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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.

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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 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.

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Most popular questions from this chapter

Assume that \(T\) is a linear operator from a normed space \(X\) into a normed space \(Y\) such that \(\left\\{T\left(x_{n}\right)\right\\}\) is bounded for every sequence \(\left\\{x_{n}\right\\} \subset X\) satisfying \(\left\|x_{n}\right\| \rightarrow 0 .\) Is \(T\) necessarily continuous? Hint: Yes. Assuming the contrary, consider a sequence \(\left\\{x_{n} / \sqrt{\left\|x_{n}\right\|}\right\\}\) for \(\left\\{x_{n}\right\\}\) such that \(x_{n} \rightarrow 0\) and \(T\left(x_{n}\right) \neq 0\)

Let \(M\) be a dense subset (not necessarily countable) of a Banach space \(X\). Show that for every \(x \in X \backslash\\{0\\}\) there are \(x_{k} \in M\) such that \(x=\sum x_{k}\) and \(\left\|x_{k}\right\| \leq \frac{3\|x\|}{2^{k}} .\) Hint: Find \(x_{1} \in M\) such that \(\left\|x-x_{1}\right\| \leq \frac{\|x\|}{2} ;\) then by induction find \(x_{k} \in M\) such that \(\left\|x-\left(x_{1}+\ldots+x_{k-1}\right)-x_{k}\right\| \leq \frac{\|x\|}{2^{k}} .\) Then \(x=\sum x_{k}\) and \(\left\|x_{k}\right\|=\left\|\left(x-\sum_{n=1}^{k-1} x_{n}\right)-\left(x-\sum_{n=1}^{k} x_{n}\right)\right\| \leq \frac{\|x\|}{2^{k-1}}+\frac{\|x\|}{2^{k}}=\frac{3\|x\|}{2^{k}}\)

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 \(\|\cdot\|_{1},\|\cdot\|_{2}\) be two norms on a vector space \(X\). Let \(B_{1}\) and \(B_{2}\) be the closed unit balls of \(\left(X,\|\cdot\|_{1}\right)\) and \(\left(X,\|\cdot\|_{2}\right)\), respectively. Prove that \(\|\cdot\|_{1} \leq C\|\cdot\|_{2}\) (that is, \(\|x\|_{1} \leq C\|x\|_{2}\) for all \(x \in X\) ) if and only if \(\frac{1}{C} B_{2} \subset B_{1}\)

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 \(1.23\) ).
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