Question

It is not known whether an infinite-dimensional Banach space \(X\) exists such that \(\mathcal{B}(X)=\operatorname{span}\left(\left\\{I_{X}\right\\} \cup \mathcal{K}(X)\right)\). Show that a Hilbert space does not satisfy this property. It is also not known whether there is a Banach space \(X\) such that \(\mathcal{K}(X)\) is complemented in \(\mathcal{B}(X)\). It is, however, known that there is a Banach space \(X\) such that \(\mathcal{B}(X)=\operatorname{span}\left(\left\\{I_{X}\right\\} \cup \mathcal{S}(X)\right)([\mathrm{GoMa}])\), where \(\mathcal{S}(X)\) is the closed subspace of \(\mathcal{B}(X)\) formed by strictly singular operators.

Short answer

An infinite-dimensional Hilbert space does not satisfy \(\mathcal{B}(X)=\operatorname{span}\left(\left\{I_{X}\right\} \cup \mathcal{K}(X)\right)\) because there exist bounded operators that are neither compact nor scalar multiples of \(I_X\).

Step-by-step solution

Title - Understand the Problem Statement

First, understand what needs to be shown: a Hilbert space does not satisfy the property that \(\backslashmathcal{B}(X)=\operatorname{span}(\left\backslash\backslash{I_{X}\right\backslash\backslash} \cup \backslashmathcal{K}(X))\) where \(\backslashmathcal{B}(X)\) is the space of bounded operators on X, \(\backslashoperatorname{span}(S)\) is the span of a set S, and \(\backslashmathcal{K}(X)\) is the ideal of compact operators on X.

Title - Recall Properties of Banach and Hilbert Spaces

Recall that in a Banach space \(\backslashmathcal{B}(X)\) includes all bounded linear operators, and for Hilbert spaces, compact operators are those that map bounded sets to relatively compact sets. Note the definition differences for infinite-dimensional spaces.

Title - Analyze Property of Hilbert Space

To show that a Hilbert space \(H\) does not satisfy the property, we need to show that \(\backslashmathcal{B}(H)\) cannot be spanned by \(\backslashleft{I_{H}\right}\) and \(\backslashmathcal{K}(H)\).

Title - Use Known Results

Use the known theorem stating \(\textbackslashmathcal{B}(H)\) on an infinite-dimensional Hilbert space \(H\) contains operators that are neither scalar multiples of \(I_{H}\) nor compact. This directly disproves the property for \(H\). Example: Consider the unilateral shift operator which is bounded but not compact nor is it a scalar multiple of \(I_{H}\).

Title - Conclusively Prove the Property

By the previous step, it is conclusively shown that there exist operators in \(\mathcal{B}(H)\) that do not belong to \(\operatorname{span}(\left\{I_{H}\right\} \cup\mathcal{K}(H))\). Thus, an infinite-dimensional Hilbert space does not satisfy the given property.

Key concepts

Banach Space

A Banach space is a complete normed vector space. This means that every Cauchy sequence in the space converges to a limit within the space. Banach spaces are very important in functional analysis and have a wide range of applications.
One key aspect of Banach spaces is that they can be either finite-dimensional or infinite-dimensional. Most interesting phenomena occur in the infinite-dimensional case.
Banach spaces include all bounded linear operators. \(\backslashmathcal{B}(X)\) denotes the set of bounded operators on a Banach space \(\backslashmathcal{B}(X)\).
It is important to remember that while every Hilbert space is a Banach space, not every Banach space is a Hilbert space. Hilbert spaces have an inner product, which provides a richer structure.

Bounded Operators

In functional analysis, an operator is a rule that assigns a function or vector to another function or vector. Operators can be bounded or unbounded.
Bounded operators are those for which there exists a constant \(C\) such that for all vectors \(x\), the norm of the operator applied to \(x\) is less than or equal to \(C\) times the norm of \(x\).
Mathematically, this is written as: \[ \backslash|Tx \backslash| \backslashleq C \backslash|x \backslash|, \] for all \(x\).
Bounded operators are essential because they guarantee that the operator's action is controlled and does not produce arbitrarily large outputs. In Banach spaces, \(\backslashmathcal{B}(X)\) represents the space of all bounded linear operators.

Compact Operators

Compact operators are a special class of bounded operators. An operator \(K\) is called compact if it maps bounded sets to relatively compact sets, meaning the closure of the image of any bounded set is compact.
Compact operators are significant in both Banach and Hilbert spaces because they generalize the notion of finite-dimensional operators to infinite-dimensional settings.
One key property of compact operators \(K\) on Hilbert spaces is that they can be approximated by finite-rank operators. This is useful for many analytical techniques and proofs.
Remember, \(\backslashmathcal{K}(X)\) denotes the set of all compact operators on a Banach space \(X\).

Operator Spanning

In functional analysis, spanning is a concept where a set of vectors (or operators) is used in linear combinations to produce every element of a space.
A set of operators is said to span a space if any operator in the space can be expressed as a linear combination of operators from this set.
For instance, in the given exercise, the task was to show that a Hilbert space \(H\) does not satisfy the property where \(\backslashmathcal{B}(H)\) can be spanned by \(\backslash{I_{H}\backslash\} \cup\backslashmathcal{K}(H)\).
Spanning is critical as it helps to understand the structure and limitations of spaces of operators. In this particular case, the fact that \(\backslashmathcal{B}(H)\) contains elements that are neither in \(\backslash{I_{H}\backslash\}\) nor \(\backslashmathcal{K}(H)\) indicates the complexity of operator spaces in infinite-dimensional settings.