A nonzero continuous map orthogonal to all polynomials

Let’s consider the vector space \(\mathcal{C}^0([a,b],\mathbb R)\) of continuous real functions defined on a compact interval \([a,b]\). We can define an inner product on pairs of elements \(f,g\) of \(\mathcal{C}^0([a,b],\mathbb R)\) by \[
\langle f,g \rangle = \int_a^b f(x) g(x) \ dx.\]

It is known that \(f \in \mathcal{C}^0([a,b],\mathbb R)\) is the always vanishing function if we have \(\langle x^n,f \rangle = \int_a^b x^n f(x) \ dx = 0\) for all integers \(n \ge 0\). Let’s recall the proof. According to Stone-Weierstrass theorem, for all \(\epsilon >0\) if exists a polynomial \(P\) such that \(\Vert f – P \Vert_\infty \le \epsilon\). Then \[
\begin{aligned}
0 &\le \int_a^b f^2 = \int_a^b f(f-P) + \int_a^b fP\\
&= \int_a^b f(f-P) \le \Vert f \Vert_\infty \epsilon(b-a)
\end{aligned}\] As this is true for all \(\epsilon > 0\), we get \(\int_a^b f^2 = 0\) and \(f = 0\).

We now prove that the result becomes false if we change the interval \([a,b]\) into \([0, \infty)\), i.e. that one can find a continuous function \(f \in \mathcal{C}^0([0,\infty),\mathbb R)\) such that \(\int_0^\infty x^n f(x) \ dx\) for all integers \(n \ge 0\). In that direction, let’s consider the complex integral \[
I_n = \int_0^\infty x^n e^{-(1-i)x} \ dx.\] \(I_n\) is well defined as for \(x \in [0,\infty)\) we have \(\vert x^n e^{-(1-i)x} \vert = x^n e^{-x}\) and \(\int_0^\infty x^n e^{-x} \ dx\) converges. By integration by parts, one can prove that \[
I_n = \frac{n!}{(1-i)^{n+1}} = \frac{(1+i)^{n+1}}{2^{n+1}} n! = \frac{e^{i \frac{\pi}{4}(n+1)}}{2^{\frac{n+1}{2}}}n!.\] Consequently, \(I_{4p+3} \in \mathbb R\) for all \(p \ge 0\) which means \[
\int_0^\infty x^{4p+3} \sin(x) e^{-x} \ dx =0\] and finally \[
\int_0^\infty u^p \sin(u^{1/4}) e^{-u^{1/4}} \ dx =0\] for all integers \(p \ge 0\) using integration by substitution with \(x = u^{1/4}\). The function \(u \mapsto \sin(u^{1/4}) e^{-u^{1/4}}\) is one we were looking for.

A group G isomorph to the product group G x G

Let’s provide an example of a nontrivial group \(G\) such that \(G \cong G \times G\). For a finite group \(G\) of order \(\vert G \vert =n > 1\), the order of \(G \times G\) is equal to \(n^2\). Hence we have to look at infinite groups in order to get the example we’re seeking for.

We take for \(G\) the infinite direct product \[
G = \prod_{n \in \mathbb N} \mathbb Z_2 = \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \dots,\] where \(\mathbb Z_2\) is endowed with the addition. Now let’s consider the map \[
\begin{array}{l|rcl}
\phi : & G & \longrightarrow & G \times G \\
& (g_1,g_2,g_3, \dots) & \longmapsto & ((g_1,g_3, \dots ),(g_2, g_4, \dots)) \end{array}\]

From the definition of the addition in \(G\) it follows that \(\phi\) is a group homomorphism. \(\phi\) is onto as for any element \(\overline{g}=((g_1, g_2, g_3, \dots),(g_1^\prime, g_2^\prime, g_3^\prime, \dots))\) in \(G \times G\), \(g = (g_1, g_1^\prime, g_2, g_2^\prime, \dots)\) is an inverse image of \(\overline{g}\) under \(\phi\). Also the identity element \(e=(\overline{0},\overline{0}, \dots)\) of \(G\) is the only element of the kernel of \(G\). Hence \(\phi\) is also one-to-one. Finally \(\phi\) is a group isomorphism between \(G\) and \(G \times G\).

Counterexamples around series (part 1)

The purpose of this article is to provide some basic counterexamples on real series. Counterexamples are provided as answers to questions.

Unless otherwise stated, \((u_n)_{n \in \mathbb{N}}\) and \((v_n)_{n \in \mathbb{N}}\) are two real sequences.

If \((u_n)\) is non-increasing and converges to zero then \(\sum u_n\) converges?

Is not true. A famous counterexample is the harmonic series \(\sum \frac{1}{n}\) which doesn’t converge as \[
\displaystyle \sum_{k=p+1}^{2p} \frac{1}{k} \ge \sum_{k=p+1}^{2p} \frac{1}{2p} = 1/2,\] for all \(p \in \mathbb N\).

If \(u_n = o(1/n)\) then \(\sum u_n\) converges?

Does not hold as can be seen considering \(u_n=\frac{1}{n \ln n}\) for \(n \ge 2\). Indeed \(\int_2^x \frac{dt}{t \ln t} = \ln(\ln x) – \ln (\ln 2)\) and therefore \(\int_2^\infty \frac{dt}{t \ln t}\) diverges. We conclude that \(\sum \frac{1}{n \ln n}\) diverges using the integral test. However \(n u_n = \frac{1}{\ln n}\) converges to zero. Continue reading Counterexamples around series (part 1)

Isomorphism of factors does not imply isomorphism of quotient groups

Let \(G\) be a group and \(H, K\) two isomorphic subgroups. We provide an example where the quotient groups \(G / H\) and \(G / K\) are not isomorphic.

Let \(G = \mathbb{Z}_4 \times \mathbb{Z}_2\), with \(H = \langle (\overline{2}, \overline{0}) \rangle\) and \(K = \langle (\overline{0}, \overline{1}) \rangle\). We have \[
H \cong K \cong \mathbb{Z}_2.\] The left cosets of \(H\) in \(G\) are \[
G / H=\{(\overline{0}, \overline{0}) + H, (\overline{1}, \overline{0}) + H, (\overline{0}, \overline{1}) + H, (\overline{1}, \overline{1}) + H\},\] a group having \(4\) elements and for all elements \(x \in G/H\), one can verify that \(2x = H\). Hence \(G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2\). The left cosets of \(K\) in \(G\) are \[
G / K=\{(\overline{0}, \overline{0}) + K, (\overline{1}, \overline{0}) + K, (\overline{2}, \overline{0}) + K, (\overline{3}, \overline{0}) + K\},\] which is a cyclic group of order \(4\) isomorphic to \(\mathbb{Z}_4\). We finally get the desired conclusion \[
G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \ncong \mathbb{Z}_4 \cong G / K.\]