Sets of finite perimeter which do not contain any open ball

Some time ago, I wondered myself if I could find a very small ball included in a set of finite perimeter (see definition for set of finite perimeter here and here). It turns out, that the answer is no, there mustn’t be any open ball included in a set of finite perimeter. My teacher gave me the following nice example.

There isn’t such a function

Prove that there is no continuous function such that and .

Functional equation on positive integers

Let be a fixed positive integer. Find all functions such that we have .

.

Arcs on a circle

Suppose that we have a finite set of arcs on a circle, with the property that every two of them intersect. Prove that there exists a diameter which intersects all arcs.

Pointwise convergence implies other type of convergence

Let be a measure space for which . Let . Suppose that is a sequence in such that and exists for -a.e. . Prove that .

PHD 4324 (Indiana)

Shape Optimization Course – Day 2

Speaker – Giuseppe Buttazzo

The problem of finding a minimal resistance body (due to Newton) consists of finding the shape of a body which travels in a straight line through a fluid when we are given a certain fixed section of it, orthogonal to the flow of the fluid. The classical problem presented here makes some assumptions about the fluid and about the movement, which are not really accurate, taking into account the physics of the fluids, but which turns to be a good approximation in the case the liquid is rare, such as the movement of the airplanes. The assumptions made are:

• the single shock property: every particle which hits the body is reflected and it doesn’t influence the behavior of other particles in the fluid, moreover, if a particle hits the body, it never touches the body after that moment.
• the part of the body below the fixed orthogonal section is neglected, which means that it is considered that its resistance is zero.

New definition for the perimeter of a set.

As you can see in this post we can define the perimeter of a Lebesgue measurable set relative to an open set (if it is the usual perimeter) of a set by using the formula

It is important that this definition would agree with the classical definition for sets, namely, if is a bounded open set of class then , where represents the surface element on .

Existence Result for the Isoperimetric Problems

The tricky part is how to define the perimeter of a Lebesgue measurable set with finite perimeter. This can be done considering the space of bounded variation functions, denoted . By definition we have for an open set that .  Here we denoted by the space of continuously differentiable functions with compact support in . Because of the density of the space of infinitely differentiable functions with compact support in in the space , we could have replaced by in the above definition. You could take a look at this blog post for a detailed description of or at the Wikipedia page.

We say that a set of finite Lebesgue measure is a set of finite perimeter in if its characteristic function belongs to . This means that the distributional gradient is a vector valued measure with finite total variation. The total variation is called the perimeter of .

In the same way we can define the perimeter of a Lebesgue measurable set relative to an open set . We say that is a set of finite perimeter relative to if the characteristic function belongs to the space .

Shape Optimization Course – Day 1

The main speakers of the course were Giuseppe Buttazzo and Edouard Oudet. See more details in the Shape Optimization page.

Day 1. Speaker – Giuseppe Buttazzo

Optimization problems have the following form:   where is a functional (sometimes called cost) and is the set of admissible objects. A Shape Optimization Problem has the following form: , where again is a functional (e.g. area or perimeter) and is a class of admissible domains (e.g. bounded area, convex, connected). There are a few aspects of a shape optimization problem, each important in its own way:

1. Existence of a solution. This is not a trivial question, because sometimes optimal forms do not exist. A general method is to provide a topology for such that the map is lower semicontinuous and the sublevels of are compact. (i.e. is compact for every ). This is not easy in general, because the two facts are in contradiction. For the compacity we need fewer open sets, but for the lower continuity of we need more open sets. The key is to find a balance between the two. There is not a general topology for ; changing the functional we may need to change the topology we use, or the class of admissible domains.
2. Uniqueness. This is not generally the case for shape optimization problems, because sometimes, if we have a solution, its translates or rigid motions of the shape are are also a solution.
3. Regularity. In some problems, we may get existence, and we may wonder if the shapes we found are regularly enough (e.g of class , etc).
4. Necessary conditions of optimality. These are conditions for which we have the following implication: is optimal implies satisfies . Maybe sometimes not all objects which satisfy are optimal.
5. Numerical approximation. This is is an important tool, since in many cases it turns out that the optimal shape is not what we would expect. Numerical approximation can give us some idea of what we are looking for and what should we try and prove theoretically. See for example the discussion on the Newton problem, where many people tried to prove that the optimal solution in case of a disk is radial. After seeing numerically that this is not the case, the theoretical proof of the existence of a better non-radial solution appeared.

A lemma of J. L. Lions

Let and be three Banach spaces with norms and . Assume that with compact injection and that with continuous injection. Prove that

satisfying .

Applications:

1. Prove that for every there exists satisfying.
2. Pick . Prove that for every there exists such that .

Source: Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, 2011

Sobolev space impossible extension

Let . And for (!!! Correction: ) consider . Prove that:

1) does not have Lipschitz boundary (i.e. its boundary is not locally the graph of Lipschitz functions).

2) .

3) For every ball which contains , there is no function in which extends .

This problem gives a counter example which states that if doesn’t have Lipschitz boundary, then there may be no extension to some functions in to greater Sobolev spaces.

Nice characterization of convergence

Suppose is a topological space, and consider the sequence with the following property:

• every subsequence has a subsequence converging to .

Then .

Compact operator maps weakly convergent sequences into strong convergent sequences

Suppose is a compact operator and is a sequence in such that (i.e. converges weakly). Prove that strongly in .

Generalized Poincare Inequality

Consider a bounded domain with Lipschitz boundary. If is a non-zero, closed subspace in , which does not contain the non-zero constant functions, then there is a constant , depending on , such that .

Note that this generalizes the usual Poincare inequality, which says that the above inequality holds for some on the space , a space which does not contain the non-zero constant functions.

Interesting inequality involving Banach spaces and an operator

Let be two Banach spaces with norms . Let (space of linear bounded operators ) be such that is closed and . Let denote another norm on which is weaker than , i.e. .

Prove that there exists a constant such that .

Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Chapter 2

Max Dimension for a linear space of singular matrices

Denote by a vector space of singular () matrices. What is the maximal dimension of such a space.

Denote by a vector space of matrices with rank smaller or equal to . What is the maximal dimension of such a space?

Slick condition for boundedness of an operator

Consider a Banach space, and the space of real functionals on . Suppose that is a linear operator such that for all . Prove that is bounded.

Infinitely Countable Sigma Algebra

A famous result in measure theory is the following

There is no -algebra on a set which is infinitely countable.

This plainly states that if is a -algebra on a space , then is finite or .

IMC 2011 Day 2 Problem 5

Let be a convex polygon in the plane. Define for all the operation which replaces with a new polygon where is the symmetric of with respect to the perpendicular bisector of . Prove that .

IMC 2011 Day 2 Problem 4

Let be a polynomial with real coefficients of degree . Suppose that is an integer for all . Prove that for all distinct integers .