# 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: $\begin{cases} T[(x,0) + (0,y)] = T[(x,y)] &= (x,y)\\ \text{while}\\ T[(x,0)] + T[(0,y)] = (0,x) + (y,0) &= (y,x) \end{cases}$

# The line with two origins

Let’s introduce and describe some properties of the line with two origins.

Let $$X$$ be the union of the set $$\mathbb R \setminus \{0\}$$ and the two-point set $$\{p,q\}$$. The line with two origins is the set $$X$$ topologized by taking as base the collection $$\mathcal B$$ of all open intervals in $$\mathbb R$$ that do not contain $$0$$, along with all sets of the form $$(-a,0) \cup \{p\} \cup (0,a)$$ and all sets of the form $$(-a,0) \cup \{q\} \cup (0,a)$$, for $$a > 0$$.

### $$\mathcal B$$ is a base for a topology $$\mathcal T$$ of $$X$$

Indeed, one can verify that the elements of $$\mathcal B$$ cover $$X$$ as $X = \left( \bigcup_{a > 0} (-a,0) \cup \{p\} \cup (0,a) \right) \cup \left( \bigcup_{a > 0} (-a,0) \cup \{q\} \cup (0,a) \right)$ and that the intersection of two elements of $$\mathcal B$$ is the union of elements of $$\mathcal B$$ (verification left to the reader).

### Each of the spaces $$X \setminus \{p\}$$ and $$X \setminus \{q\}$$ is homeomorphic to $$\mathbb R$$

Let’s prove it for $$X \setminus \{p\}$$. The map $\begin{array}{l|rcll} f : & X \setminus \{p\} & \longrightarrow & \mathbb R \\ & x & \longmapsto & x & \text{for } x \neq q\\ & q & \longmapsto & 0 \end{array}$ is a bijection. $$f$$ is continuous as the inverse image of an open interval $$I$$ of $$\mathbb R$$ is an open subset of $$X$$. For example taking $$I=(-b,c)$$ with $$0 < b < c$$, we have \begin{align*} f^{-1}[I] &= (-b,0) \cup \{q\} \cup (0,c)\\ &= \left( (-b,0) \cup \{q\} \cup (0,b) \right) \cup (b/2,c) \end{align*} One can also prove that $$f^{-1}$$ is continuous. Continue reading The line with two origins

# A power series converging everywhere on its circle of convergence defining a non-continuous function

Consider a complex power series $$\displaystyle \sum_{k=0}^\infty a_k z^k$$ with radius of convergence $$0 \lt R \lt \infty$$ and suppose that for every $$w$$ with $$\vert w \vert = R$$, $$\displaystyle \sum_{k=0}^\infty a_k w^k$$ converges.

We provide an example where the power expansion at the origin $\displaystyle f(z) = \sum_{k=0}^\infty a_k z^k$ is discontinuous on the closed disk $$\vert z \vert \le R$$.

The function $$f$$ is constructed as an infinite sum $\displaystyle f(z) = \sum_{n=1}^\infty f_n(z)$ with $$f_n(z) = \frac{\delta_n}{a_n-z}$$ where $$(\delta_n)_{n \in \mathbb N}$$ is a sequence of positive real numbers and $$(a_n)$$ a sequence of complex numbers of modulus larger than one and converging to one. Let $$f_n^{(r)}(z)$$ denote the sum of the first $$r$$ terms in the power series expansion of $$f_n(z)$$ and $$\displaystyle f^{(r)}(z) \equiv \sum_{n=1}^\infty f_n^{(r)}(z)$$.

We’ll prove that:

1. If $$\sum_n \delta_n \lt \infty$$ then $$\sum_{n=1}^\infty f_n^{(r)}(z)$$ converges and $$f(z) = \lim\limits_{r \to \infty} \sum_{n=1}^\infty f_n^{(r)}(z)$$ for $$\vert z \vert \le 1$$ and $$z \neq 1$$.
2. If $$a_n=1+i \epsilon_n$$ and $$\sum_n \delta_n/\epsilon_n < \infty$$ then $$\sum_{n=1}^\infty f_n^{(r)}(1)$$ converges and $$f(1) = \lim\limits_{r \to \infty} \sum_{n=1}^\infty f_n^{(r)}(1)$$
3. If $$\delta_n/\epsilon_n^2 \to \infty$$ then $$f(z)$$ is unbounded on the disk $$\vert z \vert \le 1$$.

First, let’s recall this corollary of Lebesgue’s dominated convergence theorem:

Let $$(u_{n,i})_{(n,i) \in \mathbb N \times \mathbb N}$$ be a double sequence of complex numbers. Suppose that $$u_{n,i} \to v_i$$ for all $$i$$ as $$n \to \infty$$, and that $$\vert u_{n,i} \vert \le w_i$$ for all $$n$$ with $$\sum_i w_i < \infty$$. Then for all $$n$$ the series $$\sum_i u_{n,i}$$ is absolutely convergent and $$\lim_n \sum_i u_{n,i} = \sum_i v_i$$.
Continue reading A power series converging everywhere on its circle of convergence defining a non-continuous function

# 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.