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 2: Problem 42

Show that \(c_{0}\) is not isomorphic to \(C[0,1]\). Hint: Check the separability of their duals.

Short Answer

Expert verified
The dual of \(c_{0}\) is separable; the dual of \(C[0,1]\) is not separable. Hence, \(c_{0}\) is not isomorphic to \(C[0,1]\).

Step by step solution

01

Understand the Problem

The task is to show that the space of sequences converging to zero, denoted by \(c_{0}\), is not isomorphic to the space of continuous functions on the interval \([0,1]\), denoted by \(C[0,1]\). A good approach is to examine the properties of their dual spaces, specifically their separability.
02

Define \(c_{0}\) and \(C[0,1]\)

\(c_{0}\) is the space of all sequences \((x_n)\) such that \(x_n \to 0\). \(C[0,1]\) is the space of all continuous functions defined on the interval \([0,1]\).
03

Determine the Dual of \(c_{0}\)

The dual space of \(c_{0}\), denoted by \((c_{0})'\), is \(\text{l}^1\), which is the space of absolutely summable sequences. \(\text{l}^1\) is separable.
04

Determine the Dual of \(C[0,1]\)

The dual space of \(C[0,1]\) is the space of signed regular Borel measures on \([0,1]\), which is not separable.
05

Compare the Separability

Since \(c_{0}\) has a separable dual \(\text{l}^1\), and \(C[0,1]\) has a non-separable dual, \(c_{0}\) cannot be isomorphic to \(C[0,1]\) as a Banach space. Isomorphic spaces must have isomorphic duals, implying both should either be separable or non-separable.
06

Conclude the Result

Therefore, since the dual spaces have different separability properties, \(c_{0}\) is not isomorphic to \(C[0,1]\).

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.

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 Spaces
A Banach space is a complete normed vector space. This means that every Cauchy sequence in the space converges to an element in the space.
One can think of it as a space where the 'distance' between elements is well-defined, and limits of sequences within the space are guaranteed to exist.
Examples of Banach spaces include \( c_{0} \) which comprises sequences converging to zero, and \( C[0,1] \), the space of continuous functions on the interval \[0,1\].
Understanding Banach spaces helps in studying many functional analysis problems and is crucial for deeper insights into spaces like \( c_{0} \) and \( C[0,1] \).
Dual Spaces
The dual space of a vector space consists of all continuous linear functionals defined on that space. Essentially, it is a way to 'measure' or 'probe' the vectors in the original space.
For a Banach space \( X \), its dual space is often denoted by \( X' \).
In our context:
  • The dual space of \( c_{0} \) is \( \text{l}^{1} \), the space of absolutely summable sequences.
  • The dual space of \( C[0,1] \) is the space of signed regular Borel measures on \[0,1\].
These dual spaces have different properties, particularly in terms of separability, which is key in proving that \( c_{0} \) is not isomorphic to \( C[0,1] \).
Separability
Separability is a property of a space where a countable dense subset exists. Simply put, it's possible to approximate any element of the space arbitrarily well using only a countable subset.
For Banach spaces:
  • \( \text{l}^{1} \) (the dual of \( c_{0} \)) is separable.
  • The dual space of \( C[0,1] \) (signed regular Borel measures) is not separable.
The contrast in the separability of these dual spaces confirms that \( c_{0} \) and \( C[0,1] \) cannot be isomorphic, as isomorphic spaces must have isomorphic duals.
Isomorphism
Isomorphic spaces are those between which there exists a bijective linear map that is continuous and whose inverse is also continuous. In simpler terms, isomorphic spaces are structurally the same in terms of their linear properties.
To prove that \( c_{0} \) is not isomorphic to \( C[0,1] \), we considered their dual spaces' separability:
  • \( c_{0} \) has a separable dual (\text{l}^{1}).
  • \( C[0,1] \) has a non-separable dual.
Since their duals have different separability, \( c_{0} \) and \( C[0,1] \) cannot be isomorphic Banach spaces.
Sequence Spaces
Sequence spaces are spaces consisting of sequences with particular properties. They are significant in functional analysis because they provide concrete examples of Banach spaces.
\( c_{0} \) is a sequence space comprising all sequences converging to zero. Its dual is \( \text{l}^{1} \), a space of absolutely summable sequences.
The study of these spaces helps in understanding more complicated functional spaces, such as \( C[0,1] \), and provides insights into different types of functional behavior within Banach 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: (i) \(\overline{\operatorname{span}}(A)=\left(A^{\perp}\right)_{\perp}\) for \(A \subset X\). (ii) \(\operatorname{span}(B) \subset\left(B_{\perp}\right)^{\perp}\) for \(B \subset X^{*}\). Note that in general we cannot put equality. (iii) \(A^{\perp}=\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp}\) for \(A \subset X\) and \(B_{\perp}=\left(\left(B_{\perp}\right)^{\perp}\right)_{\perp}\) for \(B \subset X^{*}\). Hint: (i): Using the definition, show that \(A \subset\left(A^{\perp}\right)_{\perp}\). Then use that \(B_{\perp}\) is a closed subspace for any \(B \subset X^{*}\), proving that \(\frac{\overline{\operatorname{span}}}(A) \subset\left(A^{\perp}\right)_{\perp}\). Take any \(x \notin \overline{\operatorname{span}}(A)\). Since \(\overline{\operatorname{span}}(A)\) is a closed subspace, by the separation \(\left.f\right|_{A}=0\), hence \(f \in A^{\perp}\); also \(f(x)>0\), so \(x \notin\left(A^{\perp}\right)_{\perp}\) (ii): Similar to (i). (iii): Applying (i) to \(A^{\perp}\), we get \(A^{\perp} \subset\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp} .\) On the other hand, using \(A \subset\left(A^{\perp}\right)_{\perp}\) and the previous exercise, we get \(\left(\left(A^{\perp}\right)_{\perp}\right)^{\perp} \subset A^{\perp} .\) The dual statement is proved in the same way.

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 closed subspace of \(C[0,1]\) such that every element of \(X\) is a continuously differentiable function on \([0,1]\). Show that \(X\) is finitedimensional. Hint: Let \(T: X \rightarrow C[0,1]\) be defined for \(f \in X\) by \(T(f)=f^{\prime} .\) Show that the graph of \(T\) is closed: If \(f_{n} \rightarrow f\) uniformly and \(f_{n}^{\prime} \rightarrow g\) uniformly, then \(f^{\prime}=g .\) Therefore \(T\) is continuous by the closed graph theorem. Thus for some \(n \in \mathbf{N}\) we have \(\left\|f^{\prime}\right\|_{\infty} \leq n\) whenever \(f \in X\) satisfies \(\|f\|_{\infty} \leq 1\). Let \(x_{i}=\frac{i}{4 n}\) for \(i=0,1, \ldots, 4 n\). Define an operator \(S: X \rightarrow\) \(\mathbf{R}^{4 n+1}\) by \(S(f)=\left\\{f\left(x_{i}\right)\right\\} .\) We claim that \(S\) is one-to-one. It is enough to show that if \(\|f\|_{\infty}=1\), then for some \(i, S(f)\left(x_{i}\right) \neq 0 .\) Assume that this is not true. If \(f(x)=1\) and \(x \in\left(\frac{i}{4 n}, \frac{i+1}{4 n}\right)\), then by the Lagrange mean value theorem we have \(\left|f(x)-f\left(\frac{i}{4 n}\right)\right|=\left|f^{\prime}(\xi)\right|\left|x-\frac{i}{4 n}\right| \leq n \cdot \frac{1}{4 n}\), a contradiction. Therefore \(\operatorname{dim}(X) \leq 4 n+1\)

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 \((X,\|\cdot\|)\) be a Banach space. Show that \(\mu_{B_{X}}(x)=\|x\|\). Hint: Use continuity of the norm.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure 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.

Sign-up for free