Question

Let \(X\) be an infinite-dimensional Banach space. Show that there is a bounded linear non-compact operator from \(X\) into \(c_{0}\).

Short answer

The operator \(T: X \to c_{0}\) defined by \(T(x) = (\langle x, x_n \rangle )_{n \in \mathbb{N}}\) is bounded and non-compact.

Step-by-step solution

Understand the Problem

We need to show the existence of a bounded linear operator from an infinite-dimensional Banach space, which is not compact, into the sequence space c₀.

Recall Definitions

Recall that a bounded linear operator is a map between Banach spaces that satisfies linearity and boundedness. A compact operator is one that maps bounded sets to relatively compact sets (sets whose closure is compact). The space c₀ consists of all sequences converging to zero with the sup norm.

Construct the Operator

Consider the standard basis in c₀, denoted as \(\{e_n\}\). Define an operator \(T: X \to c_{0}\) by selecting a sequence \(\{x_n\}\) in \(X\) such that \(\|x_n\| = 1\) and \(x_n \to 0\) weakly.

Check for Boundedness

Make sure the operator \(T\) defined by \(T(x) = (\langle x, x_n \rangle )_{n \in \mathbb{N}}\) is bounded. Verify that \(\|T(x)\| < \infty\) for all \(x \in X\).

Verify Non-Compactness

Show that \(T\) is not compact. For this, observe that the sequence \(\{T(x_n)\}\), where \(x_n \in X\), does not have any subsequence that converges in \(c_{0}\), using the weak convergence of \(\{x_n\}\).

Conclusion

Since \(T\) is a bounded linear operator and not compact, we have demonstrated the existence of the desired operator.

Key concepts

Banach Space

Banach spaces are a fundamental concept in functional analysis. A Banach space is a vector space with a norm, and it is complete with respect to this norm. This means every Cauchy sequence in the space converges to a limit within the space. Banach spaces generalize the notion of Euclidean spaces and allow for more complex and infinite-dimensional spaces. Examples of Banach spaces include \(\text{L}^p\) spaces and sequence spaces like \( \text{c}_0 \).

In the given exercise, we start with an infinite-dimensional Banach space \( X \). The goal is to find a bounded and linear operator from \( X \) to the sequence space \( \text{c}_0 \), which is not compact. This requirement takes advantage of the completeness property of Banach spaces.

Compact Operator

Understanding what makes an operator compact is crucial to solving the exercise. A compact operator is a linear map from one Banach space to another that maps bounded sets to relatively compact sets. A set is relatively compact if its closure is compact. In simpler terms, applying a compact operator to a bounded set results in a set that can be 'squeezed' into a compact form.

Compact operators have specific properties that are heavily utilized in analysis and differential equations. For instance, in an infinite-dimensional space, a compact operator cannot be the identity operator. This distinction helps to define the characteristics of the operator we seek, which must be bounded but not compact.

Sequence Space c₀

The sequence space \( \text{c}_0 \) consists of all sequences \( \text{x}_n \) of real (or complex) numbers that converge to zero. Mathematically, \( \text{c}_0 = \big \{ (a_n) \in \text{l}^\text{∞} : \lim_{n \rightarrow ∞} a_n = 0 \big \} \).

A key feature of \( \text{c}_0 \) is that it is equipped with the sup norm, which is the maximum absolute value within the sequence. Utilizing \( \text{c}_0 \) allows us to create operators with particular properties, essential in the exercise to show the bounded linear operator that is not compact.

Weak Convergence

Weak convergence is a type of convergence in Banach spaces that is less strict than norm convergence. A sequence \( (x_n) \) in a Banach space \(X \) converges weakly to \( x \) if and only if for every continuous linear functional \( f \) in the dual space of \( X \), the sequence of real (or complex) numbers \( f(x_n) \) converges to \( f(x) \).

In our exercise, we consider a sequence \( (x_n) \) in \( X \) such that \( \text{\big \| x_n \big \| = 1} \) and \( x_n \rightarrow 0 \) weakly. This is essential for defining the operator \( T \) and establishing that it meets our requirements for boundedness and non-compactness.

Sup Norm

The sup norm, also known as the uniform norm, is defined for a sequence \( (a_n) \) as \( \big \| (a_n) \big \|_\text{∞} = \sup|a_n| \). This norm takes the maximum absolute value of the sequence elements. It is critical in studying spaces like \( \text{c}_0 \), where we often want to consider how 'large' the elements of our sequences are in a maximal sense.

The sup norm is utilized in defining boundedness for operators. In the exercise, the definition of the operator \( T \) relies on ensuring \( \text{\big \| T(x) \big \|_\text{∞}} < \text{\big \| \text{x} \big \| } \) for all \( x \in X \), confirming that our constructed operator remains within the necessary constraints and behaves as expected when applied to sequences converging to zero.