/*! 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 1 If \(X\) is a Banach space, show... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 1

If \(X\) is a Banach space, show that the dual space \(X^{*}\) is itself a Banach space under the operator norm: (a) Show that \(X^{*}\) is a vector space. (b) Show that the operator norm gives a norm on \(X^{*}\). (c) Show that \(X^{*}\) is complete.

Short Answer

Expert verified
Considering that \(X\) is a Banach space, to show that the dual space \(X^{*}\) is itself a Banach space we evidenced that \(X^{*}\) is a vector space. This was established by asserting that the set of all bounded linear functionals on \(X\) forms a vector space. We then showed that the operator norm, defined as \(\|f\| = \sup_{\|x\| \leq 1} |f(x)|\), forms a valid norm on \(X^{*}\). Finally, we confirmed that \(X^{*}\) is complete by illustrating that every Cauchy sequence in \(X^{*}\) converges to a member of \(X^{*}\).

Step by step solution

01

Show that \(X^{*}\) is a vector space

When \(X\) is a Banach space, the dual space \(X^{*}\) consists of all bounded linear functional on \(X\). A functional is called linear if it satisfies the two properties: \(f(a*x+b*y) = a*f(x) + b*f(y)\) where \(a,b\) are scalars and \(x,y\) are elements from the space \(X\). It is important to know that the set of all linear functionals on \(X\), when added and scalar multiplied in the usual way, forms a vector space.
02

Show that the operator norm gives a norm on \(X^{*}\)

The operator norm on the dual space \(X^{*}\) is defined as: \(\|f\| = \sup_{\|x\| \leq 1} |f(x)|\). This \(f\) is a bounded linear functional from \(X\) to \(\mathbb{R}\) or \(\mathbb{C}\). Any scalar multiple \(cf\) of \(f\) and \(f + g\), where \(g\) is another bounded linear functional, can be shown to follow the basic properties of norm, thereby establishing the operator norm as a valid norm on \(X^{*}\).
03

Show that \(X^{*}\) is complete

The completion of a normed space \(X^{*}\) refers to its being a Banach space. So, we need to show that every Cauchy sequence in \(X^{*}\) converges to an element in \(X^{*}\). Suppose \(\{ f_n \}\) is a Cauchy sequence in \(X^{*}\). We can show that \(\{ f_n \}\) converges pointwise to a limit \(f\), which is also in \(X^{*}\). So, this means \(X^{*}\) is complete.

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
When talking about Banach spaces, the concept of the dual space is important. The dual space, denoted as \(X^*\), consists of all bounded linear functionals defined on the original space \(X\). Bounded linear functionals are special types of functions. They map elements from \(X\) to the real or complex numbers. But what's really neat is that they are "linear" and "bounded". Linear means they respect addition and scalar multiplication. Mathematically, if \(f\) is a linear functional, then for any scalars \(a\) and \(b\), and any elements \(x\) and \(y\) in \(X\), we have \(f(ax + by) = af(x) + bf(y)\). Bounded means there's a constant \(C\) such that for every \(x \in X\), \(|f(x)| \leq C\|x\|\). The dual space forms its own vector space. This means that if you add two bounded linear functionals or multiply them by a scalar, you're still within the dual space!
Operator Norm
The operator norm helps measure the "size" of a bounded linear functional in the dual space. For a functional \(f\) in \(X^*\), the operator norm is defined as \(\|f\| = \sup_{\|x\| \leq 1} |f(x)|\). This definition looks at the greatest value \(f(x)\) could take when \(x\) has a norm of 1 or less. This norm has some important properties:
  • It's always non-negative.
  • The norm of \(f\) is zero if and only if \(f\) is the zero functional.
  • For any scalar \(c\) and functional \(f\), \(\|cf\| = |c|\|f\|\).
  • For two functionals \(f\) and \(g\), \(\|f + g\| \leq \|f\| + \|g\|\).
These traits confirm that the operator norm truly acts as a norm on the dual space \(X^*\).
Bounded Linear Functional
Bounded linear functionals are key players in the field of functional analysis. A linear functional \(f\) is a map from a vector space \(X\) to the real or complex numbers. However, not all linear functionals are bounded. A linear functional is considered bounded if there exists a constant \(M\) such that for all \(x\) in \(X\), \(|f(x)| \leq M\|x\|\). This property of boundedness is crucial because it ensures that the functional behaves nicely and doesn't "blow up" for large inputs. It also ensures continuity. If a linear functional is bounded, it means small changes in the input \(x\) lead to small changes in the output \(f(x)\), which is a very desirable property in mathematical analysis.
Vector Space
A vector space is a foundational concept in mathematics, especially in linear algebra and functional analysis. This structure consists of a set of elements, called vectors, which can be added together and multiplied by scalars. Scalars are elements from a field, like the real or complex numbers. To qualify as a vector space, a set must satisfy several properties:
  • Closure under addition and scalar multiplication.
  • Existence of a zero vector, which acts like zero in addition.
  • Every vector has an additive inverse.
  • Compatibility of scalar multiplication with field multiplication.
  • Distributivity: \(a(v + w) = av + aw\) and \((a+b)v = av + bv\).
  • Associativity: \(a(bv) = (ab)v\).
These properties need to hold for more than just vectors in simple numerical spaces. They also apply to functionals, like the ones in dual spaces, making dual spaces themselves vector spaces. This allows mathematicians to use tools from linear algebra when analyzing functionals.

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 \(X\) is a Hilbert space and \(S\) is any subset of \(X\), define $$ S^{\perp}=\\{y \in X:\langle x, y\rangle=0 \text { for all } x \in S\\} . $$ (a) Show that \(S^{\perp}\) is a closed subspace of \(X\). (b) Show that \(S \cap S^{\perp}\) can contain only the zero vector. (c) If \(S \subset T\) are both subsets of \(X\), show that \(T^{\perp} \subset S^{\perp} .\) (d) If \(\bar{S}\) is the closure of \(S\) in \(X\), show that \(S^{\perp}=\bar{S}^{\perp}\).

Show that \(T: D \rightarrow X\) is closable if and only if \(\overline{G(T)}\) is a graph, and in this case \(G(\bar{T})=\overline{G(T)}\).

If \(F: X \rightarrow Y\) is a bounded linear operator between normed vector spaces \(X\) and \(Y\), show that the nullspace \(N=\\{x \in X: F(x)=0\\}\) is a closed subspace of \(X\).

Prove that \(H_{0}^{1, p}\left(\mathbf{R}^{n}\right)=H^{1, p}\left(\mathbf{R}^{n}\right)\)

A bounded bilinear form on a real Hilbert space \(X\) is a map \(B: X \times X \rightarrow\) R satisfying (i) \(B(\alpha x+\beta y, z)=\alpha B(x, z)+\beta B(y, z)\), (ii) \(B(x, \alpha y+\beta z)=\alpha B(x, y)+\beta B(x, z)\), (iii) \(|B(x, y)| \leq C\|x\|\|y\|\) for all \(x, y, z \in X\) and \(\alpha, \beta \in \mathbf{R}\). Under these assumptions, show that there is a unique bounded linear operator \(A: X \rightarrow X\) such that \(B(x, y)=\langle A x, y\rangle\) for all \(x, y \in X\), where \(\langle,\),\(rangle denotes the inner product on X\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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.