/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Let \(X\) be a Banach space. (... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 20

Let \(X\) be a Banach space. (i) Show that in \(X^{*}\) we have \(X^{\perp}=\\{0\\}\) and \(\\{0\\}^{\perp}=X^{*}\). Show that in \(X\) we have \(\left(X^{*}\right)_{\perp}=\\{0\\}\) and \(\\{0\\}_{\perp}=X\). (ii) Let \(A \subset B\) be subsets of \(X\). Show that \(B^{\perp}\) is a subspace of \(A^{\perp}\). Hint: Follows from the definition.

Short Answer

Expert verified
\(X^{\bot}=\text{\textlbrace}0\text{\textrbrace}\), \(\text{\textlbrace}0\text{\textrbrace}^{\bot}=X^{*}\), \((X^{*})_{\bot}=\text{\textlbrace}0\text{\textrbrace}\), \(\text{\textlbrace}0\text{\textrbrace}_{\bot}=X\), \(B^{\bot}\text{ is a subspace of }A^{\bot}\)

Step by step solution

01

Show that in \(X^{*}\) we have \(X^{\bot}=\text{\textlbrace}0\text{\textrbrace}\)

By definition, \(X^{\bot}\) is the set of all functionals \(f \text{ in } X^{*}\) such that \(f(x) = 0\) for all \(x \text{ in } X\). Since the only functional that satisfies this is the zero functional, we have \(X^{\bot} = \text{\textlbrace}0\text{\textrbrace}\).
02

Show that in \(X^{*}\) we have \(\text{\textlbrace}0\text{\textrbrace}^{\bot}=X^{*}\)

By definition, \(\text{\textlbrace}0\text{\textrbrace}^{\bot}\) is the set of all functionals \(f\) in \(X^{*}\) such that \(f(0) = 0\). Since every functional in \(X^{*}\) satisfies this, we have \(\text{\textlbrace}0\text{\textrbrace}^{\bot} = X^{*}\).
03

Show that in \(X\) we have \((X^{*})_{\bot}=\text{\textlbrace}0\text{\textrbrace}\)

By definition, \((X^{*})_{\bot}\) is the set of all \(x \text{ in } X\) such that \(f(x) = 0\) for all \(f \text{ in } X^{*}\). Since the only element that satisfies this is \(x = 0\), we have \((X^{*})_{\bot} = \text{\textlbrace}0\text{\textrbrace}\).
04

Show that in \(X\) we have \(\text{\textlbrace}0\text{\textrbrace}_{\bot}=X\)

By definition, \(\text{\textlbrace}0\text{\textrbrace}_{\bot}\) is the set of all \(x \text{ in } X\) such that \(f(x) = 0\) for the zero functional. Since every \(x \text{ in } X\) satisfies this, we have \(\text{\textlbrace}0\text{\textrbrace}_{\bot} = X\).
05

Assume \(A \text{ and } B \text{ are subsets of } X\) with \(A \text{ in } B\)

Given the subsets \(A \text{ and } B\) such that \(A \text{ in } B\), consider the orthogonal complements \(A^{\bot}\) and \(B^{\bot}\). By definition, \(B^{\bot}\) is the set of all functionals \(f \text{ in } X^{*}\) such that \(f(b) = 0\) for all \(b \text{ in } B\). Similarly, \(A^{\bot}\) is the set of all functionals \(f \text{ in } X^{*}\) such that \(f(a) = 0\) for all \(a \text{ in } A\).
06

Show that \(B^{\bot}\) is a subspace of \(A^{\bot}\)

Since \(A \text{ and } B\) with \(A \text{ in } B\), every functional in \(B^{\bot}\) will also be in \(A^{\bot}\) because the conditions on the functionals are even stricter for \(B^{\bot}\). Thus, \(B^{\bot}\) is a subspace of \(A^{\bot}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dual Space
In functional analysis, one of the key concepts to understand is the dual space. If we have a Banach space denoted by \(X\), its dual space, \(X*\), consists of all continuous linear functionals on \(X\).
These functionals are essentially functions that take an element from \(X\) and return a scalar, usually real or complex. The dual space is itself a Banach space when equipped with an appropriate norm, usually the operator norm.
The importance of the dual space lies in its role in understanding the structure of \(X\) and its subspaces. For example, we use the dual space to define orthogonal complements.
Orthogonal Complement
The orthogonal complement is a crucial concept in both linear algebra and functional analysis. Given a subspace \(A\) of a Banach space \(X\), the orthogonal complement of \(A\), denoted as \(A^{\perp}\), is defined in the context of the dual space.
Specifically, \(A^{\perp}\) consists of all functionals in \(X*\) that annihilate every element of \(A\). In other words, a functional \(f\) is in \(A^{\perp}\) if and only if \( f(a) = 0 \) for every \( a\) in \( A\).
Similarly, the orthogonal complement in the primal space \(X\) can be defined using elements in the dual space. Notably, this helps in understanding the relationships and properties between different subspaces and their complements.
Subspace Properties
Subspaces in functional analysis come with various interesting properties, particularly when considering their orthogonal complements. In the given exercise, we had to show that for any subspaces \(A \) and \( B \) of \( X* \), if \( A \subset B \), then \( B^{\perp} \) is a subspace of \( A^{\perp} \).
This follows from the definition: orthogonal complements restrict the functionals to those that annihilate every element in a given subspace.
Since \( B \) contains \( A \), any functional that annihilates all elements of \( B \) will also annihilate all elements of \( A \), making \( B^{\perp} \subset A^{\perp}\). This property is fundamental in understanding how different subspaces relate to each other.
Functional Analysis
Functional analysis is a branch of mathematics that studies vector spaces with a topology, typically focusing on infinite-dimensional spaces. Banach spaces, Hilbert spaces, and their duals are central objects of study.
Dual spaces and orthogonal complements are essential tools within functional analysis, providing deep insights into the structure of these spaces.
By examining how functionals act on Banach spaces, mathematicians can derive various properties and theorems that have applications in different areas such as differential equations, quantum mechanics, and optimization.
In the context of the given exercise, functional analysis principles were used to explore the relationship between subspaces and their orthogonal complements within the framework of Banach spaces and their duals.

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

Find a discontinuous linear map \(T\) from some Banach space \(X\) into \(X\) such that \(\operatorname{Ker}(T)\) is closed. Hint: Let \(X=c_{0}\) and \(T(x)=\left(f(x), x_{1}, x_{2}, \ldots\right)\) for \(x=\left(x_{i}\right)\), where \(f\) is a discontinuous linear functional on \(X\).

Let \(X, Y\) be normed spaces, \(T \in \mathcal{B}(X, Y)\). Consider \(\widehat{T}(\hat{x})=T(x)\) as an operator from \(X / \operatorname{Ker}(T)\) into \(\overline{T(X)}\). Then we get \(\widehat{T}^{*}: \overline{T(X)}^{*} \rightarrow\) \((X / \operatorname{Ker}(T))^{*} .\) Using Proposition \(2.7\) and \(\overline{T(X)}^{\perp}=T(X)^{\perp}=\operatorname{Ker}\left(T^{*}\right)\), we may assume that \(\widehat{T}^{*}\) is a bounded linear operator from \(Y^{*} / \operatorname{Ker}\left(T^{*}\right)\) into \(\operatorname{Ker}(T)^{\perp} \subset X^{*} .\) On the other hand, for \(T^{*}: Y^{*} \rightarrow X^{*}\) we may consider \(\widehat{T^{*}}: Y^{*} / \operatorname{Ker}\left(T^{*}\right) \rightarrow X^{*}\). Show that \(\widehat{T}^{*}=\widehat{T^{*}}\) Hint: Take any \(\hat{y} \in Y^{*} / \operatorname{Ker}\left(T^{*}\right)\) and \(x \in X\). Then, using the above identifications, we obtain $$ \widehat{T}^{*}\left(\widehat{y^{*}}\right)(\hat{x})=\widehat{y^{*}}(\widehat{T}(\hat{x}))=y^{*}(T(x))=T^{*}\left(y^{*}\right)(x)=\widehat{T^{*}}\left(\hat{y^{*}}\right)(\hat{x}) $$

Show that \(c^{*}\) is isometric to \(\ell_{1}\). Hint: We observe that \(c=c_{0} \oplus \operatorname{span}\\{e\\}\), where \(e=(1,1, \ldots)\) (express \(x=\left(\xi_{i}\right) \in c\) in the form \(x=\xi_{0} e+x_{0}\) with \(\xi_{0}=\lim _{i \rightarrow \infty}\left(\xi_{i}\right)\) and \(x_{0} \in\) \(\left.c_{0}\right)\). If \(u \in c^{*}\), put \(v_{0}^{\prime}=u(e)\) and \(v_{i}=u\left(e_{i}\right)\) for \(i \geq 1 .\) Then we have \(u(x)=u\left(\xi_{0} e\right)+u\left(x_{0}\right)=\xi_{0} v_{0}^{\prime}+\sum_{i=1}^{\infty} v_{i}\left(\xi_{i}-\xi_{0}\right)\) and \(\left(v_{1}, v_{2}, \ldots\right) \in \ell_{1}\) as in Proposition 2.14. Put \(\tilde{u}=\left(v_{0}, v_{1}, \ldots\right)\), where \(v_{0}=v_{0}^{\prime}-\sum_{i=1}^{\infty} v_{i}\), and write \(\tilde{x}=\left(\xi_{0}, \xi_{1}, \ldots\right) .\) We have \(u(x)=\xi_{0} v_{0}+\sum_{i=1}^{\infty} v_{i} \xi_{i}=\tilde{u}(\tilde{x})\) Conversely, if \(\tilde{u} \in \ell_{1}\), then the above rule gives a continuous linear functional \(u\) on \(c\) with \(\|u\| \leq\|\tilde{u}\|\), because \(|\tilde{u}(\tilde{x})| \leq\left(\sum_{i=0}^{\infty}\left|v_{i}\right|\right) \sup _{i \geq 0}\left|\xi_{i}\right|=\) \(\left|\tilde{u}\left\|\sup _{i \geq 0}\left|\xi_{i}\right|=\right\| \tilde{u}\left\|_{1}\right\| x \|_{\infty} .\right.\) The inequality \(\|\tilde{u}\| \leq\|u\|\) follows like this: Let \(\xi_{i}\) be such that \(\left|v_{i}\right|=\xi_{i} v_{i}\) if \(v_{i} \neq 0\) and \(\xi_{i}=1\) otherwise, \(i=0,1, \ldots\) Set \(x^{n}=\left(\xi_{1}, \ldots, \xi_{n}, \xi_{0}, \xi_{0}, \ldots\right) .\) Then \(\left\|x^{n}\right\|_{\infty}=1\) and \(\left|u\left(x^{n}\right)\right|=\left|\tilde{u}\left(\tilde{x}^{n}\right)\right| \geq\left|v_{0}\right|+\sum_{i=1}^{n}\left|v_{i}\right|-\sum_{i=n+1}^{\infty}\left|v_{i}\right| .\) Since \(\left|u\left(x^{n}\right)\right| \leq\|u\|\), we have \(\|u\| \geq\left|v_{0}\right|+\sum_{i=1}^{n}\left|v_{i}\right|-\sum_{i=n+1}^{\infty}\left|v_{i}\right| .\) By letting \(n \rightarrow \infty\), we get \(\|\tilde{u}\| \leq\|u\|\).

Let \(X\) be a normed space with two norms \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) such that \(X\) in both of them is a complete space. Assume that \(\|\cdot\|_{1}\) is not equivalent to \(\|\cdot\|_{2}\). Let \(I_{1}\) be the identity map from \(\left(X,\|\cdot\|_{1}\right)\) onto \(\left(X,\|\cdot\|_{2}\right)\) and \(I_{2}\) be the identity map from \(\left(X,\|\cdot\|_{2}\right)\) onto \(\left(X,\|\cdot\|_{1}\right) .\) Show that neither \(I_{1}\) nor \(I_{2}\) are continuous. Hint: The Banach open mapping theorem.

We proved the closed graph theorem using the open mapping theorem. Now prove the open mapping principle using the closed graph theorem. Hint: First, prove it for one-to-one maps using the fact that \(\left\\{\left(y, T^{-1}(y)\right)\right\\}\) is closed. For a general case, note that the quotient map is an open map by the definition of the quotient topology.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

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.