Counterexamples around Fubini’s theorem

October 31, 2015 Jean-Pierre Merx Leave a comment

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 →

Topology

Counterexamples around Banach-Steinhaus theorem

October 24, 2015 Jean-Pierre Merx Leave a comment

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 →

Algebra

A finite extension that contains infinitely many subfields

October 18, 2015 Jean-Pierre Merx Leave a comment

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

Theorem: the following conditions are equivalent:

The extension contains a primitive element.
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 →

Topology

Counterexamples around connected spaces

October 11, 2015 Jean-Pierre Merx Leave a comment

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.

Analysis

Differentiability of multivariable real functions (part2)

October 4, 2015 Jean-Pierre Merx Leave a comment

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) →

Algebra

A group isomorphic to its automorphism group

September 27, 2015 Jean-Pierre Merx Leave a comment

We consider a group \(G\) and we look at its automorphism group \(\text{Aut}(G)\). Can \(G\) be isomorphic to
\(\text{Aut}(G)\)? The answer is positive and we’ll prove that it is the case for the symmetric group \(S_3\).

Consider the morphism \[
\begin{array}{l|rcl}
\Phi : & S_3 & \longrightarrow & \text{Aut}(S_3) \\
& a & \longmapsto & \varphi_a \end{array}\]
where \(\varphi_a\) is the inner automorphism \(\varphi_a : x \mapsto a^{-1}xa\). It is easy to verify that \(\Phi\) is indeed a group morphism. The kernel of \(\Phi\) is the center of \(S_3\) which is having the identity for only element. Hence \(\Phi\) is one-to-one and \(S_3 \simeq \Phi(S_3)\). Therefore it is sufficient to prove that \(\Phi\) is onto. As \(|S_3|=6\), we’ll be finished if we prove that \(|\text{Aut}(S_3)|=6\).

Generally, for \(G_1,G_2\) groups and \(f : G_1 \to G_2\) a one-to-one group morphism, the image of an element \(x\) of order \(k\) is an element \(f(x)\) having the same order \(k\). So for \(\varphi \in \text{Aut}(S_3)\) the image of a transposition is a transposition. As the transpositions \(\{(1 \ 2), (1 \ 3), (2 \ 3)\}\) generate \((S_3)\), \(\varphi\) is completely defined by \(\{\varphi((1 \ 2)), \varphi((1 \ 3)), \varphi((2 \ 3))\}\). We have 3 choices to define the image of \((1 \ 2)\) under \(\varphi\) and then 2 choices for the image of \((1 \ 3)\) under \(\varphi\). The image of \((2 \ 3)\) under \(\varphi\) is the remaining transposition.

Finally, we have proven that \(|\text{Aut}(S_3)|=6\) as desired and \(S_3 \simeq \text{Aut}(S_3)\).

Topology

Playing with interior and closure

September 20, 2015 Jean-Pierre Merx Leave a comment

Let’s play with the closure and the interior of sets.

To start the play, we consider a topological space \(E\) and denote for any subspace \(A \subset E\): \(\overline{A}\) the closure of \(A\) and \(\overset{\circ}{A}\) the interior of \(A\).

Warm up with the closure operator

For \(A,B\) subsets of \(E\), the following results hold: \(\overline{\overline{A}}=\overline{A}\), \(A \subset B \Rightarrow \overline{A} \subset \overline{B}\), \(\overline{A \cup B} = \overline{A} \cup \overline{B}\) and \(\overline{A \cap B} \subset \overline{A} \cap \overline{B}\).

Let’s prove it.
\(\overline{A}\) being closed, it is equal to its closure and \(\overline{\overline{A}}=\overline{A}\).

Suppose that \(A \subset B\). As \(B \subset \overline{B}\), we have \(A \subset \overline{B}\). Also, \(\overline{B}\) is closed so it contains \(\overline{A}\), which proves \(\overline{A} \subset \overline{B}\).

Let’s consider \(A,B \in E\) two subsets. As \(A \subset A \cup B\), we have \(\overline{A} \subset \overline{A \cup B}\) and similarly \(\overline{B} \subset \overline{A \cup B}\). Hence \(\overline{A} \cup \overline{B} \subset \overline{A \cup B}\). Conversely, \(A \cup B \subset \overline{A} \cup \overline{B}\) and \(\overline{A} \cup \overline{B}\) is closed. So \(\overline{A \cup B} \subset \overline{A} \cup \overline{B}\) and finally \(\overline{A \cup B} = \overline{A} \cup \overline{B}\).

Regarding the inclusion \(\overline{A \cap B} \subset \overline{A} \cap \overline{B}\), we notice that \(A \cap B \subset \overline{A} \cap \overline{B}\) and that \(\overline{A} \cap \overline{B}\) is closed to get the conclusion.

However, the implication \(\overline{A} \subset \overline{B} \Rightarrow A \subset B\) doesn’t hold. For a counterexample, consider the space \(E=\mathbb R\) equipped with the topology induced by the absolute value distance and take \(A=[0,1)\), \(B=(0,1]\). We have \(\overline{A}=\overline{B}=[0,1]\).

The equality \(\overline{A} \cap \overline{B} = \overline{A \cap B}\) doesn’t hold as well. For the proof, just consider \(A=[0,1)\) and \(B=(1,2]\). Continue reading Playing with interior and closure →

Analysis

A discontinuous real convex function

September 13, 2015 Jean-Pierre Merx Leave a comment

Consider a function \(f\) defined on a real interval \(I \subset \mathbb R\). \(f\) is called convex if: \[\forall x, y \in I \ \forall \lambda \in [0,1]: \ f((1-\lambda)x+\lambda y) \le (1-\lambda) f(x) + \lambda f(y)\]

Suppose that \(I\) is a closed interval: \(I=[a,b]\) with \(a < b\). For \(a < s < t < u < b\) one can prove that: \[\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}\le\frac{f(u)-f(t)}{u-t}.\] It follows from those relations that \(f\) has left-hand and right-hand derivatives at each point of the interior of \(I\). And therefore that \(f\) is continuous at each point of the interior of \(I\).
Is a convex function defined on an interval \(I\) continuous at all points of the interval? That might not be the case and a simple example is the function: \[\begin{array}{l|rcl}
f : & [0,1] & \longrightarrow & \mathbb R \\
& x & \longmapsto & 0 \text{ for } x \in (0,1) \\
& x & \longmapsto & 1 \text{ else}\end{array}\]

It can be easily verified that \(f\) is convex. However, \(f\) is not continuous at \(0\) and \(1\).

Topology

Counterexamples to Banach fixed-point theorem

August 30, 2015 Jean-Pierre Merx Leave a comment

Let \((X,d)\) be a metric space. Then a map \(T : X \to X\) is called a contraction map if it exists \(0 \le k < 1\) such that \[d(T(x),T(y)) \le k d(x,y)\] for all \(x,y \in X\). According to Banach fixed-point theorem, if \((X,d)\) is a complete metric space and \(T\) a contraction map, then \(T\) admits a fixed-point \(x^* \in X\), i.e. \(T(x^*)=x^*\).

We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled.

First, let’s consider \[\begin{array}{l|rcl}
f : & \mathbb R & \longrightarrow & \mathbb R \\
& x & \longmapsto & x+1 \end{array}\] For all \(x,y \in \mathbb R\) we have \(\vert f(x)-f(y) \vert = \vert x- y \vert\). \(f\) is not a contraction, but an isometry. Obviously, \(f\) has no fixed-point.

We now prove that a map satisfying \[d(g(x),g(y)) < d(x,y)\] might also not have a fixed-point. A counterexample is the following map \[\begin{array}{l|rcl} g : & [0,+\infty) & \longrightarrow & [0,+\infty) \\ & x & \longmapsto & \sqrt{1+x^2} \end{array}\] Since \[g^\prime(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for all } \xi \in [0, +\infty),\] by the mean value theorem \[\vert g(x) - g(y)| = \vert g^\prime(\xi)\vert |x-y| < |x-y| \text{ for all } x, y \in [0, +\infty).\] However \(g\) has no fixed-point. Finally, let's have a look to a space \((X,d)\) which is not complete. We take \(a,b \in \mathbb R\) with \(0 < a < 1\) and for \((X,d)\) the space \(X = \mathbb R \setminus \{\frac{b}{1-a}\}\) equipped with absolute value distance. \(X\) is not complete. Consider the map \[\begin{array}{l|rcl} h : & X & \longrightarrow & X \\ & x & \longmapsto & ax + b \end{array}\] \(h\) is well defined as for \(x \neq \frac{b}{1-a}\), \(h(x) \neq \frac{b}{1-a}\). \(h\) is a contraction map as for \(x,y \in \mathbb R\) \[\vert h(x)-h(y) \vert = a \vert x - y \vert \] However, \(h\) doesn't have a fixed-point in \(X\) as \(\frac{b}{1-a}\) is the only real for which \(h(x)=x\).

Math Counterexamples

Counterexamples around Fubini’s theorem

Counterexamples around Banach-Steinhaus theorem

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

A finite extension that contains infinitely many subfields

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