# Counterexamples to Banach fixed-point theorem

Let $$(X,d)$$ be a metric space. Then a map $$T : X \to X$$ is called a contraction map if it exists $$0 \le k < 1$$ such that $d(T(x),T(y)) \le k d(x,y)$ for all $$x,y \in X$$. According to Banach fixed-point theorem, if $$(X,d)$$ is a complete metric space and $$T$$ a contraction map, then $$T$$ admits a fixed-point $$x^* \in X$$, i.e. $$T(x^*)=x^*$$.

We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled.

First, let’s consider $\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & x+1 \end{array}$ For all $$x,y \in \mathbb R$$ we have $$\vert f(x)-f(y) \vert = \vert x- y \vert$$. $$f$$ is not a contraction, but an isometry. Obviously, $$f$$ has no fixed-point.

We now prove that a map satisfying $d(g(x),g(y)) < d(x,y)$ might also not have a fixed-point. A counterexample is the following map $\begin{array}{l|rcl} g : & [0,+\infty) & \longrightarrow & [0,+\infty) \\ & x & \longmapsto & \sqrt{1+x^2} \end{array}$ Since $g^\prime(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for all } \xi \in [0, +\infty),$ by the mean value theorem $\vert g(x) - g(y)| = \vert g^\prime(\xi)\vert |x-y| < |x-y| \text{ for all } x, y \in [0, +\infty).$ However $$g$$ has no fixed-point. Finally, let's have a look to a space $$(X,d)$$ which is not complete. We take $$a,b \in \mathbb R$$ with $$0 < a < 1$$ and for $$(X,d)$$ the space $$X = \mathbb R \setminus \{\frac{b}{1-a}\}$$ equipped with absolute value distance. $$X$$ is not complete. Consider the map $\begin{array}{l|rcl} h : & X & \longrightarrow & X \\ & x & \longmapsto & ax + b \end{array}$ $$h$$ is well defined as for $$x \neq \frac{b}{1-a}$$, $$h(x) \neq \frac{b}{1-a}$$. $$h$$ is a contraction map as for $$x,y \in \mathbb R$$ $\vert h(x)-h(y) \vert = a \vert x - y \vert$ However, $$h$$ doesn't have a fixed-point in $$X$$ as $$\frac{b}{1-a}$$ is the only real for which $$h(x)=x$$.

# Differentiability of multivariable real functions (part1)

This article provides counterexamples about differentiability of functions of several real variables. We focus on real functions of two real variables (defined on $$\mathbb R^2$$). $$\mathbb R^2$$ and $$\mathbb R$$ are equipped with their respective Euclidean norms denoted by $$\Vert \cdot \Vert$$ and $$\vert \cdot \vert$$, i.e. the absolute value for $$\mathbb R$$.

We recall some definitions and theorems about differentiability of functions of several real variables.

Definition 1 We say that a function $$f : \mathbb R^2 \to \mathbb R$$ is differentiable at $$\mathbf{a} \in \mathbb R^2$$ if it exists a (continuous) linear map $$\nabla f(\mathbf{a}) : \mathbb R^2 \to \mathbb R$$ with $\lim\limits_{\mathbf{h} \to 0} \frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})-\nabla f(\mathbf{a}).\mathbf{h}}{\Vert \mathbf{h} \Vert} = 0$

Definition 2 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. Then the $$\mathbf{i^{th}}$$ partial derivative at point $$\mathbf{a}$$ is the real number
\begin{align*}
\frac{\partial f}{\partial x_i}(\mathbf{a}) &= \lim\limits_{h \to 0} \frac{f(\mathbf{a}+h \mathbf{e_i})- f(\mathbf{a})}{h}\\
&= \lim\limits_{h \to 0} \frac{f(a_1,\dots,a_{i-1},a_i+h,a_{i+1},\dots,a_n) – f(a_1,\dots,a_{i-1},a_i,a_{i+1},\dots,a_n)}{h}
\end{align*} For two real variable functions, $$\frac{\partial f}{\partial x}(x,y)$$ and $$\frac{\partial f}{\partial y}(x,y)$$ will denote the partial derivatives.

Definition 3 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. The directional derivative of $$f$$ along vector $$\mathbf{v}$$ at point $$\mathbf{a}$$ is the real $\nabla_{\mathbf{v}}f(\mathbf{a}) = \lim\limits_{h \to 0} \frac{f(\mathbf{a}+h \mathbf{v})- f(\mathbf{a})}{h}$ Continue reading Differentiability of multivariable real functions (part1)

# Two non similar matrices having same minimal and characteristic polynomials

Consider a square matrix $$A$$ of dimension $$n \ge 1$$ over a field $$\mathbb F$$, i.e. $$A \in \mathcal M_n(\mathbb F)$$. Results discuss below are true for any field $$\mathbb F$$, in particular for $$\mathbb F = \mathbb R$$ or $$\mathbb F = \mathbb C$$.

A polynomial $$P \in \mathbb F[X]$$ is called a vanishing polynomial for $$A$$ if $$P(A) = 0$$. If the matrix $$B$$ is similar to $$B$$ (which means that $$B=Q^{-1} A Q$$ for some invertible matrix $$Q$$), and the polynomial $$P$$ vanishes at $$A$$ then $$P$$ also vanishes at $$B$$. This is easy to prove as we have $$P(B)=P(Q^{-1} A Q)=Q^{-1} P(A) Q$$.

In particular, two similar matrices have the same minimal and characteristic polynomials.

Is the converse true? Are two matrices having the same minimal and characteristic polynomials similar? Continue reading Two non similar matrices having same minimal and characteristic polynomials

# A counterexample to Krein-Milman theorem

In the theory of functional analysis, the Krein-Milman theorem states that for a separated locally convex topological vector space $$X$$, a compact convex subset $$K$$ is the closed convex hull of its extreme points.

For the reminder, an extreme point of a convex set $$S$$ is a point in $$S$$ which does not lie in any open line segment joining two points of S. A point $$p \in S$$ is an extreme point of $$S$$ if and only if $$S \setminus \{p\}$$ is still convex.

In particular, according to the Krein-Milman theorem, a non-empty compact convex set has a non-empty set of extreme points. Let see what happens if we weaken some hypothesis of Krein-Milman theorem. Continue reading A counterexample to Krein-Milman theorem

# Continuity of multivariable real functions

This article provides counterexamples about continuity of functions of several real variables. In addition the article discusses the cases of functions of two real variables (defined on $$\mathbb R^2$$ having real values. $$\mathbb R^2$$ and $$\mathbb R$$ are equipped with their respective Euclidean norms denoted by $$\Vert \cdot \Vert$$ and $$\vert \cdot \vert$$, i.e. the absolute value for $$\mathbb R$$.

We recall that a function $$f$$ defined from $$\mathbb R^2$$ to $$\mathbb R$$ is continuous at $$(x_0,y_0) \in \mathbb R^2$$ if for any $$\epsilon > 0$$, there exists $$\delta > 0$$, such that $$\Vert (x,y) -(x_0,y_0) \Vert < \delta \Rightarrow \vert f(x,y) - f(x_0,y_0) \vert < \epsilon$$. Continue reading Continuity of multivariable real functions