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Chapter 13: Problem 32

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

Short Answer

Expert verified
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 by step solution

01

Understanding the problem

You should understand the operation of adjoint and closure.
02

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 \(x= B^{*} A^{*} y = B^{*} z\).
03

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 \(A B\), \((A B)^{\circ}\).

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Most popular questions from this chapter

Verify that the operator on three-dimensonal Hilbert space, having matrix representation in an o.n. basis $$ \left(\begin{array}{ccc} \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \end{array}\right) $$ is a projection operator, and find a basis of the subspace it projects onto.

If \(A: \mathcal{H} \rightarrow \mathcal{H}\) is an operator such that \(A u \perp u\) for all \(u \in \mathcal{H}\), show that \(A=0\).

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