All posts by Jean-Pierre Merx

Set Theory

A non-measurable set

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We describe here a non-measurable subset of the segment \(I=[0,1] \subset \mathbb R\).

Let’s define on \(I\) an equivalence relation by \(x \sim y\) if and only if \(x-y \in \mathbb Q\). The equivalence relation \(\sim\) induces equivalence classes on \(I\). For \(x \in I\), it’s equivalence class \([x]\) is \([x] = \{y \in I \ : \ y-x \in \mathbb Q\}\). By the Axiom of Choice, we can form a set \(A\) by selecting a single point from each equivalence class.

We claim that the set \(A\) is not Lebesgue measurable.

For all \(q \in \mathbb Q\) we denote \(A_q = \{q+x \ : x \in A\}\). Let’s take \(p,q \in \mathbb Q\). If it exists \(z \in A_p \cap A_q\), it means that there exist \(u,v \in A\) such that
\[z= p+u=q+v\] hence \(u-v=q-p=0\) as \(u,v\) are supposed to be unique representatives of the classes of the equivalence relation \(\sim\). Finally if \(p,q\) are distincts, \(A_p \cap A_q = \emptyset\).

As Lebesgue measure \(\mu\) is translation invariant, we have for \(q \in \mathbb Q \cap [0,1]\) : \(\mu(A) = \mu(A_q)\) and also \(A_q \subset [0,2]\). Hence if we denote
\[B = \bigcup_{q \in \mathbb Q \cap [0,1]} A_q\] we have \(B \subset [0,2]\). If we suppose that \(A\) is measurable, we get
\[\mu(B) = \sum_{q \in \mathbb Q \cap [0,1]} \mu(A_q) = \sum_{q \in \mathbb Q \cap [0,1]} \mu(A) \le 2\] by countable additivity of Lebesgue measure (the set \(\mathbb Q \cap [0,1]\) being countable infinite). This implies \(\mu(A) = 0\).

Let’s prove now that
\[[0,1] \subset \bigcup_{q \in \mathbb Q \cap [-1,1]} A_q\] For \(z \in [0,1]\), there exists \(u \in A\) such that \(z \in [u]\). As \(A \subset [0,1]\), we have \(q = z-u \in \mathbb Q\) and \(-1 \le q \le 1\). And \(z=q+u\) means that \(z \in A_q\). This proves the inclusion. However the inclusion implies the contradiction
\[1 = \mu([0,1]) \le \sum_{q \in \mathbb Q \cap [-1,1]} \mu(A_q) = \sum_{q \in \mathbb Q \cap [-1,1]} \mu(A) =0\]

Finally \(A\) is not Lebesgue measurable.

Topology

Isometric versus affine

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Throughout this article we let \(E\) and \(F\) denote real normed vector spaces. A map \(f : E \rightarrow F\) is an isometry if \(\Vert f(x) – f(y) \Vert = \Vert x – y \Vert\) for all \(x, y \in E\), and \(f\) is affine if \[
f((1-t) a + t b ) = (1-t) f(a) + t f(b) \] for all \(a,b \in E\) and \(t \in [0,1]\). Equivalently, \(f\) is affine if the map \(T : E \rightarrow F\), defined by \(T(x)=f(x)-f(0)\) is linear.

First note that an isometry \(f\) is always one-to-one as \(f(x) = f(y)\) implies \[
0 = \Vert f(x) – f(y) \Vert = \Vert x- y \Vert\] hence \(x=y\).

There are two important cases when every isometry is affine:

  1. \(f\) is bijective (equivalently surjective). This is Mazur-Ulam theorem, which was proven in 1932.
  2. \(F\) is a strictly convex space. Recall that a normed vector space \((S, \Vert \cdot \Vert)\) is strictly convex if and only if for all distinct \(x,y \in S\), \(\Vert x \Vert = \Vert y \Vert =1\) implies \(\Vert \frac{x+y}{2} \Vert <1\). For example, an inner product space is strictly convex. The sequence spaces \(\ell_p\) for \(1 < p < \infty\) are also strictly convex.

Continue reading Isometric versus affine →

Analysis

Counterexample around Morera’s theorem

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Let’s recall Morera’s theorem.

Morera’s theorem Suppose that \(f\) is a continuous complex-valued function in a connected open set \(\Omega \subset \mathbb C\) such that
\[\int_{\partial \Delta} f(z) \ dz = 0\] for every closed triangle \(\Delta \subset \Omega \setminus \{p\}\) where \(p \in \Omega\). Then \(f\) is holomorphic in \(\Omega\).

Does the conclusion of Morera’s theorem still hold if \(f\) is supposed to be continuous only in \(\Omega \setminus \{p\}\)? The answer is negative and we provide a counterexample.

Let \(\Omega\) be the entire complex plane, \(f\) defined as follows
\[f(z)=\begin{cases}
\frac{1}{z^2} & \text{if } z \neq 0\\
0 & \text{otherwise}
\end{cases}\] and \(p\) the origin.

For \(a,b \in \Omega \setminus \{0\}\) we have
\[\begin{aligned}
\int_{[a,b]} f(z) \ dz &= \int_{[a,b]} \frac{dz}{z^2}\\
&= \int_0^1 \frac{b-a}{[a+t(b-a)]^2} \ dt\\
&=\left[ -\frac{1}{a+t(b-a)} \right]_0^1 = \frac{1}{a} – \frac{1}{b}
\end{aligned}\]

Hence for a triangle \(\Delta\) with vertices at \(a,b ,c \in \Omega \setminus \{0\}\):
\[\int_{\partial \Delta} f(z) \ dz = \left( \frac{1}{a} – \frac{1}{b} \right) + \left( \frac{1}{b} – \frac{1}{c} \right) + \left( \frac{1}{c} – \frac{1}{a} \right)=0\]

However, \(f\) is not holomorphic in \(\Omega\) as it is even not continuous at \(0\).

Analysis

Continuity versus uniform continuity

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We consider real-valued functions.

A real-valued function \(f : I \to \mathbb R\) (where \(I \subseteq\) is an interval) is continuous at \(x_0 \in I\) when: \[(\forall \epsilon > 0) (\exists \delta > 0)(\forall x \in I)(\vert x- x_0 \vert \le \delta \Rightarrow \vert f(x)- f(x_0) \vert \le \epsilon).\] When \(f\) is continuous at all \(x \in I\), we say that \(f\) is continuous on \(I\).

\(f : I \to \mathbb R\) is said to be uniform continuity on \(I\) if \[(\forall \epsilon > 0) (\exists \delta > 0)(\forall x,y \in I)(\vert x- y \vert \le \delta \Rightarrow \vert f(x)- f(y) \vert \le \epsilon).\]

Obviously, a function which is uniform continuous on \(I\) is continuous on \(I\). Is the converse true? The answer is negative.

An (unbounded) continuous function which is not uniform continuous

The map \[
\begin{array}{l|rcl}
f : & \mathbb R & \longrightarrow & \mathbb R \\
& x & \longmapsto & x^2 \end{array}\] is continuous. Let’s prove that it is not uniform continuous. For \(0 < x < y\) we have \[\vert f(x)-f(y) \vert = y^2-x^2 = (y-x)(y+x) \ge 2x (y-x)\] Hence for \(y-x= \delta >0\) and \(x = \frac{1}{\delta}\) we get
\[\vert f(x) -f(y) \vert \ge 2x (y-x) =2 > 1\] which means that the definition of uniform continuity is not fulfilled for \(\epsilon = 1\).

For this example, the function is unbounded as \(\lim\limits_{x \to \infty} x^2 = \infty\). Continue reading Continuity versus uniform continuity →

Set Theory

Around binary relations on sets

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We are considering here binary relations on a set \(A\). Let’s recall that a binary relation \(R\) on \(A\) is a subset of the cartesian product \(R \subseteq A \times A\). The statement \((x,y) \in R\) is read as \(x\) is \(R\)-related to \(y\) and also denoted by \(x R y \).

Some importants properties of a binary relation \(R\) are:

reflexive
For all \(x \in A\) it holds \(x R y\)
irreflexive
For all \(x \in A\) it holds not \(x R y\)
symmetric
For all \(x,y \in A\) it holds that if \(x R y\) then \(y R x\)
antisymmetric
For all \(x,y \in A\) if \(x R y\) and \(y R x\) then \(x=y\)
transitive
For all \(x,y,z \in A\) it holds that if \(x R y\) and \(y R z\) then \(x R z\)

A relation that is reflexive, symmetric and transitive is called an equivalence relation. Let’s see that being reflexive, symmetric and transitive are independent properties.

Symmetric and transitive but not reflexive

We provide two examples of such relations. For the first one, we take for \(A\) the set of the real numbers \(\mathbb R\) and the relation \[R = \{(x,y) \in \mathbb R^2 \, | \, xy >0\}.\] \(R\) is symmetric as the multiplication is also symmetric. \(R\) is also transitive as if \(xy > 0\) and \(yz > 0\) you get \(xy^2 z >0\). And as \(y^2 > 0\), we have \(xz > 0\) which means that \(x R z\). However, \(R\) is not reflexive as \(0 R 0\) doesn’t hold.

For our second example, we take \(A= \mathbb N\) and \(R=\{(1,1)\}\). It is easy to verify that \(R\) is symmetric and transitive. However \(R\) is not reflexive as \(n R n\) doesn’t hold for \(n \neq 1\). Continue reading Around binary relations on sets →

Algebra

A proper subspace without an orthogonal complement

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We consider an inner product space \(V\) over the field of real numbers \(\mathbb R\). The inner product is denoted by \(\langle \cdot , \cdot \rangle\).

When \(V\) is a finite dimensional space, every proper subspace \(F \subset V\) has an orthogonal complement \(F^\perp\) with \(V = F \oplus F^\perp\). This is no more true for infinite dimensional spaces and we present here an example.

Consider the space \(V=\mathcal C([0,1],\mathbb R)\) of the continuous real functions defined on the segment \([0,1]\). The bilinear map
\[\begin{array}{l|rcl}
\langle \cdot , \cdot \rangle : & V \times V & \longrightarrow & \mathbb R \\
& (f,g) & \longmapsto & \langle f , g \rangle = \displaystyle \int_0^1 f(t)g(t) \, dt \end{array}
\] is an inner product on \(V\).

Let’s consider the proper subspace \(H = \{f \in V \, ; \, f(0)=0\}\). \(H\) is an hyperplane of \(V\) as \(H\) is the kernel of the linear form \(\varphi : f \mapsto f(0)\) defined on \(V\). \(H\) is a proper subspace as \(\varphi\) is not always vanishing. Let’s prove that \(H^\perp = \{0\}\).

Take \(g \in H^\perp\). By definition of \(H^\perp\) we have \(\int_0^1 f(t) g(t) \, dt = 0\) for all \(f \in H\). In particular the function \(h : t \mapsto t g(t)\) belongs to \(H\). Hence
\[0 = \langle h , g \rangle = \displaystyle \int_0^1 t g(t)g(t) \, dt\] The map \(t \mapsto t g^2(t)\) is continuous, non-negative on \([0,1]\) and its integral on this segment vanishes. Hence \(t g^2(t)\) is always vanishing on \([0,1]\), and \(g\) is always vanishing on \((0,1]\). As \(g\) is continuous, we finally get that \(g = 0\).

\(H\) doesn’t have an orthogonal complement.

Moreover we have
\[(H^\perp)^\perp = \{0\}^\perp = V \neq H\]

Topology

A non-compact closed ball

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Consider a normed vector space \((X, \Vert \cdot \Vert)\). If \(X\) is finite-dimensional, then a subset \(Y \subset X\) is compact if and only if it is closed and bounded. In particular a closed ball \(B_r[a] = \{x \in X \, ; \, \Vert x – a \Vert \le r\}\) is always compact if \(X\) is finite-dimensional.

What about infinite-dimensional spaces?

The space \(A=C([0,1],\mathbb R)\)

Consider the space \(A=C([0,1],\mathbb R)\) of the real continuous functions defined on the interval \([0,1]\) endowed with the sup norm:
\[\Vert f \Vert = \sup\limits_{x \in [0,1]} \vert f(x) \vert\]
Is the closed unit ball \(B_1[0]\) compact? The answer is negative and we provide two proofs.

The first one is based on open covers. For \(n \ge 1\), we denote by \(f_n\) the piecewise linear map defined by \[
\begin{cases}
f_n(0)=f_n(\frac{1}{2^n}-\frac{1}{2^{n+2}})=0 \\
f_n(\frac{1}{2^n})=1 \\
f_n(\frac{1}{2^n}+\frac{1}{2^{n+2}})=f_n(1)=0
\end{cases}\] All the \(f_n\) belong to \(B_1[0]\). Moreover for \(1 \le n < m\) we have \(\frac{1}{2^n}+\frac{1}{2^{n+2}} < \frac{1}{2^m}-\frac{1}{2^{m+2}}\). Hence the supports of the \(f_n\) are disjoint and \(\Vert f_n – f_m \Vert = 1\).

Now consider the open cover \(\mathcal U=\{B_{\frac{1}{2}}(x) \, ; \, x \in B_1[0]\}\). For \(x \in B_1[0]\}\) and \(u,v \in B_{\frac{1}{2}}(x)\), \(\Vert u -v \Vert < 1\). Therefore, each \(B_{\frac{1}{2}}(x)\) contains at most one \(f_n\) and a finite subcover of \(\mathcal U\) will contain only a finite number of \(f_n\) proving that \(A\) is not compact.

Second proof based on convergent subsequence. As \(A\) is a metric space, it is enough to prove that \(A\) is not sequentially compact. Consider the sequence of functions \(g_n : x \mapsto x^n\). The sequence is bounded as for all \(n \in \mathbb N\), \(\Vert g_n \Vert = 1\). If \((g_n)\) would have a convergent subsequence, the subsequence would converge pointwise to the function equal to \(0\) on \([0,1)\) and to \(1\) at \(1\). As this function is not continuous, \((g_n)\) cannot have a subsequence converging to a map \(g \in A\).

Riesz’s theorem

The non-compactness of \(A=C([0,1],\mathbb R)\) is not so strange. Based on Riesz’s lemma one can show that the unit ball of an infinite-dimensional normed space \(X\) is never compact. This is sometimes known as the Riesz’s theorem.

The non-compactness of \(A=C([0,1],\mathbb R)\) is just standard for infinite-dimensional normed vector spaces!

Analysis

Continuity under the integral sign

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We consider here a measure space \((\Omega, \mathcal A, \mu)\) and \(T \subset \mathbb R\) a topological subspace. For a map \(f : T \times \Omega \to \mathbb R\) such that for all \(t \in T\) the map \[
\begin{array}{l|rcl}
f(t, \cdot) : & \Omega & \longrightarrow & \mathbb R \\
& \omega & \longmapsto & f(t,\omega) \end{array}
\] is integrable, one can define the function \[
\begin{array}{l|rcl}
F : & T & \longrightarrow & \mathbb R \\
& t & \longmapsto & \int_\Omega f(t,\omega) \ d\mu(\omega) \end{array}
\]

Following theorem is well known (and can be proven using dominated convergence theorem):

THEOREM for an adherent point \(x \in T\), if

  • \(\forall \omega \in \Omega \lim\limits_{t \to x} f(t,\omega) = \varphi(\omega)\)
  • There exists a map \(g : \Omega \to \mathbb R\) such that \(\forall t \in T, \, \forall \omega \in \Omega, \ \vert f(t,\omega) \vert \le g(\omega)\)

then \(\varphi\) is integrable and \[
\lim\limits_{t \to x} F(t) = \int_\Omega \varphi(\omega) \ d\mu(\omega)\]
In other words, one can switch \(\lim\) and \(\int\) signs.

We provide here a counterexample showing that the conclusion of the theorem might not hold if \(f\) is not bounded by a function \(g\) as supposed in the premises of the theorem. Continue reading Continuity under the integral sign →

Analysis

A trigonometric series that is not a Fourier series (Lebesgue-integration)

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We already provided here an example of a trigonometric series that is not the Fourier series of a Riemann-integrable function (namely the function \(\displaystyle x \mapsto \sum_{n=1}^\infty \frac{\sin nx}{\sqrt n}\)).

Applying an Abel-transformation (like mentioned in the link above), one can see that the function \[f(x)=\sum_{n=2}^\infty \frac{\sin nx}{\ln n}\] is everywhere convergent. We now prove that \(f\) cannot be the Fourier series of a Lebesgue-integrable function. The proof is based on the fact that for a \(2 \pi\)-periodic function \(g\), Lebesgue-integrable on \([0,2 \pi]\), the sum \[\sum_{n=1}^\infty \frac{c_n-c_{-n}}{n}\] is convergent where \((c_n)_{n \in \mathbb Z}\) are the complex Fourier coefficients of \(g\): \[c_n = \frac{1}{2 \pi} \int_0^{2 \pi} g(t)e^{-ikt} \ dt.\] As the series \(\displaystyle \sum_{n=2}^\infty \frac{1}{n \ln n}\) is divergent, we will be able to conclude that the sequence defined by \[\gamma_0=\gamma_1=\gamma_{-1} = 0, \, \gamma_n=- \gamma_{-n} = \frac{1}{\ln n} \ (n \ge 2)\] cannot be the Fourier coefficients of a Lebesgue-integrable function, hence that \(f\) is not the Fourier series of any Lebesgue-integrable function. Continue reading A trigonometric series that is not a Fourier series (Lebesgue-integration) →