# Bounded functions and infimum, supremum

According to the extreme value theorem, a continuous real-valued function $$f$$ in the closed and bounded interval $$[a,b]$$ must attain a maximum and a minimum, each at least once.

Let’s see what can happen for non-continuous functions. We consider below maps defined on $$[0,1]$$.

First let’s look at $f(x)=\begin{cases} x &\text{ if } x \in (0,1)\\ 1/2 &\text{otherwise} \end{cases}$ $$f$$ is bounded on $$[0,1]$$, continuous on the interval $$(0,1)$$ but neither at $$0$$ nor at $$1$$. The infimum of $$f$$ is $$0$$, its supremum $$1$$, and $$f$$ doesn’t attain those values. However, for $$0 < a < b < 1$$, $$f$$ attains its supremum and infimum on $$[a,b]$$ as $$f$$ is continuous on this interval.

### Bounded function that doesn’t attain its infimum and supremum on all $$[a,b] \subseteq [0,1]$$

The function $$g$$ defined on $$[0,1]$$ by $g(x)=\begin{cases} 0 & \text{ if } x \notin \mathbb Q \text{ or if } x = 0\\ \frac{(-1)^q (q-1)}{q} & \text{ if } x = \frac{p}{q} \neq 0 \text{, with } p, q \text{ relatively prime} \end{cases}$ is bounded, as for $$x \in \mathbb Q \cap [0,1]$$ we have $\left\vert g(x) \right\vert < 1.$ Hence $$g$$ takes values in the interval $$[-1,1]$$. We prove that the infimum of $$g$$ is $$-1$$ and its supremum $$1$$ on all intervals $$[a,b]$$ with $$0 < a < b <1$$. Consider $$\varepsilon > 0$$ and an odd prime $$q$$ such that $q > \max(\frac{1}{\varepsilon}, \frac{1}{b-a}).$ This is possible as there are infinitely many prime numbers. By the pigeonhole principle and as $$0 < \frac{1}{q} < b-a$$, there exists a natural number $$p$$ such that $$\frac{p}{q} \in (a,b)$$. We have $-1 < g \left(\frac{p}{q} \right) = \frac{(-1)^q (q-1)}{q} = - \frac{q-1}{q} <-1 +\varepsilon$ as $$q$$ is supposed to be an odd prime with $$q > \frac{1}{\varepsilon}$$. This proves that the infimum of $$g$$ is $$-1$$. By similar arguments, one can prove that the supremum of $$g$$ on $$[a,b]$$ is $$1$$.

# On limit at infinity of functions and their derivatives

We consider continuously differentiable real functions defined on $$(0,\infty)$$ and the limits $\lim\limits_{x \to \infty} f(x) \text{ and } \lim\limits_{x \to \infty} f^\prime(x).$

### A map $$f$$ such that $$\lim\limits_{x \to \infty} f(x) = \infty$$ and $$\lim\limits_{x \to \infty} f^\prime(x) = 0$$

Consider the map $$f : x \mapsto \sqrt{x}$$. It is clear that $$\lim\limits_{x \to \infty} f(x) = \infty$$. As $$f^\prime(x) = \frac{1}{2 \sqrt{x}}$$, we have as announced $$\lim\limits_{x \to \infty} f^\prime(x) = 0$$

### A bounded map $$g$$ having no limit at infinity such that $$\lim\limits_{x \to \infty} g^\prime(x) = 0$$

One idea is to take an oscillating map whose wavelength is increasing to $$\infty$$. Let’s take the map $$g : x \mapsto \cos \sqrt{x}$$. $$g$$ doesn’t have a limit at $$\infty$$ as for $$n \in \mathbb N$$, we have $$g(n^2 \pi^2) = \cos n \pi = (-1)^n$$. However, the derivative of $$g$$ is $g^\prime(x) = – \frac{\sin \sqrt{x}}{2 \sqrt{x}},$ and as $$\vert g^\prime(x) \vert \le \frac{1}{2 \sqrt{x}}$$ for all $$x \in (0,\infty)$$, we have $$\lim\limits_{x \to \infty} g^\prime(x) = 0$$.

# Limit points of real sequences

Let’s start by recalling an important theorem of real analysis:

THEOREM. A necessary and sufficient condition for the convergence of a real sequence is that it is bounded and has a unique limit point.

As a consequence of the theorem, a sequence having a unique limit point is divergent if it is unbounded. An example of such a sequence is the sequence $u_n = \frac{n}{2}(1+(-1)^n),$ whose initial values are $0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 6, \dots$ $$(u_n)$$ is an unbounded sequence whose unique limit point is $$0$$.

Let’s now look at sequences having more complicated limit points sets.

### A sequence whose set of limit points is the set of natural numbers

Consider the sequence $$(v_n)$$ whose initial terms are $1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, \dots$ $$(v_n)$$ is defined as follows $v_n=\begin{cases} 1 &\text{ for } n= 1\\ n – \frac{k(k+1)}{2} &\text{ for } \frac{k(k+1)}{2} \lt n \le \frac{(k+1)(k+2)}{2} \end{cases}$ $$(v_n)$$ is well defined as the sequence $$(\frac{k(k+1)}{2})_{k \in \mathbb N}$$ is strictly increasing with first term equal to $$1$$. $$(v_n)$$ is a sequence of natural numbers. As $$\mathbb N$$ is a set of isolated points of $$\mathbb R$$, we have $$V \subseteq \mathbb N$$, where $$V$$ is the set of limit points of $$(v_n)$$. Conversely, let’s take $$m \in \mathbb N$$. For $$k + 1 \ge m$$, we have $$\frac{k(k+1)}{2} + m \le \frac{(k+1)(k+2)}{2}$$, hence $u_{\frac{k(k+1)}{2} + m} = m$ which proves that $$m$$ is a limit point of $$(v_n)$$. Finally the set of limit points of $$(v_n)$$ is the set of natural numbers.

# A linear map without any minimal polynomial

Given an endomorphism $$T$$ on a finite-dimensional vector space $$V$$ over a field $$\mathbb F$$, the minimal polynomial $$\mu_T$$ of $$T$$ is well defined as the generator (unique up to units in $$\mathbb F$$) of the ideal:$I_T= \{p \in \mathbb F[t]\ ; \ p(T)=0\}.$

For infinite-dimensional vector spaces, the minimal polynomial might not be defined. Let’s provide an example.

We take the real polynomials $$V = \mathbb R [t]$$ as a real vector space and consider the derivative map $$D : P \mapsto P^\prime$$. Let’s prove that $$D$$ doesn’t have any minimal polynomial. By contradiction, suppose that $\mu_D(t) = a_0 + a_1 t + \dots + a_n t^n \text{ with } a_n \neq 0$ is the minimal polynomial of $$D$$, which means that for all $$P \in \mathbb R[t]$$ we have $a_0 + a_1 P^\prime + \dots + a_n P^{(n)} = 0.$ Taking for $$P$$ the polynomial $$t^n$$ we get $a_0 t^n + n a_1 t^{n-1} + \dots + n! a_n = 0,$ which doesn’t make sense as $$n! a_n \neq 0$$, hence $$a_0 t^n + n a_1 t^{n-1} + \dots + n! a_n$$ cannot be the zero polynomial.

We conclude that $$D$$ doesn’t have any minimal polynomial.

# Non linear map preserving Euclidean norm

Let $$V$$ be a real vector space endowed with an Euclidean norm $$\Vert \cdot \Vert$$.

A bijective map $$T : V \to V$$ that preserves inner product $$\langle \cdot, \cdot \rangle$$ is linear. Also, Mazur-Ulam theorem states that an onto map $$T : V \to V$$ which is an isometry ($$\Vert T(x)-T(y) \Vert = \Vert x-y \Vert$$ for all $$x,y \in V$$) and fixes the origin ($$T(0) = 0$$) is linear.

What about an application that preserves the norm ($$\Vert T(x) \Vert = \Vert x \Vert$$ for all $$x \in V$$)? $$T$$ might not be linear as we show with following example:$\begin{array}{l|rcll} T : & V & \longrightarrow & V \\ & x & \longmapsto & x & \text{if } \Vert x \Vert \neq 1\\ & x & \longmapsto & -x & \text{if } \Vert x \Vert = 1\end{array}$

It is clear that $$T$$ preserves the norm. However $$T$$ is not linear as soon as $$V$$ is not the zero vector space. In that case, consider $$x_0$$ such that $$\Vert x_0 \Vert = 1$$. We have:$\begin{cases} T(2 x_0) &= 2 x_0 \text{ as } \Vert 2 x_0 \Vert = 2\\ \text{while}\\ T(x_0) + T(x_0) = -x_0 + (-x_0) &= – 2 x_0 \end{cases}$