## Relaxation of the Anisotropic Perimeter – Part 2

(Read this previous post to see the context and the statement of the problem)

While it is usually difficult to construct a recovery sequence for an arbitrary (or all ), the construction is often much simpler if the target function has some special structure.

It is useful to consider a subset which is dense in such that for every there exists such that in and .

A common choice for is the class of characteristic functions of smooth sets, which is dense in in the sense that for every finite perimeter set there exists a sequence of smooth sets such that and as . Note that this is different than the BV-norm convergence, but it is the notion of convergence very well suited for BV functions. In Approximation of Free-Discontinuity Problems by A. Braides a proof of the limsup estimate is given using the dense class of smooth functions.

When dealing with the anisotropic perimeter, a more natural way of working would be choosing sets whose boundary is polyhedral (piecewise affine). The main reason for doing that is the fact that polyhedrons have constant normals corresponding to one face. To be able to choose as the class of characteristic functions of polyhedral subsets of (with finite perimeter, of course), we must have a density result similar to the smooth case.

G. Cortesani and R. Toader prove the following result in their article A density result in SBV with respect to non-isotropic energies:

First, let’s write some preliminaries. Let be an open and bounded subset of and denote the space of all functions which satisfy the following properties:

(i) is essentially closed ().

(ii) is a polyhedral set (the intersection of with the union of a finite number of simplexes.

(iii) for every .

Theorem Assume that is Lipschitz, and let . Then there exists a sequence such that

strongly in ,

strongly in

for every and for every upper semicontinuous function such that for every and .

This is not quite exactly what we want here, because note that the approximate sequence is not made of characteristic functions. Still, we can construct a sequence of characteristic functions starting from in the following way: choose such that and have the same jump set (which is polyhedral). Moreover, we choose the side of the jump set on which is such that is smaller (i.e. choose the side on which most of the set defined by lies).

What we really need in our problem is the convergence and the last part with the integral on the jump set. Note that since we do not use in our problem and the rest of the expressions in the fourth property above depend only on the jump set, it follows that the property remains true. As for the convergence, we have:

which tends to zero, since the symmetric differente will appear only in a small neighbourhood of the jump set whose measure goes to ans goes to .

The above argument allows us to consider the subclass of of functions with polyhedral jump set.

In order to prove the upper estimate for the -convergence problem we can consider the optimal profile problem

It is immediate to see that if we apply Young’s inequality we find that

and the equality holds if satisfies the differential equation . We will consider a solution of the problem

Note that truncating between and makes the functional smaller, which means that is contained between and (this is logical… to go from a value to another in an optimal way, you don’t go oscilating). This also proves that is bounded and therefore is Lipschitz continuous.

Note that the optimal profile realizes the minimal cost of the functional while passing from to on the real line. We would like to get from to in finite time, and one way to do this is to consider slightly dilatated profiles:

Note that if we denote

then we have

Now we assume that is the characteristic function of a polyhedral set . Fix and denote . Choose such that passes from to in the interval . For each we will consider rectangular neighborhoods of the -dimensional facets of the points at distance of smaller than , but we choose these rectangular boxes so that boxes corresponding to different facets do not intersect. It remains a zone at distance smaller than of the jump set which is not covered by the rectangular boxes, and this region has a volume of order . We also consider the signed distance .

Define the recovery sequence on by

and extend it on the entire with the same Lipschitz constant of order using Kirszbraun’s Theorem.

Let’s estimate first the part of the integral on :

It is easy to see that because of the fact that if we take the limsup above we get zero so is negligible in the limsup estimate.

On we have for small enough:

Taking limsup as , and then making gives us the desired conclusion.