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Chapter 1: Problem 17

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 \(\left\|\sum_{n=1}^{m} y_{n}-\sum_{k=1}^{M} x_{k}\right\| \leq 2^{-m}\).

Short Answer

Expert verified
A normed space \(X\) is a Banach space if \(\sum y_{n}\) converges whenever \(\|y_{n}\| \leq 2^{-n}\).

Step by step solution

01

Understanding the problem

Given a normed space \(X\), we need to prove that it is a Banach space if and only if the series \(\sum y_{n}\) converges whenever \(\|y_{n}\| \leq 2^{-n}\) for every \(n\). Use the hint provided to structure the proof.
02

Direct implication (\(X\) is Banach implies convergence condition)

Assume \(X\) is a Banach space. Consider the series \(\sum y_{n}\) with \(\|y_{n}\| \leq 2^{-n}\). To show that the series converges, consider the partial sums \(S_{N} = \sum_{n=1}^{N} y_{n}\). Each \(y_{n}\) is bounded by \( 2^{-n}\), thus their series converges absolutely. Since \(X\) is complete, every absolutely convergent series in \(X\) converges, so \(\sum y_{n}\) converges.
03

Converse implication (Convergence condition implies \(X\) is Banach)

Assume the series \(\sum y_{n}\) converges whenever \(\|y_{n}\| \leq 2^{-n}\). We need to prove that \(X\) is a Banach space. Consider a Cauchy series \(\sum x_{k}\) in \(X\). By Lemma 1.15, for each \(n\), there exists an \(N_{n}\) such that \(\sum_{k > N_{n}}\|x_{k}\| \leq 2^{-n}\). Define \(y_{n} = \sum_{k=N_{n}+1}^{N_{n+1}} x_{k}\) so that \(\|y_{n}\| \leq 2^{-n}\). By assumption, \(\sum y_{n}\) converges.
04

Ensuring consistency with Cauchy criterion

Now, observe that if \(M > N_{m}\) then \(\left\|\sum_{n=1}^{m} y_{n} - \sum_{k=1}^{M} x_{k}\right\| \leq 2^{-m}\). This follows directly from the assumption that the tails of \(\sum x_{k}\) become arbitrarily small. Since \(\sum y_{n}\) converges, \(\sum_{k=1}^{\infty} x_{k}\) must converge as well, implying that \(X\) is complete, hence a Banach space.

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Key Concepts

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

Normed Space
Normed spaces are spaces equipped with a function called a norm, denoted \(orm{.}\). A norm measures the 'length' or 'size' of elements in the space. A norm must satisfy these properties:
  • Positivity: \(orm{x} \geq 0\) for all \(x \eq 0\), and \( orm{x} = 0 \) only if \(x = 0\).
  • Scalar Multiplication: \(orm{\beta x} = |\beta| orm{x}\) for any scalar \(\beta\) and vector \(x\).
  • Triangle Inequality: \(orm{x + y} \leq orm{x} + orm{y}\) for all \(x, y\).

Norms allow us to discuss concepts like distance and convergence in vector spaces, laying the foundation for advanced topics in functional analysis and linear algebra.
Cauchy Series
A Cauchy series (or sequence) in a normed space is a sequence where the elements get arbitrarily close to each other as the sequence progresses. More formally, a sequence \( (x_n) \) in a normed space \(X\) is Cauchy if for every \(\frac{ \varepsilon \gt 0}{\forall}\), there exists an \( N \in \mathbb{N}\) such that, for all \( m, n \geq N\), \(\|x_m - x_n\| < \varepsilon\).
Understanding Cauchy sequences is crucial as they help determine the completeness of a space. In a normed space, if every Cauchy sequence converges to a limit within the space, the space is complete, meaning it's a Banach space.
Absolutely Convergent Series
An absolutely convergent series is a series where the sum of the absolute values of its terms converges. That is, given a sequence \((x_n)\) in a normed space, the series \(\sum x_n\) is absolutely convergent if \(\sum orm{x_n} < \infty\).
Absolute convergence guarantees the convergence of the series. This property is vital in normed spaces because if a series is absolutely convergent in a normed space, the series itself converges in that space. This aspect is often used to test convergence in Banach spaces.
Completeness
A normed space \(X\) is complete if every Cauchy series in \(X\) converges to a limit that is also within \(X\). Completeness is a key property that defines Banach spaces. When a space is complete, it ensures that calculations and operations done within the space do not lead out of it.
For a practical understanding, consider the series \(\sum y_{n}\) in the given exercise. If the series converges under the condition \(\|y_{n}\| \leq 2^{-n}\), every Cauchy series must also converge in the normed space, implying the space is complete. Hence, the normed space is a Banach space.
Completeness is crucial for ensuring the stability and reliability of mathematical and functional operations in applied mathematics and functional analysis.

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Most popular questions from this chapter

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). If there is \(\delta>0\) such that \(\|T(x)\| \geq \delta\|x\|\) for all \(x \in X\), then \(T(X)\) is closed in \(Y .\) Moreover, \(T\) is an isomorphism from \(X\) into \(Y\). Hint: Let \(\left\\{T\left(x_{n}\right)\right\\} \subset T(X)\) be such that \(T\left(x_{n}\right) \rightarrow y\). Then \(\left\\{T\left(x_{n}\right)\right\\}\) is Cauchy and hence also \(\left\\{x_{n}\right\\}\) is Cauchy. By completeness, \(x_{n} \rightarrow x\), so \(T\left(x_{n}\right) \rightarrow T(x)\), by the uniqueness of the limit we get \(y=T(x)\). The inequality we assume implies that \(T\) is one-to-one, so the inverse \(T^{-1}: T(X) \rightarrow X\) is defined and \(\left\|T^{-1}\right\| \leq \frac{1}{\delta}\)

Show that a Banach space \(X\) is separable if and only if \(S_{X}\) is separable. Hint: If \(\left\\{x_{n}\right\\}\) is dense in \(S_{X}\), consider \(\left\\{r_{k} x_{n}\right\\}_{k, n}\) for some dense sequence \(\left\\{r_{k}\right\\}\) in \(\mathbf{K}\). If \(\mathcal{S}\) is countable and dense in \(X\), consider \(\left\\{\frac{x}{\|x\|} ; x \in \mathcal{S}\right\\}\).

Let \(K, C\) be subsets of a normed space \(X\). (i) Show that if \(K, C\) are closed, \(K+C\) need not be closed. (ii) Show that if \(K\) is compact and \(C\) is closed, then \(K+C\) is closed. Is \(\operatorname{con}(K \cup C)\) closed? (iii) Show that if \(K\) is compact and \(C\) is bounded and closed, then the set \(\operatorname{conv}(K \cup C)\) is closed. Hint: (i): Consider \(K=\\{(x, 0) ; x \in \mathbf{R}\\}\) and \(C=\left\\{\left(x, \frac{1}{x}\right) ; x>0\right\\}\). (ii): If \(x_{n}=k_{n}+c_{n} \rightarrow y\) for \(k_{n} \in K, c_{n} \in C\), then by compactness assume \(k_{n} \rightarrow k\); then also \(c_{n}=x_{n}-k_{n} \rightarrow(y-k)\) and use that \(C\) is closed. For a negative answer to \(\operatorname{conv}(K \cup C)\), see the previous exercise. (iii): If \(x_{n}=\lambda_{n} k_{n}+\left(1-\lambda_{n}\right) c_{n} \rightarrow x\) for \(k_{n} \in K, c_{n} \in C, \lambda \in[0,1]\) find a subsequence \(n_{i}\) such that \(k_{n,} \rightarrow k\) and \(\lambda_{n_{i}} \rightarrow \lambda\). If \(\lambda=1\), then by boundedness \(\left(1-\lambda_{n_{i}}\right) c_{n_{1}} \rightarrow 0\), and \(x=k \in K .\) If \(\lambda \neq 1, c_{n_{i}} \rightarrow \frac{x-\lambda k}{1-\lambda} \in C\) by closedness of \(C\).

Let \(H\) be a Hilbert space. Prove the generalized parallelogram equality: If \(x_{1}, \ldots, x_{n} \in H\), then $$ \sum_{\epsilon_{i}=\pm 1}\left\|\sum_{i=1}^{n} \varepsilon_{i} x_{i}\right\|^{2}=2^{n} \sum_{i=1}^{n}\left\|x_{i}\right\|^{2} $$ Hint: Induction on \(n\).

Let \(T\) be a one-to-one bounded linear operator from a normed space \(X\) into a normed space \(Y\). Show that \(T\) is an isometry onto \(Y\) if and only if \(T\left(B_{X}\right)=B_{Y}\) if and only if \(T\left(S_{X}\right)=S_{Y}\) if and only if \(T\left(B_{X}^{O}\right)=B_{Y}^{O}\), where \(B_{X}^{O}\) is the open unit ball in \(X\). Hint: By homogeneity, \(T\) is an isometry onto \(Y\) if and only if \(T\left(S_{X}\right)=S_{Y}\). Assume that \(T\left(B_{X}\right)=B_{Y}\). If there is \(x \in S_{X}\) such that \(\|T(x)\|=C<1\) then \(\|x / C\|>1\) and \(\|T(x / C)\|=1 .\) But there must be \(y \in B_{X}\) such that \(T(y)=T(x / C)\), a contradiction with \(T\) being one-to-one.
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