/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Let the functional \(f\) on \(\m... [FREE SOLUTION] | 91Ó°ÊÓ

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Let the functional \(f\) on \(\mathbf{R}^{2}\) be defined by \(f(x)=4 \xi_{1}-3 \xi_{2}\), where \(x=\left(\xi_{1}, \xi_{2}\right)\), Regard \(\mathbf{R}^{2}\) as the subspace of \(\mathbf{R}^{3}\) given by \(\xi_{3}=0 .\) Determine all linear extensions \(\tilde{f}\) of \(f\) from \(\mathbf{R}^{2}\) to \(\mathbf{R}^{3}\).

Short Answer

Expert verified
The extensions are \(\tilde{f}(\xi_1, \xi_2, \xi_3) = 4\xi_1 - 3\xi_2 + a\xi_3\) for any real \(a\).

Step by step solution

01

Understand the Problem Statement

We have a function \(f\) defined on \(\mathbb{R}^2\) as \(f(x) = 4\xi_1 - 3\xi_2\), where \(x = (\xi_1, \xi_2)\). We need to find a linear extension \(\tilde{f}\) from \(\mathbb{R}^2\) to \(\mathbb{R}^3\). This means extending the function \(f\) to include one more variable, \(\xi_3\), while maintaining linearity.
02

Write General Form for Extended Function

To find the extension \(\tilde{f}\), we express it as a linear combination of three variables: \(\tilde{f}(\xi_1, \xi_2, \xi_3) = 4\xi_1 - 3\xi_2 + a\xi_3\), where \(a\) is a constant to be determined.
03

Respect the Subspace Condition

For every vector in \(\mathbb{R}^2\), \(\xi_3 = 0\). Therefore, \(\tilde{f}(\xi_1, \xi_2, 0) = 4\xi_1 - 3\xi_2\), which matches the original function \(f(x) = 4\xi_1 - 3\xi_2\). This condition is automatically satisfied regardless of the value of \(a\).
04

Determine All Possible Extensions

Since the function must satisfy \(\tilde{f}(\xi_1, \xi_2, 0) = 4\xi_1 - 3\xi_2\) regardless of \(a\), the choice of \(a\) does not affect this condition. Hence, \(\tilde{f}(\xi_1, \xi_2, \xi_3) = 4\xi_1 - 3\xi_2 + a\xi_3\) is valid for any real number \(a\).

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Key Concepts

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

Functional Analysis
Functional analysis is a branch of mathematics focused on studying vector spaces and linear functions, particularly in the context of infinite-dimensional spaces. In this exercise, we are working with the finite-dimensional space \(\mathbb{R}^3\) to understand how a linear functional can be extended from a subspace, \(\mathbb{R}^2\), to a larger domain.
Functional analysis often investigates properties of functions that map between vector spaces, such as continuity, linearity, and boundedness. In our exercise, the function \(f(x) = 4\xi_1 - 3\xi_2\) is a linear functional defined on \(\mathbb{R}^2\). The task is to extend this functional linearly to another vector space, \(\mathbb{R}^3\), while preserving its essential linear characteristics.
This concept is deeply rooted in functional analysis, as it requires understanding how functions behave across dimensions. By considering linear extensions, we can gain insights into how vector spaces can be linked through linear mappings.
Subspace
A subspace is a vector space that is contained within another vector space. In our exercise, we consider a specific subspace of \(\mathbb{R}^3\), where vectors are constrained by the condition \(\xi_3 = 0\).
The subspace \(\mathbb{R}^2\) can be visualized as a plane within the three-dimensional space \(\mathbb{R}^3\). All vectors within this subspace have the form \((\xi_1, \xi_2, 0)\), meaning they do not extend in the direction of the \(\xi_3\) axis.
Subspace properties are crucial as they maintain all vector operations, such as addition and scalar multiplication. This means any linearly extended function must act the same on the subspace as the original function. Thus, the functional \(f(x)\) defined on \(\mathbb{R}^2\) should seamlessly integrate into the larger space \(\mathbb{R}^3\) without loss of its properties.
Linearity
Linearity is a fundamental property that describes a function or transformation that satisfies two main properties: additivity and homogeneity. In simpler terms, a linear function will preserve vector addition and scalar multiplication.
For our exercise, linearity implies that the function \(f(x) = 4\xi_1 - 3\xi_2\) satisfies:
  • Additivity: \(f(x + y) = f(x) + f(y)\) for all vectors \(x\) and \(y\).
  • Homogeneity: \(f(\alpha x) = \alpha f(x)\) for any scalar \(\alpha\).
When extending this function to \(\mathbb{R}^3\), with \(\tilde{f}(\xi_1, \xi_2, \xi_3) = 4\xi_1 - 3\xi_2 + a\xi_3\), the linearity of \(f\) is retained because the function continues to respect these properties for all choices of \(a\).
This linearity ensures that even though \(a\) can vary, the extended function \(\tilde{f}\) behaves consistently with the original function's definitions, reinforcing the seamless transition from \(\mathbb{R}^2\) to \(\mathbb{R}^3\).

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Most popular questions from this chapter

Let \(X\) be the normed space of all bounded real-valued functions on \(\mathbf{R}\) with norm defined by $$ \|x\|=\sup _{t \in \mathbf{R}}|x(t)| $$ and let \(T: X \longrightarrow X\) be defined by $$ y(t)=T_{x}(t)=x(t-\Delta) $$ where \(\Delta>0\) is a constant. (This is a model of a delay line, which is an electric device whose output \(y\) is a delayed version of the input \(x\), the time delay being \(\Delta\); see Fig. 22.) Is \(T\) linear? Bounded?

If \(Z\) is an \((n-1)\)-dimensional subspace of an \(n\)-dimensional vector space \(X\), show that \(Z\) is the null space of a suitable linear functional \(f\) on \(X\), which is uniquely determined to within a scalar multiple.

If \(d\) is a metric on a vector space \(X \neq\\{0\\}\) which is obtained from a norm, and \(\bar{d}\) is defined by $$ \tilde{d}(x, x)=0, \quad \vec{d}(x, y)=d(x, y)+1 \quad(x \neq y) $$ show that \(\bar{d}\) cannot be obtained from a norm.

Show that the operators \(T_{1}, \cdots, T_{4}\) from \(\mathbf{R}^{2}\) into \(\mathbf{R}^{2}\) defined by $$ \begin{aligned} &\left(\xi_{1}, \xi_{2}\right) \longmapsto\left(\xi_{1}, 0\right) \\ &\left(\xi_{1}, \xi_{2}\right) \longmapsto\left(0, \xi_{2}\right) \\ &\left(\xi_{1}, \xi_{2}\right) \longmapsto\left(\xi_{2}, \xi_{1}\right) \\ &\left(\xi_{1}, \xi_{2}\right) \longmapsto\left(\gamma \xi_{1}, \gamma \xi_{2}\right) \end{aligned} $$ respectively, are linear, and interpret these operators geometrically.

Which of the following subsets of \(\mathbf{R}^{3}\) constitute a subspace of \(\mathbf{R}^{3}\) ? [Here, \(\left.x=\left(\xi_{1}, \xi_{2}, \xi_{3}\right) .\right]\) (a) All \(x\) with \(\xi_{1}=\xi_{2}\) and \(\xi_{3}=0\). (b) All \(x\) with \(\xi_{1}=\xi_{2}+1\). (c) All \(x\) with positive \(\xi_{1}, \xi_{2}, \xi_{3}\). (d) All \(x\) with \(\xi_{1}-\xi_{2}+\xi_{3}=k=\) const.

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