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Chapter 2: Problem 2

Suppose \(M\) is a dense subspace in a Banach space \(X\) (meaning that the closure of \(M\) is all of \(X\) ) and suppose that \(T: M \rightarrow Y\) is linear, where \(Y\) is a Banach space, with \(\|T m\|_{Y} \leq K\|m\|_{X}\) for some \(K<\infty\) and all \(m \in M\). Show that \(T\) extends, in a unique way, to a bounded linear operator from \(X\) into \(Y\).

Short Answer

Expert verified
The operator \(T\) uniquely extends to a bounded linear operator from \(X\) to \(Y\).

Step by step solution

01

Understand the Problem

We are given a dense subspace \(M\) within a Banach space \(X\) with a linear operator \(T: M \rightarrow Y\) such that \(\|T m\|_{Y} \leq K\|m\|_{X}\) for all \(m \in M\). The task is to show that \(T\) can be extended uniquely to a bounded linear operator on the whole space \(X\).
02

Use the Density of M

Since \(M\) is dense in \(X\), for every \(x \in X\), there exists a sequence \((m_n) \subset M\) such that \(m_n \rightarrow x\) in \(X\). This means for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), \(\|m_n - x\|_X < \epsilon\).
03

Define the Extension

Define \(T: X \rightarrow Y\) for any \(x \in X\) as \(T(x) = \lim_{n \to \infty} T(m_n)\) where \(m_n \rightarrow x\). We need to show that this limit exists and does not depend on the chosen approximating sequence \((m_n)\).
04

Verify the Limit Exists

Given \(T\) is bounded on \(M\), we have \(\|T(m_n) - T(m_k)\|_Y \leq K \|m_n - m_k\|_X\). Since \(m_n\) is Cauchy and \(T\) is bounded, \(T(m_n)\) is also Cauchy in \(Y\) and converges in \(Y\) as \(Y\) is complete.
05

Show Uniqueness and Independence

Suppose \((m_n)\) and \((m'_n)\) are two sequences such that \(m_n \rightarrow x\) and \(m'_n \rightarrow x\). Then, \(\|T(m_n) - T(m'_n)\|_Y \leq K\|m_n - m'_n\|_X\). As both sequences converge to \(x\), \(\|m_n - m'_n\|_X \rightarrow 0\), hence \(\|T(m_n) - T(m'_n)\|_Y \rightarrow 0\). Thus, the limit defining \(T(x)\) is unique, independent of the choice of sequence.
06

Check Linearity and Boundedness

For any \(x, y \in X\) and scalars \(a, b\), choose sequences \(m_n \rightarrow x\) and \(n_n \rightarrow y\). We have \(T(ax + by) = \lim (aT(m_n) + bT(n_n)) = a\lim T(m_n) + b\lim T(n_n) = aT(x) + bT(y)\), so \(T\) is linear. Boundedness follows as \(\|T(x)\|_Y = \lim \|T(m_n)\|_Y \leq K\lim \|m_n\|_X = K\|x\|_X\).
07

Conclusion

Thus, the extension \(T: X \rightarrow Y\) is a bounded linear operator, and due to the uniqueness in extension, it is the only such extension from \(M\) to \(X\).

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Key Concepts

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

Banach Space
In the realm of functional analysis, a **Banach space** is a vector space equipped with a notion of distance, via a norm, in which every Cauchy sequence converges to a limit within the space. This means a Banach space is complete 鈥� a crucial property that ensures all limits of convergent sequences are also contained within the space. Banach spaces are named after the Polish mathematician Stefan Banach, who also contributed significantly to the development of functional analysis.

Key characteristics of Banach spaces include:
  • **Normed Structure**: Every Banach space is a normed vector space, where the norm provides a way to measure the size or length of a vector.
  • **Completeness**: This property guarantees that sequences of vectors that "seem" to have limits actually do so within the space.
  • **Versatility**: Banach spaces encompass a wide variety of function spaces, including spaces of continuous functions, sequences, and even other more complex function settings.
Understanding Banach spaces is fundamental in the study of functional analysis and various applications in mathematical physics and differential equations.
Dense Subspace
A **dense subspace** within a Banach space is a subset whose closure equals the entire space. In simpler terms, for any point in the Banach space, you can find points arbitrarily close from the dense subspace. This concept is important as it allows properties and results proven for the dense subspace to extend to the whole space.

When working with dense subspaces, keep these points in mind:
  • Any element of the Banach space can be approximated as closely as desired by elements from the dense subspace.
  • A set of functions is dense in a Banach space if linear combinations of those functions can approximately create any function in the space to any degree of accuracy desired.
  • Density plays a crucial role in the extension of linear operators, allowing operators defined on a dense subset to be extended to the entire space, as seen in the given exercise.
By appreciating the role of dense subspaces, you gain insight into how operations within a smaller subset may influence the entire Banach space.
Linear Operators
**Linear operators** are mappings between vector spaces that preserve the operations of addition and scalar multiplication. In mathematical notation, a linear operator \( T \) between two vector spaces \( V \) and \( W \) satisfies:

1. \( T(u + v) = T(u) + T(v) \) for all \( u, v \in V \)
2. \( T(cu) = cT(u) \) for all \( u \in V \) and scalars \( c \)

Key features to remember about linear operators:
  • A linear operator establishes a direct proportionality between inputs and outputs, maintaining the fundamental structure of the spaces involved.
  • Linear operators are foundational in understanding transformations in data, signals, and systems, making them crucial in computations involving matrices or transformations.
  • The study of linear operators involves evaluating their characteristics such as kernel, range, and invertibility.
Understanding linear operators helps in visualizing and computing effects on transformations across different types of vector spaces.
Bounded Linear Operators
A **bounded linear operator** is a linear operator between two normed vector spaces that maps bounded sets to bounded sets. Formally, a linear operator \( T: X \rightarrow Y \) is bounded if there exists a constant \( K \) such that for all \( x \in X \), \( \|T(x)\|_Y \leq K\|x\|_X \). This property ensures the operator is continuous; continuity is guaranteed in normed spaces if boundedness is present.

Essentials of bounded linear operators include:
  • **Continuity**: These operators are continuous throughout their domain because of their bounded nature.
  • **Operator Norm**: The smallest such \( K \) such that \( \|T(x)\|_Y \leq K\|x\|_X \) for all \( x \) is sometimes referred to as the operator's norm. This gives a measure of the "size" of the operator.
  • **Importance in Analysis**: Bounded linear operators are significant in functional analysis as they enable easier manipulation and study due to their continuity properties.
The concept of bounded linear operators is pivotal in extending linear operators from dense subspaces to entire spaces, a theme central to the exercise provided.

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Most popular questions from this chapter

For an operator \(T\) in \(\mathscr{B}(\mathscr{H})\), where \(\mathscr{H}\) is a Hilbert space, show that ker \(T=\) \(\left(\operatorname{ran} T^{*}\right)^{\perp}\), where \(\operatorname{ker} T=\\{h: T h=0\\}\) and \(\operatorname{ran} T^{*} \equiv \operatorname{range} T^{*}=\left\\{T^{*} h: h \in \mathscr{H}\right\\} .\)

Let \(X=[0,1]\) or more generally any compact Hausdorff space, and let \(\mathscr{Y}=\) \(C(X)\), the Banach space of continuous, complex-valued functions on \(X\), in the supremum norm. For any continuous function \(\varphi\) mapping \(X\) into itself, define the composition operator \(C_{\varphi}\) on \(\mathscr{Y}\) by \(C_{\varphi}(f)=f \circ \varphi\). Prove that \(C_{\varphi}\) is a bounded linear operator on \(\mathscr{Y}\). For which \(\varphi\) is \(C_{\varphi}\) invertible?

Show that a diagonal operator on a Hilbert space is an orthogonal projection if and only if its diagonal consists of 0 's and 1 's.

Show that a linear surjective isometry from one Hilbert space \(\mathscr{H}\) to another Hilbert space \(\mathscr{X}\) is unitary.

Suppose \(T\) is a bounded linear operator on a Hilbert space \(\mathscr{H}\) and suppose further that the range of \(T\) is one-dimensional. Show that there are vectors \(x\) and \(y\) in \(\mathscr{H}\) so that $$ T z=\langle z, x\rangle y $$ for all \(z \in \mathscr{H}\). This operator is sometimes written as \(y \otimes x\). Identify \(T^{*}\) in this case.
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