/*! 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 63 Let \(X\) be an infinite-dimensi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 63

Let \(X\) be an infinite-dimensional Banach space. Show that there is no translation-invariant Borel measure \(\mu\) on \(X\) such that \(\mu(U)>0\) for every open set \(U\) and such that \(\mu\left(U_{1}\right)<\infty\) for some open \(U_{1}\). Hint: Every open ball contains an infinite number of disjoint open balls of equal radii (see Lemma \(1.23\) ).

Short Answer

Expert verified
No such translation-invariant Borel measure can exist due to the contradiction arising from the infinite sum of measures of disjoint open balls.

Step by step solution

01

Understand the problem statement

We need to prove that there is no translation-invariant Borel measure \( \mu \) on an infinite-dimensional Banach space \( X \) which assigns a positive measure to every open set \( U \) and has a finite measure for some open set \( U_1 \).
02

Translation-invariant Borel measure

A Borel measure \( \mu \) on \( X \) is translation-invariant if \( \mu(A + x) = \mu(A) \) for any Borel set \( A \) and any \( x \) in \( X \). This means the measure assigned to any set is unaffected by translating the entire set by any vector in the space.
03

Infinite-dimensional Banach spaces and open sets

Recall that in an infinite-dimensional Banach space, every open set contains an infinite number of disjoint open balls of equal radii, as per Lemma 1.23.
04

Implication of Lemma 1.23

Given any open set \( U_1 \), it contains infinitely many disjoint open balls of equal radii. Since these balls are disjoint, we can denote them as \( B_1, B_2, B_3, \ldots \).
05

Consider the measure of disjoint open balls

By translation invariance, each of these disjoint open balls \( B_i \) should have the same measure \( \mu(B_i) = c \), where \( c \) is some constant. Since measure \( \mu \) is additive on disjoint sets, we get: \[ \mu\left( U_1 \right) \geq \mu\left( \bigcup_{i=1}^{\infty} B_i \right) = \sum_{i=1}^{\infty} \mu(B_i) = \sum_{i=1}^{\infty} c = \infty \]
06

Contradiction with finite measure assumption

This contradicts the assumption that \( \mu(U_1) < \infty \). Therefore, we cannot have a translation-invariant Borel measure on \( X \) that assigns a positive measure to every open set and is finite for some open set.
07

Final conclusion

Thus, it is shown that there is no translation-invariant Borel measure \( \mu \) on an infinite-dimensional Banach space \( X \) where \( \mu(U) > 0 \) for every open set \( U \) and such that \( \mu(U_1) < \infty \) for some open \( U_1 \).

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.

infinite-dimensional Banach space
An infinite-dimensional Banach space is a complete normed vector space with an infinite basis. Unlike finite-dimensional spaces where vectors can be represented by a finite number of coordinates, infinite-dimensional spaces require an infinite number of vectors to form a basis. This property makes them significantly more complex.

These spaces are important in functional analysis and can represent various function spaces. For instance, the space of all square-integrable functions, denoted by \(L^2\) space, is an example of an infinite-dimensional Banach space.

One critical property of these spaces is that every open set within them contains infinitely many disjoint open balls of equal radii. This fact is fundamental to understanding why certain measures cannot exist in these spaces. The disjoint nature of the balls means that they do not overlap, which leads to unique implications when we consider translation-invariant measures.
open set
In the context of topology and Banach spaces, an open set is a set where each point within it has a neighborhood entirely contained within the set. Visually, you can imagine an open set in 3D space as a hollow ball where every point inside can 'move around' a bit without ever leaving the ball.

Formally, a subset \(U\) of a Banach space \(X\) is open if for every point \(x \in U\), there exists an \(\epsilon > 0\) such that the ball of radius \(\epsilon\) centered at \(x\) (denoted by \(B(x, \epsilon)\)) is entirely within \(U\). This characteristic is what allows an infinite number of disjoint open balls to exist within any given open set in an infinite-dimensional Banach space.

Understanding open sets is vital in analyzing how measures (like our Borel measure) interact with the space, particularly when considering translation invariance and measure's properties on different sets within the space.
disjoint open balls
Disjoint open balls are open balls (sets) that do not overlap. This means for any two disjoint open balls \(B_1\) and \(B_2\), the intersection \(B_1 \cap B_2\) is empty.

In infinite-dimensional Banach spaces, each open set contains an infinite number of such disjoint open balls of equal radii. This fact is captured in Lemma 1.23 and plays a crucial role in understanding why no finite translation-invariant Borel measure can exist in these spaces.

The translation invariance of a measure \(\mu\), where \(\mu(A + x) = \mu(A)\) for any Borel set \(A\) and vector \(x\) in \(X\), means each of these disjoint balls should have the same measure. Consequently, adding the measures of infinitely many disjoint open balls leads to an infinite sum, contradicting the assumption that \(\mu(U_1) < \infty\) for some open set \(U_1\). This contradiction proves the impossibility of such a measure in these spaces.

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

Let \(X\) be a Banach space. Show that if \(A \subset X\) is totally bounded, then there is a sequence \(\left\\{x_{n}\right\\} \in X\) such that \(x_{n} \rightarrow 0\) in \(X\) and \(A \subset\) sconv \(\left\\{x_{n}\right\\}\) (see Exercise \(1.7)\) In particular, for every compact subset \(A\) of \(X\) there exists a sequence \(\left\\{x_{n}\right\\}\) such that \(x_{n} \rightarrow 0\) and \(A \subset \overline{\operatorname{conv} v}\left\\{x_{n}\right\\}\) (Grothendieck). Hint: We set \(A_{1}=A\), let \(B_{1}\) be a finite \(2^{-2}\) -net in \(A_{1}\). If \(A_{i}\) and \(B_{i}\) were defined for \(i \leq n\), let \(A_{n+1}=\left(A_{n}-B_{n}\right) \cap 2^{-2 n} B_{X} ;\) note that every \(a_{n} \in A_{n}\) is of the form \(a_{n}=a_{n+1}+b_{n}\), where \(a_{n+1} \in A_{n+1}, b_{n} \in B_{n}\). Let \(B_{n+1}\) be a finite \(2^{-2 n}\) -net in \(A_{n+1}\). Therefore, every element \(a \in A=A_{1}\) is of the form \(a=b_{1}+a_{2}=b_{1}+b_{2}+a_{3}=\ldots=\sum_{1}^{n} b_{n}+a_{n+1} .\) Since \(a_{n+1} \in 2^{-2 n} B_{X_{1}}\)

Let \(X, Y\) be normed spaces and \(T \in \mathcal{B}(X, Y)\). Show that $$ \|T\|=\sup \left\\{\|T(x)\|_{Y} ;\|x\|_{X}<1\right\\}=\sup \left\\{\|T(x)\|_{Y} ;\|x\|_{X}=1\right\\} $$ Hint: Clearly, both suprema are not greater than \(\|T\|\). Given \(\varepsilon>0\), find \(x \in B_{X}\) such that \(\|T(x)\|_{Y} \geq \sqrt{1-\varepsilon}\|T\| .\) Then \(\|\sqrt{1-\varepsilon} x\|<1\) and \(\frac{x}{\|x\|_{x}} \in S_{X}\), and both vectors give \(\|T(y)\|_{Y} \geq(1-\varepsilon)\|T\|\).

Show that \(\overline{\operatorname{span}}(L)=\overline{\operatorname{span}(L)}\) and \(\overline{\operatorname{conv}}(M)=\overline{\operatorname{conv}(M)}\). Hint: By the definition, \(\overline{\operatorname{span}}(L)\) is the intersection of all closed subspaces containing \(L\), and \(\overline{\operatorname{span}(L)}\) is one such subspace. From this, one inclusion follows. The other inclusion follows similarly because \(\operatorname{span}(L)\) is the intersection of all subspaces containing \(L .\)

Let \(A\) be a set in a real Banach space \(X\). Show that the symmetric convex hull of \(A\) (i.e., the intersection of all symmetric convex sets containing \(A\) ) is equal to \(\left\\{\sum_{i=1}^{n} \lambda_{i} x_{i} ; x_{i} \in A, \sum_{i=1}^{n}\left|\lambda_{i}\right| \leq 1\right\\}\). Hint: Show first that this set is convex and symmetric.

Let \(f \in L_{p_{0}}[0,1]\) for some \(p_{0}>1 .\) Show that \(\lim _{p \rightarrow 1^{+}}\|f\|_{L_{p}}=\|f\|_{L_{1}} .\) If \(f \in L_{\infty}[0,1]\), then \(\lim _{p \rightarrow \infty}\|f\|_{L_{p}}=\|f\|_{L_{\infty}}\) Let \(x \in \ell_{q}\) for some \(q \geq 1\). Show that \(\lim _{p \rightarrow \infty}\|x\|_{\ell_{p}}=\|x\|_{\ell_{\infty}}\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Discrete 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.