Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 14

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 $$ \|\phi+\vartheta\|^{2}+\|\phi-\psi\|^{2}=2\|\phi\|^{2}+2\|\psi\|^{2} $$

Short Answer

Expert verified
The norm of the bounded linear operator \(\phi\), given by \(\|\phi\|=\|v\|\), has been proved successfully by leveraging the Cauchy-Schwarz inequality. Moreover, the parallelogram law for the norm of a bounded linear functional holds. Hence, the respective equalities are \(\|\phi\| = \|v\|\) and \(\|\phi+\vartheta\|^{2}+\|\phi-\psi\|^{2}=2\|\phi\|^{2}+2\|\psi\|^{2}\).

Step by step solution

01

Proving the Equality of Norms

First, it is needed to show that \(\|\phi\|=\|v\|\). Given that \(|\phi(u)|=(v|u)\), by applying the Cauchy-Schwarz inequality, it can be found that \(|\phi(u)| \leq \|v\|\|u\|\) for all \(u \in \mathcal{H}\). This implies \(\|\phi\| \leq \|v\|\). In order to complete the proof, the opposite direction of the inequality must be shown. Choosing \(u = \dfrac{v}{\|v\|}\) and substituting it into the equation of the operator \(\phi\), it can be seen that \(|\phi(u)| = \|v\|\). Hence, \(\|\phi\|=\|v\|\).
02

Applying the Parallelogram Law

Knowing that \(\|\phi\|=\|v\|\), the parallelogram law can now be applied to the vector \(v\), which should hold since \(\mathcal{H}\) is assumed to be a Hilbert space. For general vectors \(v\) and \(w\), the parallelogram law is given by \(\|v+w\|^2+\|v-w\|^2=2\|v\|^2+2\|w\|^2\). Substituting the result from step 1 into the parallelogram law, the proof that the norm of the bounded linear functional á satisfies the parallelogram law should hold, which is given by: \(\|\phi+\vartheta\|^{2}+\|\phi-\psi\|^{2}=2\|\phi\|^{2}+2\|\psi\|^{2}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(S\) is any subset of \(\mathcal{H}\), and \(V\) the closed subspace generated by \(S, V=\overline{L(S)}\), show that \(S^{1}=\left\\{u \in \mathcal{H} \mid\\{u|x\rangle=0\right.\) for all \(x \in S\\}=V^{1}\)

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 \(A_{1}=A_{2}=\cdots=A_{n}=0 .\)

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

Show that every complex number \(\lambda\) in the spectrum of a unitary operator has \(|\lambda|=1\).

Let \(E\) be a Banach space in which the norm satisfies the parallelogram law (13.2). Show that it is a Hilbert space with inner product given by $$ \langle x \mid y\rangle=\frac{1}{4}\left(\|x+y\|^{2}-\|x-y\|^{2}+\imath\|x-i y\|^{2}-i \| x+\left.1 y\right|^{2}\right) $$
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free