# 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 $$\epsilon >0$$ it 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)

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 $G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \ncong \mathbb{Z}_4 \cong G / K.$