An operator A on a Hilbert space H over K is said to be compactif for every bounded sequence (x₂) in H, the sequence (A(x₂)) containsa subsequence which converges in H. (This definition is equivalent tothe one given earlier. See 17.1(a).)If A is a compact operator on H, then there is some a > 0 suchthat ||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. Thenthe sequence (A(x)) has no convergent subsequence although (x₂)is a bounded sequence. This shows that every compact operator isbounded. However, the converse is not true. For example, if H isinfinite dimensional, then the identity operator I is clearly bounded,but it is not compact. For if (un) is an infinite orthonormal sequencein H, then ||un|| = 1 but ||un − Um|| = √√2 for all n ‡ m, so that thebounded sequence (I(un)) has no convergent subsequence.Requestexplain

Kindly refer to Step 2 for the solution.

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

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

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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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

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