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Zngg909

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Q. #2

Compact operators in functional Analysis from the book a Course in Functional Analysis Second Edition

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Homework 4 1. Let K : L2 [0, 1] ! L2 [0, 1] be given by (Kf )(x) = (a) (b) (c) (d) Z 1 f (y)(x y)2 dy. 0 Show that K is self-adjoint. Show that ran K ✓ span{1, x, x2 }. Find an orthonormal basis for the range of K consisting of eigenvectors of K. Find kKk. 2. Let ' 2 L1 [0, 1]. Show that the multiplication operator M' : L2 [0, 1] ! L2 [0, 1] is compact if and only if '(x) = 0 for almost ever x 2 [0, 1]. [Hint: Find an orthonormal sequence in the range of M' .] 3. Let T be a compact operator on an infinite dimensional Hilbert space H. Prove the following: (a) If {en }1 n=1 is an orthonormal sequence, then lim kT en k = 0 n!1 (b) Show that T is not invertible, that is, there is no operator S 2 B(H) such that ST h = T Sh = h for all h 2 H. (c) Show that 0 is not necessarily in the point spectrum, that is, find a compact operator T so that 0 2 / p (T ). 4. Let H be a Hilbert space, and let M ⇢ H be a linear manifold. Let f : M ! C be a linear functional such that |f (x)|  kxk. Without using the Hahn-Banach theorem, show that there is a linear functional F : H ! C such that F (x) = f (x) for all h 2 M and |F (h)|  khk for all h 2 H. [Hint: First, define F on M . Show that F has the desired properties, then use the projection onto M to define F on all of H.] 1
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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...


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