# Characterising “jumps” of functions in $\mathbb R^n$ (Distribution Theory, Generalised functions)

Let $$f:\mathbb R\to \mathbb R$$ be a function differentiable on $$\mathbb R\backslash\{0\}$$. Define the jump of the derivatives of $$f$$ by $$\sigma_k =\lim_{\epsilon\to 0} (f^{(k)}(\epsilon)-f^{(k)}(-\epsilon)).$$ $$f$$ defines a distribution in $$\mathcal D'(\mathbb R)$$. Let $$f'$$ be the derivative of $$f$$ in the sense of distributions, and let $$\{f'\}$$ be the distribution given by the derivative $$\frac{df}{dx}$$ in the sense of functions. (This is awkward - please suggest better notation if possible.)

Now, by direct computation, $$f'=\{f'\}+\sigma_0\delta.$$

This question concerns the generalisation of this to higher dimensions. Let's now work in $$\mathbb R^n$$. Let $$S$$ be a surface in $$\mathbb R^n$$ defined by $$S=\{\mathbf x\in\mathbb R^n: F(\mathbf x)=0\}$$, where $$\nabla F\neq 0$$. Let $$f$$ be a function defined on $$\mathbb R^n$$ and differentiable on $$\mathbb R^n\backslash S$$. For $$\mathbf x\in S$$ and a multi-index $$\alpha$$, define $$\sigma_\alpha(x)=\lim_{\mathbf y\to \mathbf x\\ F(\mathbf x)>0} \partial^\alpha f(\mathbf y)-\lim_{\mathbf y\to \mathbf x\\ F(\mathbf x)<0} \partial^\alpha f(\mathbf y)$$

Now, consider the derivative $$\frac{\partial f}{\partial x_1}$$ in the sense of distributions. Let $$\phi\in \mathcal D$$. Under the assumption that on $$S$$, $$x_1=x_1(x_2,...,x_n)$$, $$\langle\frac{\partial f}{\partial x_1},\phi\rangle =-\int \mathbb d x_2...\mathbb d x_n\int f(x_1,..., x_n) \frac{\partial \phi}{\partial x_1} \mathbb dx_1\\ =\int \mathbb d x_2...\mathbb d x_n\int \left(\frac{\partial f(x_1,..., x_n)}{\partial x_1}+\delta_{x_1(x_2,...,x_n)}\sigma_0(x_1(x_2,...,x_n))\right) \phi(x_1,\ldots, x_n) \mathbb dx_1\\ =\int \mathbb d x_2\ldots\mathbb d x_n\left[\int \left(\frac{\partial f(x_1,..., x_n)}{\partial x_1}\phi(x_1,\ldots, x_n)\right) \mathbb dx_1+\phi(x_1(x_2,...,x_n),\ldots, x_n)\sigma_0(x_1(x_2,...,x_n))\right]\\ =\langle \{\frac{\partial f}{\partial x_1}\},\phi \rangle+\int \phi(x_1(x_2,...,x_n),\ldots, x_n)\sigma_0(x_1(x_2,...,x_n)) \mathbb d x_2\ldots\mathbb d x_n$$ Note that the second term of the last expression is $$\int_S \sigma_0 \phi \cos \theta_1 ds$$, where $$\theta_1$$ is the angle between $$x_1$$-axis and the normal of the surface, because the surface element $$ds=\cos \theta_1 \mathbb d x_2\ldots\mathbb d x_n$$.

Now, here is my question: can I establish the result $$\langle\frac{\partial f}{\partial x_1},\phi\rangle =\langle \{\frac{\partial f}{\partial x_1}\},\phi \rangle+\int_S \sigma_0 \phi \cos \theta_1 ds$$ rigorously and formally, without the assumption that $$x_1$$ is a function of other $$x_i$$?

Also: $$\int_S \sigma_0 \phi \cos \theta_1 ds$$ also gives a distribution. Is there a special notation for this distribution?

With $$g,h$$ smooth $$\Bbb{R^n\to R}$$ $$f(x) = g(x) 1_{h(x) > 0}$$ is a distribution, $$\partial_{x_1} f =(\partial_{x_1} g) 1_{h > 0}+g\ (\partial_{x_1}1_{h>0})=(\partial_{x_1} g) 1_{h > 0}+g\ (\partial_{x_1}h)\delta(h)$$ where $$\partial_{x_1}1_{h>0}= (\partial_{x_1}h)\delta(h)$$ is always a distribution but for $$\delta(h)= \underset{\text{in the sense of distributions}}{\lim_{t\to 0}} \frac{1_{h \in [0,t]}}{t}$$ to be a distribution on its own we need that $$\|\nabla h\|$$ doesn't vanish on $$h=0$$ and $$h$$ has locally finitely many vanishing hypersurfaces. If so the second term of
$$\langle \partial_{x_1} f,\phi\rangle= \langle (\partial_{x_1} g) 1_{h > 0},\phi\rangle-\langle 1_{h>0} , \partial_{x_1} (g\phi)\rangle$$
can be evaluated as $$-\langle 1_{h>0} , \partial_{x_1} (g\phi)\rangle =\langle \delta(h) , (\partial_{x_1} h)g\phi\rangle=\int_{h = 0} (\partial_{x_1} h)g\phi d\nu$$ $$\nu$$ is the measure such that for $$h(x)=0$$ as $$r\to 0$$ it approximates $$\frac{1_{h\in [0,r]}}{r}$$ ie. $$\nu(B_r(x)\cap (h=0))\sim \frac{Vol(B_r(x)\cap h\in [0,r])}{r}\sim \frac{S_{n-1}r^{n-1}}{\|\nabla h(x)\|}$$
• Are you assuming that $f$ is zero when $h(x)<0$? What about the general case (the $f$ described in my question)? Thank you. – Ma Joad Dec 31 '19 at 5:06
• Also, what is $1_{h\in [0,r]}$? I think it is an indicator function, but in the last line, it seems to be a set. – Ma Joad Dec 31 '19 at 5:08
• With $g(x) 1_{h(x) > 0}+g_2(x) 1_{h(x) < 0}$ (if $\| \nabla h\|$ doesn't vanish) you have all the functions with a jump discontinuity at $h=0$ – reuns Dec 31 '19 at 5:09
• Thanks. What about the measure $\nu$? Why the second expression in the last line (with $B_r(x)$ in it) is the surface element? – Ma Joad Dec 31 '19 at 5:14
• Also, $g$ doesn't need to be $\mathbb R^n\to \mathbb R$, does it? $g$ just need to be defined in the region where $h>0$. – Ma Joad Dec 31 '19 at 7:11