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Chapter 10: Problem 29

Show that if \(x_{n} \rightarrow x\) in a normed vector space then $$ \frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \rightarrow x $$

Short Answer

Expert verified
The key point is to express the difference between the arithmetic mean and x, and make each term less than \( \frac{\varepsilon}{2} \) by carefully choosing N and M based on the definition of sequence convergence in a normed vector space. The steps of the proof show that both terms can be made arbitrarily small, thus the arithmetic mean of the sequence converges to x.

Step by step solution

01

Definition of sequence convergence in a normed vector space

Given a sequence \( x_n \) in a normed vector space that converges to x, this means that for any \( \varepsilon > 0 \), there exists a natural number N such that for all \( n > N \), \( ||x_n - x|| < \varepsilon \).
02

Define the arithmetic mean

The arithmetic mean of the sequence \( x_n \) up to the nth term is defined as \( \frac{x_1 + x_2 + \cdots + x_n}{n} \). In order to prove that the arithmetic mean converges to x, we need to show that for any \( \varepsilon > 0 \), there exists a natural number M such that for all \( n > M \) we have \( ||\frac{x_1 + x_2 + \cdots + x_n}{n} - x|| < \varepsilon \).
03

Express the norm of the arithmetic mean minus x

Consider \( ||\frac{x_1 + x_2 + \cdots + x_n}{n} - x|| \). This can be rewritten as \( ||\frac{x_1 + x_2 + \cdots + x_N + x_{N + 1} + \cdots + x_n}{n} - x|| \). Expand this and apply triangle inequality to obtain \( ||\frac{x_1 + x_2 + \cdots + x_N}{n} - \frac{Nx}{n} + \frac{x_{N + 1} + \cdots + x_n}{n} - \frac{(n - N)x}{n}|| \leq \frac{1}{n}||x_1 + x_2 + \cdots + x_N - Nx|| + \frac{1}{n}||x_{N + 1} + \cdots + x_n - (n - N)x||.
04

Apply the convergence definition of \( x_n \)

Since \( x_n \rightarrow x \) for \( n > N \), by choosing \( N > \frac{2}{\varepsilon}||x_1 + x_2 + \cdots + x_N - Nx|| \) the first term of the inequality in the previous step can be made less than \( \frac{\varepsilon}{2} \). For the second term, choose \( M > \max {N} \) such that for \( n > M \) and for \( n > N \), we have \( ||x_n - x|| < \frac{\varepsilon}{2(n - N)} \). Therefore, the second term can also be made less than \( \frac{\varepsilon}{2} \).
05

Conclude the proof

Since both terms from the inequality of step 3 can be made less than \( \frac{\varepsilon}{2} \), we then have \( ||\frac{x_1 + x_2 + \cdots + x_n}{n} - x|| = ||mean_{n=1}^{N}{x_n} - x|| < \varepsilon \). This means that as \( n \) gets very large, the means \( mean_{n=1}^{n}{x_n} \) converge to \( x \), hence the proof is completed.

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Most popular questions from this chapter

Let \(V\) be a Banach space and \(W\) te a vector subspace of \(V .\) Define its closture \(\bar{W}\) to be the union of \(W\) and all hmuts of Cauchy sequences of elements of \(W\). Show that \(\bar{W}\) is a closed vector subspace of \(V\) in the sense that the limit points of all Cauchy scquences in \(\bar{W}\) lie in \(\bar{W}\) (note that the Cauchy sequences may include the newly added limit points of \(W\) ).

Show that if \(x_{n}\) is a sequence in anormed vector space \(V\) such that every subsequence has a subsequence comergent to \(x\), then \(x_{n} \rightarrow x\).

If \(G_{0}\) is the component of the identity of a locally connected topological group \(G\), the factor group \(G / G_{0}\) is called the group of components of \(G .\) Show that the group of components is a discrete topological group with respect to the topology induced by the natural projection map \(\pi: g \mapsto g G_{0}\)

Show that a map \(f \cdot X \rightarrow Y\) between two topological spaces \(X\) and \(Y\) is contintous if and only if \(f(\bar{U}) \subseteq \overline{f(U)}\) for all sets \(U \subseteq X\) Show that \(f\) is a lwomeomorplism only if \(f(\bar{U})=\overline{f(U)}\) for all sets \(U \subseteq X\)

Show that the set \(V^{\prime}\) consisting of bounded linear functionals on a Banach space \(V\) is a normed vector space with respect to the norm $$ \|\varphi\|=\sup [M|| \varphi(x) \mid \leq M\|x\| \text { for all } x \in V \mid $$ Show that this norm is complete on \(V^{\prime}\).
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