/*! 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 60 Let \(X\) be a Banach space. Sho... [FREE SOLUTION] | 91影视

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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 \(a_{n+1} \in 2^{-2 n} B_{X_{1}}\)

Short Answer

Expert verified
Construct a sequence {b鈧檥 from finite nets covering A. The decomposition ensures convergence to 0 and containment in the closed convex hull.

Step by step solution

01

Understanding Total Boundedness

A set is total bounded in a Banach space if for every 蔚 > 0, there exists a finite number of points such that the set can be covered by balls of radius 蔚 centered at these points.
02

Defining Subsequences

Set the initial subset as A鈧 = A. Let B鈧 be a finite 2鈦宦-net in A鈧. A 2鈦宦-net is a set of points such that every point in A鈧 is within a distance of 2鈦宦 from at least one point in B鈧.
03

Inductive Definition

Assume A岬 and B岬 for i 鈮 n have been defined. Let A鈧欌倞鈧 = (A鈧 - B鈧) 鈭 2鈦宦测伩 B_X, where B_X is the unit ball in the Banach space. This means each element in A鈧欌倞鈧 is within 2鈦宦测伩 distance from some finite set of points in B_X.
04

Finite Net

Let B鈧欌倞鈧 be a finite 2鈦宦测伩-net in A鈧欌倞鈧. This construction ensures that every element in A鈧欌倞鈧 is of the form a鈧 = a鈧欌倞鈧 + b鈧, where a鈧欌倞鈧 鈭 A鈧欌倞鈧 and b鈧 鈭 B鈧.
05

Decomposition of Elements

Every element a 鈭 A can be written as a sum of elements from the sequence B岬 and residual sequences from A岬. Specifically, a = b鈧 + a鈧 = b鈧 + b鈧 + a鈧 = ... = 危鈧佲伩 b鈧 + a鈧欌倞鈧. Here, a鈧欌倞鈧 belongs to 2鈦宦测伩 B_X.
06

Convergence to Zero

As n 鈫 鈭, the residual term a鈧欌倞鈧 in the formulation a = 危鈧佲伩 b鈧 + a鈧欌倞鈧 becomes arbitrarily small since they are in 2鈦宦测伩 B_X. Thus, the sequence {x鈧檥 converges to 0 in X.
07

Convex Combinations

From the construction, since each element a is expressed in terms of the finite nets B岬, A is contained in the closed convex hull sconv {x鈧檥.

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Key Concepts

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

Banach space
A Banach space is a type of complete normed vector space. The term 'complete' means that any Cauchy sequence (a sequence where the elements get arbitrarily close to each other) in the space has a limit that is also within the space. This property is critical because it ensures that every sequence that appears to converge actually does.
To visualize, imagine you have a collection of vectors in a multi-dimensional space where every vector has a length (distance from the origin). The concept of a 'norm' provides a way to measure this length. Formally, a norm on a vector space X is a function \(orm{.} \) from X to the non-negative real numbers \(orm{.} \).
Here are a few essential properties of a norm:
  • \(orm{a} \) is always non-negative, and \(orm{a} = 0 \) only when the vector a is the zero vector
  • For any scalar \(\beta \) and any vector a, \(orm{\beta a} = |\beta| orm{a} \)
  • The norm satisfies the triangle inequality: \(orm{a+b} \) 鈮 \(orm{a} + orm{b} \)
Additionally, a Banach space allows mathematicians to handle infinite-dimensional spaces while still preserving the structure and properties found in finite-dimensional vector spaces. This is especially useful in functional analysis, where Banach spaces often serve as the foundational framework.
Total Boundedness
Total boundedness is a stronger condition than just boundedness. A set A in a metric space is totally bounded if, for any given distance 蔚 > 0, you can cover the set with a finite number of balls of radius 蔚. This means that no matter how small 蔚 gets, you can still cover the set with a limited number of tiny balls.
This property is important because it implies that the set can be approximated very closely with a finite subset. In essence, a totally bounded set doesn't spread out indefinitely; it stays 'compact' even when you zoom in. For instance, consider a set of points on a line that keeps clustering closer and closer together. If it is totally bounded, you can always find a finite number of small intervals that cover all points.
In the context of Banach spaces, total boundedness helps in dealing with limits and convergence. When trying to show that each element of a totally bounded set can be represented as sums of sequences converging to zero, total boundedness ensures that these sequences can be carefully controlled and manipulated. Thus, proving results like the one in the exercise becomes feasible by ensuring that our sequences and nets stay within manageable bounds, no matter how finely we need to cover the set.
Sequence Convergence
Sequence convergence is a foundational concept in analysis. A sequence \(orm{a_n} \) in a Banach space converges to an element \(orm{a} \) if the distance between the sequence elements and \(orm{a} \) gets arbitrarily small as n grows larger. Mathematically, \(orm{a_n \rightarrow a} \) means that for any 蔚 > 0, there's an N such that for all n 鈮 N, \(orm{a_n - a} < 蔚 \).
Convergence in Banach spaces has some useful implications:
  • It ensures that sequences can be approximated as closely as needed by other sequences within the space.
  • It makes it possible to isolate and work with limit points and helps to approximate more complex functions with simpler ones.
In the provided exercise, sequence convergence is employed to show that every element of a totally bounded set can be written as sums of sequences converging to zero. This shows that elements of these sets don't 'escape' the space but instead can be tightly controlled in their behavior. So, the sequences in a Banach space not only converge but they can be manipulated to help represent elements of totally bounded or compact subsets effectively.

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Most popular questions from this chapter

Show that \(c_{0}(\Gamma)\) is the closure of \(c_{00}(\Gamma)\) in \(\ell_{\infty}(\Gamma)\).

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 \(\|T(y)\|_{Y} \geq(1-\varepsilon)\|T\|\).

Let \(C\) be a convex set in a normed space \(X\), assume \(\operatorname{Int}(C) \neq \emptyset\) (recall that \(\operatorname{Int}(C)\) denotes the interior of \(C\) ). Show that \(\overline{\operatorname{Int}(C)}=\bar{C}\) and \(\operatorname{Int}(\bar{C})=\operatorname{Int}(C)\) Hint: Use the "cone" argument. There is a point \(x_{0}\) and an open ball \(B_{b}^{\circ}\) such that \(x_{0}+B_{\delta}^{O} \subset C .\) Note that if we take any \(c \in C\), then by the convexity of \(C\), the cone with the vertex \(c\) and the base \(x_{0}+B_{\delta}^{O}\) is a subset of \(C\), and all points in this cone but the vertex \(c\) are its interior points (draw a picture!). Therefore, given \(c \in C\), we can find points from \(\operatorname{Int}(C)\) arbitrarily close to \(c\), showing that \(\bar{C} \subset \overline{\operatorname{Int}(C)}\) Given \(c \in \operatorname{Int}(\bar{C})\), there is \(\varepsilon>0\) such that \(c+\varepsilon B_{X} \subset \bar{C} .\) Let \(d=\) \(c+\varepsilon\left(c-x_{0}\right) .\) Then \(d \in \bar{C}\), so for every \(\nu>0\) there is \(y \in C\) with \(\|d-y\|<\nu\) If \(\nu\) is small (namely \(\left.\nu \leq \frac{\varepsilon \delta}{\sqrt{\delta^{2}+\left\|x_{0}-c\right\|^{2}}}\right), c\) is in the cone with the base \(x_{0}+B_{\delta}^{O}\) and vertex \(y\), so \(c \in \operatorname{Int}(C)\).

Find two (obviously not closed) subspaces \(F_{1}\) and \(F_{2}\) of a Banach space \(X\) such that \(F_{1} \cap F_{2}=\\{0\\}\) and both \(F_{1}\) and \(F_{2}\) are dense in \(X\). Hint: \(X=C[0,2 \pi]\), let \(F_{1}\) be all polynomials on \([0,2 \pi]\) and \(F_{2}\) be all trigonometric polynomials on \([0,2 \pi]\).

Let \(X\) be the normed space obtained by taking \(c_{0}\) with the norm \(\|x\|_{0}=\sum 2^{-i}\left|x_{i}\right| .\) Show that \(X\) is not a Banach space. Note that this shows that \(\|\cdot\|_{0}\) is not an equivalent norm on \(c_{0}\). Hint: The sequence \(\\{(1,1, \ldots, 1,0, \ldots)\\}_{n=1}^{\infty}\) is Cauchy and not convergent since the only candidate for the limit would be \((1,1, \ldots) \notin c_{0}\).

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