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Consider the characteristic function of $[0, \infty)^2$ (i.e. a function $u: \mathbb{R^2} \to \mathbb{R}$ given by $u(x, y) = 1$ if $x \geq 0$, $y \geq 0$ and $u(x, y) = 0$ otherwise). I want to compute $\partial_x \partial_y u$, where the derivatives are distributional, to avoid blowup at jump discontinuities. Since $u$ projected onto the $y$-axis is just the Heaviside function, $\partial_y u(x, y) = \delta_0(y)$ if $x \geq 0$ and $\partial_y u(x, y) = 0$ otherwise.

I visualize this $\partial_y u$ as a map which sends the plane to itself, except on the ray $\{(x, 0) \in \mathbb{R}^2: x \geq 0\}$, in which case all points are "mapped" to $+\infty$. The trouble is, this gives me no intuition how to approach the computation of its second derivative $\partial_x$. My best guess is that I should expect $\partial_x \partial_y u$ to be identically $0$ outside of the origin, where it is $+\infty$; i.e. that $\partial_x \partial_y u = \delta_0$. But this also feels like "cheating" somehow: I would expect $\delta_0$ to only be the derivative of a finite multiple of the Heaviside function, not of a distribution which is already infinite!

Is this method of approaching differentiation of distributions appropriate? If I bashed out convolutions of $C_0^\infty$ functions I could probably (dis?)prove that $\partial_x \partial_y u = \delta_0$, but I'd still feel unsure why it's true/false, because I don't know how to imagine the partial derivatives of a distribution.

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You have $u(x,y) = H(x) \otimes H(y)$ so $$\partial_x \partial_y u(x,y) = \partial_x H(x) \, \partial_y H(y) = \delta(x) \delta(y) = \delta(x,y).$$ (Here $x$ and $y$ should be viewed as placeholders rather than values.)

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  • $\begingroup$ To visualize a distribution $T$ simply look at the sequence of $C^\infty$ functions $T \ast \phi_n$ with $\phi_n(x) = n^d \phi(nx), \phi \in C^\infty_c,\int_{\mathbb{R}^d} \phi(x)dx = 1$ ! $\endgroup$
    – reuns
    Commented Sep 15, 2017 at 17:09
  • $\begingroup$ @reuns. Why did you put your comment under my answer? And isn't your answer like saying "To visualize $\mathbb R^4$ simply look at $\mathbb R^n$ and then take $n=4$"? $\endgroup$
    – md2perpe
    Commented Sep 15, 2017 at 20:43
  • $\begingroup$ What do you mean with $\mathbb{R}^4$ ? It is like saying real numbers are limits of sequences of rationals. Do you realize $T \ast \phi_n \to T$ in the sense of distributions where $T \ast \phi_n$ is a sequence of $C^\infty$ functions, and $\partial^\alpha T = \lim_{n \to \infty} \partial^\alpha (T \ast \phi_n)$ ? Why do you refuse to explain distributions this way to those asking for an intuitive explanation ? $\endgroup$
    – reuns
    Commented Sep 15, 2017 at 21:01
  • $\begingroup$ How do you visualize $T * \phi_n$ when you don't even know how to visualize $T$? $\endgroup$
    – md2perpe
    Commented Sep 15, 2017 at 21:20
  • $\begingroup$ Here $T = \partial_x \partial_y u$ thus $T \ast \phi_n = \partial_x \partial_y \Phi_n$ where $\Phi_n = u \ast \phi_n$ is just a smoothed version of $u(x,y) = 1_{x \ge 0, y \ge 0}$ $\endgroup$
    – reuns
    Commented Sep 15, 2017 at 21:24

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