Non-Basic Sequence In Banach Space Exercise 4.34 Megginson

In the realm of functional analysis, understanding the structure and properties of Banach spaces is paramount. These spaces, which are complete normed vector spaces, serve as the foundation for numerous advanced mathematical concepts and applications. One crucial aspect of studying Banach spaces involves the notion of Schauder bases, which provide a way to represent elements of the space as infinite linear combinations of basis vectors. However, not all sequences in a Banach space qualify as Schauder bases. This article delves into Exercise 4.34 from Robert E. Megginson's "Introduction to Banach Space Theory," which challenges us to construct a sequence in a Banach space that, despite seemingly promising properties, fails to be a basic sequence.

This article aims to provide a comprehensive understanding of the solution to Exercise 4.34. We will begin by revisiting the fundamental definitions and concepts necessary to tackle the problem, including Banach spaces, Schauder bases, and basic sequences. Then, we will discuss the subtleties involved in constructing a sequence that satisfies the given conditions but does not form a basic sequence. By carefully examining the properties of such sequences, we can gain deeper insights into the intricacies of Banach space theory.

Before diving into the solution, it is crucial to establish a solid understanding of the core concepts involved. Let's start by defining a Banach space.

Banach Space: A Banach space is a complete normed vector space. This means it is a vector space equipped with a norm (a function that assigns a non-negative length or size to each vector) and satisfies the property that every Cauchy sequence in the space converges to a limit within the space. Completeness is a crucial characteristic of Banach spaces, allowing for the development of powerful analytical tools.

Examples of Banach spaces include the familiar Euclidean spaces ( ${ \mathbb{R}^n }$ or ${ \mathbb{C}^n }$ with the usual Euclidean norm), the space of continuous functions on a closed interval ( ${ C[a, b] }$ with the supremum norm), and the sequence spaces ${ \ell^p }$ (where ${ 1 \leq p \leq \infty }$).

Next, let's define a Schauder basis, which is a generalization of the concept of a basis from finite-dimensional vector spaces to infinite-dimensional spaces.

Schauder Basis: A sequence ${ (x_n) }$ in a Banach space ${ X }$ is called a Schauder basis if every element ${ x \in X }$ can be uniquely represented as an infinite series:

${ x = \sum_{n=1}^{\infty} a_n x_n }$

where the coefficients ${ a_n }$ are scalars. This means that the series converges to ${ x }$ in the norm of the Banach space, and the coefficients ${ a_n }$ are uniquely determined by ${ x }$. A Schauder basis provides a fundamental way to decompose and represent elements within a Banach space.

Finally, we introduce the concept of a basic sequence, which is closely related to Schauder bases.

Basic Sequence: A sequence ${ (x_n) }$ in a Banach space ${ X }$ is called a basic sequence if it forms a Schauder basis for its closed linear span, denoted as ${ [x_n] }$. The closed linear span ${ [x_n] }$ is the smallest closed subspace of ${ X }$ that contains all the vectors ${ x_n }$. In other words, a basic sequence is a sequence that behaves like a Schauder basis within the subspace it generates.

The key distinction between a Schauder basis and a basic sequence is that a Schauder basis spans the entire Banach space, while a basic sequence only spans its closed linear span, which may be a proper subspace of the Banach space.

Now, let's formally state Exercise 4.34 from Megginson's book:

Exercise 4.34: Give an example of a sequence ${ (x_n) }$ in a Banach space ${ X }$ such that ${ (x_n) }$ is not a basic sequence, even though every subsequence of ${ (x_n) }$ converges weakly to 0, and ${ \inf \|x_n\| > 0 }$.

This exercise presents a fascinating challenge. We are tasked with constructing a sequence that possesses certain desirable properties—namely, weak convergence to 0 for every subsequence and a positive lower bound for the norms—yet fails to be a basic sequence. This highlights the subtle differences between various notions of convergence and the structural requirements for a sequence to form a basis.

To tackle this problem, we need to carefully consider what conditions might prevent a sequence from being basic, even if it exhibits these seemingly well-behaved convergence properties. Recall that a sequence is basic if it forms a Schauder basis for its closed linear span. This implies that every element in the closed linear span can be uniquely represented as an infinite linear combination of the sequence elements. Therefore, to construct a non-basic sequence, we need to find a sequence where this uniqueness property fails.

The conditions given in the exercise provide clues about the type of sequence we should look for. Weak convergence to 0 suggests that the sequence elements, in some sense, become "orthogonal" to bounded linear functionals as ${ n }$ tends to infinity. The positive lower bound on the norms ensures that the sequence elements do not simply vanish. These conditions together hint at a sequence that oscillates in some way, preventing a stable linear combination representation.

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One effective approach to solving this exercise is to construct a sequence in the Banach space ${ c_0 }$, which is the space of all sequences of scalars that converge to 0, equipped with the supremum norm:

${ \| (a_n) \| = \sup_{n \in \mathbb{N}} |a_n| }$

The space ${ c_0 }$ is a well-understood Banach space, making it a suitable setting for our construction. Let's define the sequence ${ (x_n) }$ in ${ c_0 }$ as follows:

${ x_n = e_n + \frac{1}{n} e_1 }$

where ${ e_n }$ denotes the sequence with a 1 in the ${ n }$-th position and 0 elsewhere (the standard basis vectors in ${ c_0 }$). In other words:

${ x_1 = (1, 1, 0, 0, ...) }$

${ x_2 = (\frac{1}{2}, 0, 1, 0, ...) }$

${ x_3 = (\frac{1}{3}, 0, 0, 1, ...) }$

and so on. Now, let's verify that this sequence satisfies the conditions of the exercise and that it is indeed not a basic sequence.

Verifying the Conditions

First, let's show that every subsequence of ${ (x_n) }$ converges weakly to 0. Recall that a sequence ${ (x_n) }$ converges weakly to ${ x }$ in a Banach space ${ X }$ if for every bounded linear functional ${ f \in X^* }$ (where ${ X^* }$ is the dual space of ${ X }$), we have:

${ f(x_n) \rightarrow f(x) \text{ as } n \rightarrow \infty }$

In our case, we want to show that for any subsequence ${ (x_{n_k}) }$ and any ${ f \in c_0^* }$, we have ${ f(x_{n_k}) \rightarrow 0 }$ as ${ k \rightarrow \infty }$.

The dual space of ${ c_0 }$ is ${ \ell^1 }$, the space of absolutely summable sequences. Thus, any ${ f \in c_0^* }$ can be represented as ${ f(x) = \sum_{i=1}^{\infty} a_i x_i }$ for some ${ (a_i) \in \ell^1 }$. Now, consider the subsequence ${ (x_{n_k}) }$:

${ x_{n_k} = e_{n_k} + \frac{1}{n_k} e_1 }$

Applying the functional ${ f }$ to ${ x_{n_k} }$, we get:

${ f(x_{n_k}) = f(e_{n_k} + \frac{1}{n_k} e_1) = f(e_{n_k}) + \frac{1}{n_k} f(e_1) = a_{n_k} + \frac{1}{n_k} a_1 }$

Since ${ (a_i) \in \ell^1 }$, we know that ${ a_{n_k} \rightarrow 0 }$ as ${ k \rightarrow \infty }$. Also, ${ \frac{1}{n_k} a_1 \rightarrow 0 }$ as ${ k \rightarrow \infty }$. Therefore, ${ f(x_{n_k}) \rightarrow 0 }$ as ${ k \rightarrow \infty }$, which confirms that every subsequence of ${ (x_n) }$ converges weakly to 0.

Next, we need to show that ${ \inf \|x_n\| > 0 }$. Recall that the norm in ${ c_0 }$ is the supremum norm. For each ${ n }$, we have:

${ \|x_n\| = \|e_n + \frac{1}{n} e_1\| = \sup \{ |1/n|, 1 \} = 1 }$

Thus, ${ \|x_n\| = 1 }$ for all ${ n }$, and ${ \inf \|x_n\| = 1 > 0 }$.

Proving the Sequence is Not Basic

Now comes the crucial part: proving that the sequence ${ (x_n) }$ is not a basic sequence. To do this, we need to show that there exists an element in the closed linear span of ${ (x_n) }$ that does not have a unique representation as an infinite linear combination of the ${ x_n }$'s. Let's consider the vector ${ e_1 }$. We will demonstrate that ${ e_1 }$ can be represented in two different ways as a limit of linear combinations of the ${ x_n }$'s.

First, we can write:

${ e_1 = \lim_{n \rightarrow \infty} n (\frac{1}{n} e_1) = \lim_{n \rightarrow \infty} n (x_n - e_n) }$

This shows that ${ e_1 }$ is in the closed linear span of ${ (x_n) }$.

Now, suppose that ${ (x_n) }$ is a basic sequence. Then, there should be a unique sequence of scalars ${ (a_n) }$ such that:

${ e_1 = \sum_{n=1}^{\infty} a_n x_n = \sum_{n=1}^{\infty} a_n (e_n + \frac{1}{n} e_1) }$

If such a unique representation exists, we can rewrite the sum as:

${ e_1 = \left( \sum_{n=1}^{\infty} \frac{a_n}{n} \right) e_1 + \sum_{n=1}^{\infty} a_n e_n }$

For this equality to hold, the coefficients of the basis vectors ${ e_n }$ must be equal. Therefore, we must have:

${ 1 = \sum_{n=1}^{\infty} \frac{a_n}{n} }$

and

${ 0 = a_n \text{ for all } n > 1 }$

This implies that ${ a_n = 0 }$ for all ${ n > 1 }$, and the first equation becomes:

${ 1 = a_1 }$

However, we can also represent ${ e_1 }$ as an infinite linear combination where all coefficients are zero. Consider the trivial representation:

${ e_1 = 0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 + ... = \sum_{n=1}^{\infty} 0 \cdot x_n }$

This is a different representation of ${ e_1 }$ as a linear combination of the ${ x_n }$'s. Since we have found two different representations of ${ e_1 }$ in the closed linear span of ${ (x_n) }$, the uniqueness condition for a basic sequence is violated. Therefore, the sequence ${ (x_n) }$ is not a basic sequence.

In conclusion, Exercise 4.34 from Megginson's "Introduction to Banach Space Theory" provides a compelling example of a sequence ${ (x_n) }$ in the Banach space ${ c_0 }$ that is not a basic sequence, even though every subsequence converges weakly to 0 and the norms are bounded away from 0. The sequence ${ x_n = e_n + \frac{1}{n} e_1 }$ satisfies these conditions but fails to be basic because the vector ${ e_1 }$ in its closed linear span has multiple representations as infinite linear combinations of the ${ x_n }$'s. This exercise underscores the importance of the uniqueness condition in the definition of basic sequences and highlights the subtle nuances of convergence in Banach spaces.

By carefully constructing this counterexample, we have deepened our understanding of the properties and limitations of basic sequences, reinforcing the fundamental principles of Banach space theory. This exploration serves as a valuable exercise for anyone studying functional analysis and seeking a more profound grasp of the structure and behavior of infinite-dimensional vector spaces.