# 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 $$\lim\limits_{x \to \infty} g^\prime(x) = 0$$.

# 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:$\begin{cases} T(2 x_0) &= 2 x_0 \text{ as } \Vert 2 x_0 \Vert = 2\\ \text{while}\\ T(x_0) + T(x_0) = -x_0 + (-x_0) &= – 2 x_0 \end{cases}$

# Non linear map preserving orthogonality

Let $$V$$ be a real vector space endowed with an inner product $$\langle \cdot, \cdot \rangle$$.

It is known that a bijective map $$T : V \to V$$ that preserves the inner product $$\langle \cdot, \cdot \rangle$$ is linear.

That might not be the case if $$T$$ is supposed to only preserve orthogonality. Let’s consider for $$V$$ the real plane $$\mathbb R^2$$ and the map $\begin{array}{l|rcll} T : & \mathbb R^2 & \longrightarrow & \mathbb R^2 \\ & (x,y) & \longmapsto & (x,y) & \text{for } xy \neq 0\\ & (x,0) & \longmapsto & (0,x)\\ & (0,y) & \longmapsto & (y,0) \end{array}$

The restriction of $$T$$ to the plane less the x-axis and the y-axis is the identity and therefore is bijective on this set. Moreover $$T$$ is a bijection from the x-axis onto the y-axis, and a bijection from the y-axis onto the x-axis. This proves that $$T$$ is bijective on the real plane.

$$T$$ preserves the orthogonality on the plane less x-axis and y-axis as it is the identity there. As $$T$$ swaps the x-axis and the y-axis, it also preserves orthogonality of the coordinate axes. However, $$T$$ is not linear as for non zero $$x \neq y$$ we have: $\begin{cases} T[(x,0) + (0,y)] = T[(x,y)] &= (x,y)\\ \text{while}\\ T[(x,0)] + T[(0,y)] = (0,x) + (y,0) &= (y,x) \end{cases}$

# A linear map having all numbers as eigenvalue

Consider a linear map $$\varphi : E \to E$$ where $$E$$ is a linear space over the field $$\mathbb C$$ of the complex numbers. When $$E$$ is a finite dimensional vector space of dimension $$n \ge 1$$, the number of eigenvalues is finite. The eigenvalues are the roots of the characteristic polynomial $$\chi_\varphi$$ of $$\varphi$$. $$\chi_\varphi$$ is a complex polynomial of degree $$n \ge 1$$. Therefore the set of eigenvalues of $$\varphi$$ is non-empty and its cardinal is less than $$n$$.

Things are different when $$E$$ is an infinite dimensional space.

### A linear map having all numbers as eigenvalue

Let’s consider the linear space $$E=\mathcal C^\infty([0,1])$$ of smooth complex functions having derivatives of all orders and defined on the segment $$[0,1]$$. $$E$$ is an infinite dimensional space: it contains all the polynomial maps.

On $$E$$, we define the linear map $\begin{array}{l|rcl} \varphi : & \mathcal C^\infty([0,1]) & \longrightarrow & \mathcal C^\infty([0,1]) \\ & f & \longmapsto & f^\prime \end{array}$

The set of eigenvalues of $$\varphi$$ is all $$\mathbb C$$. Indeed, for $$\lambda \in \mathbb C$$ the map $$t \mapsto e^{\lambda t}$$ is an eigenvector associated to the eigenvalue $$\lambda$$.

### A linear map having no eigenvalue

On the same linear space $$E=\mathcal C^\infty([0,1])$$, we now consider the linear map $\begin{array}{l|rcl} \psi : & \mathcal C^\infty([0,1]) & \longrightarrow & \mathcal C^\infty([0,1]) \\ & f & \longmapsto & x f \end{array}$

Suppose that $$\lambda \in \mathbb C$$ is an eigenvalue of $$\psi$$ and $$h \in E$$ an eigenvector associated to $$\lambda$$. By hypothesis, there exists $$x_0 \in [0,1]$$ such that $$h(x_0) \neq 0$$. Even better, as $$h$$ is continuous, $$h$$ is non-vanishing on $$J \cap [0,1]$$ where $$J$$ is an open interval containing $$x_0$$. On $$J \cap [0,1]$$ we have the equality $(\psi(h))(x) = x h(x) = \lambda h(x)$ Hence $$x=\lambda$$ for all $$x \in J \cap [0,1]$$. A contradiction proving that $$\psi$$ has no eigenvalue.

# A strictly increasing map that is not one-to-one

Consider two partially ordered sets $$(E,\le)$$ and $$(F,\le)$$ and a strictly increasing map $$f : E \to F$$. If the order $$(E,\le)$$ is total, then $$f$$ is one-to-one. Indeed for distinct elements $$x,y \in E$$, we have either $$x < y$$ or $$y < x$$ and consequently $$f(x) < f(y)$$ or $$f(y) < f(x)$$. Therefore $$f(x)$$ and $$f(y)$$ are different. This is not true anymore for a partial order $$(E,\le)$$. We give a counterexample.

Consider a finite set $$E$$ having at least two elements and partially ordered by the inclusion. Let $$f$$ be the map defined on the powerset $$\wp(E)$$ that maps $$A \subseteq E$$ to its cardinal $$\vert A \vert$$. $$f$$ is obviously strictly increasing. However $$f$$ is not one-to-one as for distincts elements $$a,b \in E$$ we have $f(\{a\}) = 1 = f(\{b\})$

# Radius of convergence of power series

We look here at the radius of convergence of the sum and product of power series.

Let’s recall that for a power series $$\displaystyle \sum_{n=0}^\infty a_n x^n$$ where $$0$$ is not the only convergence point, the radius of convergence is the unique real $$0 < R \le \infty$$ such that the series converges whenever $$\vert x \vert < R$$ and diverges whenever $$\vert x \vert > R$$.

Given two power series with radii of convergence $$R_1$$ and $$R_2$$, i.e.
\begin{align*}
\displaystyle f_1(x) = \sum_{n=0}^\infty a_n x^n, \ \vert x \vert < R_1 \\ \displaystyle f_2(x) = \sum_{n=0}^\infty b_n x^n, \ \vert x \vert < R_2 \end{align*} The sum of the power series \begin{align*} \displaystyle f_1(x) + f_2(x) &= \sum_{n=0}^\infty a_n x^n + \sum_{n=0}^\infty b_n x^n \\ &=\sum_{n=0}^\infty (a_n + b_n) x^n \end{align*} and its Cauchy product:
\begin{align*}
\displaystyle f_1(x) \cdot f_2(x) &= \left(\sum_{n=0}^\infty a_n x^n\right) \cdot \left(\sum_{n=0}^\infty b_n x^n \right) \\
&=\sum_{n=0}^\infty \left( \sum_{l=0}^n a_l b_{n-l}\right) x^n
\end{align*}
both have radii of convergence greater than or equal to $$\min \{R_1,R_2\}$$.

The radii can indeed be greater than $$\min \{R_1,R_2\}$$. Let’s give examples.

# Isometric versus affine

Throughout this article we let $$E$$ and $$F$$ denote real normed vector spaces. A map $$f : E \rightarrow F$$ is an isometry if $$\Vert f(x) – f(y) \Vert = \Vert x – y \Vert$$ for all $$x, y \in E$$, and $$f$$ is affine if $f((1-t) a + t b ) = (1-t) f(a) + t f(b)$ for all $$a,b \in E$$ and $$t \in [0,1]$$. Equivalently, $$f$$ is affine if the map $$T : E \rightarrow F$$, defined by $$T(x)=f(x)-f(0)$$ is linear.

First note that an isometry $$f$$ is always one-to-one as $$f(x) = f(y)$$ implies $0 = \Vert f(x) – f(y) \Vert = \Vert x- y \Vert$ hence $$x=y$$.

There are two important cases when every isometry is affine:

1. $$f$$ is bijective (equivalently surjective). This is Mazur-Ulam theorem, which was proven in 1932.
2. $$F$$ is a strictly convex space. Recall that a normed vector space $$(S, \Vert \cdot \Vert)$$ is strictly convex if and only if for all distinct $$x,y \in S$$, $$\Vert x \Vert = \Vert y \Vert =1$$ implies $$\Vert \frac{x+y}{2} \Vert <1$$. For example, an inner product space is strictly convex. The sequence spaces $$\ell_p$$ for $$1 < p < \infty$$ are also strictly convex.

# A trigonometric series that is not a Fourier series (Lebesgue-integration)

We already provided here an example of a trigonometric series that is not the Fourier series of a Riemann-integrable function (namely the function $$\displaystyle x \mapsto \sum_{n=1}^\infty \frac{\sin nx}{\sqrt n}$$).

Applying an Abel-transformation (like mentioned in the link above), one can see that the function $f(x)=\sum_{n=2}^\infty \frac{\sin nx}{\ln n}$ is everywhere convergent. We now prove that $$f$$ cannot be the Fourier series of a Lebesgue-integrable function. The proof is based on the fact that for a $$2 \pi$$-periodic function $$g$$, Lebesgue-integrable on $$[0,2 \pi]$$, the sum $\sum_{n=1}^\infty \frac{c_n-c_{-n}}{n}$ is convergent where $$(c_n)_{n \in \mathbb Z}$$ are the complex Fourier coefficients of $$g$$: $c_n = \frac{1}{2 \pi} \int_0^{2 \pi} g(t)e^{-ikt} \ dt.$ As the series $$\displaystyle \sum_{n=2}^\infty \frac{1}{n \ln n}$$ is divergent, we will be able to conclude that the sequence defined by $\gamma_0=\gamma_1=\gamma_{-1} = 0, \, \gamma_n=- \gamma_{-n} = \frac{1}{\ln n} \ (n \ge 2)$ cannot be the Fourier coefficients of a Lebesgue-integrable function, hence that $$f$$ is not the Fourier series of any Lebesgue-integrable function. Continue reading A trigonometric series that is not a Fourier series (Lebesgue-integration)

# A trigonometric series that is not a Fourier series (Riemann-integration)

We’re looking here at convergent trigonometric series like $f(x) = a_0 + \sum_{k=1}^\infty (a_n \cos nx + b_n \sin nx)$ which are convergent but are not Fourier series. Which means that the terms $$a_n$$ and $$b_n$$ cannot be written$\begin{array}{ll} a_n = \frac{1}{\pi} \int_0^{2 \pi} g(t) \cos nt \, dt & (n= 0, 1, \dots) \\ b_n = \frac{1}{\pi} \int_0^{2 \pi} g(t) \sin nt \, dt & (n= 1, 2, \dots) \end{array}$ where $$g$$ is any integrable function.

This raises the question of the type of integral used. We cover here an example based on Riemann integral. I’ll cover a Lebesgue integral example later on.

We prove here that the function $f(x)= \sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ is a convergent trigonometric series but is not a Fourier series. Continue reading A trigonometric series that is not a Fourier series (Riemann-integration)

# Counterexamples around Dini’s theorem

In this article we look at counterexamples around Dini’s theorem. Let’s recall:

Dini’s theorem: If $$K$$ is a compact topological space, and $$(f_n)_{n \in \mathbb N}$$ is a monotonically decreasing sequence (meaning $$f_{n+1}(x) \le f_n(x)$$ for all $$n \in \mathbb N$$ and $$x \in K$$) of continuous real-valued functions on $$K$$ which converges pointwise to a continuous function $$f$$, then the convergence is uniform.

We look at what happens to the conclusion if we drop some of the hypothesis.

### Cases if $$K$$ is not compact

We take $$K=(0,1)$$, which is not closed equipped with the common distance. The sequence $$f_n(x)=x^n$$ of continuous functions decreases pointwise to the always vanishing function. But the convergence is not uniform because for all $$n \in \mathbb N$$ $\sup\limits_{x \in (0,1)} x^n = 1$

The set $$K=\mathbb R$$ is closed but unbounded, hence also not compact. The sequence defined by $f_n(x)=\begin{cases} 0 & \text{for } x < n\\ \frac{x-n}{n} & \text{for } n \le x < 2n\\ 1 & \text{for } x \ge 2n \end{cases}$ is continuous and monotonically decreasing. It converges to $$0$$. However, the convergence is not uniform as for all $$n \in \mathbb N$$: $$\sup\{f_n(x) : x \in \mathbb R\} =1$$. Continue reading Counterexamples around Dini’s theorem