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Chapter 13: Problem 34

If \(\left(A, D_{A}\right)\) is a densely defined openator and \(D_{A}\) is dense in \(\mathcal{H}_{1}\) show that \(A \subseteq A^{* *}\).

Short Answer

Expert verified
The operator \(A\) is a subspace of its double adjoint \(A^{**}\) because for each \(x\) in its domain, \(Ax\) is in the domain of \(A^{**}\) and \(Ax = A^{**}x\).

Step by step solution

01

Introduction to Elements and Spaces

Here, \(A\) is an operator and \(D_A\) is the domain of \(A\) which is densely defined in the Hilbert space \(\mathcal{H}_1\). This means that \(D_A\) is a subset of \(\mathcal{H}_1\) and that the closure of \(D_A\) is \(\mathcal{H}_1\). The problem statement is also saying that \(A\) is densely defined because its domain \(D_A\) is dense in \(\mathcal{H}_1\). The goal is to show that \(A \subseteq A^{**}\).
02

Defining the Adjoint

The adjoint of an operator \(A\), denoted \(A^*\), is an operator such that for all \(x\) in \(D_A\) and \(y\) in the domain of \(A^*\), we have \(\langle Ax, y \rangle = \langle x, A^* y \rangle\). Here, \(\langle ., . \rangle\) denotes the inner product in the Hilbert space.
03

Defining the Second Adjoint

The second adjoint of \(A\), denoted \(A^{**}\), is defined as the adjoint of \(A^*\). For all \(x\) in the domain of \(A^*\) and \(y\) in the domain of \(A^{**}\), we have \(\langle A^* x, y \rangle = \langle x, A^{**}y \rangle\).
04

Showing Inclusion

To show that \(A \subseteq A^{**}\), we need to show that for each \(x\) in \(D_A\), \(Ax\) is also in the domain of \(A^{**}\) and \(Ax = A^{**}x\). This can be shown using the definitions of adjoints. For each \(x\) in \(D_A\) and each \(y\) in the domain of \(A^*\), since \(A \subseteq A^*\), we have \(\langle Ax, y \rangle = \langle x, A^*y \rangle = \langle x, A^{**}y \rangle\). Thus, we find that \(Ax\) is in the domain of \(A^{**}\) and \(Ax = A^{**}x\). Therefore, \(A \subseteq A^{**}\).

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Most popular questions from this chapter

Show that a non-zero vector \(u\) is an cigenvector of an operator \(A\) if and only if \(|\langle u \mid A u\rangle|=\| A u|| u \mid .\)

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