Tag Archives: banach-spaces

Separability of a vector space and its dual

November 11, 2019 Jean-Pierre Merx Leave a comment

Recall that a topological space is considered separable when it contains a countable dense set. The following theorem establishes a significant connection between separability and dual spaces:

Theorem: If the dual $X^*$ of a normed vector space $X$ is separable, then so is the space $X$ itself.

Proof outline: let ${f_n}$ be a countable dense set in $X^*$ unit sphere $S_*$ . For any $n \in \mathbb{N}$ one can find $x_n$ in $X$ unit ball such that $f_n(x_n) \ge \frac{1}{2}$ . We claim that the countable set $F = \mathrm{Span}_{\mathbb{Q}}(x_0,x_1,…)$ is dense in $X$ . If not, we would find $x \in X \setminus \overline{F}$ and according to the Hahn-Banach theorem, there would exist a linear functional $f \in X^*$ such that $f_{\overline{F}} = 0$ and $\Vert f \Vert=1$ . But then for all $n \in \mathbb{N}$ , $\Vert f_n-f \Vert \ge \vert f_n(x_n)-f(x_n)\vert = \vert f_n(x_n) \vert \ge \frac{1}{2}$ . A contradiction since ${f_n}$ is supposed to be dense in $S_*$ .

We prove that the converse is not true, i.e., a dual space can be separable, while the space itself may not be.

Introducing some normed vector spaces

Given a closed interval $K \subset \mathbb{R}$ and a set $A \subset \mathbb{R}$ , we define the $4$ following spaces. The first three are endowed with the supremum norm and the last with the $\ell^1$ norm.

$\mathcal{C}(K,\mathbb{R})$ , the space of continuous functions from $K$ to $\mathbb{R}$ , is separable as the polynomial functions with coefficients in $\mathbb{Q}$ are dense and countable.
$\ell^{\infty}(A, \mathbb{R})$ is the space of real bounded functions defined on $A$ with countable support.
$c_0(A, \mathbb{R}) \subset \ell^{\infty}(A, \mathbb{R})$ is the subspace of elements of $\ell^{\infty}(A)$ going to $0$ at $\infty$ .
$\ell^1(A, \mathbb{R})$ is the space of summable functions on $A$ : $u \in \mathbb{R}^{A}$ is in $\ell^1(A, \mathbb{R})$ iff $\sum \limits_{a \in A} |u_x| < +\infty$ .

We find the usual sequence spaces when $A = \mathbb{N}$ . It should be noted that $c_0(A, \mathbb{R})$ and $\ell^1(A, \mathbb{R})$ are separable iff $A$ is countable (otherwise the subset $\big\{x \mapsto 1_{\{a\}}(x),\ a \in A \big\}$ is uncountable, and discrete), and that $\ell^{\infty}(A, \mathbb{R})$ is separable iff $A$ is finite (otherwise the subset $\{0,1\}^A$ is uncountable, and discrete).

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Topology

A non complete normed vector space

March 12, 2017 Jean-Pierre Merx Leave a comment

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.

Analysis

Existence of a continuous function with divergent Fourier series

October 29, 2016 Jean-Pierre Merx 1 Comment

In that article, I provided an example of a continuous function with divergent Fourier series. We prove here the existence of such a function using Banach-Steinhaus theorem, also called uniform boundedness principle.

Theorem (Uniform Boundedness Theorem) Let $(X, \Vert \cdot \Vert_X)$ be a Banach space and $(Y, \Vert \cdot \Vert_Y)$ be a normed vector space. Suppose that $F$ is a set of continuous linear operators from $X$ to $Y$ . If for all $x \in X$ one has $\sup\limits_{T \in F} \Vert T(x) \Vert_Y \lt \infty$ then $\sup\limits_{T \in F, \ \Vert x \Vert = 1} \Vert T(x) \Vert_Y \lt \infty$

Let’s take for $X$ the vector space $\mathcal C_{2 \pi}$ of continuous functions from $\mathbb R$ to $\mathbb C$ which are periodic with period $2 \pi$ endowed with the norm $\Vert f \Vert_\infty = \sup\limits_{- \pi \le t \le \pi} \vert f(t) \vert$ . $(\mathcal C_{2 \pi}, \Vert \cdot \Vert_\infty)$ is a Banach space. For the vector space $Y$ , we take the complex numbers $\mathbb C$ endowed with the modulus.

For $n \in \mathbb N$ , the map $\begin{array}{l|rcl} \ell_n : & \mathcal C_{2 \pi} & \longrightarrow & \mathbb C \\ & f & \longmapsto & \displaystyle \sum_{p=-n}^n c_p(f) \end{array}$ is a linear operator, where for $p \in \mathbb Z$ , $c_p(f)$ denotes the complex Fourier coefficient $c_p(f) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(t) e^{-i p t} \ dt$

We now prove that
$\begin{align*} \Lambda_n &= \sup\limits_{f \in \mathcal C_{2 \pi}, \Vert f \Vert_\infty=1} \vert \ell_n(f) \vert\\ &= \frac{1}{2 \pi} \int_{- \pi}^{\pi} \left\vert \frac{\sin (2n+1)\frac{t}{2}}{\sin \frac{t}{2}} \right\vert \ dt = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \left\vert h_n(t) \right\vert \ dt, \end{align*}$ where one can notice that the function $\begin{array}{l|rcll} h_n : & [- \pi, \pi] & \longrightarrow & \mathbb C \\ & t & \longmapsto & \frac{\sin (2n+1)\frac{t}{2}}{\sin \frac{t}{2}} &\text{for } t \neq 0\\ & 0 & \longmapsto & 2n+1 \end{array}$ is continuous.
Continue reading Existence of a continuous function with divergent Fourier series →

Topology

Counterexamples around Banach-Steinhaus theorem

October 24, 2015 Jean-Pierre Merx Leave a comment

In this article we look at what happens to Banach-Steinhaus theorem when the completness hypothesis is not fulfilled. One form of Banach-Steinhaus theorem is the following one.

Banach-Steinhaus Theorem Let $T_n : E \to F$ be a sequence of continuous linear maps from a Banach space $E$ to a normed space $F$ . If for all $x \in E$ the sequence $T_n x$ is convergent to $Tx$ , then $T$ is a continuous linear map.

A sequence of continuous linear maps converging to an unbounded linear map

Let $c_{00}$ be the vector space of real sequences $x=(x_n)$ eventually vanishing, equipped with the norm $\Vert x \Vert = \sup_{n \in \mathbb N} \vert x_n \vert$ For $n \in \mathbb N$ , $T_n : E \to E$ denotes the linear map defined by $T_n x = (x_1,2 x_2, \dots, n x_n,0,0, \dots).$ $T_n$ is continuous as for $\Vert x \Vert \le 1$ , we have
$\begin{align*} \Vert T_n x \Vert &= \Vert (x_1,2 x_2, \dots, n x_n,0,0, \dots) \Vert\\ & = \sup_{1 \le k \le n} \vert k x_k \vert \le n \Vert x \Vert \le n \end{align*}$ Continue reading Counterexamples around Banach-Steinhaus theorem →

Topology

Counterexamples to Banach fixed-point theorem

August 30, 2015 Jean-Pierre Merx Leave a comment

Let $(X,d)$ be a metric space. Then a map $T : X \to X$ is called a contraction map if it exists $0 \le k < 1$ such that $d(T(x),T(y)) \le k d(x,y)$ for all $x,y \in X$ . According to Banach fixed-point theorem, if $(X,d)$ is a complete metric space and $T$ a contraction map, then $T$ admits a fixed-point $x^* \in X$ , i.e. $T(x^*)=x^*$ .

We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled.

First, let’s consider $\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & x+1 \end{array}$ For all $x,y \in \mathbb R$ we have $\vert f(x)-f(y) \vert = \vert x- y \vert$ . $f$ is not a contraction, but an isometry. Obviously, $f$ has no fixed-point.

We now prove that a map satisfying $d(g(x),g(y)) < d(x,y)$ might also not have a fixed-point. A counterexample is the following map $\begin{array}{l|rcl} g : & [0,+\infty) & \longrightarrow & [0,+\infty) \\ & x & \longmapsto & \sqrt{1+x^2} \end{array}$ Since $g^\prime(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for all } \xi \in [0, +\infty),$ by the mean value theorem $\vert g(x) - g(y)| = \vert g^\prime(\xi)\vert |x-y| < |x-y| \text{ for all } x, y \in [0, +\infty).$ However $g$ has no fixed-point. Finally, let's have a look to a space $(X,d)$ which is not complete. We take $a,b \in \mathbb R$ with $0 < a < 1$ and for $(X,d)$ the space $X = \mathbb R \setminus \{\frac{b}{1-a}\}$ equipped with absolute value distance. $X$ is not complete. Consider the map $\begin{array}{l|rcl} h : & X & \longrightarrow & X \\ & x & \longmapsto & ax + b \end{array}$ $h$ is well defined as for $x \neq \frac{b}{1-a}$ , $h(x) \neq \frac{b}{1-a}$ . $h$ is a contraction map as for $x,y \in \mathbb R$ $\vert h(x)-h(y) \vert = a \vert x - y \vert$ However, $h$ doesn't have a fixed-point in $X$ as $\frac{b}{1-a}$ is the only real for which $h(x)=x$ .

Topology

Counterexample around Arzela-Ascoli theorem

June 7, 2015 Jean-Pierre Merx Leave a comment

Let’s recall Arzelà–Ascoli theorem:

Suppose that $F$ is a Banach space and $E$ a compact metric space. A subset $\mathcal{H}$ of the Banach space $\mathcal{C}_F(E)$ is relatively compact in the topology induced by the uniform norm if and only if it is equicontinuous and and for all $x \in E$ , the set $\mathcal{H}(x)=\{f(x) \ | \ f \in \mathcal{H}\}$ is relatively compact.

We look here at what happens if we drop the requirement on space $E$ to be compact and provide a counterexample where the conclusion of Arzelà–Ascoli theorem doesn’t hold anymore.

We take for $E$ the real interval $[0,+\infty)$ and for all $n \in \mathbb{N} \setminus \{0\}$ the real function
$f_n(t)= \sin \sqrt{t+4 n^2 \pi^2}$ We prove that $(f_n)$ is equicontinuous, converges pointwise to $0$ but is not relatively compact.

According to the mean value theorem, for all $x,y \in \mathbb{R}$
$\vert \sin x – \sin y \vert \le \vert x – y \vert$ Hence for $n \ge 1$ and $x,y \in [0,+\infty)$
$\begin{align*} \vert f_n(x)-f_n(y) \vert &\le \vert \sqrt{x+4 n^2 \pi^2} -\sqrt{y+4 n^2 \pi^2} \vert \\ &= \frac{\vert x – y \vert}{\sqrt{x+4 n^2 \pi^2} +\sqrt{y+4 n^2 \pi^2}} \\ &\le \frac{\vert x – y \vert}{4 \pi} \end{align*}$ using multiplication by the conjugate.

Which enables to prove that $(f_n)$ is equicontinuous.

We also have for $n \ge 1$ and $x \in [0,+\infty)$
$\begin{align*} \vert f_n(x) \vert &= \vert f_n(x) – f_n(0) \vert \le \vert \sqrt{x+4 n^2 \pi^2} -\sqrt{4 n^2 \pi^2} \vert \\ &= \frac{\vert x \vert}{\sqrt{x+4 n^2 \pi^2} +\sqrt{4 n^2 \pi^2}} \\ &\le \frac{\vert x \vert}{4 n \pi} \end{align*}$

Hence $(f_n)$ converges pointwise to $0$ and for $t \in [0,+\infty), \mathcal{H}(t)=\{f_n(t) \ | \ n \in \mathbb{N} \setminus \{0\}\}$ is relatively compact

Finally we prove that $\mathcal{H}=\{f_n \ | \ n \in \mathbb{N} \setminus \{0\}\}$ is not relatively compact. While $(f_n)$ converges pointwise to $0$ , $(f_n)$ does not converge uniformly to $f=0$ . Actually for $n \ge 1$ and $t_n=\frac{\pi^2}{4} + 2n \pi^2$ we have
$f_n(t_n)= \sin \sqrt{\frac{\pi^2}{4} + 2n \pi^2 +4 n^2 \pi^2}=\sin \sqrt{\left(\frac{\pi}{2} + 2 n \pi\right)^2}=1$ Consequently for all $n \ge 1$ $\Vert f_n – f \Vert_\infty \ge 1$ . If $\mathcal{H}$ was relatively compact, $(f_n)$ would have a convergent subsequence with $f=0$ for limit. And that cannot be as for all $n \ge 1$ $\Vert f_n – f \Vert_\infty \ge 1$ .

Topology

Distance between a point and a hyperplane not reached

April 26, 2015 Jean-Pierre Merx Leave a comment

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 →

Analysis

A solution of a differential equation not exploding in finite time

January 11, 2015 Jean-Pierre Merx 2 Comments

In this post, I mention that Peano existence theorem is valid for finite dimensional vector spaces, but not for Banach spaces of infinite dimension. I highlight here a second property of ordinary differential equations which is valid for finite dimensional vector spaces but not for infinite dimensional Banach spaces. Continue reading A solution of a differential equation not exploding in finite time →

Topology

An empty intersection of nested closed convex subsets in a Banach space

August 4, 2014 Jean-Pierre Merx Leave a comment

We consider a decreasing sequence $(C_n)_{n \in \mathbb{N}}$ of non empty closed convex subsets of a Banach space $E$ .

If the convex subsets are closed balls, their intersection is not empty. To see this let $x_n$ be the center and $r_n > 0$ the radius of the ball $C_n$ . For $0 \leq n < m$ we have $\Vert x_m-x_n\Vert \leq r_n – r_m$ which proves that $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence. As the space $E$ is Banach, $(x_n)_{n \in \mathbb{N}}$ converges to a limit $x$ and $x \in \bigcap_{n=0}^{+\infty} C_n$ . Continue reading An empty intersection of nested closed convex subsets in a Banach space →

Math Counterexamples

Tag Archives: banach-spaces

Separability of a vector space and its dual

Introducing some normed vector spaces

A non complete normed vector space

Existence of a continuous function with divergent Fourier series

Counterexamples around Banach-Steinhaus theorem

A sequence of continuous linear maps converging to an unbounded linear map

Counterexamples to Banach fixed-point theorem

Counterexample around Arzela-Ascoli theorem

Distance between a point and a hyperplane not reached

The case of finite dimensional vector spaces

A solution of a differential equation not exploding in finite time

An empty intersection of nested closed convex subsets in a Banach space

Mathematical exceptions to the rules or intuition