What are examples of two BV functions $u:\mathbb{R}^2 \to \mathbb{R}^2$ with singular derivative?

More precisely, I'd like to see an example of

  • a function $$u_1 \in BV(\mathbb R^2; \mathbb R^2)$$ with only jump part in the derivative $$Du_1 = D^{jump} u_1$$
  • and of a function with only Cantor part in the derivative: $$u_2 \in BV(\mathbb R^2; \mathbb R^2)$$ with $$Du_2 = D^{cantor} u_2$$

A related more general question is on MathOverflow.


You can turn a one-dimensional BV functions $f$ into higher-dimensional examples by defining a radial function $u(x) = f(\|x\|)$. E.g., if $f(x)=-1$ for $x \le 1$ and $f(x)=0$ for $x>1$, then $u(x) = f(\|x\|)$ as a function from $\mathbb{R}^n$ to $\mathbb{R}$ has as distributional derivative the surface measure on the unit sphere, by the divergence theorem, so it is a BV function with $Du$ consisting only of a "jump" part. Explicitly, if $\phi:\mathbb{R}^2 \to \mathbb{R}^2$ is a smooth test function with compact support, then $$ \int_{\mathbb{R}^2} \phi \cdot \nabla u= -\int_{\mathbb{R}^2} u \,\, \textrm{div}\, \phi = \int_{\|x\|\le 1} \textrm{div}\, \phi = \int_{\|x\|=1} \phi \cdot n \, dS $$ where the first integral is "formal" in the sense that $\nabla u$ does not exist as a function, with the precise meaning given by the second integral (using integration by parts/Green's formula over a large disk), and the last equality is the divergence theorem, with $n$ the normal vector and $dS$ the boundary measure, in this case 1-dimensional length measure on the circle. Since this is true for all test functions, we get that $\nabla u = n \, dS$ in the distributional sense.

A similar example with the Cantor staircase will give you a function whose derivative only has a singularly continuous ("Cantor") part.

  • $\begingroup$ This is interesting. 1. Could you please add a picture of the radial $2$-dimensional version of Cantor Staircase (with Mathematica or Matlab)? 2. Do you have any other genuinely 2-dimensional examples in mind? $\endgroup$
    – Riku
    Apr 26 '19 at 16:51
  • $\begingroup$ Also, I'm not sure that $f(\|x\|)$ satisfies Alberti rank-one theorem. What would the distributional derivative of $f(\|x\|)$ be? $\endgroup$
    – Riku
    Apr 26 '19 at 18:05
  • $\begingroup$ @Riku: Just take a picture of the 1-dimensional Cantor staircase and rotate it about the $y$-axis... The distributional gradient of $u$ is given by the flux through the unit sphere, and it is trivially of rank one here, because the codomain is one-dimensional. (I realize you asked for an example from $\mathbb{R}^2$ to $\mathbb{R}^2$, but you can just compose the given $u$ with a trivial embedding of the real line into the plane, or more generally some Lipschitz embedding.) $\endgroup$ Apr 26 '19 at 20:51
  • $\begingroup$ What do you mean when you say that "The distributional gradient of u is given by the flux through the unit sphere"? $\endgroup$
    – Riku
    Apr 26 '19 at 20:54
  • $\begingroup$ Also, the singularity appears to be on concentric circles around the origin, which does not seem "unidirectional" in the sense of Alberti rank-one theorem. What am I missing? $\endgroup$
    – Riku
    Apr 26 '19 at 20:58

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