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Chapter 5: Problem 47

Let \(\mathscr{H}\) be a Hilbert space. An operator \(T\) in \(\mathscr{B}(\mathscr{H})\) is said to be a contraction if \(\|T\| \leq 1\). (a) Show that \(T\) is a contraction if and only if \(I-T^{*} T \geq 0\). (b) Suppose that \(A\) and \(B\) are bounded linear operators on \(\mathscr{H}\) with \(B\) invertible. Show that \(A B^{-1}\) is a contraction if and only if \(A^{*} A \leq B^{*} B\).

Short Answer

Expert verified
(a) \(T\) is a contraction if and only if \(I-T^{*} T \geq 0\). (b) \(A B^{-1}\) is a contraction if and only if \(A^{*} A \leq B^{*} B\).

Step by step solution

01

Understanding a Contraction

Recall that an operator \(T\) on a Hilbert space \(\mathscr{H}\) is a contraction if its operator norm satisfies \(\|T\| \leq 1\). This means that for all \(x \in \mathscr{H}\) with \(\|x\| = 1\), it holds that \(\|Tx\| \leq \|x\|\).
02

Showing \( T \) Contraction Equivalence

To prove part (a), we need to show \( T \) is a contraction if and only if \( I - T^*T \geq 0 \). Start by noting that \(\|T\| \leq 1\) implies for any \( x \in \mathscr{H} \), we have \( \langle Tx, Tx \rangle \leq \|x\|^2 \). Expanding, this becomes \( \langle T^*Tx, x \rangle \leq \langle x, x \rangle \), or \( \langle (I - T^*T)x, x \rangle \geq 0 \) for all \( x \), showing \( I - T^*T \geq 0 \). The reverse direction follows similarly.
03

Conditions on Bounded Invertible Operators

Now consider part (b). Recall that if \(B\) is bounded and invertible, \(B^{-1}\) also exists and is bounded. We want \(A B^{-1}\) to be a contraction, i.e., \(\|AB^{-1}\| \leq 1\). This implies the condition \((AB^{-1})^*(AB^{-1}) \leq I\), where \(I\) is the identity operator.
04

Using Equivalent Expression

For \(A B^{-1}\) to be a contraction, \((AB^{-1})^*(AB^{-1}) = B^{-1*}A^*AB^{-1} \leq I\) must hold, which is equivalent to \(A^*A \leq B^*B\). This gives the necessary and sufficient condition for \(A B^{-1}\) to be a contraction.

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Key Concepts

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

Hilbert Space
Hilbert spaces are fundamental in functional analysis. They are complete inner product spaces, which means that the space is not only equipped with an inner product that allows measurement of angles and lengths, but also every Cauchy sequence converges within the space. This makes Hilbert spaces a natural generalization of Euclidean spaces to potentially infinite-dimensional spaces. In a Hilbert space \(\mathscr{H}\), we can analyze the properties of vectors and operators with ease, thanks to the structure provided by the inner product.
  • The inner product \(\langle x, y \rangle\) for vectors \(x\) and \(y\) offers notions of orthogonality and the concept of projection, similar to those in 2D or 3D Euclidean spaces.
  • Completeness ensures that if a sequence of vectors is tending to a limit, that limit indeed exists within the space.
Hilbert spaces find applications in quantum mechanics, signal processing, and much more, providing a framework to study linear operators' behavior.
Bounded Linear Operators
Bounded linear operators are operators \(T: \mathscr{H} \to \mathscr{H}\) that remain bounded in their action. This means there exists a constant \(C\) such that for all vectors \(x\) in the Hilbert space, the inequality \(\|Tx\| \leq C\|x\|\) holds.Characteristics of bounded linear operators include:
  • Linearity: The operator respects vector addition and scalar multiplication, meaning \(T(ax + by) = aT(x) + bT(y)\).
  • Boundedness: There is a maximum "stretching" factor \(C\), beyond which the operator cannot increase a vector's norm, demonstrating stability.
These operators are essential in analysis because they ensure continuity of operations. In a way, boundedness can be compared to a "disciplined" operation on vectors that maintains predictability in many mathematical settings.
Operator Norm
The operator norm is a measure of how much an operator can "stretch" a vector in a Hilbert space. For an operator \(T\), the operator norm is defined as \(\|T\| = \sup_{\|x\| = 1} \|Tx\|\), meaning it is the maximum length \(T\) can stretch any unit vector.This measure plays a key role in analyzing operators because:
  • It gives a clear maximum bound for the change in vector lengths, guiding stability analysis.
  • An operator is a contraction if its operator norm \(\|T\|\) is \(\leq 1\), meaning it does not increase the length of any vector.
Understanding the operator norm is crucial in many areas, including systems theory and numerical analysis, where stability and control predict outcomes in dynamic systems.
Contraction Mapping
A contraction mapping is an operator that brings points closer together, ensuring the existence of a unique fixed point in certain contexts. These mappings, defined for an operator \(T\) with \(\|T\| \leq 1\), guarantee that applying \(T\) always results in a contraction of distances between any two points.Key features and implications include:
  • A fixed point of a contraction is a point \(x\) for which \(Tx = x\), existing uniquely under certain conditions.
  • The contraction mapping principle helps in proving the existence and uniqueness of solutions to various equations, such as differential equations via the Banach fixed-point theorem.
Given its robustness, contraction mappings are applied in numerous mathematical and practical problems to assure stability and convergence.

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Most popular questions from this chapter

Suppose that \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) are two norms on a \(*\)-algebra \(\mathscr{A}\), each of which make \(\mathscr{A}\) into a \(C^{*}\)-algebra. Show \(\|\cdot\|_{1}=\|\cdot\|_{2}\).

Suppose that \(P\) and \(Q\) are orthogonal projections onto closed subspaces \(M\) and \(N\) in \(\mathscr{H}\), respectively. Show that \(P \geq Q\) if and only if \(N \subseteq M\).

Suppose \(\mathscr{A}\) and \(\mathscr{B}\) are Banach algebras with common identity and \(\mathscr{B} \subseteq \mathscr{A}\). Show \(\sigma_{\mathscr{I}}(A) \subseteq \sigma_{\mathscr{B}}(A)\) and \(\partial \sigma_{\mathscr{D}}(A) \subseteq \partial \sigma_{\mathscr{I}}(A)\), for any \(A \in \mathscr{B}\). Hint for the second part: Since the first part implies that the interior of \(\sigma_{\sigma f}(A)\) is contained in the interior of \(\sigma_{\mathscr{B}}(A)\) for any \(A\) in \(\mathscr{B}\), argue first that it suffices to show that if \(\lambda \in \partial \sigma_{\mathscr{D}}(A)\), then \(\lambda \in \sigma_{\sigma d}(A)\).

Suppose that \(0 \leq A \leq B\) for self-adjoint elements \(A, B\) in a \(C^{*}\) algebra. (a) Show that \(B \leq\|B\| I\). Hint: Consider \(C^{*}(B) \cong C(\sigma(B))\) where \(B\) corresponds to the identity function on the spectrum of \(B\). Use the functional calculus, with the function \(f(x)=\|B\|-x\) on \(\sigma(B)\). (b) Show \(\|A\| \leq\|B\| .\) Hint: Consider \(C^{*}(A) \cong C(\sigma(A))\) with \(A\) corresponding to the identity function on the spectrum of \(A\). Use the functional calculus with the function \(f(x)=\|B\|-x\). (c) Show that \(0 \leq A \leq B\) need not imply \(A^{2} \leq B^{2}\) by considering $$ X=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] $$ and $$ Y=\left[\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array}\right] $$ Show \(0 \leq X \leq X+Y\). Is \(X^{2} \leq(X+Y)^{2}\) ? (d) Show that if \(0 \leq A \leq B\) and \(A\) and \(B\) commute, then \(A^{n} \leq B^{n}\) for every positive integer \(n\). More generally, show that if there are positive elements \(C_{j}, 1 \leq j \leq k\), with $$ 0 \leq A \leq C_{1} \leq C_{2} \leq \cdots \leq C_{k} \leq B $$ so that any two neighbors in this list commute, then \(A^{n} \leq B^{n}\) for any positive integer \(n\).

Find the norm of the operator on \(\mathscr{B}(\mathscr{H})\) where \(\mathscr{H}=\mathbb{C}^{2}\) which is given by the matrix $$ \left[\begin{array}{ll} a & b \\ c & d \end{array}\right] $$ Give your answer in terms of the numbers \(S=|a|^{2}+|b|^{2}+|c|^{2}+|d|^{2}\) and \(D=\) \(a d-b c\).
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