Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 1.
FREDHOLM THEORY
1.2 Preliminaries
Let X and Y be complex Banach spaces. Write B(X, Y ) for the set of bounded linear
operators from X to Y and abbreviate B(X, X) to B(X). If T ∈ B(X) write ρ(T ) for
the resolvent set of T ; σ(T ) for the spectrum of T ; π 0 (T ) for the set of eigenvalues
of T .
We begin with:
Definition 1.2.1. Let X be a normed space and let X ∗ be the dual space of X. If
Y is a subset of X, then
Y ⊥ = {f ∈ X ∗ : f(x) = 0 for all x ∈ Y } = {f ∈ X ∗ : Y ⊂ f −1 (0)}
is called the annihilator of Y . If Z is a subset of X ∗ then
. ⊥ Z = {x ∈ X : f(x) = 0 for all f ∈ Z} = ∩
is called the back annihilator of Z.
f∈Z
f −1 (0)
Even if Y and Z are not subspaces, and Y ⊥ and . ⊥ Z are closed subspaces.
Lemma 1.2.2. Let Y, Y ′ ⊂ X and Z, Z ′ ⊂ X ∗ . Then
(a) Y ⊂ . ⊥ (Y ⊥ ), Z ⊂ ( ⊥ Z) ⊥ ;
(b) Y ⊂ Y ′ =⇒ (Y ′ ) ⊥ ⊂ Y ⊥ ; Z ⊂ Z ′ =⇒ . ⊥ (Z ′ ) ⊂ . ⊥ Z;
(c) ( ⊥ (Y ⊥ )) ⊥ = Y ⊥ , . ⊥ (( ⊥ Z) ⊥ ) = ⊥ Z;
(d) {0} ⊥ = X ∗ , X ⊥ = {0}, . ⊥ {0} = X.
Proof. This is straightforward.
Theorem 1.2.3. Let M be a subspace of X. Then
(a) X ∗ /M ⊥ ∼ = M ∗ ;
(b) If M is closed then (X/M) ∗ ∼ = M ⊥ ;
(c) . ⊥ (M ⊥ ) = cl M.
Proof. See [Go, p.25].
Theorem 1.2.4. If T ∈ B(X, Y ) then
(a) T (X) ⊥ = (T ∗ ) −1 (0);
(b) cl T (X) = . ⊥ (T ∗−1 (0));
(c) T −1 (0) ⊂ . ⊥ T ∗ (Y ∗ );
(d) cl T ∗ (Y ∗ ) ⊂ T −1 (0) ⊥ .
Proof. See [Go, p.59].
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