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?


1 Answer 1


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. $$\nu(B_r(x)\cap (h=0))\sim \frac{Vol(B_r(x)\cap h\in [0,r])}{r}\sim \frac{S_{n-1}r^{n-1}}{\|\nabla h(x)\|}$$

  • $\begingroup$ Are you assuming that $f$ is zero when $h(x)<0$? What about the general case (the $f$ described in my question)? Thank you. $\endgroup$
    – Ma Joad
    Dec 31, 2019 at 5:06
  • $\begingroup$ 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. $\endgroup$
    – Ma Joad
    Dec 31, 2019 at 5:08
  • $\begingroup$ 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 $h=0$ $\endgroup$
    – reuns
    Dec 31, 2019 at 5:09
  • $\begingroup$ Thanks. What about the measure $\nu$? Why the second expression in the last line (with $B_r(x)$ in it) is the surface element? $\endgroup$
    – Ma Joad
    Dec 31, 2019 at 5:14
  • $\begingroup$ 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 $h>0$. $\endgroup$
    – Ma Joad
    Dec 31, 2019 at 7:11

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