/*! 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 21 Suppose \(X, Y\) are Banach spac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose \(X, Y\) are Banach spaces and \(T: X \rightarrow Y\) is linear. Suppose further that whenever \(x_{n} \rightarrow 0\) and \(T x_{n} \rightarrow y\) then \(y=0\). Show that \(T\) is continuous.

Short Answer

Expert verified
The operator \( T \) is continuous because it maps converging sequences to zero.

Step by step solution

01

Understand the Criterion for Continuity

A linear operator between Banach spaces is continuous if and only if it is bounded. An equivalent condition for continuity is that for any sequence \( x_n \rightarrow 0 \) in \( X \), the sequence \( T x_n \rightarrow 0 \) in \( Y \).
02

Analyze the Given Condition

The problem states that if \( x_n \rightarrow 0 \) in \( X \) and \( T x_n \rightarrow y \) for some \( y \) in \( Y \), then \( y = 0 \). This is a version of the sequential criterion for continuity: \( T \) maps sequences converging to zero in \( X \) to sequences converging to zero in \( Y \).
03

Apply the Criterion to Show Continuity

Given that \( T x_n \rightarrow y \) and by the problem's condition, \( y = 0 \), it follows for any \( x_n \rightarrow 0 \), that \( T x_n \rightarrow 0 \). This satisfies the criterion that \( T \) takes sequences converging to zero to sequences converging to zero in the image space, implying \( T \) is continuous.

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.

Linear Operator
A linear operator is a fundamental concept in functional analysis and an extension of the idea of matrix multiplication to infinite-dimensional spaces. The operator maps elements from one vector space to another, while respecting the linear structure. To be specific, a linear operator between two Banach spaces, say from space \(X\) to space \(Y\), satisfies two important properties:
  • **Additivity**: For any two elements \(x_1, x_2\) in \(X\), \(T(x_1 + x_2) = T(x_1) + T(x_2)\).
  • **Homogeneity**: For any scalar \(a\) and any element \(x\) in \(X\), \(T(ax) = aT(x)\).
These properties ensure that the operator aligns with how linear transformations behave in finite-dimensional settings. Understanding linear operators is crucial since they form the building blocks for more complex operations in functional analysis.
Continuity
In the realm of Banach spaces, continuity of an operator ensures that small changes in the input result in small changes in the output. For a linear operator \(T: X \rightarrow Y\) to be continuous, particularly between Banach spaces, it must satisfy the condition that for any sequence \(x_n\) converging to zero in \(X\), the sequence \(Tx_n\) converges to zero in \(Y\).
This continuous behavior guarantees predictability in the transformation process. While continuity might seem intuitive, verifying it requires specific tests, especially in infinite-dimensional spaces, where our regular notion of inspection by graphs is not feasible. Thus, in practice, the boundedness of the operator is used as a mutual criterion for continuity in Banach spaces.
Sequential Criterion
The sequential criterion offers a practical approach to verifying continuity. It states that an operator is continuous if it maps sequences in the domain that converge to a point, to sequences in the codomain that converge to the operator's image of that point.
In simpler terms, if \(x_n \rightarrow 0\) in \(X\), then \(Tx_n \rightarrow 0\) in \(Y\). This criterion can often simplify the task of proving continuity, especially within the context of linear operators between Banach spaces. When dealing with sequences, we leverage their convergence properties to establish a link between input sequences converging to zero and their transformed counterparts also approaching zero. This provides a strong and versatile tool to demonstrate continuity in practice.
Boundedness
Boundedness is a key property in understanding the continuity of linear operators between Banach spaces. An operator \(T: X \rightarrow Y\) is bounded if there exists a constant \(C\) such that for all \(x\) in \(X\), \(\|Tx\| \leq C\|x\|\). Boundedness implies that the operator does not distort vectors too heavily, meaning the growth of the output is controlled relative to the input. This property is intimately linked with continuity—indeed, a linear operator is continuous if and only if it is bounded.
This equivalence provides a practical method for proving continuity of operators by showing boundedness. In the setting of Banach spaces, this approach simplifies the examination of operators and facilitates analyzing their properties rigorously.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

We introduce some terminology for the purpose of this problem: If \(X\) is either a real or complex vector space (meaning that the scalars used in scalar multiplication are real, or, respectively, complex), we say that a real-valued \(\varphi\) is a real-linear functional if \(\varphi(x+y)=\varphi(x)+\varphi(y)\) and \(\varphi(\alpha x)=\alpha \varphi(x)\) holds for all \(x, y \in X\) and \(\alpha\) real. For \(X\) a complex vector space, we say that (a complex-valued) \(\varphi\) is a complexlinear functional if these relationships hold for all \(x, y \in X\) and \(\alpha\) complex. (a) Show that for any complex number \(z, z=\operatorname{Re} z-i \operatorname{Re}(i z)\). (b) Suppose \(X\) is a complex vector space and \(\varphi\) is a complex-linear functional on \(X\). Define \(u: X \rightarrow \mathbb{R}\) by \(u(x)=\operatorname{Re} \varphi(x)\). Show that \(\varphi(x)=u(x)-i u(i x)\) for all \(x \in X\). (c) Suppose \(u\) is a real-linear functional on \(X\). Define \(\varphi: X \rightarrow \mathbb{C}\) by \(\varphi(x)=\) \(u(x)-i u(i x)\). Show that \(\varphi\) is a complex-linear functional on \(X\). (Hint: Check the condition \(\varphi(\alpha x)=\alpha \varphi(x)\) first for \(\alpha\) real, then for \(\alpha=i\), then for \(\alpha\) complex.) (d) Now suppose \(X\) is a normed linear space. For \(\varphi\) and \(u\) related as above, show that \(\|\varphi\|=\|u\|\).

Show that the quotient \(X / M\) of a Banach space \(X\) by a closed subspace \(M\) is a Banach space. (Begin by showing that $$ \|x+M\| \equiv \inf \\{\|x+m\|: m \in M\\} $$ is a norm on \(X / M\).)

Let \(L_{\mathbb{R}}^{\infty}\) be the space of real-valued essentially bounded functions on \([0,1]\) with respect to Lebesgue measure. Let \(M\) be the subspace of constant functions. Define \(f: M \rightarrow \mathbb{R}\) by \(f(c)=c\), where on the left hand side, \(c\) denotes the constant function with value \(c\). Let \(g_{0}(x)=x\), and set \(N \equiv\left\\{c+\operatorname{tg}_{0}: c \in M, t \in \mathbb{R}\right\\}\). The proof of the one-step extension process in the Hahn- Banach theorem tells you how to find all linear \(F: N \rightarrow \mathbb{R}\) so that \(F\) extends \(f\) and \(\|F\|=\|f\|\). Find all such \(F\).

This problem outlines a proof of the statement: If \(X\) is a Banach space and \(T \in \mathscr{B}(X)\) is such that \(X / T X\), as a vector space, is finite- dimensional, then \(T X\) is closed. (a) Argue that since the map \(A: X /\) ker \(T \rightarrow X\) defined by \(A(x+\operatorname{ker} T)=T x\) is oneto-one and has the same range as \(T\), we may assume without loss of generality that \(T\) is one-to-one. (b) Suppose that \(X / T X\) has dimension 1 , so that there exists \(y \in X\) such that \(X=\) \(\\{T x+\alpha y: x \in X, \alpha \in \mathbb{C}\\}\). Show that the map \(S\) defined on \(X \times \mathbb{C}\) in the onenorm by \(S(x, \alpha)=T x+\alpha y\) is continuous and bijective. Use this to show that \(T\) is bounded below and thus has closed range. (c) Prove the full result.

Give an example of a diagonal operator \(T: \mathscr{H} \rightarrow \mathscr{H}\) whose range is not closed.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.