# A non complete normed vector space

Consider a real normed vector space $$V$$. $$V$$ is called complete if every Cauchy sequence in $$V$$ converges in $$V$$. A complete normed vector space is also called a Banach space.

A finite dimensional vector space is complete. This is a consequence of a theorem stating that all norms on finite dimensional vector spaces are equivalent.

There are many examples of Banach spaces with infinite dimension like $$(\ell_p, \Vert \cdot \Vert_p)$$ the space of real sequences endowed with the norm $$\displaystyle \Vert x \Vert_p = \left( \sum_{i=1}^\infty \vert x_i \vert^p \right)^{1/p}$$ for $$p \ge 1$$, the space $$(C(X), \Vert \cdot \Vert)$$ of real continuous functions on a compact Hausdorff space $$X$$ endowed with the norm $$\displaystyle \Vert f \Vert = \sup\limits_{x \in X} \vert f(x) \vert$$ or the Lebesgue space $$(L^1(\mathbb R), \Vert \cdot \Vert_1)$$ of Lebesgue real integrable functions endowed with the norm $$\displaystyle \Vert f \Vert = \int_{\mathbb R} \vert f(x) \vert \ dx$$.

Let’s give an example of a non complete normed vector space. Let $$(P, \Vert \cdot \Vert_\infty)$$ be the normed vector space of real polynomials endowed with the norm $$\displaystyle \Vert p \Vert_\infty = \sup\limits_{x \in [0,1]} \vert p(x) \vert$$. Consider the sequence of polynomials $$(p_n)$$ defined by
$p_n(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \cdots + \frac{x^n}{2^n} = \sum_{k=0}^{n} \frac{x^k}{2^k}.$ For $$m < n$$ and $$x \in [0,1]$$, we have $\vert p_n(x) - p_m(x) \vert = \left\vert \sum_{i=m+1}^n \frac{x^i}{2^i} \right\vert \le \sum_{i=m+1}^n \frac{1}{2^i} \le \frac{1}{2^m}$ which proves that $$(p_n)$$ is a Cauchy sequence. Also for $$x \in [0,1]$$ $\lim\limits_{n \to \infty} p_n(x) = p(x) \text{ where } p(x) = \frac{1}{1 - \frac{x}{2}}.$ As uniform converge implies pointwise convergence, if $$(p_n)$$ was convergent in $$P$$, it would be towards $$p$$. But $$p$$ is not a polynomial function as none of its $$n$$th-derivative always vanishes. Hence $$(p_n)$$ is a Cauchy sequence that doesn't converge in $$(P, \Vert \cdot \Vert_\infty)$$, proving as desired that this normed vector space is not complete. More generally, a normed vector space with countable dimension is never complete. This can be proven using Baire category theorem which states that a non-empty complete metric space is not the countable union of nowhere-dense closed sets.

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

# 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 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

# A homeomorphism of the unit ball having no fixed point

Let’s recall Brouwer fixed-point theorem.

Theorem (Brouwer): Every continuous function from a convex compact subset $$K$$ of a Euclidean space to $$K$$ itself has a fixed point.

We here describe an example of a homeomorphism of the unit ball of a Hilbert space having no fixed point. Let $$E$$ be a separable Hilbert space with $$(e_n)_{n \in \mathbb{Z}}$$ as a Hilbert basis. $$B$$ and $$S$$ are respectively $$E$$ closed unit ball and unit sphere.

There is a unique linear map $$u : E \to E$$ for which $$u(e_n)=e_{n+1}$$ for all $$n \in \mathbb{Z}$$. For $$x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$$ we have $$u(x)= \sum_{n \in \mathbb{Z}} \xi_n e_{n+1}$$. $$u$$ is isometric as $\Vert u(x) \Vert^2 = \sum_{n \in \mathbb{Z}} \vert \xi_n \vert^2 = \Vert x \Vert^2$ hence one-to-one. $$u$$ is also onto as for $$x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$$, $$\sum_{n \in \mathbb{Z}} \xi_n e_{n-1} \in E$$ is an inverse image of $$x$$. Finally $$u$$ is an homeomorphism. Continue reading A homeomorphism of the unit ball having no fixed point

# Distance between a point and a hyperplane not reached

Let’s investigate the following question: “Is the distance between a point and a hyperplane always reached?”

In order to provide answers to the question, we consider a normed vector space $$(E, \Vert \cdot \Vert)$$ and a hyperplane $$H$$ of $$E$$. $$H$$ is the kernel of a non-zero linear form. Namely, $$H=\{x \in E \text{ | } u(x)=0\}$$.

## The case of finite dimensional vector spaces

When $$E$$ is of finite dimension, the distance $$d(a,H)=\inf\{\Vert h-a \Vert \text{ | } h \in H\}$$ between any point $$a \in E$$ and a hyperplane $$H$$ is reached at a point $$b \in H$$. The proof is rather simple. Consider a point $$c \in H$$. The set $$S = \{h \in H \text{ | } \Vert a- h \Vert \le \Vert a-c \Vert \}$$ is bounded as for $$h \in S$$ we have $$\Vert h \Vert \le \Vert a-c \Vert + \Vert a \Vert$$. $$S$$ is equal to $$D \cap H$$ where $$D$$ is the inverse image of the closed real segment $$[0,\Vert a-c \Vert]$$ by the continuous map $$f: x \mapsto \Vert a- x \Vert$$. Therefore $$D$$ is closed. $$H$$ is also closed as any linear subspace of a finite dimensional vector space. $$S$$ being the intersection of two closed subsets of $$E$$ is also closed. Hence $$S$$ is compact and the restriction of $$f$$ to $$S$$ reaches its infimum at some point $$b \in S \subset H$$ where $$d(a,H)=\Vert a-b \Vert$$. Continue reading Distance between a point and a hyperplane not reached

# An unbounded convex not containing a ray

We consider a normed vector space $$E$$ over the field of the reals $$\mathbb{R}$$ and a convex subset $$C \subset E$$.

We suppose that $$0 \in C$$ and that $$C$$ is unbounded, i.e. there exists points in $$C$$ at distance as big as we wish from $$0$$.

The following question arises: “does $$C$$ contains a ray?”. It turns out that the answer depends on the dimension of the space $$E$$. If $$E$$ is of finite dimension, then $$C$$ always contains a ray, while if $$E$$ is of infinite dimension $$C$$ may not contain a ray. Continue reading An unbounded convex not containing a ray