Question

Let \(X, Y\) be Banach spaces, \(T \in \mathcal{B}(X, Y)\). Show that if \(T\) is one-to-one and \(B_{Y}^{O} \subset T\left(B_{X}\right) \subset B_{Y}\), then \(T\) is an isometry onto \(Y\). Hint: Since \(B_{Y}^{O} \subset T\left(B_{X}\right), T\) is onto (Exercise \(\left.2.29\right)\) and hence invertible. From \(T\left(B_{X}\right) \subset B_{Y}\) we get \(\|T\| \leq 1\). Assume that there is \(x \in S_{X}\) such that \(\|T(x)\|<\|x\| .\) Pick \(\delta>1\) such that \(\delta\|T(x)\|<1 .\) Then \(T(\delta x) \in B_{Y}^{O} \subset\) \(T\left(B_{X}\right)\). Thus, there must be \(z \in B_{X}\) such that \(T(z)=T(\delta x)\), but it cannot be \(\delta x \notin B_{X}\), a contradiction with \(T\) being one-to-one.

Short answer

Using the assumptions and contradiction method, \(T\) preserving norms and being onto makes it an isometry onto \(Y\).

Step-by-step solution

Understand the problem statement

Given Banach spaces \(X\) and \(Y\), and a linear operator \(T \in \mathcal{B}(X, Y)\), we need to show that \(T\) is an isometry onto \(Y\) given that \(T\) is one-to-one and that the open unit ball of \(Y\), \(B_{Y}^{O}\), is contained in the image of the closed unit ball of \(X\) under \(T\), \(T(B_{X})\), which in turn is contained in the closed unit ball of \(Y\).

Show that \(T\) is onto

Since \(B_{Y}^{O} \subset T(B_{X})\), for every element in the open unit ball of \(Y\), there exists an element in \(B_{X}\) such that under \(T\), this element maps into \(B_{Y}^{O}\). From Exercise 2.29, it follows that \(T\) is onto.

Show that \(T\) is invertible

Since \(T\) is one-to-one and we have shown that \(T\) is onto, \(T\) is therefore invertible. This means there exists a continuous linear operator \(T^{-1} : Y \to X\) such that \(T(T^{-1}(y)) = y\) for all \(y \in Y\) and \(T^{-1}(T(x)) = x\) for all \(x \in X\).

Show that \(\|T\| \leq 1\)

Given that \(T(B_{X}) \subset B_{Y}\), for any \(x \in B_{X}\), \(\|T(x)\| \leq 1\). Hence, the norm of \(T\) is \(\|T\| \leq 1\).

Assume a contradiction

Assume there exists an \(x \in S_{X}\) where \(\|T(x)\| < \|x\| = 1\). Pick a \(\delta > 1\) such that \(\delta \|T(x)\| < 1\).

Use the contradiction to show inconsistency

Since \(\delta \|T(x)\| < 1\), \(T(\delta x) \in B_{Y}^{O}\). From the assumption \(B_{Y}^{O} \subset T(B_{X})\), there must be some \(z \in B_{X}\) such that \(T(z) = T(\delta x)\). However, \(\delta x ot\in B_{X}\) contradicts \(T\) being one-to-one. This contradiction implies \(\|T(x)\| = \|x\|\) for all \(x \in S_{X}\).

Conclude that \(T\) is an isometry

Since \(\|T(x)\| = \|x\|\) holds for all \(x \in S_{X}\), \(T\) preserves the norm. Therefore, \(T\) is an isometry. As shown before, \(T\) is onto, making \(T\) an isometry onto \(Y\).

Key concepts

Banach Spaces

A Banach space is a type of vector space that has a norm and is complete with respect to that norm. This means that every Cauchy sequence in the space converges to a point within the space.

Here are some important properties of Banach spaces:

• **Vector Space:** Like any vector space, it consists of vectors which can be added and scaled by numbers.
• **Normed:** Each vector has a length or 'norm' which is a non-negative number.
• **Complete:** Its most important characteristic is completeness. In practical terms, this ensures that series of vectors that get infinitely closer together will converge to a point within the space.

Examples include spaces of continuous functions, sequences, and integrable functions. Understanding Banach spaces helps us study how functions and vectors behave under different operations.

Linear Operators

A linear operator is a function between two vector spaces that respects the operations of vector addition and scalar multiplication. Formally, for a linear operator \(T : X \to Y\), it must satisfy:

• \( T(x_1 + x_2) = T(x_1) + T(x_2) \)
• \( T(\beta x) = \beta T(x) \)

where \(x_1, x_2\) are elements of Banach space \(X\) and \(\beta\) is a scalar.

These operators are fundamental in the study of functional analysis. They allow us to generalize concepts from finite-dimensional vector spaces to infinite-dimensional ones. They can represent various transformations or changes applied to functions and are key to solving differential and integral equations in Banach spaces.

Isometry

An isometry is a mapping that preserves distances between all pairs of points. In the context of our problem:

• For a Banach space \(X\) and a linear operator \(T\), \(T\) is an isometry if for every \(x\) and \(y\) in \(X\), the distance between \(T(x)\) and \(T(y)\) in \(Y\) is the same as in \(X\).

In simpler terms, \(T\) does not distort the length of vectors. If \(T\) is isometry, then for all vectors \(x\) in \(X\), we have \(\|T(x)\| = \|x\|\). This property is particularly useful because it ensures that geometrical and topological properties of the space are preserved under \(T\).

Norm Preservation

Norm preservation is a crucial property for linear operators in functional analysis. A linear operator \(T\) from a Banach space \(X\) to another Banach space \(Y\) is said to preserve the norm if for all \(x\) in \(X\), \(\|T(x)\| = \|x\|\). This means the length of the vector does not change when the operator \(T\) is applied.

Norm preservation is a key characteristic of isometries. In our context, showing \(T\) preserves the norm means proving \(\|T(x)\|\) equals \(\|x\|\) for every \(x\) in the unit sphere of \(X\). This helps us conclude that \(T\) is indeed an isometry, maintaining the structure and properties of the vectors in the space.