How does $\Delta f$ behave where $f$ has jump discontinuities in first-order partials? In the wake of a prior question, I've solidified my understanding that for a one-dimensional function $f(x)$, when its first derivative has a jump discontinuity of height $h$ at $x_0$, we can regard the second derivative (for the purpose of eventually integrating it) as having a $\delta(x-x_0)$ factor plus a locally continuous function. What if, as in my former question, a continuous function is nonzero everywhere outside of the bounded region $\Omega \subset \mathbb{R}^2$, and its first-order partials have jump discontinuities at the boundary: how then should we derive the form of $\Delta f$ ?
My guess is that if we define $\nabla f$ on the boundary by taking the limit from inside $\Omega$, then we'll get something akin to $\Delta f = \delta(0) \| \nabla f \|$ on the boundary, and $0$ otherwise. Or is it possible that when integrating $\Delta f$ over a path hitting a single boundary point $x_0$, the resulting "$\delta$-factor" depends on what direction the path hits $x_0$?
 A: Let $\Omega\subset\mathbb{R}^2$ be a bounded domain with a $C^1$ boundary. Let $f$ be zero outside of $\Omega$ and $C^2$ inside, with $f=0$ on $\partial\Omega$. Also assume that all of the partial derivatives of $f$ extend continuously to $\partial\Omega$. Let us define
  $$
  g(x) = \begin{cases}
    \Delta f(x) &\text{if $x\in\Omega$},\\\\
    0 &\text{otherwise}.
    \end{cases}
  $$
Now, $g$ is not equal to $\Delta f$ on all of $\mathbb{R}^2$. In fact, $\Delta f$ may not even exist as a function on $\mathbb{R}^2$. But it does exist as a distribution. Let us see how $\Delta f$, the distribution, compares with $g$.
Let $\varphi$ be a smooth test function. Then
  $$
  \langle\Delta f,\varphi\rangle = \langle f,\Delta\varphi\rangle
    = \int_{\mathbb{R}^2}f(x)\Delta\varphi(x)\,dx.
  $$
Since $f=0$ outside of $\Omega$, we need only integrate over $\Omega$. We can then use Green's second identity to obtain
  $$
  \int_\Omega f(x)\Delta\varphi(x)\,dx
    = \int_\Omega g(x)\varphi(x)\,dx
    + \int_{\partial\Omega}
    (f\partial_{\bf n}\varphi - \varphi\partial_{\bf n}f)\,d\sigma,
  $$
where $\partial_{\bf n}$ is the derivative in the direction of the outward normal, and $\sigma$ is the surface measure on $\partial\Omega$. (Since we are in $\mathbb{R}^2$, this is just the arclength measure.) Since $f=0$ on $\partial\Omega$, we get
  \begin{align*}
  \langle\Delta f,\varphi\rangle
    &= \int_\Omega g\varphi\\,dx
    -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\\,d\sigma\\\\
  &= \int_{\mathbb{R}^2} g\varphi\\,dx
    -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\\,d\sigma\\\\
  &= \langle g,\varphi\rangle
    -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\\,d\sigma.
  \end{align*}
What this shows is that $\Delta f = g + \nu$, where $\nu$ is the distribution that maps $\varphi$ to
  $$
  \langle\nu,\varphi\rangle
    = -\int_{\partial\Omega} (\partial_{\bf n}f)\varphi\,d\sigma.
  $$
For example, if $\partial_{\bf n}f(x)=-1$ for all $x\in\partial\Omega$, then $\nu$ is just the surface measure on $\partial\Omega$. This is the case in the example I gave in my answer to the other question. And in that case, the surface measure on the boundary of $[0,\pi]$ in $\mathbb{R}$ is just two point masses, one at $0$ and one at $\pi$.
Lastly, to address a comment from the other question, the Laplacian is still rotationally invariant when interpreted as a distributional derivative. To prove this, we apply $\Delta f$ to a smooth test function, then move the Laplacian over to the test function (where it acts in the classical way), and then utilize the rotational invariance of the classical Laplacian.
A fairly short and accessible reference for tempered distributions, which is free online, is Chapter 11 of Applied Analysis by Hunter and Nachtergaele. Also, there is a chapter on the Laplace operator in Folland.
Edit:
To address the question in the comments, the Radon transform of a tempered distribution is defined by $\langle Rf,\varphi\rangle=\langle f,R^*\varphi\rangle$. More precisely, it is $\langle Rf,\varphi\rangle_{S^1\times\mathbb{R}}=\langle f,R^*\varphi\rangle_{\mathbb{R}^2}$. The inner product on the right is the usual $L^2$ inner product on $\mathbb{R}^2$; the inner product on the left is defined by
  $$
  \langle f,g\rangle_{S^1\times\mathbb{R}}
    = \frac1{2\pi}\int_0^{2\pi}\int_{\mathbb{R}}
    f(\theta,s)g(\theta,s)\,ds\,d\theta.
  $$
It follows that
  \begin{align*}
  \langle R(\Delta f),\varphi\rangle
    &= \langle \Delta f,R^\*\varphi\rangle
    = \langle f,\Delta(R^\*\varphi)\rangle\\\\
  &= \langle f,R^\*(\partial_s^2\varphi)\rangle
    = \langle Rf,\partial_s^2\varphi\rangle
    = \langle \partial_s^2(Rf),\varphi\rangle,
  \end{align*}
and so $R(\Delta f)=\partial_s^2(Rf)$, even in the distributional sense.
What is relevant for your other question is how to compute $R\nu$. My guess is that you are asking if
  $$
  R\nu = -\sum_{x\in\partial\Omega\cap L}\partial_{\bf n}f(x).
  $$
The intuition behind this formula does not work, because it does not account for the angle between $L$ and $\partial\Omega$ at the point of intersection. I will leave it as an exercise to show that if $\sigma$ is the arclength measure on $S^1$, that is, if
  $$
  \langle\sigma,\varphi\rangle = \int_0^{2\pi}\varphi(e^{i\theta})\,d\theta,
  $$
then
  $$
  R\sigma(\theta,s) = 2\frac1{\sqrt{1 - s^2}}\chi_{[-1,1]}(s).
  $$
The $2$ comes from the two points where the line intersects the circle, and the factor of $|1-s^2|^{-1/2}$ comes from the angle of intersection. If $L$ is the line corresponding to $(\theta,s)$ and $\alpha(L)\in[0,\pi/2]$ is the angle with which $L$ intersects $S^1$, then $|s|=\cos\alpha(L)$. We therefore have
  $$
  R\sigma(\theta,s) = 2\csc\alpha(L)\chi_{[-1,1]}(s).
  $$
The natural generalization to $\nu$ would be
  $$
  R\nu(L) = -\sum_{x\in\partial\Omega\cap L}
    \partial_{\bf n}f(x)\csc\alpha(x),
  $$
where, for $x\in\partial\Omega\cap L$, we define $\alpha(x)\in[0,\pi/2]$ to be the angle of intersection between $L$ and the tangent line to $\partial\Omega$ at $x$.
A: Let the boundary $S=\partial \Omega$ be smooth enough. Applying the second Green's formula for test function $\varphi$ we get
$$
\int_{\mathbb R^n}(f\Delta \varphi -\Delta f\varphi)\mathrm dx=
\int_{\Omega}(f\Delta \varphi -\Delta f\varphi)\mathrm dx=
$$
$$
\int_{S}\left(f\frac{\partial \varphi}{\partial\bar n}-
\varphi\frac{\partial f}{\partial \bar n}\right)\mathrm dx=
-\int_{S}\frac{\partial f}{\partial\bar n}\mathrm dx,
$$
where $\bar n$ is the unit outword normal to $S$. So in the sense of distributions $\Delta f=\{\Delta f\}-\frac{\partial\Delta f^+}{\partial\bar n}\delta_S$, where $\{\Delta f\}$ is  $\Delta f$ outside $S$. The derivative in the last term is the limite values of $ \frac{\partial\Delta f}{\partial\bar n}$ inside the domain.
