Exercise 4.34(a) Megginson An Introduction To Banach Space Theory Example Of A Non Basic Sequence

In this comprehensive exploration, we delve into Exercise 4.34(a) from Robert E. Megginson's esteemed book, "An Introduction to Banach Space Theory." This exercise challenges our understanding of basic sequences within the framework of Banach spaces. Specifically, it asks us to construct a sequence $(x_n)$ in a Banach space $X$ that, despite certain properties, fails to be a basic sequence. This seemingly simple request opens up a fascinating avenue for exploring the intricacies of functional analysis and the subtle interplay between sequences and series in infinite-dimensional spaces. Understanding the nuances of Schauder bases and their connection to basic sequences is crucial for tackling this problem effectively. Our discussion will not only present a solution to the exercise but also provide a detailed exposition of the underlying concepts, ensuring a thorough grasp of the material. This journey will involve navigating the landscape of Banach spaces, carefully constructing a sequence with specific characteristics, and rigorously proving that it indeed satisfies the conditions of the problem. By the end of this exploration, you will gain a deeper appreciation for the subtleties of basic sequences and their role in the broader context of Banach space theory.

Understanding Basic Sequences and Banach Spaces

To effectively address Exercise 4.34(a), it's crucial to have a solid grasp of the fundamental definitions and concepts. Let's begin by defining a Banach space: A Banach space is a complete normed vector space. This means it's a vector space equipped with a norm (a function that assigns a non-negative length or size to each vector) and that every Cauchy sequence in the space converges to a limit within the space. Completeness is a crucial property that distinguishes Banach spaces and allows for powerful analytical tools to be applied. Examples of Banach spaces abound in mathematics, including the familiar Euclidean spaces ($\mathbb{R}^n$, $\mathbb{C}^n$), the spaces of continuous functions on a closed interval ($C[a, b]$), and the Lebesgue spaces ($L^p$ spaces). These spaces serve as the foundation for many areas of analysis and its applications.

Next, we turn our attention to the concept of a basic sequence. A sequence $(x_n)$ in a Banach space $X$ is said to be basic if it is linearly independent and if the closed linear span of $(x_n)$, denoted by $\overline{\text{span}}\{x_n\}$, is a Banach space under the norm inherited from $X$. Furthermore, for every $x$ in $\overline{\text{span}}\{x_n\}$, there exists a unique sequence of scalars $(a_n)$ such that the series $\sum_{n=1}^{\infty} a_n x_n$ converges to $x$. This uniqueness property is a defining characteristic of basic sequences and highlights their role as building blocks for the subspace they span. The concept of a Schauder basis is closely related to that of a basic sequence. A Schauder basis for a Banach space $X$ is a sequence $(x_n)$ such that every $x$ in $X$ can be uniquely represented as the limit of a series $\sum_{n=1}^{\infty} a_n x_n$. In other words, a Schauder basis is a basic sequence whose closed linear span is the entire space $X$. However, not every Banach space possesses a Schauder basis, making the study of basic sequences and their properties even more significant.

Key Properties and Implications

Several key properties are associated with basic sequences. One important result is that a basic sequence in a Banach space generates a sequence of bounded linear functionals, known as the biorthogonal functionals, which satisfy certain orthogonality conditions with respect to the basic sequence. These biorthogonal functionals play a crucial role in the representation of elements in the closed linear span of the basic sequence. Another important aspect is the stability of basic sequences under small perturbations. This means that if we slightly perturb the elements of a basic sequence, the resulting sequence may still be basic, provided the perturbations are sufficiently small in a suitable sense. This stability property is essential for applications in approximation theory and numerical analysis. Furthermore, the concept of basic sequences is closely tied to the notion of complemented subspaces in Banach spaces. A subspace $Y$ of a Banach space $X$ is said to be complemented if there exists another subspace $Z$ of $X$ such that $X$ is the direct sum of $Y$ and $Z$. If $(x_n)$ is a basic sequence in $X$, then its closed linear span $\overline{\text{span}}\{x_n\}$ is a complemented subspace if and only if there exists a bounded linear projection from $X$ onto $\overline{\text{span}}\{x_n\}$. This connection between basic sequences and complemented subspaces sheds light on the geometric structure of Banach spaces.

Deconstructing Exercise 4.34(a): The Challenge

Exercise 4.34(a) from Megginson's book presents a specific challenge: to construct a sequence $(x_n)$ in a Banach space $X$ that is not basic, despite possessing certain characteristics. This counterexample highlights that not every sequence that satisfies some seemingly necessary conditions is automatically a basic sequence. This underscores the importance of a precise understanding of the definition of a basic sequence and the subtleties involved. The exercise implies that there exist sequences that may exhibit some properties reminiscent of basic sequences but ultimately fail to meet all the criteria. This failure can stem from various reasons, such as the lack of linear independence, the failure of the closed linear span to be a Banach space, or the non-uniqueness of the representation of elements in the span as infinite series. The core of the exercise lies in carefully crafting a sequence that intentionally violates one or more of these conditions while potentially satisfying others. This requires a deep understanding of the definition of a basic sequence and the ability to manipulate sequences in infinite-dimensional spaces. The challenge is not merely to find any non-basic sequence, but to construct one that elucidates the nuances of the definition and the potential pitfalls in assuming a sequence is basic based on limited information.

Understanding the Implicit Conditions

To tackle this exercise effectively, we must carefully consider the implicit conditions that might make a sequence appear to be basic at first glance. For instance, one might initially think that a sequence that converges to zero or a sequence whose elements are linearly independent would automatically be basic. However, these conditions alone are not sufficient. The crux of the problem lies in the uniqueness of the representation of elements in the closed linear span as infinite series. This uniqueness condition is often the most challenging to verify and is the key to constructing a counterexample. We need to find a sequence where, even if we can express elements in its closed linear span as infinite series, this representation is not unique. This non-uniqueness can arise if there are different ways to combine the elements of the sequence to approximate the same vector. The construction of such a sequence requires a clever choice of elements and a careful analysis of their linear combinations and limits. We need to think about how to introduce dependencies or relationships between the elements of the sequence that lead to non-unique representations without making the sequence trivially non-basic (e.g., by making it linearly dependent from the outset). The subtle interplay between linear independence, convergence, and uniqueness is at the heart of this exercise.

Constructing the Counterexample: A Step-by-Step Approach

Now, let's embark on the construction of a sequence $(x_n)$ in a Banach space $X$ that satisfies the conditions of Exercise 4.34(a) but is not basic. We will work in the Banach space $\ell^1$, the space of absolutely summable sequences, equipped with the norm $\|(a_n)\|_{\ell^1} = \sum_{n=1}^{\infty} |a_n|$. This space provides a convenient setting for our construction due to its well-understood properties and the ease with which we can manipulate sequences within it. Our strategy will be to define a sequence $(x_n)$ in $\ell^1$ such that its elements exhibit a specific dependency, leading to non-unique representations in its closed linear span. This dependency will be carefully designed to violate the uniqueness condition required for a basic sequence.

Defining the Sequence (x_n)

We define the sequence $(x_n)$ in $\ell^1$ as follows:

• $x_1 = (1, 0, 0, 0, ...)$
• $x_2 = (1, 1, 0, 0, ...)$
• $x_3 = (1, 1, 1, 0, ...)$
• ...
• $x_n = (1, 1, ..., 1, 0, 0, ...)$ (where there are $n$ ones)
• ...

In other words, $x_n$ is a sequence of 1s followed by 0s, where the number of 1s is equal to $n$. Each $x_n$ is clearly an element of $\ell^1$ since the sum of the absolute values of its components is simply $n$, which is finite. It's also important to observe that the elements of this sequence are linearly independent. To see this, suppose we have a finite linear combination of the form $\sum_{i=1}^{k} a_i x_i = 0$, where the $a_i$ are scalars. By examining the components of this linear combination, we can deduce that all the coefficients $a_i$ must be zero, thus establishing linear independence.

Proving Non-Basicity

Now, we need to demonstrate that this sequence $(x_n)$ is not basic. To do this, we will show that the representation of elements in its closed linear span is not unique. Consider the vector $x = (1, 1/2, 1/3, 1/4, ...)$, which is an element of $\ell^1$ since the harmonic series $\sum_{n=1}^{\infty} 1/n$ diverges, but the series $\sum_{n=1}^{\infty} |1/n|$ also diverges. However the alternating harmonic series $\sum_{n=1}^{\infty} (-1)^{n+1}/n$ converges. Although $x$ itself is not in the span of $(x_n)$, we can modify our approach slightly to create a similar situation within the closed linear span. Let's consider the differences between consecutive elements of our sequence:

• $x_2 - x_1 = (0, 1, 0, 0, ...)$
• $x_3 - x_2 = (0, 0, 1, 0, ...)$
• $x_4 - x_3 = (0, 0, 0, 1, ...)$
• ...

These differences form the standard basis vectors in $\ell^1$, which we denote as $e_n$, where $e_n$ has a 1 in the $n$-th position and 0s elsewhere. Now, consider the vector $y = \sum_{n=1}^{\infty} \frac{1}{n} e_n = (0, 1, 1/2, 1/3, ...)$. This vector is not in $\ell^1$ because the sum of the absolute values of its components diverges. However, we can work with finite truncations of this series. Let's define the partial sums:

$s_k = \sum_{n=1}^{k} \frac{1}{n} e_n$

Each $s_k$ is in the span of the $e_n$, and therefore in the span of $(x_n)$. Now, we can express each $e_n$ as a difference of $x_i$s:

$e_n = x_{n+1} - x_n$ for $n \geq 1$

Therefore, we can rewrite the partial sums $s_k$ in terms of the $x_i$:

$s_k = \sum_{n=1}^{k} \frac{1}{n} (x_{n+1} - x_n)$

Rearranging the terms, we get:

$s_k = -x_1 + \sum_{n=2}^{k} x_n (\frac{1}{n-1} - \frac{1}{n}) + \frac{1}{k} x_{k+1}$

This expression shows that each $s_k$ is a linear combination of the $x_i$s. However, if the sequence $(x_n)$ were basic, the representation of each $s_k$ as a linear combination of the $x_i$s would be unique. To demonstrate non-uniqueness, we need to find a different way to represent these vectors (or a limit of these vectors) as a series involving the $x_i$s. This is where the careful construction of the $x_i$s comes into play.

Consider the sequence $z_n = x_n / n$. This sequence converges to 0 in $\ell^1$ since $\|z_n\| = \|x_n / n\| = \|x_n\| / n = 1$. If the sequence $(x_n)$ were basic, then the only representation of the zero vector as a series involving the $x_n$s would be the trivial one (all coefficients equal to zero). However, we can construct a non-trivial representation of zero using the $z_n$s:

$0 = \lim_{k \to \infty} (\sum_{n=1}^{k} \frac{1}{n} x_n - x_{k+1})$

This equation demonstrates that there exists a non-trivial representation of zero as a limit of linear combinations of the $x_n$s, which contradicts the uniqueness requirement for basic sequences. Therefore, the sequence $(x_n)$ is not basic.

Conclusion: Key Takeaways from Exercise 4.34(a)

In conclusion, by constructing the sequence $(x_n)$ in $\ell^1$ as described above, we have successfully demonstrated an example of a sequence that is not basic, even though it may initially appear to possess some characteristics of basic sequences. This exercise highlights the crucial role of the uniqueness condition in the definition of a basic sequence and the subtle ways in which this condition can be violated. The key to our construction was the introduction of a dependency between the elements of the sequence, allowing for non-unique representations of vectors in its closed linear span. This exploration reinforces the importance of a thorough understanding of the definitions and properties of basic sequences and their relationship to Schauder bases in the context of Banach space theory. Understanding these nuances is essential for further studies in functional analysis and its applications.