \\$$ I chose $\frac{1}{\sqrt{x}}$ because this makes $g$ a integrable function, and the distributional derivative of $g$ can be explicitly written out as an ingetral $$\langle T_g ',\phi \rangle := -\langle T_g, \phi' \rangle = \int_{-1}^1 g \phi' dx$$ However integration by parts formula does not work here, because $\frac{1}{\sqrt{x}}$ is not absolute continuous since the derivative of $\frac{1}{\sqrt{x}}$ is no longer integrable in $(0,1]$. Here I tried the standard approach, for $\epsilon >0$ we define a new function $$g_\epsilon(x) = \left\{ \begin{array}{l l} \frac{1}{\sqrt{x}} & \quad \text{if } x\in (\epsilon,1]\\ \frac{1}{\sqrt{\epsilon}} & \quad \text{if } x\in (0, \epsilon]\\ 0 & \quad \text{if } x\in [-1,0)\\ \end{array} \right. \\$$ which gives us the following, $$\int_\epsilon^1 \frac{1}{\sqrt x} \phi' dx = -\frac{1}{\sqrt \epsilon} \phi(\epsilon) -\int_\epsilon^1 \left(\frac{1}{\sqrt x}\right)' \phi dx $$ $$\int_0^\epsilon \frac{1}{\sqrt \epsilon} \phi' dx = \frac{1}{\sqrt \epsilon} \phi(\epsilon) - \frac{1}{\sqrt \epsilon} \phi(0).$$ Combine them we get $$\langle T_{g_\epsilon} ', \phi\rangle:=\langle T_{g_\epsilon} , \phi'\rangle = \frac{1}{\sqrt \epsilon} \phi(0) + \int_\epsilon^1 \left(\frac{1}{\sqrt x}\right)' \phi dx $$ But taking $\epsilon \rightarrow 0$ does not give us anything meaningful. Could you guys give me some example calculations for this type of distributional derivative.

  • $\begingroup$ Distributional derivatives are not always $L^1$ functions or nice functions, that's one of the key points. I feel like you are assuming you should somehow get a nice function in the end? $\endgroup$ – Thompson Aug 2 '16 at 15:29
  • $\begingroup$ What kind of representation are you looking for? $f \mapsto -\int_0^1 \frac{f'(x)}{\sqrt{x}} dx$ seems like a reasonable way to write a distribution to me. If you wanted to integrate by parts further, you would want to put another derivative on $f$, not on $x^{-1/2}$. $\endgroup$ – Ian Aug 2 '16 at 15:30
  • $\begingroup$ In $\mathbb{R}^3$, the laplacian of $\frac{1}{|x|}$ in the sense of distribution is $-4\pi\delta_0$, I thought we could get something looks as simple as this one, maybe it is not the case. $\endgroup$ – Xiao Aug 2 '16 at 15:35
  • $\begingroup$ (and essential singularities are a completely different thing) $\endgroup$ – reuns Aug 2 '16 at 16:34

for $a > 0$, let $g_a(x) = x^{-1/2}1_{x \in (a,1)}$. It is clear that $\lim_{a \to 0^+}g_a = g$ in the sense of distributions, so $g' = \lim_{a \to 0^+} g_a'$.

you have $$g_a' = - \frac{1}2 x^{-3/2}1_{x \in (a,1)} + a^{-1/2}\delta(x-a)-\delta(x-1)$$

Let $\phi \in C^\infty_c$

  • if $\phi(0) = 0$ then $\phi(x) \sim \phi'(0) x$ as $x \to 0$, so

$$\langle \phi, g' \rangle = \lim_{a \to 0^+}\langle \phi, g_a' \rangle = \langle \phi, - \frac{1}2 x^{-3/2}1_{x \in (0,1)} -\delta(x-1) \rangle$$

  • now if $\varphi \in C^\infty_c$ is constant on $[0,1]$ then

$$\lim_{a \to 0} \langle \varphi, g_a' \rangle = -\lim_{a \to 0} \langle \varphi', g_a \rangle = 0$$

Hence, taking any $\varphi \in C^\infty_c$ constant on $[0,1]$ and $\varphi(0) = 1$, using that $\phi = \phi - \phi(0)\varphi+\phi(0)\varphi$ you get $$\langle \phi,g' \rangle = \langle \phi-\phi(0)\varphi,- \frac{1}2 x^{-3/2}1_{x \in (0,1)} -\delta(x-1) \rangle$$

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  • $\begingroup$ Thank you very much. I have never seen this kind of calculation in the examples. So in your last line, the second pairing is still integration right? So $\phi - \phi(0)\psi$ will remove the non integrable part in $x^{-3/2}$. $\endgroup$ – Xiao Aug 2 '16 at 17:08
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    $\begingroup$ @Xiao the idea is that you have to decompose $C^\infty_c$ into two subspaces : the one where $\phi(0) = 0$, and the one where $\varphi$ is constant on $[0,1]$. on the former, $g'$ is the distribution $-1/2 x^{-3/2} 1_{x \in (0,1)}-\delta(x-1)$, on the latter $g'$ is the zero distribution. and decomposing $\phi$ as $(\phi-\phi(0)\varphi) +\phi(0)\varphi$ you get the result $\endgroup$ – reuns Aug 2 '16 at 17:11
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    $\begingroup$ @Xiao it is not so different to using the principal value for seing $\langle ., \frac{1}{x} \rangle$ as a distribution, where again you consider the subspace $\phi(0) = 0$, and the subspace $\varphi$ even $\endgroup$ – reuns Aug 2 '16 at 17:14

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