Tag Archives: normed-vector-spaces

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:\[
T : & V & \longrightarrow & V \\
& x & \longmapsto & x & \text{if } \Vert x \Vert \neq 1\\
& x & \longmapsto & -x & \text{if } \Vert x \Vert = 1\end{array}\]

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:\[
T(2 x_0) &= 2 x_0 \text{ as } \Vert 2 x_0 \Vert = 2\\
T(x_0) + T(x_0) = -x_0 + (-x_0) &= – 2 x_0

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.

Continue reading Isometric versus affine

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