# A continuous function not differentiable at the rationals but differentiable elsewhere

We build here a continuous function of one real variable whose derivative exists on $$\mathbb{R} \setminus \mathbb{Q}$$ and doesn’t have a left or right derivative on each point of $$\mathbb{Q}$$.

As $$\mathbb{Q}$$ is (infinitely) countable, we can find a bijection $$n \mapsto r_n$$ from $$\mathbb{N}$$ to $$\mathbb{Q}$$. We now reuse the function $$f$$ defined here. Recall $$f$$ main properties: Continue reading A continuous function not differentiable at the rationals but differentiable elsewhere

# A differentiable function except at one point with a bounded derivative

We build here a continuous function of one real variable whose derivative exists except at $$0$$ and is bounded on $$\mathbb{R^*}$$.

We start with the even and piecewise linear function $$g$$ defined on $$[0,+\infty)$$ with following values:
$g(x)= \left\{ \begin{array}{ll} 0 & \mbox{if } x =0\\ 0 & \mbox{if } x \in \{\frac{k}{4^n};(k,n) \in \{1,2,4\} \times \mathbb{N^*}\}\\ 1 & \mbox{if } x \in \{\frac{3}{4^n};n \in \mathbb{N^*}\}\\ \end{array} \right.$
The picture below gives an idea of the graph of $$g$$ for positive values. Continue reading A differentiable function except at one point with a bounded derivative

# Converse of Lagrange’s theorem does not hold

Lagrange’s theorem, states that for any finite group $$G$$, the order (number of elements) of every subgroup $$H$$ of $$G$$ divides the order of $$G$$ (denoted by $$\vert G \vert$$).

Lagrange’s theorem raises the converse question as to whether every divisor $$d$$ of the order of a group is the order of some subgroup. According to Cauchy’s theorem this is true when $$d$$ is a prime.

However, this does not hold in general: given a finite group $$G$$ and a divisor $$d$$ of $$\vert G \vert$$, there does not necessarily exist a subgroup of $$G$$ with order $$d$$. The alternating group $$G = A_4$$, which has $$12$$ elements has no subgroup of order $$6$$. We prove it below. Continue reading Converse of Lagrange’s theorem does not hold

# A continuous function which is not of bounded variation

## Introduction on total variation of functions

Recall that a function of bounded variation, also known as a BV-function, is a real-valued function whose total variation is bounded (finite).

Being more formal, the total variation of a real-valued function $$f$$, defined on an interval $$[a,b] \subset \mathbb{R}$$ is the quantity:
$V_a^b(f) = \sup\limits_{P \in \mathcal{P}} \sum_{i=0}^{n_P-1} \left\vert f(x_{i+1}) – f(x_i) \right\vert$ where the supremum is taken over the set $$\mathcal{P}$$ of all partitions of the interval considered. Continue reading A continuous function which is not of bounded variation

# A curve filling a square – Lebesgue example

## Introduction

We aim at defining a continuous function $$\varphi : [0,1] \rightarrow [0,1]^2$$. At first sight this looks quite strange.

Indeed, $$\varphi$$ cannot be a bijection. If $$\varphi$$ would be bijective, it would also be an homeomorphism as a continuous bijective function from a compact space to a Haussdorff space is an homeomorphism. But an homeomorphism preserves connectedness and $$[0,1] \setminus \{1/2\}$$ is not connected while $$[0,1]^2 \setminus \{\varphi(1/2)\}$$ is.

Nor can $$\varphi$$ be piecewise continuously differentiable as the Lebesgue measure of $$\varphi([0,1])$$ would be equal to $$0$$.

$$\varphi$$ is defined in two steps using the Cantor space $$K$$. Continue reading A curve filling a square – Lebesgue example