/*! 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 27 Let \(X, Y\) be Banach spaces an... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 27

Let \(X, Y\) be Banach spaces and \(T \in \mathcal{B}(X, Y)\). If there is \(\delta>0\) such that \(\|T(x)\| \geq \delta\|x\|\) for all \(x \in X\), then \(T(X)\) is closed in \(Y .\) Moreover, \(T\) is an isomorphism from \(X\) into \(Y\). Hint: Let \(\left\\{T\left(x_{n}\right)\right\\} \subset T(X)\) be such that \(T\left(x_{n}\right) \rightarrow y\). Then \(\left\\{T\left(x_{n}\right)\right\\}\) is Cauchy and hence also \(\left\\{x_{n}\right\\}\) is Cauchy. By completeness, \(x_{n} \rightarrow x\), so \(T\left(x_{n}\right) \rightarrow T(x)\), by the uniqueness of the limit we get \(y=T(x)\). The inequality we assume implies that \(T\) is one-to-one, so the inverse \(T^{-1}: T(X) \rightarrow X\) is defined and \(\left\|T^{-1}\right\| \leq \frac{1}{\delta}\)

Short Answer

Expert verified
If \( \| T(x) \| \geq \delta \| x \| \), then \(T(X)\) is closed, and \(T\) is an isomorphism with \( \left\| T^{-1} \right\| \leq \frac{1}{\delta} \).

Step by step solution

01

Define conditions and given information

Given Banach spaces \(X\) and \(Y\), and a bounded linear operator \(T \, \in \, \mathcal{B}(X, Y)\), such that there exists \(\delta > 0\) with \(\| T(x) \| \geq \delta \| x \|\) for all \(x \in X\).
02

Show \(T(X)\) is closed

Given \( \left\{ T(x_n) \right\} \subset T(X) \) and \(T(x_n) \rightarrow y\), prove that \(y \in T(X)\). Since \( \left\{ T(x_n) \right\} \) is Cauchy, \( \left\{ x_n \right\} \) is also Cauchy due to the inequality \( \| T(x) \| \geq \delta \| x \| \). Completeness of \(X\) implies \( x_n \rightarrow x \). Therefore, \( T(x_n) \rightarrow T(x) \). By the uniqueness of limits, \( y = T(x) \), so \( y \in T(X) \). Hence, \(T(X)\) is closed.
03

Prove \(T\) is an isomorphism

To show \(T\) is one-to-one, consider the inequality \(\| T(x) \| \geq \delta \| x \|\). Suppose \( T(x) = 0 \). Then, \( 0 \geq \delta \| x \| \), which implies \( x = 0 \). Hence, \(T\) is injective (one-to-one).
04

Define the inverse and bound its norm

Given \(T(X)\) is closed and \(T\) is injective, define the inverse operator \(T^{-1}: T(X) \to X\). The norm of the inverse can be bounded as \( \left\| T^{-1} \right\| \leq \frac{1}{\delta} \) by the definition of the inverse and the given inequality.
05

Conclude the isomorphism

Since \(T\) is a bounded linear operator, injective, with closed image and a bounded inverse, \(T\) is an isomorphism from \(X\) into \(Y\).

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Key Concepts

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

Banach Spaces
A Banach space is a complete normed vector space, meaning it is a vector space equipped with a norm, and every Cauchy sequence converges within the space. The completeness criterion is crucial for many analysis results.
For example, the sequence \{x_n\} in a Banach space X, which satisfies \(| x_{n+1} - x_n | \to 0\) for n \to \infty\ is called a Cauchy sequence.
If every such Cauchy sequence \( x_n \) converges to some limit x within the space, then X is a Banach space. Common examples are:
  • \( \mathbb{R}^n \) with the Euclidean norm
  • \( L^p \) spaces for \( 1 \leq p \leq \infty \)
  • the space of continuous functions over a closed interval with the maximum norm
Bounded Linear Operator
A linear operator \(T\) between two normed spaces X and Y is called bounded if there exists a constant C such that \( \|T(x)\| \leq C \|x\| \) for all \(x\) in X. The smallest such C is called the operator norm of T and is denoted by \( \|T\| \).
For a bounded linear operator T, the value \(T(x)\) scales appropriately with the magnitude of \(x\), ensuring stability and preventing 'blow-ups'.
This property is significant because bounded operators map bounded sets to bounded sets, which is vital in various functional analysis contexts.
In the given exercise, T being in \( \mathcal{B}(X, Y) \) signifies it is a bounded linear operator.
Closed Set
In a topological space, a set is closed if it contains all its limit points. In simpler terms, if a sequence within this set converges, its limit is also included in the set. This property ensures that certain sequences or functions do not 'escape' the set upon convergence.
For instance, in the exercise, proving that \(T(X)\) is closed ensures that the image of the Banach space X under T remains within a well-defined boundary within Y.
The closed nature of T(X) within Y implies any limit of a convergent sequence of elements mapped by T from X remains in T(X). This property is critical when demonstrating the structured behavior of the operator T.
Isomorphism
An isomorphism is a bijective (one-to-one and onto) linear map between two vector spaces that preserves the operations of vector addition and scalar multiplication. This concept ensures a form of equivalence between the spaces, suggesting they are structurally the same concerning their linear properties.
In the context of the exercise, proving that \(T\) is an isomorphism shows that it establishes a strong correspondence between X and Y, allowing elements of X to be translated into Y uniquely and vice-versa.
  • One-to-one: \( T(x) = T(y) \) implies \( x = y \)
  • Onto: For every \( y \in Y\), there exists \( x \in X\) such that \( T(x) = y \)
The boundedness of T and its inverse reinforces stability and maintains controlled behavior of the spaces under this mapping.

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Most popular questions from this chapter

Find a vector space \(X\) with two norms on it such that both of them are complete norms and they are not equivalent. Hint: Take a vector space \(V\) of linear dimension \(c\), and let \(T_{1}\) and \(T_{2}\) be linear bijections of \(V\) onto \(\ell_{2}\) and \(\ell_{4}\), respectively. Define norms on \(V\) by \(\|x\|_{1}=\left\|T_{1}(x)\right\|_{2}\) and \(\|x\|_{2}=\left\|T_{2}(x)\right\|_{4}\). Then \(\left(V_{1}\|\cdot\|_{1}\right)\) is isomorphic to \(\ell_{2}\) and \(\left(V_{i}\|\cdot\|_{2}\right)\) is isomorphic to \(\ell_{4} .\) Since \(\ell_{2}\) is not isomorphic to \(\ell_{4}\) (see exercises above), \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) are not equivalent.

Let \(X, Y\) be normed spaces and \(T \in \mathcal{B}(X, Y)\). Show that $$ \|T\|=\sup \left\\{\|T(x)\|_{Y} ;\|x\|_{X}<1\right\\}=\sup \left\\{\|T(x)\|_{Y} ;\|x\|_{X}=1\right\\} $$ Hint: Clearly, both suprema are not greater than \(\|T\|\). Given \(\varepsilon>0\), find \(x \in B_{X}\) such that \(\|T(x)\|_{Y} \geq \sqrt{1-\varepsilon}\|T\| .\) Then \(\|\sqrt{1-\varepsilon} x\|<1\) and \(\frac{x}{\|x\|_{x}} \in S_{X}\), and both vectors give \(\|T(y)\|_{Y} \geq(1-\varepsilon)\|T\|\).

Show that the following are equivalent: (i) \(\sum x_{i}\) is unconditionally convergent. (ii) \(\sum x_{n}\), is convergent for every increasing sequence \(\left\\{n_{i}\right\\}_{i=1}^{\infty}\). (iii) \(\sum \varepsilon_{i} x_{i}\) converges for every \(\varepsilon_{i}=\pm 1\). Hint: (i) \Longrightarrow (ii): Given \(\varepsilon>0\), there is \(n_{0}\) such that \(\left\|\sum_{\sigma} x_{i}\right\|<\varepsilon\) whenever a finite \(\sigma\) has the property that \(\min (\sigma)>n_{0} .\) If \(k, l>n_{0}\), then \(n_{k}, n_{l}>n_{0}\) and thus \(\left\|\sum_{i=k}^{l} x_{n_{i}}\right\|<\varepsilon .\) Thus \(\sum x_{n_{i}}\) is convergent. (ii) \(\Rightarrow(\mathrm{i}):\) If \(\sum x_{i}\) is not unconditionally Cauchy, then there is \(\varepsilon>0\) and a sequence \(F_{i}\) of finite sets such that \(\max \left(F_{k}\right)<\min \left(F_{k+1}\right)\) and \(\left\|\sum_{F_{k}} x_{i}\right\| \geq\) \varepsilon. Then enumerate the sets \(F_{k}\) into a sequence \(\left\\{n_{i}\right\\}\) to get that \(\sum x_{n_{1}}\) is not Cauchy. (i) \(\Longrightarrow\) (iii): Given \(\varepsilon>0\), get a finite \(F\) such that \(\left\|\sum_{\sigma} x_{i}\right\|<\varepsilon\) whenever \(\sigma \cap F=\emptyset\). Let \(n_{0}>\max (F)\). Then, for \(m>n>n_{0}\) we have that \(\\{n, n+1, \ldots, m\\} \cap F=\emptyset\) and therefore $$ \left\|\sum_{i=n}^{m} \varepsilon_{i} x_{i}\right\| \leq\left\|\sum_{n \leq i \leq m, e_{i}=1} x_{i}\right\|+\left\|\sum_{n \leq i \leq m, e_{i}=-1}-x_{i}\right\| \leq 2 \varepsilon $$ Thus, \(\sum \varepsilon_{i} x_{i}\) is Cauchy and hence convergent. \((i i) \Longrightarrow\) (i): By contradiction, assume that \(\sum x_{i}\) is not unconditionally Cauchy. Thus, there is \(\varepsilon>0\) and finite sets \(F_{k}\) such that \(\max \left(F_{k}\right)<\) \(\min \left(F_{k+1}\right)\) and \(\left\|\sum_{F_{k}} x_{i}\right\|>\varepsilon .\) Define \(\varepsilon_{i}=1\) for \(i \in \bigcup F_{k}\) and \(\varepsilon_{i}=-1\) for \(i \notin \bigcup F_{k}\), let \(n_{k}=\min \left(F_{k}\right) .\) Then \(\left\|\sum_{i=n_{k}+1}^{n_{k+1}} x_{i}+\sum_{i=n_{k}+1}^{n_{k+1}} \varepsilon_{i} x_{i}\right\|=\) \(2\left\|\sum_{i \in F_{k+1}} x_{i}\right\| \geq 2 \varepsilon\), so either \(\left\|\sum_{i=n_{k}+1}^{n_{k+1}} x_{i}\right\| \geq \varepsilon\) for infinitely many \(k\), contra- dicting the convergence of \(\sum x_{i}\), or \(\left\|\sum_{i=n_{k}+1}^{n_{k+1}} \varepsilon_{i} x_{i}\right\| \geq \varepsilon\) for infinitely many \(k\), contradicting the convergence of \(\sum \varepsilon_{i} x_{i}\)

Show that \(C\) is a convex set in a vector space if and only if \(\sum \lambda_{i} x_{i} \in C\) whenever \(x_{1}, \ldots, x_{n} \in C\) and \(\lambda_{1}, \ldots, \lambda_{n} \geq 0\) satisfy \(\sum \lambda_{i}=1\) Hint: (a) \(\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3}=\left(\lambda_{1}+\lambda_{2}\right)\left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}} x_{1}+\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}} x_{2}\right)+\lambda_{3} x_{3}\) and induction.

We say that \(\sum x_{i}\) is unconditionally Cauchy if, given \(\varepsilon>0\), there is a finite set \(F\) in \(\mathbf{N}\) such that \(\left\|\sum_{F^{\prime}} x_{i}\right\|<\varepsilon\) whenever \(F^{\prime}\) is a finite set in \(\mathbf{N}\) satisfying \(F^{\prime} \cap F=\emptyset\) Show that a series \(\sum x_{i}\) in a Banach space \(X\) is unconditionally Cauchy if and only if it is unconditionally convergent. Hint: If \(\sum x_{i}\) is unconditionally Cauchy, then it is Cauchy and thus it converges to some \(x \in X\). Given \(\varepsilon>0\), find a finite \(F_{1}\) such that \(\left\|\sum_{\vec{F}^{\prime}} x_{i}\right\|<\varepsilon\) for every finite \(F^{\prime}\) satisfying \(F^{\prime} \cap F_{1}=\emptyset\). Then find \(n_{0}>\max \left(F_{1}\right)\) such that \(\left\|\sum_{i=1}^{n_{0}} x_{i}-x\right\|<\varepsilon\) and set \(F=\left\\{1, \ldots, n_{0}\right\\}\). If \(F^{\prime} \supset F\), then \(\left\\{i \in F^{\prime} ; i\right\rangle\) \(\left.n_{0}\right\\} \cap F_{1}=\emptyset\), so \(\left\|\sum_{F^{\prime}} x_{i}-x\right\| \leq\left\|_{i \in F^{\prime}, i>n_{0}} x_{i}-x\right\|+\left\|\sum_{i=1}^{n_{0}} x_{i}\right\|<2 \varepsilon\)
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