Properties of an infinite dimensional Banach space over the field with two elements

TL;DR

This paper demonstrates that over the field with two elements, there exists an infinite dimensional Banach space where every bounded linear operator attains its norm, addressing a longstanding open question.

Contribution

It proves the existence of an infinite dimensional Banach space over GF(2) in which all bounded linear operators attain their norm, a previously unresolved problem.

Findings

Existence of such Banach space over GF(2)

All bounded linear operators on this space attain their norm

Addresses a question posed by M.I. Ostrovskii

Abstract

A banach space X is a normed vector space, which is complete with respect to the metric induced by the norm. Given a bounded linear operator T acting on a banach space X, T is said to attain its norm if there is a unit vector z in X, such that the norm of Tz equals the norm of T. The existence of an infinite dimensional banach space X, in which each bounded linear operator acting on X attains its norm, is still undetermined. This question was posed by M.I. Ostrovskii at St. John's University. In this paper we show that if an infinite dimensional banach space is considered over GF(2), then it is possible for every bounded linear operator to attain its norm.

Tables

Table 1. Table 1 : Addition in GF(2).

+	0	1
0	0	1
1	1	0

Table 2. Table 2 : Multiplication in GF(2).

$\cdot$	0	1
0	0	0
1	0	1

Equations

$$|0|=0,|1|=1$$

$$S=\{(s_1,s_2,\dots )|s_i\ne 0\text{ for finitely many }i\in \mathbb{N}\}$$

$$\mathbf{x}+\mathbf{y}=(x_1+y_1,x_2+y_2,\dots )$$

$$\Vert \mathbf{x}{\Vert }_S=\{\begin{array}{ll}0,& \mathbf{x}=0\\ 1,& \mathbf{x}\ne 0\end{array}$$

$${\mathbf{e}}_n=(0,\dots ,0,1,0,\dots )$$

$$\alpha (\beta \mathbf{x})=\{\begin{array}{ll}0,& \alpha =0\text{ or }\beta =0\\ \mathbf{x},& \alpha =1\text{ and }\beta =1\end{array}$$

$$\Vert \alpha \mathbf{x}{\Vert }_S=\Vert \mathbf{x}{\Vert }_S=1\Vert \mathbf{x}{\Vert }_S=|\alpha |\Vert \mathbf{x}{\Vert }_S$$

$$\Vert \alpha \mathbf{x}{\Vert }_S=\Vert 0{\Vert }_S=0=0\Vert \mathbf{x}{\Vert }_S=|\alpha |\Vert \mathbf{x}{\Vert }_S$$

$$\Vert \mathbf{x}-\mathbf{y}{\Vert }_S=\Vert \mathbf{x}+(-\mathbf{y}){\Vert }_S=\Vert \mathbf{x}+\mathbf{y}{\Vert }_S=\{\begin{array}{ll}1,& \mathbf{x}\ne \mathbf{y}\\ 0,& \mathbf{x}=\mathbf{y}\end{array}$$

$$\Vert T\Vert =\sup \{\Vert T\mathbf{x}{\Vert }_S|\mathbf{x}\in S,\Vert \mathbf{x}{\Vert }_S=1\}$$

$$\Vert T\Vert =1$$

$$\Vert \mathbf{x}+\mathbf{y}{\Vert }_S^2+\Vert \mathbf{x}-\mathbf{y}{\Vert }_S^2=2\Vert \mathbf{x}{\Vert }_S^2+2\Vert \mathbf{y}{\Vert }_S^2$$

$$\bigcup _{m=0}^p{S}_m^p$$

$$S=\bigcup _{p=0}^{\infty }\bigcup _{m=0}^p{S}_m^p.$$

Full text

Introduction

The Galois Field of two elements, denoted GF(2), is the field containing 0 (zero) and 1 (one). The operations of addition and multiplication are defined as follows:

Since GF(2) satisfies the axioms required to be a field, we may consider vector spaces over GF(2), which may be endowed with a norm. In order to meaningfully define a norm on a vector space over GF(2), we define a function $|\cdot|:\mathrm{GF(2)}\rightarrow\mathbb{R}$ which acts as an absolute value.

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This definition of absolute value trivially satisfies non-negativity, positive-definiteness, multiplicativity, as well as the triangle inequality. So it is indeed sensible to define the absolute value for elements of GF(2) in this way.

Theorem**.**

There exists an infinite dimensional banach space S over GF(2) such that each bounded linear operator on S attains its norm.

Proof.

Define an infinite dimensional banach space $S$ over GF(2) as follows:

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Vector addition and scalar multiplication are defined entry-wise.

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Note here that the operations $x_{i}+y_{i}$ and $\alpha x_{i}$ occur in GF(2). The space $S$ will be given the norm $\|\,\|_{S}:S\rightarrow\mathbb{R}$ defined by:

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Here, the zero vector is taken to be the sequence of all zeros. This space has the canonical basis, where $\mathbf{e}_{n}$ has a 1 in the $n^{th}$ spot and 0 in the rest.

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First we must verify that $S$ is a vector space. Since addition is performed entry-wise, associativity and commutativity are inherited properities of addition in the field. The identity element is the sequence of all zeros. Furthermore, since $1+1=0$ in GF(2), every element of $S$ is its own inverse with respect to addition. Now let $\alpha$ and $\beta$ be elements of GF(2), and $\mathbf{x}$ be in $S$. Then:

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Similiary for $(\alpha\beta)\mathbf{x}$, and thus scalar multiplication is compatible with field multiplication. The identity element of scalar multiplication is $1\in\text{GF(2)}$. Finally, scalar multiplication trivially distributes over vector addition as well as field addition. Thus $S$ is a vector space.

Now we verify that $S$ is a normed space. By the definition of $\|\,\|_{S}$, only the zero vector has norm zero, and all others have norm one. Thus positive-definiteness of the norm is satisfied. Now let $\alpha$ be an element of GF(2), and let $\mathbf{x}$ be a zero vector in $S$. If $\alpha$ is one then we observe:

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If $\alpha$ is zero then instead we have:

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In either case the result is that $\|\,\|_{S}$ is absolutely homogeneous. Moving forward, if $\mathbf{x}$ and $\mathbf{y}$ are distinct non-zero vectors in $S$, the norm of their sum will equal one. However the sum of their norms will be two, and thus be greater. For all other cases - one of them is the zero vector, both of them are the zero vector, or they are inverses - the triangle inequality trivially holds. Thus $\|\,\|_{S}$ is a norm on the vector space $S$.

Lastly, in order for $S$ to be a Banach space, it must be complete with respect to the metric induced by the norm. If $\mathbf{x}$ and $\mathbf{y}$ are non-zero vectors in $S$, then the distance between them is given as:

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We see that the metric induced by the norm is the discrete metric, so for a sequence in $S$ to be Cauchy, it must eventually be constant. Thus every Cauchy sequence in $S$ converges. Now that we have verified that $S$ is a Banach space, we must show that every operator $T$ in the space of bounded linear operators acting on $S$, $L(S)$ attains its norm. Let $T\in L(S)$ be a non-zero operator. The norm of $T$, $\|T\|$ is defined as:

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For any vector in $S$, the norm of its image under $T$ can only be either zero or one. Since $T$ was assumed not to be the zero operator, we obtain:

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Furthermore, there must exist some $\tilde{\mathbf{x}}\in S$ such that $\lVert T\tilde{\mathbf{x}}\rVert_{S}=1$, otherwise $T$ would have to be the zero operator. It should also be noted that the zero operator attains its norm via any point in $S$. Therefore, every operator in $L(S)$ attains its norm. This concludes the proof. ∎

Remarks

It is worth noting that by definition, $S$ cannot be a Hilbert space, since given some distinct and non-zero $\mathbf{x},\mathbf{y}\in S$, the parallelogram identity:

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Is true if and only if $\mathbf{x}=\mathbf{y}=\mathbf{0}$. Now consider the subset of sequences, $S_{m}^{p}$, which has $m$ 1’s, all of which occur before or at the $p^{th}$ position in the sequence. There are $p$-choose-$m$ such sequences.

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The union of these sets, as shown above, consists of all sequences with zeros after the $p^{th}$ entry. This union is a finite union of finite sets, and thus finite. The infinite union for all $p$ will be exactly our space $S$.

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This is a countable union of finite sets, and thus $S$ is countable. Furthermore, because $S$ has the discrete topology, the only dense subset of $S$ is $S$ itself. Thus $S$ is a countable dense subset, and $S$ is trivially separable.