Question

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). Show that \(T\) maps \(X\) onto a dense set in \(Y\) if and only if \(T^{*}\) maps \(Y^{*}\) one-to-one into \(X^{*}\). Also, if \(T^{*}\) maps onto a dense set, then \(T\) is one-to-one. Hint: If \(\overline{T(X)} \neq Y\), let \(f \in Y^{*} \backslash\\{0\\}\) be such that \(f=0\) on \(T(X)\). Then \(T^{*}(f)=0 .\) The other implications are straightforward.

Short answer

T maps X onto a dense set in Y if and only if T* is injective. If T* maps onto a dense set, T is injective.

Step-by-step solution

Understand the Definitions

Ensure that you are clear about the definitions of Banach spaces, bounded linear operators, and the adjoint operator. A Banach space is a complete normed vector space. An operator \( T \in \mathcal{B}(X, Y) \) is a bounded linear operator from \(X\) to \(Y\). The adjoint operator \( T^{*} \) maps \( Y^{*} \) to \( X^{*} \).

Show T Maps X onto a Dense Set in Y

Show that if the closure of \( T(X) \), denoted by \( \overline{T(X)} \), is equal to \( Y \), then for any \( f \in Y^{*} \) such that \( f = 0 \) on \( T(X) \), it follows that \( f = 0 \). This implies that \( T^{*}(f) = 0 \) if and only if \( f = 0 \).

Using the Hint

Suppose \( \overline{T(X)} eq Y \). Then there exists a non-zero functional \( f \in Y^{*} \) such that \( f = 0 \) on \( T(X) \). Since \( f \) is not zero but \( T^{*}(f) = 0 \), this would imply that \( T^{*} \) is not one-to-one.

Complete the 'If' Direction

Conversely, if \( T^{*} \) is one-to-one, assume \( f \in Y^{*} \) and \( f eq 0 \). Then \( T^{*}(f) eq 0 \), which implies \( f eq 0 \) on \( T(X) \). Therefore, \( T(X) \) must be dense in \( Y \).

Show T is One-to-One if T* Maps Onto a Dense Set

Finally, if \( T^{*} \) maps onto a dense set, show that \( T \) is one-to-one. Suppose \( T(x) = 0 \). Then for any \( f \in Y^{*} \), we have \( \langle T^{*}(f), x \rangle = \langle f, T(x) \rangle = 0 \). If \( T^{*} \) maps onto a dense set, then dense functionals separate points, implying \( x = 0 \). Thus, \( T \) is one-to-one.

Key concepts

Banach Spaces

A Banach space is a complete normed vector space. This means that it is a vector space equipped with a norm, and any Cauchy sequence in this space converges to a limit within the space. Understanding Banach spaces is crucial in functional analysis because they provide a structure where we can apply techniques from analysis. Examples of Banach spaces include spaces of continuous functions, denoted as \(C([a,b])\), and the \(L^p\) spaces of integrable functions.

Since the space is complete, we can perform operations on limits of sequences without stepping outside the space. This property is significant because it allows for the study of infinite-dimensional spaces in a rigorous way.

Bounded Linear Operators

A bounded linear operator is a linear transformation between two normed vector spaces that maps bounded sets to bounded sets. In simpler terms, it means that the operator won't 'blow up' values and remains controlled in how it scales inputs. Formally, an operator \(T: X \to Y\) is bounded if there exists a constant \(C\) such that \(\|T(x)\|_Y \leq C \|x\|_X\) for all \(x \in X\).

Bounded linear operators are central in functional analysis because they ensure stability and continuity in transformations. This means that small changes in the input lead to small changes in the output, which is a desirable property in both theoretical and applied contexts.

Adjoint Operator

The adjoint operator is a concept that extends the idea of linear transformations to the dual spaces of the original vector spaces. If \(T\) is a bounded linear operator from \(X\) to \(Y\), then its adjoint operator, denoted \(T^*\), is a map from \(Y^*\) to \(X^*\). Here, \(X^*\) and \(Y^*\) denote the dual spaces of \(X\) and \(Y\), which consist of all bounded linear functionals on these spaces.

The adjoint operator is defined such that for every \(y^* \in Y^*\) and \(x \in X\), we have \( \langle T^*(y^*), x \rangle = \langle y^*, T(x) \rangle \). This condition ensures that the adjoint operator preserves the inner product structure between the original space and its dual. The properties of adjoint operators are key in analyzing the behavior of the original operators, such as determining if they are injective or surjective.

Dense Sets

A subset \(A\) of a topological space \(X\) is said to be dense in \(X\) if every point in \(X\) can be approximated arbitrarily closely by points from \(A\). In mathematical terms, the closure of \(A\), denoted by \(\overline{A}\), is equal to \(X\).

In the context of Banach spaces and bounded linear operators, if \(T: X \to Y\) maps \(X\) onto a dense set in \(Y\), it means that \(\overline{T(X)} = Y\). This property is crucial because it implies that \(T\) is 'almost' surjective; we can get as close as we want to any point in \(Y\) by using points from \(T(X)\). Understanding dense sets helps in grasping how operators behave in infinite-dimensional spaces, indicating their reach and limitations.