2
$\begingroup$

Let $F:\mathbb{R} \to \mathbb{R}$ be a $L$-Lipschitz function.

Consider the function $$G(x,y) = \chi_{\{x \le F(y)\}}(x,y),$$ where $\chi$ is the indicator function.

  • How can I plot this function using MATLAB or Mathematica in the case, for example $F(y) = y$?
  • Is it true that $G$ is Lipschitz continuous (at least with respect to one of the variables?

Follow-up:

  1. Is $G$ a BV function?
  2. What is its distributional derivative?
$\endgroup$
0
0
$\begingroup$

If $G$ is continuous then it's constant because $G$ can assume only values $0$ and $1$.

The set $E=\left\{(x, y)|x\leq F(y)\right\}$ is the subgraph of function $F$ where $x$ axis is changed with $y$ axis, and its topological boundary is $$ \partial E=\left\{(x, y)|x=F(y)\right\} $$ and coincides with $F$ graph. Because $F$ is Lipscitz then $E$ boundary is also Lipscitz then $E$ has locally finite perimeter that implies $\chi_E(x, y)$ is a locally bounded variation function and its exterior normal $\nu_E$ is well defined on $\mathcal H^1$-almost every point over $\partial E$.

Also $$ \nu_E[F(y), y]=\left(-\frac{1}{\sqrt{1+\left\lvert F'(y)\right\rvert^2}}; \frac{F'(y)}{\sqrt{1+\left\lvert F'(y)\right\rvert^2}}\right) $$ because $F$ derivative is defined almost everywhere.

Due to De-Giorgi Structure theorem we have $$ D\chi_E(A)=\int_{A\cap\partial E}\nu_E(x, y)d\mathcal H^1 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.