Question

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

Short answer

If any subsequence of \(x_n\) has a subsequence convergent to \(x\), then \(x_n\) converges to \(x\). This is proven by constructing a subsequence that for each positive integer \(m\), \(\|x_{n_k} - x\| < 1/m\) for \(k > m\). The index in this sequence can be used to demonstrate \(x_n\) satisfies the conditions of convergence.

Step-by-step solution

State The Property to Be Shown

To show that \(x_n \rightarrow x\), it has to be proven that for any given \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that \(\| x_n - x \| < \epsilon\) for all \(n > N\). This is the definition of convergence in normed vector spaces.

Construct Subsequence

For each positive integer \(m\), choose a subsequence \(x_{n_k}\) such that \(\|x_{n_k} - x\| < 1/m\) for all \(k>m\). It is possible to do so since every subsequence of \(x_n\) must have a convergent subsequence, which by definition means an \(N\) exists such that \(\|x_{n_k} - x\| < 1/m\) for all \(k>N\). Choose the first element of the subsequence where this inequality holds.

Use Subsequence to Prove Convergence of Original Sequence

The constructed subsequence the properties that for a given positive integer \(m\), the index \(n_k\) associated with the constructed subsequence is such that for all \(n > n_k\), \(\|x_n - x\| < 1/m\). This shows that for any given \(\epsilon = 1/m\), an integer \(N = n_k\) can be found, such that for all \(n > N\), \(\|x_n - x\| < \epsilon\). Therefore, we have shown that \(x_n\) converges to \(x\).

Key concepts

Subsequences

In mathematics, a subsequence is a derived sequence created by deleting some or none of the elements of the original sequence, without altering the order of the remaining elements. For example, if you have a sequence of numbers like

• 1, 2, 3, 4, 5, 6

you can create a subsequence such as

• 1, 3, 4

Subsequences are useful for analyzing patterns and convergence properties in sequences. In the context of normed vector spaces, working with subsequences can help one determine the overall behavior of a sequence, especially when studying convergent sequences. Subsequence analysis is powerful because unless a subsequence exhibits certain convergence behaviors, the original sequence cannot be assumed to converge.

Normed Vector Space

A normed vector space is a vector space equipped with a function known as a "norm." The norm assigns a length to each vector, usually denoted by \(\|x\|\).For example, if \(x\) is a vector in the space, the norm might represent its magnitude or size, typically a non-negative real number.Properties of Norms:

• Non-negativity: \(\|x\| \geq 0\) and \(\|x\| = 0\) if and only if \(x = 0\).
• Scalar multiplication: \(\|\alpha x\| = |\alpha| \cdot \|x\|\) for any scalar \(\alpha\).
• Triangle inequality: \(\|x + y\| \leq \|x\| + \|y\|\).

Understanding these properties helps when proving convergence, as it involves showing that the distance between sequences becomes arbitrarily small.

Convergent Sequences

A sequence in a normed vector space is said to be convergent if it approaches a specific point as it progresses to infinity. To say that a sequence \(x_n\) converges to a point \(x\), we mean that for every small distance \(\epsilon > 0\), there exists a point \(N\) such that the distance between \(x_n\) and \(x\) becomes smaller than \(\epsilon\) for all \(n > N\).Key Points:

• Convergence implies that the terms of the sequence cluster around a specific point.
• This idea can be visualized as the terms of the sequence getting increasingly closer to the point \(x\).
• In the context of our problem, if every subsequence has a convergent subsequence to \(x\), it implies that \(x_n\) itself converges to \(x\).

Convergent sequences are foundational in analysis and many mathematical proofs because they show stability and regularity in sequences.