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Chapter 13: Problem 33

For unbounded operators, show that \(A^{*}+B^{*} \subseteq(A+B)^{\circ}\)

Short Answer

Expert verified
The statement \(A^{*}+B^{*} \subseteq(A+B)^{\circ}\) is true as shown by taking any vector from the domain of A and B and any vector from the Hilbert space, and confirming that the inner product of these vectors with \(A^{*}+B^{*}\) and \((A+B)^{\circ}\) are equal, confirming the given relation.

Step by step solution

01

Defining the terms

We first start by defining our terms. An unbounded operator \(A\) on a Hilbert space \(\mathcal{H}\) is a linear operator whose domain, denoted by \(D(A)\), is not necessarily the whole space. The adjoint \(A^*\) of an operator \(A\) is defined as the operator satisfying \(\langle Ax , y\rangle = \langle x , A^{*}y \rangle\) for all \(x\) in \(D(A)\) and all \(y\) in \(\mathcal{H}\). \(B^{*}\) is the adjoint operator of \(B\), and \(A+B\) denotes the sum of the operators \(A\) and \(B\). The statement \(A^{*}+B^{*} \subseteq(A+B)^{\circ}\) means that every element from \(A^{*}+B^{*}\) is also an element of \((A+B)^{\circ}\).
02

The Statement

The statement to prove is \(A^{*}+B^{*} \subseteq(A+B)^{\circ}\), where \(A^{*}\) and \(B^{*}\) are the adjoint operators of \(A\) and \(B\) respectively, and \((A+B)^{\circ}\) is the adjoint of the sum of the operators \(A\) and \(B\).
03

Using definition of adjoint operator

By the definition of the adjoint, we need to show that for all \(x\) in the domain of \(A+B\) and all \(y\) in \(\mathcal{H}\), \(\langle (A^{*}+B^{*})x,y \rangle = \langle x,((A+B)^{\circ})y \rangle\). We then need to take an element \(x\) in the domain of \(A\) and \(B\) and a suitable \(y\) in \(\mathcal{H}\).
04

Applying the Inner Product Properties

The inner product of two vectors satisfies \(\langle x, Ay \rangle + \langle x, By \rangle = \langle x, (A+B)y \rangle\). Hence, from the definitions of the adjoint, we have \(\langle Ax,y \rangle + \langle Bx,y \rangle = \langle x, (A^{*}+B^{*})y \rangle\). Therefore, \( \langle x, (A^{*}+B^{*})y \rangle = \langle x, ((A+B)^{\circ})y \rangle\). Therefore, \(A^{*}+B^{*} \subseteq(A+B)^{\circ}\).
05

Concluding the Proof

We have shown that for any \(x\) in the domain of \(A\) and \(B\) and any \(y\) in \(\mathcal{H}\), the inner product of \(x\) and \((A^{*}+B^{*})y\) is equal to the inner product of \(x\) and \((A+B)^{\circ}y\). Hence, every element from the set \(A^{*}+B^{*}\) is also an element from the set \(A+B)^{\circ}\). This means our proof is complete.

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Most popular questions from this chapter

For every bounded operator \(A\) on a Hilbert space \(\mathcal{H}\) show that the exponential operator $$ \mathrm{e}^{A}=\sum_{n=0}^{\infty} \frac{A^{n}}{n !} $$ is well-defined and bounded on \(\mathcal{H}\). Show that (a) \(e^{0}=1\) (b) For all posituve integers \(n,\left(c^{A}\right)^{n}=e^{n A}\). (c) \(\mathrm{e}^{A}\) is invertuble for all bounded operators \(A\) (even if \(A\) is not mivertible) and \(e^{-A}=\left(e^{4}\right)^{-1} .\) (d) If \(A\) and \(B\) are commuting operators then \(e^{A+B}=\mathrm{e}^{A} \mathrm{e}^{\theta}\) (c) If \(A\) is hermitian then \(e^{i A}\) is unitary.

If \(A_{1}, A_{2}, \ldots . A_{n}\) are operators on a dense domain such that $$ \sum_{i=1}^{n} A_{1}^{*} A_{1}=0 $$ show that \(A_{1}=A_{2}=\cdots=A_{n}=0 .\)

The norm \(\|\phi\|\) of a bounded linear operator \(\phi: \mathcal{H} \rightarrow \mathrm{C}\) is defined as the greatest lower bound of all \(M\) such that \(|\phi(u)| \leq M\|u\|\) for all \(u \in \mathcal{H}\). If \(\phi(u)=(v \mid u)\) show that \(\|\phi\|=\|v\|\). Hence show that ahe bounded lanear functional norm satisfies the parallelogram law $$ \|\phi+\vartheta\|^{2}+\|\phi-\psi\|^{2}=2\|\phi\|^{2}+2\|\psi\|^{2} $$

On the vector space \(\mathcal{F}^{\prime}[a, b]\) of complex continuous differentiable functions on the, interval \([a, b]\), set $$ \langle f| g)=\int_{a}^{b} \overline{f^{\prime}(x)} g^{\prime}(x) \mathrm{dr} \text { where } f^{\prime}=\frac{\mathrm{d} f}{\mathrm{~d} x}, \quad g^{\prime}=\frac{\mathrm{d} g}{\mathrm{~d} x} $$ Show that this is not an inner product, but becomes one if restricted to the space of functions \(f \in\) \(F^{\prime}[a, b]\) having \(f(c)=0\) for seme fixed \(a \leq c \leq b\). Is it a Hilbert space? Give a similar analysis for the case \(a=-\infty, b=\infty\), and restricting functions to those of compact support.

For any pair of hermitian operators \(A\) and \(B\) on a Hilbert space \(\mathcal{H}\), define \(A \leq B\) iff \((u \mid A u) \leq(u \mid B u)\) for all \(u \in \mathcal{H}\). Show that this is a partial order on the set of hermatian operators pay particular attention to the symmetry property, \(A \leq B\) and \(B \leq A\) umplies \(A=B\). (a) For multiplication operators on \(L^{2}(X)\) show that \(A_{0} \leq A_{B}\). Iff \(\alpha(x) \leq \beta(x)\) a.e. on \(X\). (b) For projection operators show that the definition given here reduces to that given in the text, \(P_{N} \leq P_{N}\) iff \(M \subseteq N\)
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