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 rightderivative on each point of \(\mathbb{Q}\).
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→
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→
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→
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 \(0\).
The Cantor ternary set (named Cantor set below) \(K\) is a subset of the real segment \(I=[0,1]\). It is built by induction:
Starting with \(K_0=I\)
If \(K_n\) is a finite disjoint union of segments \(K_n=\cup_k \left[a_k,b_k\right]\), \[K_{n+1}=\bigcup_k \left(\left[a_k,a_k+\frac{b_k-a_k}{3}\right] \cup \left[a_k+2\frac{b_k-a_k}{3},b_k\right]\right)\]
And finally \(K=\displaystyle \bigcap_{n \in \mathbb{N}} K_n\). The Cantor set is created by repeatedly deleting the open middle third of a set of line segments starting with the segment \(I\).
Let’s start by recalling some background about modules.
Suppose that \(R\) is a ring and \(1_R\) is its multiplicative identity. A left \(R\)-module \(M\) consists of an abelian group \((M, +)\) and an operation \(R \times M \rightarrow M\) such that for all \(r, s \in R\) and \(x, y \in M\), we have:
\(r \cdot (x+y)= r \cdot x + r \cdot y\) (\( \cdot\) is left-distributive over \(+\))
\((r +s) \cdot x= r \cdot x + s \cdot x\) (\( \cdot\) is right-distributive over \(+\))
\((rs) \cdot x= r \cdot (s \cdot x)\)
\(1_R \cdot x= x \)
\(+\) is the symbol for addition in both \(R\) and \(M\).
If \(K\) is a field, \(M\) is \(K\)-vector space. It is well known that a vector space \(V\) is having a basis, i.e. a subset of linearly independent vectors that spans \(V\). Unlike for a vector space, a module doesn’t always have a basis.Continue reading A module without a basis→
We consider a normed vector space \(E\) over the field of the reals \(\mathbb{R}\) and a convex subset \(C \subset E\).
We suppose that \(0 \in C\) and that \(C\) is unbounded, i.e. there exists points in \(C\) at distance as big as we wish from \(0\).
The following question arises: “does \(C\) contains a ray?”. It turns out that the answer depends on the dimension of the space \(E\). If \(E\) is of finite dimension, then \(C\) always contains a ray, while if \(E\) is of infinite dimension \(C\) may not contain a ray. Continue reading An unbounded convex not containing a ray→
We consider a topological vector space \(E\) over the field of the reals \(\mathbb{R}\). The convex hull of a subset \(X \subset E\) is the smallest convex set that contains \(X\).
The convex hull may also be defined as the intersection of all convex sets containing X or as the set of all convex combinations of points in X.
In this page \(\mathbb{F}\) refers to a field. Given any vector space \(V\) over \(\mathbb{F}\), the dual space \(V^*\) is defined as the set of all linear functionals \(f: V \mapsto \mathbb{F}\). The dual space \(V^*\) itself becomes a vector space over \(\mathbb{F}\) when equipped with the following addition and scalar multiplication:
\[\left\{
\begin{array}{lll}(\varphi + \psi)(x) & = & \varphi(x) + \psi(x) \\
(a \varphi)(x) & = & a (\varphi(x)) \end{array} \right. \] for all \(\phi, \psi \in V^*\), \(x \in V\), and \(a \in \mathbb{F}\).
There is a natural homomorphism \(\Phi\) from \(V\) into the double dual \(V^{**}\), defined by \((\Phi(v))(\phi) = \phi(v)\) for all \(v \in V\), \(\phi \in V^*\). This map \(\Phi\) is always injective. Continue reading A vector space not isomorphic to its double dual→
