All posts by Jean-Pierre Merx

Counterexamples around Dini’s theorem

In this article we look at counterexamples around Dini’s theorem. Let’s recall:

Dini’s theorem: If \(K\) is a compact topological space, and \((f_n)_{n \in \mathbb N}\) is a monotonically decreasing sequence (meaning \(f_{n+1}(x) \le f_n(x)\) for all \(n \in \mathbb N\) and \(x \in K\)) of continuous real-valued functions on \(K\) which converges pointwise to a continuous function \(f\), then the convergence is uniform.

We look at what happens to the conclusion if we drop some of the hypothesis.

Cases if \(K\) is not compact

We take \(K=(0,1)\), which is not closed equipped with the common distance. The sequence \(f_n(x)=x^n\) of continuous functions decreases pointwise to the always vanishing function. But the convergence is not uniform because for all \(n \in \mathbb N\) \[\sup\limits_{x \in (0,1)} x^n = 1\]

The set \(K=\mathbb R\) is closed but unbounded, hence also not compact. The sequence defined by \[f_n(x)=\begin{cases}
0 & \text{for } x < n\\ \frac{x-n}{n} & \text{for } n \le x < 2n\\ 1 & \text{for } x \ge 2n \end{cases}\] is continuous and monotonically decreasing. It converges to \(0\). However, the convergence is not uniform as for all \(n \in \mathbb N\): \(\sup\{f_n(x) : x \in \mathbb R\} =1\). Continue reading Counterexamples around Dini’s theorem

A ring whose characteristic is a prime having a zero divisor

Consider a ring \(R\) whose characteristic is a composite number \(p=ab\) with \(a,b\) integers greater than \(1\). Then \(R\) has a zero divisor as we have \[0=p.1=(a.b).1=(a.1).(b.1).\]

What can we say of a ring \(R\) having zero divisors? It is known that the rings \(\mathbb{Z}/p.\mathbb{Z}\) where \(p\) is a prime are fields and therefore do not have zero divisors. Is this a general fact? That is, does a ring whose characteristic is a prime do not have zero divisors?

The answer is negative and we give below a counterexample.

Let’s consider the field \(\mathbb{F}_p = \mathbb{Z}/p.\mathbb{Z}\) where \(p\) is a prime and the product of rings \(R=\mathbb{F}_p \times \mathbb{F}_p\). One can verify following facts:

  • \(R\) additive identity is equal to \((0,0)\).
  • \(R\) multiplicative identity is equal to \((1,1)\).
  • \(R\) is commutative.
  • The characteristic of \(R\) is equal to \(p\) as for \(n\) integer, we have \(n.(1,1)=(n.1,n.1)\) which is equal to \((0,0)\) if and only if \(p\) divides \(n\).

However, \(R\) does have zero divisors as following identity holds: \[(1,0).(0,1)=(0,0)\]

The Smith Volterra Cantor Set

In Cantor set article, I presented the Cantor set which is a null set having the cardinality of the continuum. I present here a modification of the Cantor set named the Smith-Volterra-Cantor set.

Construction of the Smith-Volterra-Cantor set

The Smith-Volterra-Cantor set (also named SVC set below) \(S\) is a subset of the real segment \(I=[0,1]\). It is built by induction:

  • Starting with \(S_0=I\)
  • \(S_1=[0,\frac{3}{8}] \cup [\frac{5}{8},1]\)
  • If \(S_n\) is a finite disjoint union of segments \(s_n=\cup_k \left[a_k,b_k\right]\), \[S_{n+1}=\bigcup_k \left(\left[a_k,\frac{a_k+b_k}{2}-\frac{1}{2^{2n+3}}\right] \cup \left[\frac{a_k+b_k}{2}+\frac{1}{2^{2n+3}},b_k\right]\right)\]

Continue reading The Smith Volterra Cantor Set

No minimum at the origin but a minimum along all lines

We look here at an example, from the Italian mathematician Giuseppe Peano of a real function \(f\) defined on \(\mathbb{R}^2\). \(f\) is having a local minimum at the origin along all lines passing through the origin, however \(f\) does not have a local minimum at the origin as a function of two variables.

The function \(f\) is defined as follows
\[\begin{array}{l|rcl}
f : & \mathbb{R}^2 & \longrightarrow & \mathbb{R} \\
& (x,y) & \longmapsto & f(x,y)=3x^4-4x^2y+y^2 \end{array}\] One can notice that \(f(x, y) = (y-3x^2)(y-x^2)\). In particular, \(f\) is strictly negative on the open set \(U=\{(x,y) \in \mathbb{R}^2 \ : \ x^2 < y < 3x^2\}\), vanishes on the parabolas \(y=x^2\) and \(y=3 x^2\) and is strictly positive elsewhere. Consider a line \(D\) passing through the origin. If \(D\) is different from the coordinate axes, the equation of \(D\) is \(y = \lambda x\) with \(\lambda > 0\). We have \[f(x, \lambda x)= x^2(\lambda-3x)(\lambda -x).\] For \(x \in (-\infty,\frac{\lambda}{3}) \setminus \{0\}\), \(f(x, \lambda x) > 0\) while \(f(0,0)=0\) which proves that \(f\) has a local minimum at the origin along the line \(D \equiv y – \lambda x=0\). Along the \(x\)-axis, we have \(f(x,0)=3 x^ 4\) which has a minimum at the origin. And finally, \(f\) also has a minimum at the origin along the \(y\)-axis as \(f(0,y)=y^2\).

However, along the parabola \(\mathcal{P} \equiv y = 2 x^2\) we have \(f(x,2 x^2)=-x^4\) which is strictly negative for \(x \neq 0\). As \(\mathcal{P}\) is passing through the origin, \(f\) assumes both positive and negative values in all neighborhood of the origin.

This proves that \(f\) does not have a minimum at \((0,0)\).

Two matrices A and B for which AB and BA have different minimal polynomials

We consider here the algebra of matrices \(\mathcal{M}_n(\mathbb F)\) of dimension \(n \ge 1\) over a field \(\mathbb F\).

It is well known that for \(A,B \in \mathcal{M}_n(\mathbb F)\), the characteristic polynomial \(p_{AB}\) of the product \(AB\) is equal to the one (namely \(p_{BA}\)) of the product of \(BA\). What about the minimal polynomial?

Unlikely for the characteristic polynomials, the minimal polynomial \(\mu_{AB}\) of \(AB\) maybe different to the one of \(BA\).

Consider the two matrices \[
A=\begin{pmatrix}
0 & 1\\
0 & 0\end{pmatrix} \text{, }
B=\begin{pmatrix}
0 & 0\\
0 & 1\end{pmatrix}\] which can be defined whatever the field we consider: \(\mathbb R, \mathbb C\) or even a field of finite characteristic.

One can verify that \[
AB=A=\begin{pmatrix}
0 & 1\\
0 & 0\end{pmatrix} \text{, }
BA=\begin{pmatrix}
0 & 0\\
0 & 0\end{pmatrix}\]

As \(BA\) is the zero matrix, its minimal polynomial is \(\mu_{BA}=X\). Regarding the one of \(AB\), we have \((AB)^2=A^2=0\) hence \(\mu_{AB}\) divides \(X^2\). Moreover \(\mu_{AB}\) cannot be equal to \(X\) as \(AB \neq 0\). Finally \(\mu_{AB}=X^2\) and we verify that \[X^2=\mu_{AB} \neq \mu_{BA}=X.\]

Counterexamples around Fubini’s theorem

We present here some counterexamples around the Fubini theorem.

We recall Fubini’s theorem for integrable functions:
let \(X\) and \(Y\) be \(\sigma\)-finite measure spaces and suppose that \(X \times Y\) is given the product measure. Let \(f\) be a measurable function for the product measure. Then if \(f\) is \(X \times Y\) integrable, which means that \(\displaystyle \int_{X \times Y} \vert f(x,y) \vert d(x,y) < \infty\), we have \[\int_X \left( \int_Y f(x,y) dy \right) dx = \int_Y \left( \int_X f(x,y) dx \right) dy = \int_{X \times Y} f(x,y) d(x,y)\] Let's see what happens when some hypothesis of Fubini's theorem are not fulfilled. Continue reading Counterexamples around Fubini’s theorem

Counterexamples around Banach-Steinhaus theorem

In this article we look at what happens to Banach-Steinhaus theorem when the completness hypothesis is not fulfilled. One form of Banach-Steinhaus theorem is the following one.

Banach-Steinhaus Theorem
Let \(T_n : E \to F\) be a sequence of continuous linear maps from a Banach space \(E\) to a normed space \(F\). If for all \(x \in E\) the sequence \(T_n x\) is convergent to \(Tx\), then \(T\) is a continuous linear map.

A sequence of continuous linear maps converging to an unbounded linear map

Let \(c_{00}\) be the vector space of real sequences \(x=(x_n)\) eventually vanishing, equipped with the norm \[\Vert x \Vert = \sup_{n \in \mathbb N} \vert x_n \vert\] For \(n \in \mathbb N\), \(T_n : E \to E\) denotes the linear map defined by \[T_n x = (x_1,2 x_2, \dots, n x_n,0,0, \dots).\] \(T_n\) is continuous as for \(\Vert x \Vert \le 1\), we have
\begin{align*}
\Vert T_n x \Vert &= \Vert (x_1,2 x_2, \dots, n x_n,0,0, \dots) \Vert\\
& = \sup_{1 \le k \le n} \vert k x_k \vert \le n \Vert x \Vert \le n
\end{align*} Continue reading Counterexamples around Banach-Steinhaus theorem

A finite extension that contains infinitely many subfields

Let’s consider \(K/k\) a finite field extension of degree \(n\). The following theorem holds.

Theorem: the following conditions are equivalent:

  1. The extension contains a primitive element.
  2. The number of intermediate fields between \(k\) and \(K\) is finite.

Our aim here is to describe a finite field extension having infinitely many subfields. Considering the theorem above, we have to look at an extension without a primitive element.

The extension \(\mathbb F_p(X,Y) / \mathbb F_p(X^p,Y^p)\) is finite

For \(p\) prime, \(\mathbb F_p\) denotes the finite field with \(p\) elements. \(\mathbb F_p(X,Y)\) is the algebraic fraction field of two variables over the field \(\mathbb F_p\). \(\mathbb F_p(X^p,Y^p)\) is the subfield of \(\mathbb F_p(X,Y)\) generated by the elements \(X^p,Y^p\). Continue reading A finite extension that contains infinitely many subfields

Counterexamples around connected spaces

A connected space is a topological space that cannot be represented as the union of two or more disjoint nonempty open subsets. We look here at unions and intersections of connected spaces.

Union of connected spaces

The union of two connected spaces \(A\) and \(B\) might not be connected “as shown” by two disconnected open disks on the plane.

union-connected-spaces-image
The union of two connected spaces might not be connected.

However if the intersection \(A \cap B\) is not empty then \(A \cup B\) is connected.

Intersection of connected spaces

The intersection of two connected spaces \(A\) and \(B\) might also not be connected. An example is provided in the plane \(\mathbb R^2\) by taking for \(A\) the circle centered at the origin with radius equal to \(1\) and for \(B\) the segment \(\{(x,0) \ : \ x \in [-1,1]\}\). The intersection \(A \cap B = \{(-1,0),(1,0)\}\) is the union of two points which is not connected.

Differentiability of multivariable real functions (part2)

Following the article on differentiability of multivariable real functions (part 1), we look here at second derivatives. We consider a function \(f : \mathbb R^n \to \mathbb R\) with \(n \ge 2\).

Schwarz’s theorem states that if \(f : \mathbb R^n \to \mathbb R\) has continuous second partial derivatives at any given point in \(\mathbb R^n\), then for \((a_1, \dots, a_n) \in \mathbb R^n\) and \(i,j \in \{1, \dots, n\}\):
\[\frac{\partial^2 f}{\partial x_i \partial x_j}(a_1, \dots, a_n)=\frac{\partial^2 f}{\partial x_j \partial x_i}(a_1, \dots, a_n)\]

A function for which \(\frac{\partial^2 f}{\partial x \partial y}(0,0) \neq \frac{\partial^2 f}{\partial y \partial x}(0,0)\)

We consider:
\[\begin{array}{l|rcl}
f : & \mathbb R^2 & \longrightarrow & \mathbb R \\
& (0,0) & \longmapsto & 0\\
& (x,y) & \longmapsto & \frac{xy(x^2-y^2)}{x^2+y^2} \text{ for } (x,y) \neq (0,0)
\end{array}\] Continue reading Differentiability of multivariable real functions (part2)