/*! 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 44 Let \(x_{n} \in S_{X}\) and \(y_... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

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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.
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.
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: \( \lim_{n \to \infty} \| z_n \| = 1 \).

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Most popular questions from this chapter

Let \(A\) be a subset of a metric space \((X, \rho)\). We say that a point \(x \in A\) is diametral if \(\sup \\{\|x-y\| ; y \in A\\}=\operatorname{diam}(A)\) Define \(A=\\{x(t) \in C[0,1] ; 0=x(0) \leq x(t) \leq x(1)=1\\}\). Show that every point of \(A\) is a diametral point.

Let \(A\) be a subset of \(\mathbf{R}\) with \(\lambda(A)=0\) (Lebesgue measure). Follow the hint to show that there is a Lipschitz function \(f\) on \(\mathbf{R}\) not differentiable at points of \(A\) ([Zah]). Since there is a residual set of measure 0 in \(\mathbf{R}\), this shows that the set of differentiability points of a Lipschitz function need not be residual. Hint: \(([\mathrm{BeLi}])\) Take open sets \(G_{1} \supset G_{2} \supset \supset \ldots \supset A\) such that \(\lambda\left(G_{n}\right) \leq 2^{-n}\) and \(\lambda\left((a, b) \cap G_{n+1}\right) \leq \frac{b-a}{3}\) for every component \((a, b)\) of \(G_{n} .\) Put \(f_{n}(x)=\) \(\lambda\left((-\infty, x) \cap G_{n}\right)\) and \(f=\sum(-1)^{n+1} f .\) Then \(f\) is Lipschitz. If \(x \in \bigcap G_{n}\), then \(f\) is not differentiable at \(x .\) Indeed, let \(\left(\alpha_{n}, \beta_{n}\right)\) be components of \(G_{n}\) containing \(x\). Fix \(n\), define \(\gamma_{j}=\frac{f\left(\beta_{n}\right)-f\left(\alpha_{n}\right)}{\beta_{n}-\alpha_{n}}\), and observe that \(\left\\{\gamma_{j}\right\\}\) is decreasing, \(\gamma_{1}=\ldots=\gamma_{n}=1\), and \(\gamma_{n+1} \leq \frac{1}{3} .\) Thus if \(n\) is even, then \(\frac{f\left(\beta_{n}\right)-f\left(\alpha_{n}\right)}{\beta_{n}-\alpha_{n}}=1-1+\ldots-1+\gamma_{n+1}-\gamma_{n+2}+\ldots \leq \frac{1}{3}\), while if \(n\) is odd, \(\frac{f\left(\beta_{n}\right)-f\left(\alpha_{n}\right)}{\beta_{n}-\alpha_{n}}=1-1+\ldots+1-\gamma_{n+1}+\gamma_{n+2}-\ldots \geq \frac{2}{3}\)

Let \(T: \ell_{2} \rightarrow \ell_{2}\) be a diagonal operator defined for \(x=\left(x_{i}\right) \in \ell_{2}\) by \(T(x)=\left(\left(1-\frac{1}{i}\right) x_{i}\right) .\) Show that \(T\) does not attain its norm.

Let \(X\) be an infinite-dimensional Banach space. Show that the weak and norm topologies do not coincide on \(B_{X}\).

\(\mathbf{}\) Let \(p \in(1, \infty) .\) Show that the norm of \(L_{p}[0,1]\) is Gâteaux differentiable and calculate its Gâteaux derivative. Hint: By the standard rules (use the monotonicity in the differential quotient), we get \(\|\cdot\|_{x}^{\prime}(h)=\|x\|^{1-p} \int|x(t)|^{p-1} \operatorname{sign}(x(t)) h(t) d t\). The convergence of the integral follows from Hölder's inequality.

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