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.
Sets of finite perimeter which do not contain any open ball
Posted: February 27, 2012 in Geometry, shape optimizationTags: shape optimization
There isn’t such a function
Posted: February 9, 2012 in Analysis, OlympiadTags: continuous, function
Prove that there is no continuous function such that
and
.
Functional equation on positive integers
Posted: February 9, 2012 in UncategorizedTags: functional equation
Let be a fixed positive integer. Find all functions
such that
we have
.
.
Arcs on a circle
Posted: February 3, 2012 in Analysis, Combinatorics, Geometry, OlympiadTags: Combinatorics, continuity, Geometry
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
Posted: January 3, 2012 in Measure Theory, Real AnalysisTags: measure space
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
Posted: November 26, 2011 in shape optimizationTags: shape optimization
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.
Posted: November 5, 2011 in Partial Differential Equations, shape optimizationTags: green formula, partial differential equations, shape optimization
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
Posted: November 5, 2011 in Partial Differential Equations, shape optimizationTags: isoperimetric, shape optimization
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
Posted: November 4, 2011 in shape optimizationTags: shape optimization
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:
- 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.
- 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.
- 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).
- 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.
- 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
Posted: October 13, 2011 in Functional Analysis, Partial Differential Equations, Sobolev SpacesTags: Banach, sobolev
Let and
be three Banach spaces with norms
and
. Assume that
with compact injection and that
with continuous injection. Prove that
satisfying
.
Applications:
- Prove that for every
there exists
satisfying
.
- 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
Posted: October 7, 2011 in Partial Differential Equations, Sobolev SpacesTags: sobolev
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
Posted: October 6, 2011 in TopologyTags: limit, sequence, topology, trick
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
Posted: October 6, 2011 in Functional AnalysisTags: compact, functional, operators
Suppose is a compact operator and
is a sequence in
such that
(i.e. converges weakly). Prove that
strongly in
.
Generalized Poincare Inequality
Posted: October 5, 2011 in Inequalities, Partial Differential EquationsTags: poincare, sobolev
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
Posted: September 28, 2011 in Functional Analysis, Linear Algebra, TopologyTags: Banach, operators
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
Posted: September 27, 2011 in Higher Algebra, Linear AlgebraTags: linear
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
Posted: September 27, 2011 in Functional AnalysisTags: bounded, functional, operators
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
Posted: September 8, 2011 in "Normal" Problem Solving, Analysis, Measure Theory, Real AnalysisTags: measurable, measure
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
.
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
.
Let be a polynomial with real coefficients of degree
. Suppose that
is an integer for all
. Prove that
for all distinct integers
.


