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For real number $a \in \mathbb{R}$, the heaviside step function $H_{a} : \mathbb{R} \to \{0,1\}$ is usually defined as \begin{equation} H_a(x) = \begin{cases} 1, & x \geq a; \\ 0, & x < a. \end{cases} \end{equation} From wikipedia I learned that using distribution theory, the Dirac delta function $\delta_a$ can be formulated as a distributional derivative of the heaviside step function $H_a$. My question is: if we extend the definition of the heaviside step functions to include the case $a = \infty$, i.e., we define an extended real-valued function $H_{\infty}: \mathbb{R} \cup \{\infty\} \to \{0,1\}$ as \begin{equation} H_{\infty}(x) = \begin{cases} 1, & x = \infty; \\ 0, & x \in \mathbb{R}, \end{cases} \end{equation} then is it still possible to extend the Dirac delta function to an extended one $\delta_{\infty}$ as a distributional derivative with respect to $H_{\infty}$? Anyone has an idea? Thanks very much.

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  • $\begingroup$ It could be thought of as a distribution in $\mathcal{E}'$ but generally that space does not have convenient properties to talk about since it's pool is so limited (It does, however, have convenient Fourier transform properties). A general heuristic notion on the usual spaces of distributions, $\mathcal{S}'$ and $\mathcal{D}'$, is that when a distribution's support runs off to infinity, it truly "vanishes". So in the those last two spaces, "$\delta_\infty$"$\equiv 0$, which you can prove rigorously by using the translation operator. $\endgroup$ – Ninad Munshi Sep 27 at 9:49
  • $\begingroup$ You need to see how the distributional derivative follows from the properties of $C^\infty_c(\Bbb{R})$, how do you define an equivalent on $\Bbb{R} \cup \infty$ ? @NinadMunshi If you say $\phi \in C^\infty_c(\Bbb{R} \cup \infty)$ iff $\phi \circ \tan \in C^\infty(\Bbb{R/Z})$ then OP's function is the zero distribution and the distributional derivative of $1_{x > 0}$ becomes $\langle 1_{x > 0}',\phi\rangle =-\langle 1_{x > 0},\phi'\rangle = \phi(\infty)-\phi(0)$. Also note $\varphi$ is Schwartz iff $\varphi \circ \tan$ is smooth and all its derivative vanish at $0$. $\endgroup$ – reuns Sep 27 at 10:52

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