Description
Q. #2
Compact operators in functional Analysis from the book a Course in Functional Analysis Second Edition
Unformatted Attachment Preview
Purchase answer to see full attachment
Explanation & Answer
I don't know what proof they meant, but I have mine:)
𝜑 ∈ 𝐿∞ [0, 1], 𝑀𝜑 : 𝐿2 [0, 1] → 𝐿2 [0, 1], (𝑀𝜑 𝑓)(𝑥) = 𝜑(𝑥)𝑓(𝑥).
Prove that 𝑀𝜑 is compact iff 𝜑 = 0 (a.e.).
Proof. It is obvious that if 𝜑 = 0 then 𝑀𝜑 is compact. Prove the opposite direction by contradiction, let
𝜑 ≠ 0. Then there is some 𝜀0 > 0: 𝜇({𝑥: |𝜑(𝑥) ≥ 𝜀0 |}) > 0.
[ if there are no such epsilon, then 𝜇({𝑥: |𝜑(𝑥) > 0|}) = ∑ 𝜇 ({𝑥:
1
2𝑛
≥ |𝜑(𝑥) >
1
|})
2𝑛+1
=0]
Denote the set {𝑥: |𝜑(𝑥) ≥ 𝜀0 |} as 𝐴 and the subspace 𝐿2 [𝐴] = {𝑓 ∈ 𝐿2 [0, 1]: 𝑓(𝐴𝐶 ) = 0}.
This entire subspace lies in the image of 𝑀𝜑 , and there is an inverse operator
𝑓(𝑥)⁄𝜑(𝑥) , 𝑥 ∈ 𝐴
𝑀1⁄𝜑 : 𝐿2 [𝐴] → 𝐿2 [0, 1], (𝑀1⁄𝜑 𝑓)(𝑥) = {
.
0, 𝑥 ∉ 𝐴
Both 𝑀𝜑 and 𝑀1⁄𝜑 are linear bounded operators, so 𝐿2 [𝐴] and 𝐿2 [0, 1] are isomorphic.
The space 𝐿2 [𝐴] is infinite-dimensional, so its closed ball is not norm-compact, so the operator 𝑀𝜑 is not
a compact operator. ∎
𝜑 ∈ 𝐿∞ [0, 1], 𝑀𝜑 : 𝐿2 [0, 1] → 𝐿2 [0, 1], (𝑀𝜑 𝑓)(𝑥) = 𝜑(𝑥)𝑓(𝑥).
Prove that 𝑀𝜑 is compact iff 𝜑 = 0 (a.e.).
Proof. It is obvious that if 𝜑 = 0 then 𝑀𝜑 is compact. Prove the opposite direction by contradiction, let
𝜑 ≠ 0. Then there is some 𝜀0 > 0: 𝜇({𝑥: |𝜑(𝑥) ≥ 𝜀0 |}) > 0.
[ if there are no such epsilon, then 𝜇({𝑥: |𝜑(𝑥...
Review
Review
24/7 Homework Help
Stuck on a homework question? Our verified tutors can answer all questions, from basic math to advanced rocket science!
Similar Content
Related Tags
Ezperanza Rising
by Pam Muñoz Ryan
A Portrait of the Artist as a Young Man
by James Joyce
Normal People
by Sally Rooney
You Are a Badass
by Jen Sincero
The Power of Habit - Why We Do What We Do in Life and Business
by Charles Duhigg
Black Beauty
by Anna Sewell
Where the Crawdads Sing
by Delia Owens
Tess of the DUrbervilles
by Thomas Hardy