# 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 $f(\{a\}) = 1 = f(\{b\})$

# A uniformly but not normally convergent function series

Consider a functions series $$\displaystyle \sum f_n$$ of functions defined on a set $$S$$ to $$\mathbb R$$ or $$\mathbb C$$. It is known that if $$\displaystyle \sum f_n$$ is normally convergent, then $$\displaystyle \sum f_n$$ is uniformly convergent.

The converse is not true and we provide two counterexamples.

Consider first the sequence of functions $$(g_n)$$ defined on $$\mathbb R$$ by:
$g_n(x) = \begin{cases} \frac{\sin^2 x}{n} & \text{for } x \in (n \pi, (n+1) \pi)\\ 0 & \text{else} \end{cases}$ The series $$\displaystyle \sum \Vert g_n \Vert_\infty$$ diverges as for all $$n \in \mathbb N$$, $$\Vert g_n \Vert_\infty = \frac{1}{n}$$ and the harmonic series $$\sum \frac{1}{n}$$ diverges. However the series $$\displaystyle \sum g_n$$ converges uniformly as for $$x \in \mathbb R$$ the sum $$\displaystyle \sum g_n(x)$$ is having only one term and $\vert R_n(x) \vert = \left\vert \sum_{k=n+1}^\infty g_k(x) \right\vert \le \frac{1}{n+1}$

For our second example, we consider the sequence of functions $$(f_n)$$ defined on $$[0,1]$$ by $$f_n(x) = (-1)^n \frac{x^n}{n}$$. For $$x \in [0,1]$$ $$\displaystyle \sum (-1)^n \frac{x^n}{n}$$ is an alternating series whose absolute value of the terms converge to $$0$$ monotonically. According to Leibniz test, $$\displaystyle \sum (-1)^n \frac{x^n}{n}$$ is well defined and we can apply the classical inequality $\displaystyle \left\vert \sum_{k=1}^\infty (-1)^k \frac{x^k}{k} – \sum_{k=1}^m (-1)^k \frac{x^k}{k} \right\vert \le \frac{x^{m+1}}{m+1} \le \frac{1}{m+1}$ for $$m \ge 1$$. Which proves that $$\displaystyle \sum (-1)^n \frac{x^n}{n}$$ converges uniformly on $$[0,1]$$.

However the convergence is not normal as $$\sup\limits_{x \in [0,1]} \frac{x^n}{n} = \frac{1}{n}$$.

# Root test

The root test is a test for the convergence of a series $\sum_{n=1}^\infty a_n$ where each term is a real or complex number. The root test was developed first by Augustin-Louis Cauchy.

We denote $l = \limsup\limits_{n \to \infty} \sqrt[n]{\vert a_n \vert}.$ $$l$$ is a non-negative real number or is possibly equal to $$\infty$$. The root test states that:

• if $$l < 1$$ then the series converges absolutely;
• if $$l > 1$$ then the series diverges.

The root test is inconclusive when $$l = 1$$.

### A case where $$l=1$$ and the series diverges

The harmonic series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$$ is divergent. However $\sqrt[n]{\frac{1}{n}} = \frac{1}{n^{\frac{1}{n}}}=e^{- \frac{1}{n} \ln n}$ and $$\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n}} = 1$$ as $$\lim\limits_{n \to \infty} \frac{\ln n}{n} = 0$$.

### A case where $$l=1$$ and the series converges

Consider the series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$$. We have $\sqrt[n]{\frac{1}{n^2}} = \frac{1}{n^{\frac{2}{n}}}=e^{- \frac{2}{n} \ln n}$ Therefore $$\limsup\limits_{n \to \infty} \sqrt[n]{\frac{1}{n^2}} = 1$$, while the series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$$ is convergent as we have seen in the ratio test article. Continue reading Root test

# Ratio test

The ratio test is a test for the convergence of a series $\sum_{n=1}^\infty a_n$ where each term is a real or complex number and is nonzero when $$n$$ is large. The test is sometimes known as d’Alembert’s ratio test.

Suppose that $\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = l$ The ratio test states that:

• if $$l < 1$$ then the series converges absolutely;
• if $$l > 1$$ then the series diverges.

What if $$l = 1$$? One cannot conclude in that case.

### Cases where $$l=1$$ and the series diverges

Consider the harmonic series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$$. We have $$\lim\limits_{n \to \infty} \frac{n+1}{n} = 1$$. It is well know that the harmonic series diverges. Recall that one proof uses the Cauchy’s convergence test based for $$k \ge 1$$ on the inequalities: $\sum_{n=2^k+1}^{2^{k+1}} \frac{1}{n} \ge \sum_{n=2^k+1}^{2^{k+1}} \frac{1}{2^{k+1}} = \frac{2^{k+1}-2^k}{2^{k+1}} \ge \frac{1}{2}$

An even simpler case is the series $$\displaystyle \sum_{n=1}^\infty 1$$.

### Cases where $$l=1$$ and the series converges

We also have $$\lim\limits_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = 1$$ for the infinite series $$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$$. The series is however convergent as for $$n \ge 1$$ we have:$0 \le \frac{1}{(n+1)^2} \le \frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}$ and the series $$\displaystyle \sum_{n=1}^\infty \left(\frac{1}{n} – \frac{1}{n+1} \right)$$ obviously converges.

Another example is the alternating series $$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n}$$.

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# A simple ring which is not a division ring

Let’s recall that a simple ring is a non-zero ring that has no two-sided ideal besides the zero ideal and itself. A division ring is a simple ring. Is the converse true? The answer is negative and we provide here a counterexample of a simple ring which is not a division ring.

We prove that for $$n \ge 1$$ the matrix ring $$M_n(F)$$ of $$n \times n$$ matrices over a field $$F$$ is simple. $$M_n(F)$$ is obviously not a division ring as the matrix with $$1$$ at position $$(1,1)$$ and $$0$$ elsewhere is not invertible.

Let’s prove first following lemma. Continue reading A simple ring which is not a division ring

# Small open sets containing the rationals

The set $$\mathbb Q$$ of the rational number is countable infinite and dense in $$\mathbb R$$. You can have a look here on a way to build a bijective map between $$\mathbb N$$ and $$\mathbb Q$$.

Now given $$\epsilon > 0$$, can one find an open set $$O_\epsilon$$ of measure less than $$\epsilon$$ with $$\mathbb Q \subseteq O_\epsilon$$?

The answer is positive. Let’s denote $O_\epsilon = \bigcup_{n \in \mathbb N} (r_n – \frac{\epsilon}{2^{n+1}},r_n + \frac{\epsilon}{2^{n+1}})$ where $$(r_n)_{n \in \mathbb N}$$ is an enumeration of the rationals. Obviously $$\mathbb Q \subseteq O_\epsilon$$. Using countable subadditivity of Lebesgue measure $$\mu$$, we get:
\begin{align*}
\mu(O_\epsilon) &\le \sum_{n \in \mathbb N} \mu((r_n – \frac{\epsilon}{2^{n+1}},r_n + \frac{\epsilon}{2^{n+1}}))\\
&= \sum_{n \in \mathbb N} \frac{2 \epsilon}{2^{n+1}} = \sum_{n \in \mathbb N} \frac{\epsilon}{2^n} = \epsilon
\end{align*}

• While Lebesgue measure of the reals is infinite and the rationals are dense in the reals, we can include the rationals in an open set of measure as small as desired!
• The open segments $$(r_n – \frac{\epsilon}{2^{n+1}},r_n + \frac{\epsilon}{2^{n+1}})$$ are overlapping. Hence $$\mu(O_\epsilon)$$ is strictly less than $$\epsilon$$.

# A continuous function with divergent Fourier series

It is known that for a piecewise continuously differentiable function $$f$$, the Fourier series of $$f$$ converges at all $$x \in \mathbb R$$ to $$\frac{f(x^-)+f(x^+)}{2}$$.

We describe Frejér example of a continuous function with divergent Fourier series. Frejér example is the even, $$(2 \pi)$$-periodic function $$f$$ defined on $$[0,\pi]$$ by: $f(x) = \sum_{p=1}^\infty \frac{1}{p^2} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right]$
According to Weierstrass M-test, $$f$$ is continuous. We denote $$f$$ Fourier series by $\frac{1}{2} a_0 + (a_1 \cos x + b_1 \sin x) + \dots + (a_n \cos nx + b_n \sin nx) + \dots.$

As $$f$$ is even, the $$b_n$$ are all vanishing. If we denote for all $$m \in \mathbb N$$:$\lambda_{n,m}=\int_0^{\pi} \sin \left[ (2m + 1) \frac{t}{2} \right] \cos nt \ dt \text{ and } \sigma_{n,m} = \sum_{k=0}^n \lambda_{k,m},$
we have:\begin{aligned} a_n &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos nt \ dt= \frac{2}{\pi} \int_0^{\pi} f(t) \cos nt \ dt\\ &= \frac{2}{\pi} \int_0^{\pi} \left(\sum_{p=1}^\infty \frac{1}{p^2} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right]\right) \cos nt \ dt\\ &=\frac{2}{\pi} \sum_{p=1}^\infty \frac{1}{p^2} \int_0^{\pi} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right] \cos nt \ dt\\ &=\frac{2}{\pi} \sum_{p=1}^\infty \frac{1}{p^2} \lambda_{n,2^{p^3-1}} \end{aligned} One can switch the $$\int$$ and $$\sum$$ signs as the series is normally convergent.

We now introduce for all $$n \in \mathbb N$$:$S_n = \frac{\pi}{2} \sum_{k=0}^n a_k = \sum_{p=1}^\infty \sum_{k=0}^n \frac{1}{p^2} \lambda_{k,2^{p^3-1}} =\sum_{p=1}^\infty \frac{1}{p^2} \sigma_{n,2^{p^3-1}}$

We will prove below that for all $$n,m \in \mathbb N$$ we have $$\sigma_{m,m} \ge \frac{1}{2} \ln m$$ and $$\sigma_{n,m} \ge 0$$. Assuming those inequalities for now, we get:$S_{2^{p^3-1}} \ge \frac{1}{p^2} \sigma_{2^{p^3-1},2^{p^3-1}} \ge \frac{1}{2p^2} \ln(2^{p^3-1}) = \frac{p^3-1}{2p^2} \ln 2$
As the right hand side diverges to $$\infty$$, we can conclude that $$(S_n)$$ diverges and consequently that the Fourier series of $$f$$ diverges at $$0$$. Continue reading A continuous function with divergent Fourier series