91Ó°ÊÓ

Let \(\mathfrak{X}\) be a real Banach space. Show that the space \(\mathfrak{Y}=\mathfrak{X} \times \mathfrak{X}\) is a complex Banach space with the operations $$ \begin{aligned} \left(x_{1}, x_{2}\right)+\left(y_{1}, y_{2}\right) &=\left(x_{1}+y_{1}, x_{2}+y_{2}\right) \\ (\alpha+\mathrm{i} \beta)\left(x_{1}, x_{2}\right) &=\left(\alpha x_{1}-\beta x_{2}, \alpha x_{2}+\beta x_{1}\right) \\ \left\|\left(x_{1}, x_{2}\right)\right\| &=\sup _{\theta}\left\|x_{1} \cos \theta+x_{2} \sin \theta\right\| \end{aligned} $$ Show that the set \(\mathfrak{Y}_{r}=\mathfrak{X} \times\\{0\\}\) is a real subspace of \(\mathfrak{y}\) such that \(\mathfrak{Y}=\mathfrak{Y}_{r}+\mathrm{i} \mathfrak{Y}_{r}\), and that the map \(x \rightarrow(x, 0)\) is an isometry of \(\mathfrak{X}\) onto \(\mathfrak{Y}_{r}\).

Step by step solution

01

Verify Addition Operations

The addition operation in \(\mathfrak{Y}\) is defined as \[ (x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2). \]To check if \(\mathfrak{Y}\) satisfies the conditions of a Banach space, confirm that standard vector addition properties such as commutativity, associativity, and having an additive identity hold in \(\mathfrak{Y}\). Given the addition operation mirrors standard vector addition in \(\mathfrak{X}\times\mathfrak{X}\), these properties hold.

02

Verify Scalar Multiplication Operations

For scalar multiplication in \(\mathfrak{Y}\), \[ (\alpha + i\beta)(x_1, x_2) = (\alpha x_1 - \beta x_2, \alpha x_2 + \beta x_1). \]This operation satisfies properties such as distributivity over vector addition, distributivity over scalar addition, compatibility with scalar multiplication, and having a multiplicative identity. Check whether these properties hold based on the complex number rules and operations defined, which aligns with standard behaviors.

03

Confirm Norm Definition

The norm in \(\mathfrak{Y}\) is given by\[ \| (x_1, x_2) \| = \sup_{\theta}\|x_1 \cos \theta + x_2 \sin \theta\|. \]This norm satisfies the properties: positivity, scalability, triangle inequality, and non-degeneracy. Use the completeness of \(\mathfrak{X}\) and analyze whether taking the supremum will also satisfy these conditions.

04

Prove Banach Space Completeness

Since \(\mathfrak{X}\) is a real Banach space, i.e., complete with respect to the norm, show that for any Cauchy sequence \((x_n, y_n)\) in \(\mathfrak{Y}\), both \((x_n)\) and \((y_n)\) are Cauchy in \(\mathfrak{X}\) and hence converge. Therefore, \(\mathfrak{Y}\), being composed of two complete spaces, is also a complete space, proving \(\mathfrak{Y}\) is a Banach space.

05

Confirm Subspace \(\mathfrak{Y}_r\)

The subspace \(\mathfrak{Y}_r = \mathfrak{X} \times \{0\}\) contains elements of the form \((x, 0)\). Check that for any elements \((x, 0), (y, 0) \in \mathfrak{Y}_r\) and real scalars \(\alpha\), their closure under addition and scalar multiplication holds, indicating it’s a real subspace.

06

Represent \(\mathfrak{Y}\) with \(\mathfrak{Y}_r+i\mathfrak{Y}_r\)

For any element \((x_1, x_2)\) in \(\mathfrak{Y}\), show that\[ (x_1, x_2) = (x_1, 0) + i(x_2, 0), \]where \((x_1, 0), (x_2, 0) \in \mathfrak{Y}_r\). This indicates that every element in \(\mathfrak{Y}\) can be expressed as a sum of an element from \(\mathfrak{Y}_r\) and \(i\mathfrak{Y}_r\).

07

Establish Isometry from \(\mathfrak{X}\) to \(\mathfrak{Y}_r\)

Define the map \(f(x) = (x, 0)\). For every \(x, y \in \mathfrak{X}\), the norm equivalence \(\|f(x)\| = \|x\|\) holds since \[ \sup_\theta \|x \cos \theta\| = \|x\|. \]This confirms that \(f\) is an isometry from \(\mathfrak{X}\) into \(\mathfrak{Y}_r\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Subspace

In the realm of Banach spaces, a **subspace** is a subset that still retains the properties of the larger space it is part of. When given a Banach space like \(\mathfrak{Y}\), checking if a subset is a subspace involves confirming it is closed under operations like addition and scalar multiplication.

For instance, consider \(\mathfrak{Y}_r = \mathfrak{X} \times \{0\}\), which is our specific subspace here. In this setup, any element takes the form \((x, 0)\). This subset will be a real subspace of \(\mathfrak{Y}\) because:

• It must contain the zero vector, which is \((0, 0)\).
• If it contains any two elements \((x, 0)\) and \((y, 0)\), then \((x, 0) + (y, 0) = (x+y, 0)\) is also included.
• It must still hold the structure when scaled by a real number – i.e., \(\alpha(x, 0) = (\alpha x, 0)\) should remain within the subspace.

Once these conditions are verified, you can confidently assert that \(\mathfrak{Y}_r\) functions as a legitimate subspace within \(\mathfrak{Y}\).

Isometry

An **isometry** is essentially a distance-preserving map between metric spaces. In simpler terms, if you map one element to another in this way, the 'size' or 'norm' of the element does not change.

Consider the mapping defined by \(f(x) = (x, 0)\) from the real Banach space \(\mathfrak{X}\) to the subspace \(\mathfrak{Y}_r\). This map is an isometry. Here's why:

• For any \(x \in \mathfrak{X}\), the norm in \(\mathfrak{Y}_r\) can be calculated as \(\|(x, 0)\|\).
• Since the element \((x, 0)\) does not change its essence through the isometric transformation, it holds that \(\|f(x)\| = \|x\|\).
• This result showcases that while \(x\) is re-expressed as \((x, 0)\), its inherent properties remain intact - a hallmark of isometries.

Understanding this concept helps to grasp how structures in space can be transformed without altering their dimensional properties.

Complex Numbers

Complex numbers extend real numbers with the inclusion of the imaginary unit \(i\), where \(i^2 = -1\). They are typically expressed as \(a + bi\) where \(a\) and \(b\) are real numbers.

This exercise uses complex numbers to define an operation in the space \(\mathfrak{Y}\). Specifically, scalar multiplication in this complex space takes the form
\[ ( \alpha + i\beta)(x_1, x_2) = ( \alpha x_1 - \beta x_2, \alpha x_2 + \beta x_1) \] which utilizes both the real and imaginary components.

• The real component \(\alpha x_1 - \beta x_2\) blends real multiplication and the rotational effect from \(\beta\) acting as \(-\beta x_2\).
• The imaginary component combines \(\alpha x_2 + \beta x_1\) to capture the transformation, showing how complex numbers influence vectors.

Understanding complex numbers and their operations is essential for grasping many advanced mathematical concepts, especially when they involve rotation or mixing of components in multidimensional spaces.

Norm Completeness

A Banach space is essentially a complete normed vector space. Here, **norm completeness** refers to whether every Cauchy sequence in the space converges within that same space.

In the problem at hand, given that \(\mathfrak{X}\) is a complete space, it was needed to show that the newly defined complex space \(\mathfrak{Y}\) retains this property through its inherit structure.

The key steps include:

• Any Cauchy sequence \((x_n, y_n)\) in \(\mathfrak{Y}\) needs individual sequences \((x_n)\) and \((y_n)\) in \(\mathfrak{X}\) to also be Cauchy.
• Since \(\mathfrak{X}\) is complete, these sequences converge in \(\mathfrak{X}\).
• As both components \((x_n)\) and \((y_n)\) converge independently, \((x_n, y_n)\) converges in \(\mathfrak{Y}\), proving its completeness.

Norm completeness is like assuring all sequences find their home base in the space, regardless of how far they stretch.