/*! 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 26 Let \(T\) be a one-to-one bounde... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 26

Let \(T\) be a one-to-one bounded linear operator from a normed space \(X\) into a normed space \(Y\). Show that \(T\) is an isometry onto \(Y\) if and only if \(T\left(B_{X}\right)=B_{Y}\) if and only if \(T\left(S_{X}\right)=S_{Y}\) if and only if \(T\left(B_{X}^{O}\right)=B_{Y}^{O}\), where \(B_{X}^{O}\) is the open unit ball in \(X\). Hint: By homogeneity, \(T\) is an isometry onto \(Y\) if and only if \(T\left(S_{X}\right)=S_{Y}\). Assume that \(T\left(B_{X}\right)=B_{Y}\). If there is \(x \in S_{X}\) such that \(\|T(x)\|=C<1\) then \(\|x / C\|>1\) and \(\|T(x / C)\|=1 .\) But there must be \(y \in B_{X}\) such that \(T(y)=T(x / C)\), a contradiction with \(T\) being one-to-one.

Short Answer

Expert verified
T is an isometry onto Y if and only if T(B_X) = B_Y, T(S_X) = S_Y, and T(B_X^O) = B_Y^O.

Step by step solution

01

- Definition and Homogeneity

First, recall that a linear operator, by definition, is a mapping between two vector spaces that preserves the operations of vector addition and scalar multiplication. We need to use the property of homogeneity which states that for a scalar \( \beta \) and vector \( x \), the operator satisfies \( T(\beta x) = \beta T(x) \). Given that \( T \) is one-to-one and bounded, we need to establish its isometry property.
02

- Isometry and Unit Sphere

To prove that \( T \) is an isometry, recall that an isometry preserves distances. For normed spaces, this means \( \|T(x)\| = \|x\| \) for all \( x \). Suppose \( T(B_X) = B_Y \), where \( B_X \) and \( B_Y \) are the unit balls in \( X \) and \( Y \) respectively. By the hint, \( T \) being an isometry onto \( Y \) is equivalent to \( T(S_X) = S_Y \, \) where \( S_X \) and \( S_Y \) are the unit spheres in \( X \) and \( Y \) respectively.
03

- Contradiction

Assume for contradiction that there exists \( x \in S_X \) such that \( \|T(x)\| = C < 1 \). Then, consider the scaled vector \( x / C \) such that \( \| x / C \| = 1/C > 1 \. \) Applying \( T \) to \( x / C \) results in \( \| T(x / C) = 1 \). By the assumption, there must exist \( y \in B_X \) such that \( T(y) = T(x / C) \. \) This contradicts \( T \) being one-to-one.
04

- Open Unit Ball

Given the results, we have established that \( T(S_X) = S_Y \). Now let us consider the open unit balls \( B_X^O \) and \( B_Y^O \. \) If \( T(B_X) = B_Y \), then for any \( y \in B_Y^O, T^{-1}(y) \in B_X^O \). Hence, \( T(B_X^O) = B_Y^O \). Therefore, \( T \) is an isometry if and only if \( T(B_X) = B_Y \), \( T(S_X) = S_Y \), and \( T(B_X^O) = B_Y^O \).

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Key Concepts

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

isometry
An isometry is a fundamental concept in mathematics, especially in the study of normed spaces and linear operators. To put it simply, an isometry is a function between two metric spaces that preserves distances. In the context of normed spaces, this means that a linear operator \(T\) preserves the norm. In equation form, this is written as \(\forall x, \|T(x)\| = \|x\|\).
To give an intuitive picture, imagine points on a rubber sheet that鈥檚 being stretched. If the stretching process keeps the distance between every pair of points the same, the process can be considered an isometry. In our given problem, to show that \(T\) is an isometry, we need to demonstrate that \(\forall x \in S_X, \|T(x)\| = 1\), which signifies that every point on the unit sphere in space \X\ is mapped to a point on the unit sphere in space \Y\.
normed spaces
A normed space is a vector space on which a norm is defined. The norm is a function that assigns a non-negative length or size to each vector in the space. The norm must follow these properties:
  • ||x|| 鈮� 0 for all x in the space, and ||x|| = 0 if and only if x is the zero vector.
  • ||伪x|| = |伪| ||x|| for any scalar 伪 and vector x (scalar multiplication property).
  • ||x + y|| 鈮� ||x|| + ||y|| for all vectors x and y (triangle inequality).

Understanding normed spaces is crucial because operators like \(T\) act on elements of these spaces. The spaces \X\ and \Y\, mentioned in the exercise, are examples of normed spaces that could be full of different types of vectors 鈥� from simple 2D vectors to complex function spaces. In mathematics, making sure that transformations and operations between such spaces maintain certain properties, like distances, ensures structured and predictable outcomes.
unit ball
The unit ball in a normed space \(X\) is the set of all vectors whose norm is less than or equal to 1. Mathematically, it is represented as \(B_X = \{ x \in X : \|x\| \leq 1 \}\). If we visualize normed spaces as physical spaces, the unit ball would be like a filled sphere (or circle) encompassing all points that are at most 1 unit away from the origin.
Similarly, the open unit ball \(B_X^O\) consists of all vectors in \X\ whose norm is strictly less than 1, i.e., \(B_X^O = \{ x \in X : \|x\| < 1 \}\).
Analyzing these unit balls helps us understand how operators transform spaces. In our problem, showing \T(B_X) = B_Y\ means that \T\ transforms every vector within the unit ball of \X\ into vectors within the unit ball of \Y\. This evidence helps to prove that \T\ preserves the norm and thus, that \T\ is indeed an isometry.

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Most popular questions from this chapter

Let \(X\) be a Banach space and \(C\) be a compact set in \(X .\) Is it true that \(\operatorname{conv}(C)\) is compact? Hint: Not in general. Consider \(C=\left\\{\frac{1}{i} e_{i}\right\\} \cup\\{0\\}\) in \(\ell_{2}\), where \(e_{i}\) are the standard unit vectors. Clearly, \(C\) is compact. The vector \(\left(2^{-i \frac{1}{i}}\right)\) is in \(\overline{\operatorname{conv}(C)}\) and it is not in \(\operatorname{conv}(C)\), since any point in \(\operatorname{conv}(C)\) is finitely supported.

Let \(\|\cdot\|_{1},\|\cdot\|_{2}\) be two norms on a vector space \(X\). Let \(B_{1}\) and \(B_{2}\) be the closed unit balls of \(\left(X,\|\cdot\|_{1}\right)\) and \(\left(X,\|\cdot\|_{2}\right)\), respectively. Prove that \(\|\cdot\|_{1} \leq C\|\cdot\|_{2}\) (that is, \(\|x\|_{1} \leq C\|x\|_{2}\) for all \(x \in X\) ) if and only if \(\frac{1}{C} B_{2} \subset B_{1}\)

Suppose \(\left\\{x^{k}\right\\}_{k=1}^{\infty}\) is an orthonormal sequence in \(\ell_{2}\), where \(x^{k}=\left(x_{i}^{k}\right)\). Show that \(\lim _{k \rightarrow \infty}\left(x_{i}^{k}\right)=0\) for every \(i \in \mathbf{N}\). Hint: Use the Bessel inequality to show that \(\left(e_{i}, x^{k}\right) \rightarrow 0\) as \(k \rightarrow \infty\).

Find a Hilbert space \(H\) and its subspace \(F\) such that \(H \neq F+F^{\perp}\). This shows that the assumption of closedness in Theorem \(1.33\) is crucial. Hint: Consider the subspace \(F\) of finitely supported vectors in \(\ell_{2} .\) Then \(F^{\perp}=\\{0\\}\) because given \(x \in H \backslash\\{0\\},\left(x, e_{i}\right) \neq 0\) for \(i \in \operatorname{supp}(x) .\)

Let \(\sum x_{i}\) be a series in a Banach space \(X, x \in X\). Show that the following are equivalent: (i) For every \(\varepsilon>0\), there is a finite set \(F \subset \mathbf{N}\) such that \(\left\|x-\sum_{i \in F^{\prime}} x_{i}\right\|<\varepsilon\) whenever \(F^{\prime}\) is a finite set in \(\mathbf{N}\) satisfying \(F^{\prime} \supset F\). (ii) If \(\pi\) is any permutation of \(\mathbf{N}\), then \(\sum x_{\pi(i)}=x\). If these conditions hold, we say that the series is unconditionally convergent to \(x .\) As in the real case, a series \(\sum x_{i}\) is unconditionally convergent if it is absolutely convergent: \(\left\|\sum_{i \in G} x_{i}\right\| \leq \sum_{i=n}^{\infty}\left\|x_{i}\right\|\) for \(G \subset \mathbf{N}\) with \(n \leq \min (G)\). Note that the unconditional convergence of a series does not in general imply its absolute convergence; consider \(\sum \frac{1}{i} e_{i}\) in \(\ell_{2}\). Hint: (i) \Longrightarrow (ii): For every \(\varepsilon>0\), get a finite \(F\) such that \(\left\|\sum_{F^{\prime}} x_{i}-x\right\|<\varepsilon\) whenever \(F^{\prime}\) is a finite set in \(\mathbf{N}\) such that \(F^{\prime} \supset F\). For some \(n_{0}\), we get \(\left\\{\pi(1), \pi(2), \ldots, \pi\left(n_{0}\right)\right\\} \supset F .\) Thus \(\left\|x-\sum_{i=1}^{n} x_{\pi(i)}\right\|<\varepsilon\) for \(n \geq n_{0}\) (ii) \(\Longrightarrow\) (i): By contradiction, get by induction a sequence \(n_{k}\) such that \(\left\|\sum_{i=1}^{n_{k}} x_{i}-x\right\|<\frac{1}{k}\) and a sequence of finite sets \(M_{k}\) such that \(n_{k+1}>\) \(\max \left(M_{k}\right) \geq \min \left(M_{k}\right)>n_{k}\) in such a way that \(\left\|\sum_{i=1}^{n_{k}} x_{i}+\sum_{M_{k}} x_{i}-x\right\| \geq \varepsilon\) Indeed, since (i) fails for \(F=\left\\{1, \ldots, n_{k}\right\\}\), we find \(F^{\prime}\) and set \(M_{k}=F^{\prime} \backslash F\) Find a permutation \(\pi\) such that \(\pi\left(\left\\{1, \ldots, n_{k}+\left|M_{k}\right|\right\\}\right)=\left\\{1, \ldots, n_{k}\right\\} \cup M_{k}\) then \(\sum x_{\pi(i)}\) does not converge. To see that \(\sum \frac{1}{i} e_{i}\) converges unconditionally in \(\ell_{2}\), note that \(\left\|\sum_{G} \frac{1}{i} e_{i}\right\|\) is small if \(\min (G)\) is large enough.
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