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 $$n \ge 1$$ $$\Vert f_n – f \Vert_\infty \ge 1$$.