Woo Young Lee Lecture Notes on Operator Theory

CHAPTER 1.
FREDHOLM THEORY
Theorem 1.7.2. (Atkinson’s Theorem) Let T ∈ B(X). Then
T is Fredholm ⇐⇒ [T ] is invertible in B(X)/K(X).
Proof. (⇒) If T is Fredholm then
∃ S ∈ B(X) such that ST − I and T S − I are compact.
Hence [S][T ] = [T ][S] = [I], so that [S] is the inverse of [T ] in the Calkin algebra.
(⇐) If [S][T ] = [T ][S] = [I] then
ST = I − K 1 and T S = I − K 2 ,
where K 1 , K 2 are compact operators. Thus T is Fredholm.
Let T ∈ B(X). The essential spectrum σ e (T ) of T is defined by
σ e (T ) = {λ ∈ C : T − λI is not Fredholm}
We thus have
σ e (T ) = σ B(X)/K(X) (T + K(X)).
Evidently σ e (T ) is compact. If dim X = ∞ then
σ e (T ) ≠ ∅
(because B(X)/K(X) ≠ ∅).
In particular, Theorem 1.6.2 implies that
σ e (T ) = σ e (T + K)
for every K ∈ K(X).
Theorem 1.7.3. If T ∈ B(X, Y ) then
T is Weyl ⇐⇒ ∃ a finite rank operator F such that T + F is invertible.
Proof. (⇒) Let T be Weyl and put
Since index T = 0, it follows that
X = T −1 (0) ⊕ X 0 and Y = T (X) ⊕ Y 0 .
dim T −1 (0) = dim Y 0 .
Thus there exists an invertible operator F 0 : T −1 (0) → Y 0 . Define F := F 0 (I − P ),
where P is the projection of X onto X 0 . Obviously, T + F is invertible.
(⇐) Assume S = T + F is invertible, where F is of finite rank. By Theorem 1.6.2,
T is Fredholm and index T = index S = 0.
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