1. If is a solution of the Cauchy functional equation which is surjective, but not injective, then has the Darboux property.
2. For every solution of the Cauchy functional equation there exist two non-trivial solutions of the same equation, such that and have the Darboux property and .
These two results were proven in this post. The version presented here is a simplified one, identifying exactly what we need in order to obtain the desired results.
We say that is an additive function if
1. Prove that there exist additive functions which are discontinuous with or without the Darboux Property.
2. Prove that for every additive function there exist two functions which are additive, have the Darboux Property, and .
The second part is similar to Sierpinski’s Theorem which states that every real function can be written as the sum of two real functions with Darboux property.
(A function has the Darboux property if for every , is an interval.)
I will present here an algorithm to find numerically the polygon with sides which minimizes the -th eigenvalue of the Dirichlet Laplacian with a volume constraint.
The first question is: how do we calculate the eigenvalues of a polygon? I adapted a variant of the method of fundamental solutions (see the general 2D case here) for the polygonal case. The method of fundamental solutions consists in finding a function which already satisfies the equation on the whole plane, and see that it is zero on the boundary of the desired shape. We choose the fundamental solutions as being the radial functions where are some well chosen source points and . We search our solution as a linear combination of the functions , so we will have to solve a system of the form
in order to find the desired eigenfunction. Since we cannot solve numerically this system for every we choose a discretization on the boundary of and we arrive at a system of equations like:
and this system has a nontrivial solution if and only if the matrix is singular. The values for which is singular are exactly the square roots of the eigenvalues of our domain .
Problem 1. Let be a point inside a given triangle and denote the feet of the perpendiculars from to the lines respectively. Find such that the quantity
Problem 2. Let and consider all subsets of elements of the set . Each of these subsets has a smallest member. Let denote the arithmetic mean of these smallest numbers. Prove that
Problem 3. Determine the maximum value of where with .
Suppose is a continuous function such that for every we have
Prove that .
Let’s start by definining the Cantor set. Define and . At each step we delete the middle third of all the intervals of to obtain . Note that we obviously have (an easy inductive argument) and . The sets are compact and descending, therefore we can define which is a compact subset of with zero measure and it is called the Cantor set.
Since at each step we remove a middle third of all the intervals in , one way to look at the Cantor set is to look at the ternary representation of the points in it. In the first step, we remove all the elements of which have on their first position in the ternary representation. In the second step we remove those (remaining) which have on the second position, and so on. In the end we are left only with elements of which have only digits in their ternary representation. Using this representation we can construct a bijection between and which maps
where if and if . This proves that the Cantor set is uncountable.
We can construct the Cantor function in the following way. Denote the set (i.e. the set removed in step ). On we let . On we have two intervals: on the left one we let and on the right one we let . We continue like this iteratively, at each step choosing constant on each of the intervals which construct such that the constant on an interval is the mean of the values of neighboring interval values.
We can look at a square matrix and see it as a table of numbers. In this case, matrices and below are completely different:
If instead we look at a square matrix as at a linear transformation things change a lot. Since the transformation is arbitrary, it seems normal that does not act in every direction in the same way, and some directions are privileged in the sense that the transformation is a simple dilatation in those special directions, i.e. there exists and a non-zero vector (the direction) such that . The values and the corresponding vectors are so important for the matrix that they almost characterize it; hence their names are eigenvalue and eigenvector which means own value and own vector (eigen = own in German). It turns out that and above both have the same eigenvalues , and because they are distinct, both the matrices are similar to the diagonal matrix ( and are similar if there exists invertible such that ).