Notes on Fredholm operators

(2) If K ∈ B(X) is compact, then for all λ ∈ C \ {0}, K − λ1 is Fredholm with index zero. (3) The shift operator S± ∈ B(`p) for 1 ≤ p ≤ ∞ defined by (S±x)n = xn±1 is Fredholm with index ±1. (4) If X,Y are finite dimensional and T ∈ B(X,Y ), then by the Rank-Nullity Theorem, ind(T ) = dim(X)− dim(Y ). Lemma 3. Suppose E,F ⊆ X are closed subspaces with F finite dimensional. (1) The subspace E + F ⊆ X is closed. (2) If in addition E is complemented, then E + F ⊆ X is complemented. Proof. Consider the canonical surjection Q : X → X/E. Then QF ⊆ X/E is finite dimensional and thus closed, and E + F = Q−1(QF ) is closed since Q is continuous. Suppose now that E is complemented, and let P ∈ B(X) be an idempotent with PX = E. Then (1− P )F ⊆ (1− P )X is finite dimensional, and thus complemented. Let Q0 ∈ B((1− P )X) be an idempotent with Q0(1−P )X = (1−P )F . Extend Q0 to Q ∈ B(X) by Q = 0 on PX. Then QP = PQ = 0, so P +Q ∈ B(X) is an idempotent with (P +Q)X = E + F . Proposition 4. Suppose T ∈ B(X,Y ) such that codim(TX) <∞. Then TX is closed. Proof. Pick y1, . . . , yn ∈ Y such that {y1+TX, . . . , yn+TX} are a basis for the vector space Y/TX, and define F = span{y1, . . . , yn} ⊂ Y . Now consider the Banach space Z = X/ kerT ⊕ F with the `1 norm ‖(x + kerT ) + f‖Z := ‖x + kerT‖X/ kerT + ‖f‖F . We define a linear map S : Z → Y by S((x + kerT ) + f) = Tx + f . It is straightforward to verify S is bounded and bijective, and thus invertible by the Open Mapping Theorem. Finally, we have S(X/ kerT ) = TX is closed as S is a closed map.