Problem 32 - A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry

Question

Show that if \(\left(A, D_{A}\right)\) and \(\left(B, D_{B}\right)\) arc operators on dense domains in \(H\) then \(B^{*} A^{*} \subseteq\) \((A B)^{\circ}\)

Step-by-Step Solution

Answer

The proof requires understanding of adjoint operators in a Hilbert space. Given \(x \in B^{*} A^{*}\), demonstrates that \(x\) is also in \((A B)^{\circ}\), thereby showing that \(B^{*} A^{*} \subseteq (A B)^{\circ}\).

Step 1: Understanding the problem

You should understand the operation of adjoint and closure.

Step 2: Assume x is in \(B^{*} A^{*}\)

Suppose an element \(x\) belongs to \(B^{*} A^{*}\). That means there exists \(y\) in \(D_{A}\) and \(z\) in \(D_{B}\) such that \(x= B^{*} A^{*} y = B^{*} z\).

Step 3: Show x is in \((A B)^{\circ}\)

To prove that \(B^{*} A^{*} \subseteq (A B)^{\circ}\), you need to show that \(x\) is also in \((A B)^{\circ}\), that is, \(x=(A B)^{\circ} y\). Using definition of closure, we know \((A B)^{\circ}\) is the smallest closed set that contains \(A B\). If \(x\in B^{*} A^{*}\), then \(x\) will also belongs to the closure of \(A B\), \((A B)^{\circ}\).

Key Concepts

Adjoint Operators

In the realm of Hilbert spaces, an adjoint operator is a foundational concept. If you have a linear operator \(A\) acting within a Hilbert space, the adjoint of \(A\), often denoted as \(A^*\), is an operator such that for every pair of vectors \(x\) and \(y\) in the space, the following holds:

• \(\langle Ax, y \rangle = \langle x, A^* y \rangle\)

Adjoint operators have several properties that prove useful in functional analysis. They essentially help define what it means for an operator to be 'close' or 'related to' its original operator \(A\).
Often, computing the actual form of \(A^*\) might not be straightforward, but understanding its existence and its relation to inner products can aid significantly when solving complex problems.

Dense Domain

When we talk about operators on Hilbert spaces, the term "dense domain" frequently comes up. Basically, a dense domain means that the operator is defined on a subset of the Hilbert space that is 'dense'.
But what does dense mean? If we say a subset \(D\) of a space \(H\) is dense, it implies that every element in \(H\) can be approximated as closely as desired by elements of \(D\). In other words, if you pick any vector in \(H\), you can find another vector from \(D\) that is arbitrarily close to it.

• It implies that the closure of \(D\) is the entire space \(H\).
• Operators with dense domains are pivotal in defining adjoint operators correctly.

Understanding dense domains is essential for working with operators in Hilbert spaces because it assures the adjoint operator can be defined over the entire space.

Closure of Operators

The concept of the closure of an operator might seem a bit abstract at first, but it's crucial in understanding the behavior of operators in a Hilbert space. Let's say we have an operator \(A\), and its closure is denoted as \((A)^{\circ}\). This closure represents the smallest closed operator that extends \(A\).
How do we comprehend this? Consider:

• A closed operator is one where the graph is closed in the product space.
• Closed operators are well-behaved regarding convergence.
• The idea of taking the closure ensures that by completing all possible limits in its action, we do not lose information when dealing with the convergence related to \(A\).

In practical terms, the closure is an operator that acts 'just like' \(A\) on its domain but might have a larger domain due to including limits of sequences within the original domain.

Hilbert Space

A Hilbert space is a complete vector space equipped with an inner product. This essentially means that every Cauchy sequence of vectors in the space has a limit that is also within the space itself, ensuring completeness of the space.
This concept extends the idea of Euclidean space to potentially infinite dimensions, making it possible to work with functions in a very flexible and powerful way.

• Hilbert spaces have structured geometric concepts such as angles and orthogonality.
• They are foundational in quantum mechanics and various branches of functional analysis.

The rich structure of Hilbert spaces allows the application of geometry to abstract vector spaces. They play a significant role in understanding bounded and unbounded operators, eigenvalue problems, and more. Essentially, they offer a robust framework for exploring and manipulating mathematical objects with infinite dimensions.