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.

RZUrE

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}.

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