An operator A on a Hilbert space H over K is said to be compact if for every bounded sequence (x₂) in H, the sequence (A(x₂)) contains a subsequence which converges in H. (This definition is equivalent to the one given earlier. See 17.1(a).) If A is a compact operator on H, then there is some a > 0 such that ||A(x)|| ≤ a for all ¤ ¤ H with ||x|| ≤ 1. For otherwise, we can find some xn ¤ H with ||xn|| ≤ 1 and ||A(x₁)|| > n. Then the sequence (A(x₂)) has no convergent subsequence although (x₂) is a bounded sequence. This shows that every compact operator is bounded. However, the converse is not true. For example, if H is infinite dimensional, then the identity operator I is clearly bounded, but it is not compact. For if (un) is an infinite orthonormal sequence in H, then ||un|| = 1 but ||un − Um|| = √√2 for all n ‡ m, so that the bounded sequence (I(un)) has no convergent subsequence. - Request explain

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 32EQ
icon
Related questions
Question
100%
An operator A on a Hilbert space H over K is said to be compact
if for every bounded sequence (x₂) in H, the sequence (A(x₂)) contains
a subsequence which converges in H. (This definition is equivalent to
the one given earlier. See 17.1(a).)
If A is a compact operator on H, then there is some a > 0 such
that ||A(x)|| ≤ a for all x ¤ H with ||x|| ≤ 1. For otherwise, we
¤
can find some în € H with ||xn|| ≤ 1 and ||A(x₁)|| > n. Then
the sequence (A(x)) has no convergent subsequence although (x₂)
is a bounded sequence. This shows that every compact operator is
bounded. However, the converse is not true. For example, if H is
infinite dimensional, then the identity operator I is clearly bounded,
but it is not compact. For if (un) is an infinite orthonormal sequence
in H, then ||un|| = 1 but ||un − Um|| = √√2 for all n ‡ m, so that the
bounded sequence (I(un)) has no convergent subsequence.
Request
explain
Transcribed Image Text:An operator A on a Hilbert space H over K is said to be compact if for every bounded sequence (x₂) in H, the sequence (A(x₂)) contains a subsequence which converges in H. (This definition is equivalent to the one given earlier. See 17.1(a).) If A is a compact operator on H, then there is some a > 0 such that ||A(x)|| ≤ a for all x ¤ H with ||x|| ≤ 1. For otherwise, we ¤ can find some în € H with ||xn|| ≤ 1 and ||A(x₁)|| > n. Then the sequence (A(x)) has no convergent subsequence although (x₂) is a bounded sequence. This shows that every compact operator is bounded. However, the converse is not true. For example, if H is infinite dimensional, then the identity operator I is clearly bounded, but it is not compact. For if (un) is an infinite orthonormal sequence in H, then ||un|| = 1 but ||un − Um|| = √√2 for all n ‡ m, so that the bounded sequence (I(un)) has no convergent subsequence. Request explain
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question

In the last step how is ||un-um|| equal to root 2?

Solution
Bartleby Expert
SEE SOLUTION
Recommended textbooks for you
Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Elements Of Modern Algebra
Elements Of Modern Algebra
Algebra
ISBN:
9781285463230
Author:
Gilbert, Linda, Jimmie
Publisher:
Cengage Learning,
College Algebra
College Algebra
Algebra
ISBN:
9781938168383
Author:
Jay Abramson
Publisher:
OpenStax