Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 1: Problem 23

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 \((1,1, \ldots) \notin c_{0}\).

Short Answer

Expert verified
The space is not complete since a Cauchy sequence does not converge within the space.

Step by step solution

01

- Understanding the Space and Norm

The space in question is the sequence space \(c_0\), which consists of all sequences that converge to zero. The norm is given by \(\|x\|_0 = \sum 2^{-i}|x_i|\). We need to show that this space with this norm is not complete.
02

- Define the Cauchy Sequence

Consider the sequence \(\{x^n\}\) where \(x^n = (1, 1, ..., 1, 0, 0, ... )\), with 1 repeated \(n\) times. Each \(x^n\) is in \(c_0\) since it eventually becomes zero.
03

- Check Cauchy Property

We need to show that \(\{x^n\}\) is Cauchy under the norm \(\|\cdot\|_0\). For m, n large, \(\|x^n - x^m\|_0 = \sum_{i=1}^{\infty}2^{-i}|x_i^n - x_i^m|\). When \(m\) and \(n\) are large enough, this norm will be small.
04

- Explicitly Calculate the Norm Difference

For \(m > n\), \(\|x^n - x^m\|_0 = \sum_{i=n+1}^m 2^{-i} |1 - 0| = \sum_{i=n+1}^m 2^{-i} = 2^{-n} - 2^{-m}\). As \(n\) and \(m\) increase, this difference becomes smaller and smaller.
05

- Sequence Not Converging

Even though \(\{x^n\}\) is Cauchy, the only candidate for its limit in \(c_0\) would be \(x = (1,1,...)\) which is not in \(c_0\). Therefore, it does not converge in \(c_0\).
06

- Conclusion

Since \(c_0\) with the norm \(\|\cdot\|_0\) contains a Cauchy sequence that does not converge to an element within the space, it is not complete. Therefore, it is not a Banach space.

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!

Key Concepts

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

Banach Space
A Banach space is a special type of normed space that is also complete. This means that every Cauchy sequence in this space converges to a limit within the same space. Banach spaces are important in functional analysis because completeness ensures stability and predictability when dealing with sequences and series. In the exercise, we saw that the normed space \(c_0\) with the norm \(\|x\|_0 = \sum 2^{-i}|x_i|\) is not a Banach space. This is because it contains a Cauchy sequence that does not converge within the space.
Cauchy Sequence
A Cauchy sequence is a sequence where the elements become arbitrarily close to each other as the sequence progresses. In other words, for any given small distance \(\epsilon\), there exists a point in the sequence beyond which all elements are within \(\epsilon\) distance of each other. This concept is used to determine completeness in normed spaces. In the provided exercise, the sequence \ \{x^n\}\, defined by \ x^n = (1, 1, ..., 1, 0, 0, ... ) \ with 1 repeated \ n \ times, is shown to be Cauchy under the norm \(\|\cdot\|_0\). However, it does not converge in \(c_0\), indicating that the space is not complete.
Equivalent Norms
Two norms \(\|\cdot\|_1\) and \(\|\cdot\|_2\) on a vector space are said to be equivalent if they induce the same topological structure. Specifically, there exist constants \(C_1\) and \(C_2\) such that \(C_1\|x\|_1 \leq \|x\|_2 \leq C_2\|x\|_1\) for all elements \(x\) in the space. Equivalent norms ensure that convergence and completeness properties are preserved. In the problem, the norm \(\|\cdot\|_0\) on \(c_0\) is shown to not be equivalent to the standard supremum norm \(\|\cdot\|_{\infty}\), because \(c_0\) with \(\|\cdot\|_0\) is not complete.
Sequence Space
Sequence spaces are vector spaces whose elements are sequences of numbers. The space \(c_0\) is an example, consisting of all sequences that converge to zero. Norms in sequence spaces provide a measure of the 'size' of the elements. In the exercise, the norm \(\|x\|_0 = \sum 2^{-i}|x_i|\) is used on \(c_0\). Understanding how norms operate in sequence spaces is crucial for analyzing their structure and properties, such as completeness.
Completeness
A normed space is said to be complete if every Cauchy sequence in the space converges to a limit that is also within the space. Completeness is a vital property of Banach spaces. In the exercise, it is demonstrated that the space \(c_0\) with the norm \(\|\cdot\|_0\) is not complete because there exists a Cauchy sequence in this space that does not converge to an element in \(c_0\). This lack of completeness confirms that \(c_0\) with \(\|\cdot\|_0\) is not a Banach space, as Banach spaces must be complete by definition.

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 vector space \(X\) with two norms on it such that both of them are complete norms and they are not equivalent. Hint: Take a vector space \(V\) of linear dimension \(c\), and let \(T_{1}\) and \(T_{2}\) be linear bijections of \(V\) onto \(\ell_{2}\) and \(\ell_{4}\), respectively. Define norms on \(V\) by \(\|x\|_{1}=\left\|T_{1}(x)\right\|_{2}\) and \(\|x\|_{2}=\left\|T_{2}(x)\right\|_{4}\). Then \(\left(V_{1}\|\cdot\|_{1}\right)\) is isomorphic to \(\ell_{2}\) and \(\left(V_{i}\|\cdot\|_{2}\right)\) is isomorphic to \(\ell_{4} .\) Since \(\ell_{2}\) is not isomorphic to \(\ell_{4}\) (see exercises above), \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) are not equivalent.

Let \(C\) be a compact set in a finite-dimensional Banach space \(X\). Show that \(\operatorname{conv}(C)\) is compact. Hint: If a point \(x\) lies in the \(\operatorname{conv}(E) \subset \mathbf{R}^{n}\), then \(x\) lies in the convex hull of some subset of \(E\) that has at most \(n+1\) points. Indeed, assume that \(r>n\) and \(x=\sum t_{i} x_{i}\) is a convex combination of some \(r+1\) vectors \(x_{i} \in E\). We will show that then \(x\) is actually a convex combination of some \(r\) of these vectors. Assume that \(t_{i}>0\) for \(1 \leq i \leq r+1\). The \(r\) vectors \(x_{i}-x_{r+1}\) for \(1 \leq i \leq r\) are linearly dependent since \(r>n\). Thus, there are real numbers \(a_{i}\) not all zero such that \(\sum_{i=1}^{r+1} a_{i} x_{i}=0\) and \(\sum_{i=1}^{r+1} a_{i}=0 .\) Choose \(m\) so that \(\left|a_{i} / t_{i}\right| \leq\left|a_{m} / t_{m}\right|\) for \(1 \leq i \leq r+1\) and define \(c_{i}=t_{i}-\frac{a_{i} t_{m}}{a_{m}}\) for \(1 \leq i \leq r+1\). Then \(c_{i} \geq 0, \sum c_{i}=\sum t_{i}=1, x=\sum c_{i} x_{i}\), and \(c_{m}=0\) Having this, we can use the fact that in \(\mathbf{R}^{n}, \operatorname{conv}(C)\) is an image of the compact set \(S \times C^{n+1}\) in \(\mathbf{R}^{n+1} \times C^{n+1}\), where \(S\) is formed by points \(\left\\{\lambda_{i}\right\\}_{1}^{n+1}\) such that \(\lambda_{i} \geq 0\) and \(\sum \lambda_{i}=1\), under the \(\operatorname{map}\left(\\{\lambda\\},\left\\{x_{i}\right\\}\right) \mapsto \sum \lambda_{i} x_{i}\).

Show that a normed space \(X\) is a Banach space if and only if \(\sum y_{n}\) converges whenever \(\left\|y_{n}\right\| \leq 2^{-n}\) for every \(n\). Hint: Use Lemma 1.15. Note that if \(\sum x_{k}\) is absolutely convergent, there are \(N_{n}\) so that \(\sum_{k>N_{n}}\left\|x_{k}\right\| \leq 2^{-k} .\) Set \(y_{n}=\sum_{k=N_{n}+1}^{N_{k+1}} x_{k}\), then \(\left\|y_{n}\right\| \leq 2^{-n}\), and if \(M>N_{m}\), then \(\left\|\sum_{n=1}^{m} y_{n}-\sum_{k=1}^{M} x_{k}\right\| \leq 2^{-m}\).

Let \(\left\\{K_{i}\right\\}\) be a finite family of convex compact sets in \(\mathbf{R}^{n}\) such that every subfamily consisting of \(n+1\) members has a nonempty intersection. Helly's theorem asserts that then \(\bigcap K_{i} \neq \emptyset\). Prove it for \(n=1\). Show an example for \(n=2\) that \(n+1\) is necessary. Hint: Consider the interval between the maximum of the left endpoints of \(K_{i}\) and the minimum of the right endpoints of \(K_{i}\). For the general case, see for example [DGK]. Example: three lines forming a triangle. 1.60 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 \(A \subset \overline{\operatorname{conv}}\left\\{x_{n}\right\\}\) (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 \(a_{n+1} \in 2^{-2 n} B_{X}\)

Show that a bounded set \(M\) in \(c_{0}\) is totally bounded if and only if for every \(\varepsilon>0\) there is \(n_{0}\) such that \(\left|x_{n}\right| \leq \varepsilon\) for every \(x \in M\) and \(n \geq n_{0}\). Formulate and prove the analogous result for \(\ell_{p}\) spaces. Hint: Every bounded subset of \(\mathbf{R}^{n_{0}}\) is totally bounded.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

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