Chapter 8: Problem 44
Let \(x_{n} \in S_{X}\) and \(y_{n} \in S_{X}\) be such that \(\lim _{n \rightarrow \infty}\left\|x_{n}+y_{n}\right\|=2\). Let \(z_{n}\) be a point on the line segment between \(x_{n}\) and \(y_{n}\) for every \(n\). Show that \(\lim _{n \rightarrow \infty}\left\|z_{n}\right\|=1\)
Short Answer
Expert verified
\(\lim_{n \rightarrow \infty} \left\|z_{n}\right\| = 1\)
Step by step solution
01
Understand the Given Information
Given sequences \(x_{n} \in S_{X}\) and \(y_{n} \in S_{X}\) such that \(\lim _{n \rightarrow \infty}\left\|x_{n}+y_{n}\right\|=2\). Also, \(z_{n}\) is a point on the line segment between \(x_{n}\) and \(y_{n}\) for each \(n\). Need to show that \(\lim _{n \rightarrow \infty}\left\|z_{n}\right\|=1\).
02
Recall Properties of Line Segments
Recall that any point on the line segment between two points \(x_{n}\) and \(y_{n}\) can be written as \(z_{n} = \lambda x_{n} + (1-\lambda)y_{n}\) for some \( \lambda \in [0,1] \).
03
Use the Limit and Norm Properties
Given \(\lim _{n \rightarrow \infty}\left\|x_{n}+y_{n}\right\|=2\), and knowing that \(x_{n} \in S_{X}\) and \(y_{n} \in S_{X}\) (which suggests \(\|x_{n}\| = 1\) and \(\|y_{n}\| = 1\)), work with the expression \(\|z_{n}\| = \|\lambda x_{n} + (1-\lambda) y_{n}\|\).
04
Apply Triangle Inequality and Convexity
Using the triangle inequality and the fact that \(\|z_{n}\|\) represents a point within the convex combination of two unit norm vectors, write \( \|z_{n}\| = \|\lambda x_{n} + (1-\lambda) y_{n}\| \leq \lambda\|x_{n}\| + (1-\lambda)\|y_{n}\| = \lambda + (1-\lambda) = 1\).
05
Analyze the Limit of \| z_n \|
Because \(z_{n}\) lies on the line segment between \(x_{n}\) and \(y_{n}\), where the norm condition holds, and given that \( \|x_{n}+y_{n}\| = 2\) suggests that \(x_{n}\) and \(y_{n}\) are close to being collinear and symmetric around the origin, \(\|z_{n}\|\) is squeezed to also approach 1 as \(n\) approaches infinity.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normed Spaces
In functional analysis, a normed space is a vector space on which a norm is defined. A norm is a function that assigns a non-negative length or size to each vector in the space. For example, if we have a vector space \(X\), a norm is usually denoted by \(\| \cdot \|\). Important properties of norms include:
- \( \| x \| \geq 0 \) (Non-negativity)
- \( \| x \| = 0 \) if and only if \( x = 0 \) (Definiteness)
- \( \| \alpha x \| = |\alpha| \cdot \| x \| \) for any scalar \( \alpha \) (Homogeneity)
- \( \| x + y \| \leq \| x \| + \| y \| \) (Triangle Inequality)
Normed spaces form a fundamental framework in which the concepts of distance and convergence can be defined. For instance, the unit sphere \( S_X \) in a normed space \( X \) is the set of all vectors with a norm of 1: \( S_X = \{ x \in X : \| x \| = 1 \} \). This concept is central to the given exercise because we work with sequences \( x_n \in S_X \) and \( y_n \in S_X \) leading to conclusions about their behavior.
- \( \| x \| \geq 0 \) (Non-negativity)
- \( \| x \| = 0 \) if and only if \( x = 0 \) (Definiteness)
- \( \| \alpha x \| = |\alpha| \cdot \| x \| \) for any scalar \( \alpha \) (Homogeneity)
- \( \| x + y \| \leq \| x \| + \| y \| \) (Triangle Inequality)
Normed spaces form a fundamental framework in which the concepts of distance and convergence can be defined. For instance, the unit sphere \( S_X \) in a normed space \( X \) is the set of all vectors with a norm of 1: \( S_X = \{ x \in X : \| x \| = 1 \} \). This concept is central to the given exercise because we work with sequences \( x_n \in S_X \) and \( y_n \in S_X \) leading to conclusions about their behavior.
Convex Combinations
A convex combination involves taking a weighted average of points where the weights sum up to 1 and are non-negative. For two points \( x \) and \( y \) on a line segment, a convex combination is given by \( z = \lambda x + (1 - \lambda) y \) where \( \lambda \in [0, 1] \). This ensures that \( z \) lies on the line segment connecting \( x \) and \( y \).
In the context of the exercise, \( z_n \) lies on the line segment between \( x_n \) and \( y_n \), which means \( z_n \) can be expressed as:
\( z_n = \lambda x_n + (1 - \lambda) y_n \),
where \( \lambda \in [0, 1] \). This representation is useful because it utilizes linearity and the properties of convex sets. For instance, the norm of a convex combination of unit vectors (our \( x_n \) and \( y_n \) are from the unit sphere) does not exceed 1. This aligns with the step-by-step logic provided: using the triangle inequality to justify that \( \| z_n \| \leq 1 \) and leveraging the specific properties of norms.
In the context of the exercise, \( z_n \) lies on the line segment between \( x_n \) and \( y_n \), which means \( z_n \) can be expressed as:
\( z_n = \lambda x_n + (1 - \lambda) y_n \),
where \( \lambda \in [0, 1] \). This representation is useful because it utilizes linearity and the properties of convex sets. For instance, the norm of a convex combination of unit vectors (our \( x_n \) and \( y_n \) are from the unit sphere) does not exceed 1. This aligns with the step-by-step logic provided: using the triangle inequality to justify that \( \| z_n \| \leq 1 \) and leveraging the specific properties of norms.
Limit Properties
Limits play a vital role in functional analysis, especially in the context of sequences in normed spaces. A sequence \( \{ x_n \} \) is said to converge to a limit \( x \) if for every \( \epsilon > 0 \), there exists a natural number \( N \) such that \( \| x_n - x \| < \epsilon \) for all \( n > N \). This means that the distance between \( x_n \) and \( x \) becomes arbitrarily small as \( n \) increases.
For the given problem, we know:
- \( \lim_{n \to \infty} \| x_n + y_n \| = 2 \).
- \( \{ x_n \} \) and \( \{ y_n \} \) are sequences in the unit sphere, implying \( \| x_n \| = 1 \) and \( \| y_n \| = 1 \).
This information leads us to understand the behavior of \( \{ z_n \} \):
- Since \( \| x_n + y_n \| = 2 \), \( x_n \) and \( y_n \) are very close to being collinear, meaning they almost point in the exact same direction.
- This close alignment and the convex combination property squeeze \( z_n \) to maintain its norm close to 1. Eventually, as \( n \to \infty \), \( z_n \rightarrow 1 \).
Thus, by understanding these limit properties, we can conclude the required result: \( \lim_{n \to \infty} \| z_n \| = 1 \).
For the given problem, we know:
- \( \lim_{n \to \infty} \| x_n + y_n \| = 2 \).
- \( \{ x_n \} \) and \( \{ y_n \} \) are sequences in the unit sphere, implying \( \| x_n \| = 1 \) and \( \| y_n \| = 1 \).
This information leads us to understand the behavior of \( \{ z_n \} \):
- Since \( \| x_n + y_n \| = 2 \), \( x_n \) and \( y_n \) are very close to being collinear, meaning they almost point in the exact same direction.
- This close alignment and the convex combination property squeeze \( z_n \) to maintain its norm close to 1. Eventually, as \( n \to \infty \), \( z_n \rightarrow 1 \).
Thus, by understanding these limit properties, we can conclude the required result: \( \lim_{n \to \infty} \| z_n \| = 1 \).