Question

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). If \(T\) is one-to- one, is \(T^{*}\) necessarily onto? Hint: No, consider the identity map from \(\ell_{1}\) into \(\ell_{2}\).

Short answer

No, as shown by the identity map from \(\ell_{1}\) into \(\ell_{2}\).

Step-by-step solution

Understand the problem

The problem is asking whether the adjoint operator of a one-to-one bounded linear operator on Banach spaces is necessarily onto.

Identify properties of the operators

Given that the operator \(T\) is one-to-one, we need to analyze whether \(T^*\), the adjoint operator, is onto.

Know the counterexample

Consider the identity map from \(\ell_{1}\) to \(\ell_{2}\). The space \(\ell_{1}\) consists of absolutely summable sequences, while \(\ell_{2}\) consists of square-summable sequences.

Understand the identity map properties

The identity map \(I: \ell_{1} \rightarrow \ell_{2}\) is one-to-one because each element in \(\ell_{1}\) maps uniquely to an element in \(\ell_{2}\).

Analyze the adjoint map

For \(T = I: \ell_{1} \rightarrow \ell_{2}\), the adjoint operator \(T^* = I^*: \ell_{2}^* \rightarrow \ell_{1}^*\). However, we need to realize that \(\ell_{2}^*\) can be identified with \(\ell_{2}\) and \(\ell_{1}^*\) with \(\ell_{\infty}\), the space of bounded sequences.

Conclusion from counterexample

Since \(T^*\) maps from \(\ell_{2}\) to \(\ell_{\infty}\), and \(\ell_{2} \) is a proper subspace of \(\ell_{\infty}\), the operator \(T^*\) is not onto.

Key concepts

adjoint operator

To get a good grasp of what an adjoint operator is, let's start with the basics. If you have a linear operator denoted by \( T \) that maps elements from one vector space, say \( X \), to another vector space, like \( Y \), the adjoint operator is an associated operator that essentially reverses the process while preserving certain properties.
Formally, if \( T \) is a bounded linear operator from a Banach space \( X \) to another Banach space \( Y \), the adjoint operator \( T^* \) takes each continuous linear functional on \( Y \) into a continuous linear functional on \( X \). This means that for each \( f \, \text{in} \, Y^* \), the dual space of \( Y \), \( T^* \) maps \( f \) to a new functional \( T^*f \) in the dual space of \( X \), denoted as \( X^* \).
Mathematically, this is expressed by the relation: \[ \forall x \, \text{in} \, X, \, \forall f \, \text{in} \, Y^*, \, (T^* f)(x) = f(Tx) \]
This relationship preserves the action of \(f\) after mapping through \( T \), ensuring the correspondence between \( X \) and \( Y \) via their respective dual spaces.

bounded linear operator

Understanding bounded linear operators is crucial when working with Banach spaces. A linear operator \( T \) from one Banach space \( X \) to another \( Y \) is termed 'bounded' if there exists a constant \( C \) such that for all elements \( x \) in \( X \), the norm of \( Tx \) in \( Y \) is less than or equal to \( C \) times the norm of \( x \) in \( X \).
Formally, \( T \) is bounded if:\[ \forall x \, \text{in} \, X, \, \text{there exists a constant} \, C \, \text{such that} \, orm{T(x)}_Y \, \text{\leq} \, C orm{x}_X \]
This property ensures that the operator does not magnify input sequences disproportionately, making it stable and well-behaved. In simpler terms, a bounded linear operator has control over the output based on the input, which is particularly useful when dealing with functional analysis.

identity map

An identity map, in the context of Banach spaces, is quite intuitive. It is the simplest type of linear operator where each element in the space maps to itself. Denoted generally as \( I \), for any Banach space \( X \), the identity map \( I: X \rightarrow X \) satisfies \( I(x) = x \) for all \( x \) in \( X \).
The identity map is always a bounded linear operator because the norm of \( Ix \) remains unchanged, i.e., it is equal to the norm of \( x \).
Considering our specific problem of spaces \( \ell_1 \) (absolutely summable sequences) and \( \ell_2 \) (square-summable sequences), the identity map from \( \ell_1 \) to \( \ell_2 \) would imply that each element in \( \ell_1 \) is also viewed as an element in \( \ell_2 \), even though \( \ell_2 \) does not include all elements of \( \ell_1 \).
Moreover, as highlighted in the step-by-step solution, considering the topological duals where \( \ell_{2}^{*}= \ell_{2} \) and \( \ell_{1}^{*}=\text{\ell}_{\text{\infty}} \), the identity map does not ensure that the adjoint map is onto, indicating the depth and subtlety in Banach space theory.