# A homeomorphism of the unit ball having no fixed point

Let’s recall Brouwer fixed-point theorem.

Theorem (Brouwer): Every continuous function from a convex compact subset $$K$$ of a Euclidean space to $$K$$ itself has a fixed point.

We here describe an example of a homeomorphism of the unit ball of a Hilbert space having no fixed point. Let $$E$$ be a separable Hilbert space with $$(e_n)_{n \in \mathbb{Z}}$$ as a Hilbert basis. $$B$$ and $$S$$ are respectively $$E$$ closed unit ball and unit sphere.

There is a unique linear map $$u : E \to E$$ for which $$u(e_n)=e_{n+1}$$ for all $$n \in \mathbb{Z}$$. For $$x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$$ we have $$u(x)= \sum_{n \in \mathbb{Z}} \xi_n e_{n+1}$$. $$u$$ is isometric as $\Vert u(x) \Vert^2 = \sum_{n \in \mathbb{Z}} \vert \xi_n \vert^2 = \Vert x \Vert^2$ hence one-to-one. $$u$$ is also onto as for $$x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$$, $$\sum_{n \in \mathbb{Z}} \xi_n e_{n-1} \in E$$ is an inverse image of $$x$$. Finally $$u$$ is an homeomorphism. Continue reading A homeomorphism of the unit ball having no fixed point

# Counterexamples around differentiation of sequences of functions

We consider here sequences of real functions defined on a closed interval. Following theorem is the main one regarding the differentiation of the limit.

Theorem: Suppose $$(f_n)$$ is a sequence of functions, differentiable on $$[a,b]$$ and such that $$(f_n(x_0))$$ converges for some point $$x_0 \in [a,b]$$. If $$(f_n^\prime)$$ converges uniformly on $$[a,b]$$, then $$(f_n)$$ converges uniformly on $$[a,b]$$ to a function $$f$$ and for all $$x \in [a,b]$$ $f^\prime(x)=\lim\limits_{n \to \infty} f_n^\prime(x)$ What happens if we drop some hypothesis of the theorem? Continue reading Counterexamples around differentiation of sequences of functions

# An infinite group whose proper subgroups are all finite

We study some properties of the Prüfer $$p$$-group $$\mathbb{Z}_{p^\infty}$$ for a prime number $$p$$. The Prüfer $$p$$-group may be identified with the subgroup of the circle group, consisting of all $$p^n$$-th roots of unity as $$n$$ ranges over all non-negative integers:
$\mathbb{Z}_{p^\infty}=\bigcup_{k=0}^\infty \mathbb{Z}_{p^k} \text{ where } \mathbb{Z}_{p^k}= \{e^{\frac{2 i \pi m}{p^k}} \ | \ 0 \le m \le p^k-1\}$

### $$\mathbb{Z}_{p^\infty}$$ is a group

First, let’s notice that for $$0 \le m \le n$$ integers we have $$\mathbb{Z}_{p^m} \subseteq \mathbb{Z}_{p^n}$$ as $$p^m | p^n$$. Also for $$m \ge 0$$ $$\mathbb{Z}_{p^m}$$ is a subgroup of the circle group. We also notice that all elements of $$\mathbb{Z}_{p^\infty}$$ have finite orders which are powers of $$p$$. Continue reading An infinite group whose proper subgroups are all finite

# Counterexample around Arzela-Ascoli theorem

Let’s recall Arzelà–Ascoli theorem:

Suppose that $$F$$ is a Banach space and $$E$$ a compact metric space. A subset $$\mathcal{H}$$ of the Banach space $$\mathcal{C}_F(E)$$ is relatively compact in the topology induced by the uniform norm if and only if it is equicontinuous and and for all $$x \in E$$, the set $$\mathcal{H}(x)=\{f(x) \ | \ f \in \mathcal{H}\}$$ is relatively compact.

We look here at what happens if we drop the requirement on space $$E$$ to be compact and provide a counterexample where the conclusion of Arzelà–Ascoli theorem doesn’t hold anymore.

We take for $$E$$ the real interval $$[0,+\infty)$$ and for all $$n \in \mathbb{N} \setminus \{0\}$$ the real function
$f_n(t)= \sin \sqrt{t+4 n^2 \pi^2}$ We prove that $$(f_n)$$ is equicontinuous, converges pointwise to $$0$$ but is not relatively compact.

According to the mean value theorem, for all $$x,y \in \mathbb{R}$$
$\vert \sin x – \sin y \vert \le \vert x – y \vert$ Hence for $$n \ge 1$$ and $$x,y \in [0,+\infty)$$
\begin{align*}
\vert f_n(x)-f_n(y) \vert &\le \vert \sqrt{x+4 n^2 \pi^2} -\sqrt{y+4 n^2 \pi^2} \vert \\
&= \frac{\vert x – y \vert}{\sqrt{x+4 n^2 \pi^2} +\sqrt{y+4 n^2 \pi^2}} \\
&\le \frac{\vert x – y \vert}{4 \pi}
\end{align*} using multiplication by the conjugate.

Which enables to prove that $$(f_n)$$ is equicontinuous.

We also have for $$n \ge 1$$ and $$x \in [0,+\infty)$$
\begin{align*}
\vert f_n(x) \vert &= \vert f_n(x) – f_n(0) \vert \le \vert \sqrt{x+4 n^2 \pi^2} -\sqrt{4 n^2 \pi^2} \vert \\
&= \frac{\vert x \vert}{\sqrt{x+4 n^2 \pi^2} +\sqrt{4 n^2 \pi^2}} \\
&\le \frac{\vert x \vert}{4 n \pi}
\end{align*}

Hence $$(f_n)$$ converges pointwise to $$0$$ and for $$t \in [0,+\infty), \mathcal{H}(t)=\{f_n(t) \ | \ n \in \mathbb{N} \setminus \{0\}\}$$ is relatively compact

Finally we prove that $$\mathcal{H}=\{f_n \ | \ n \in \mathbb{N} \setminus \{0\}\}$$ is not relatively compact. While $$(f_n)$$ converges pointwise to $$0$$, $$(f_n)$$ does not converge uniformly to $$f=0$$. Actually for $$n \ge 1$$ and $$t_n=\frac{\pi^2}{4} + 2n \pi^2$$ we have
$f_n(t_n)= \sin \sqrt{\frac{\pi^2}{4} + 2n \pi^2 +4 n^2 \pi^2}=\sin \sqrt{\left(\frac{\pi}{2} + 2 n \pi\right)^2}=1$ Consequently for all $$n \ge 1$$ $$\Vert f_n – f \Vert_\infty \ge 1$$. If $$\mathcal{H}$$ was relatively compact, $$(f_n)$$ would have a convergent subsequence with $$f=0$$ for limit. And that cannot be as for all $$n \ge 1$$ $$\Vert f_n – f \Vert_\infty \ge 1$$.