# Step functions on a plane

In this post, I give a possible definition for step functions in ${\mathbb{R}^2}$ and a related problem and its possible solution.

1. Problem

Let ${\alpha(s) = (x(s),y(s))}$ be the arc length parametrization of a plane, smooth, closed, convex curve, of length ${L}$. Let ${J:[0,L]\rightarrow\mathbb{R}}$ be a smooth and Bounded variation function, with ${J(0) = J(L)}$. Let ${\Omega}$ be the set of all interior points of the region bounded by the curve ${\alpha(s)}$.

The problem is to determine ${f:\mathbb{R}^2 \rightarrow \mathbb{R}}$, such that

1. ${f(x) = 0}$, for all ${x}$ not in the inetrior of ${\Omega}$.

2. Given any ${s_0 \in (0,L)}$, and let ${p_0 = \alpha(s_0)}$,

$\displaystyle {\lim_{p\rightarrow p_0}}_{p\in\Omega}f(p) = J(s_0)$

3. ${f\in C^{\infty}(\Omega \setminus R)}$, where ${R}$ is a set such that ${\mathcal{H}^1(R) = 0}$, meaning 1-Hausdorff measure of ${R}$ is 0.

4. The total variation ${V(f,\Omega)}$ is minimum possible, where

$\displaystyle V(f,\Omega) = \int_{\Omega}|Df|$

and ${Df}$ is the weak/distributional gradient of ${f}$.

Motivation : ${f}$ is my definition of a step function in ${\mathbb{R}^2}$

2. Solution

I roughly sketch a solution without a proof, and also a possible minimum possible value for total variation of ${f}$.

Let ${f}$ assume values of ${J}$ on the boundary curve ${\alpha}$. We construct infinite number of scaled down versions of the curve ${\alpha}$ with scale factor ranging from ${1}$ and converging to ${0}$, and on each curve we let ${f}$ take values of corresponding scaled down version of ${J}$. If curve is scaled by ${c}$, then new function is taken as

$\displaystyle J_c(x) = J(\frac{x}{c})$

where ${J_c}$ is defined on ${(0,Lc)}$. ${c\in(0,1)}$

This is illustrated roughly in the picture below.

This way we construct ${f}$ on entire ${\Omega}$ and it is smooth(as ${J}$ and ${\alpha}$ are smooth) except possibly at ${O}$, the centroid of region ${\Omega}$, where it is not continuous. But this discontinuty at ${O}$ has no effect on Variation of ${f}$ in ${\Omega}$.

Variation of ${f}$ for this case, is given as

$\displaystyle V(f,\Omega) = V_0^L(J) \mathcal{H}^2(\Omega)$

I am yet to sketch a proof that this is the minimum possible variation for ${f}$.

PS : ${V_0^L(J)}$ is the total variation of ${J}$ in ${(0,L)}$, and ${\mathcal{H}^2(\Omega)}$ is the two dimensional Hausdorff measure of the set ${\Omega}$.