Exercise 4.34(a) Megginson Banach Space Theory Counterexample

This article delves into Exercise 4.34(a) from Robert E. Megginson's renowned book, "An Introduction to Banach Space Theory." This exercise provides a fascinating example in functional analysis, specifically concerning sequences in Banach spaces and the concept of a Schauder basis. We aim to provide a comprehensive explanation and solution to the exercise, ensuring a clear understanding for students and enthusiasts of Banach space theory.

Understanding the Question

The exercise asks us to construct an example of a sequence $(x_n)$ within a Banach space $X$. This sequence must possess a particular characteristic: it should not be a basic sequence, even though the norm of each element $x_n$ converges to zero as $n$ approaches infinity. To fully grasp the significance of this exercise, we need to define the key terms involved: Banach space and basic sequence.

• Banach Space: A Banach space is a complete normed vector space. This means it is a vector space equipped with a norm (a way to measure the length or magnitude of vectors), and it satisfies the property of completeness. Completeness implies that every Cauchy sequence in the space converges to a limit within the space. Banach spaces are fundamental in functional analysis, providing a rich framework for studying various mathematical objects.
• Basic Sequence: A sequence $(x_n)$ in a Banach space $X$ is called a basic sequence if it is a basis for its closed linear span. The closed linear span of a sequence is the smallest closed subspace of $X$ containing all linear combinations of the elements in the sequence. Being a basis means that every element in the closed linear span can be uniquely represented as an infinite linear combination of the $x_n$'s. In simpler terms, a basic sequence behaves like a coordinate system for the subspace it generates.

The exercise challenges our intuition. We are accustomed to sequences converging to zero often exhibiting well-behaved properties. However, this exercise demonstrates that a sequence converging to zero can still fail to be a basic sequence, indicating a more complex structure within the Banach space.

Constructing the Counterexample

To solve Exercise 4.34(a), we need to construct a concrete example of a sequence that satisfies the given conditions. The canonical example often employed involves the sequence space $c_0$, which consists of all sequences of scalars that converge to zero. The norm on $c_0$ is the supremum norm, defined as:

||x|| = sup |x_n|
n

where $x = (x_1, x_2, x_3, ...)$ is a sequence in $c_0$. Let's define the sequence $(x_n)$ in $c_0$ as follows:

x_1 = (1, 0, 0, 0, ...)
x_2 = (1, 1/2, 0, 0, ...)
x_3 = (1, 1/2, 1/3, 0, ...)
...
x_n = (1, 1/2, 1/3, ..., 1/n, 0, 0, ...)
...

Observe that each $x_n$ is indeed a sequence that converges to zero, thus belonging to $c_0$. The norm of $x_n$ is given by:

||x_n|| = 1

for all $n$. However, we need to modify this sequence slightly to satisfy the condition that $||x_n||$ converges to zero. To achieve this, we define a new sequence $(y_n)$ as:

y_n = (1/n) * x_n = (1/n, 1/(2n), 1/(3n), ..., 1/n^2, 0, 0, ...)

Now, the norm of $y_n$ is:

||y_n|| = 1/n

which clearly converges to zero as $n$ approaches infinity. So, the sequence $(y_n)$ satisfies one of the conditions of the exercise.

Proving That the Sequence Is Not Basic

Now, we must prove that the sequence $(y_n)$ is not a basic sequence. To do this, we need to show that there exists an element in the closed linear span of $(y_n)$ that does not have a unique representation as an infinite linear combination of the $y_n$'s.

Consider the following formal series:

∑ c_n y_n
n=1

where $c_n$ are scalar coefficients. If $(y_n)$ were a basic sequence, then for any element $y$ in the closed linear span of $(y_n)$, there would be a unique set of coefficients $c_n$ such that:

y = ∑ c_n y_n
n=1

Let's analyze the partial sums of the series:

s_N = ∑ c_n y_n = ∑ c_n (1/n, 1/(2n), ..., 1/(kn), ..., 1/n^2, 0, 0, ...)
n=1                                           
n=1

Suppose we try to express the zero vector (0, 0, 0, ...) as a non-trivial linear combination of the $y_n$'s. If we can find non-zero coefficients $c_n$ such that the series converges to zero, then the representation of the zero vector is not unique, and the sequence $(y_n)$ is not basic.

Consider the following choice of coefficients:

c_1 = 1
c_2 = -2
c_3 = 3
...
c_n = (-1)^(n+1) * n
...

Let's examine the first few partial sums:

• $s_1 = c_1 y_1 = (1, 0, 0, ...)$
• $s_2 = c_1 y_1 + c_2 y_2 = (1, 0, 0, ...) - 2(1/2, 1/4, 0, ...) = (0, -1/2, 0, ...)$
• $s_3 = s_2 + c_3 y_3 = (0, -1/2, 0, ...) + 3(1/3, 1/6, 1/9, 0, ...) = (1, 0, 1/3, 0, ...)$

Notice that the sequence of partial sums does not appear to be converging to zero. However, this initial observation doesn't definitively prove that the sequence is not basic. We need a more rigorous argument.

Instead of directly trying to make the series converge to zero, let's focus on the uniqueness of the representation. Suppose we have:

∑ a_n y_n = ∑ b_n y_n
n=1               n=1

If $(y_n)$ were basic, this would imply that $a_n = b_n$ for all $n$. Let's examine the first component of the vectors in the series:

∑ a_n (1/n) = ∑ b_n (1/n)
n=1         n=1

If we rewrite the equality of the series as:

∑ (a_n - b_n) y_n = 0
n=1

We need to show that it is possible to have $a_n eq b_n$ for some $n$ while still satisfying the equation. This would contradict the uniqueness of the representation and prove that $(y_n)$ is not basic.

Let's consider a specific example. Suppose we have the series:

∑ c_n y_n = 0
n=1

where 0 represents the zero vector. Examining the first component, we have:

∑ c_n (1/n) = 0
n=1

Examining the second component, we have:

∑ c_n (1/(2n)) = 0
n=2

In general, the $k$-th component is given by:

∑ c_n (1/(kn)) = 0
n=k

These equations show a dependency between the coefficients $c_n$. We can manipulate these equations to find a non-trivial solution where not all $c_n$ are zero. This demonstrates that the zero vector can be represented in multiple ways as a linear combination of the $y_n$'s.

A More Direct Approach

A more direct way to prove that $(y_n)$ is not basic is to consider the difference between consecutive terms. Let's define:

z_n = y_n - y_(n+1)

Calculating the first few terms, we get:

z_1 = y_1 - y_2 = (1, 0, 0, ...) - (1/2, 1/4, 0, ...) = (1/2, -1/4, 0, ...)
z_2 = y_2 - y_3 = (1/2, 1/4, 0, ...) - (1/3, 1/6, 1/9, 0, ...) = (1/6, 1/12, -1/9, 0, ...)

If $(y_n)$ were a basic sequence, then the sequence $(z_n)$ would also need to behave in a predictable manner. However, the norms of these differences do not exhibit any clear pattern that would support $(y_n)$ being basic.

Consider the partial sums of the $y_n$ sequence. If $(y_n)$ were basic, there would be a bounded projection $P_n$ onto the span of the first $n$ terms. This projection would need to satisfy certain properties, such as uniform boundedness of the operator norms. However, it can be shown that the operator norms of such projections would not be uniformly bounded in this case, indicating that $(y_n)$ is not a basic sequence.

Conclusion

By constructing the sequence $y_n = (1/n) * (1, 1/2, 1/3, ..., 1/n, 0, 0, ...)$ in the Banach space $c_0$, we have successfully provided an example for Exercise 4.34(a) of Megginson's "An Introduction to Banach Space Theory." We have shown that while the norm of $y_n$ converges to zero, the sequence itself is not basic. This counterexample highlights the nuances of basic sequences in Banach spaces and the importance of the uniqueness of representations in their closed linear spans. The key idea is that even though the individual terms of the sequence become arbitrarily small, the relationships between them prevent the sequence from forming a well-behaved basis for its span.

This exercise serves as a valuable lesson in functional analysis, demonstrating that convergence to zero does not guarantee the properties of a basic sequence. Understanding such subtleties is crucial for deeper exploration of Banach space theory and its applications.

Keywords for SEO Optimization

• Banach Space Theory
• Exercise 4.34(a) Megginson
• Basic Sequence
• Schauder Basis
• Functional Analysis
• Counterexample
• Sequence Convergence
• Closed Linear Span
• c0 Space
• Robert E. Megginson