Show description

1). 1) is satisfied. 1) we have that ∞ 1 (f, gn )ϕn ϕ := µn 1 converges in the Hilbert space X. Applying A to this series we have that ∞ Aϕ = (f, gn )gn . 1 But, since f ∈ N (A∗ )⊥ , this is the singular value decomposition of f corresponding to the operator A∗ and hence Aϕ = f . Note that Picard’s theorem illustrates the ill-posed nature of the equation Aϕ = f . In particular, setting f δ = f + δgn we obtain a solution of Aϕδ = f δ given by ϕδ = ϕ + δϕn /µn . 14 we have that µn → 0. We say that Aϕ = f is mildly illposed if the singular values decay slowly to zero and severely ill-posed if they decay very rapidly (for example exponentially).

Let A : X → Y be an injective compact operator with dense range in Y , let f ∈ Y and δ > 0. Then there exists a smallest integer m such that ||ARm f − f || ≤ δ . Proof. Since A(X) = Y , A∗ is injective. Hence the singular value decomposition with the singular system (µn , gn , ϕn ) for A∗ implies that for every f ∈ Y we have that ∞ (f, gn )gn . 3) µn <µm as m → ∞. In particular, there exists a smallest integer m = m(δ) such that ||ARm f − f || ≤ δ. 3) we have that 2 2 2 ||ARm f − f || = ||f || − |(f, gn )| .

Hence µn = λn and the compact operator equation Aϕ = f is severely illposed. Picard’s theorem suggests trying to regularize Aϕ = f by damping or filtering out the influence of the higher order terms in the solution ϕ given by ∞ ϕ= 1 1 (f, gn )ϕn . µn The following theorem does exactly that. We will subsequently consider two specific regularization schemes by making specific choices of the function q that appears in the theorem. 9. Let A : X → Y be an injective compact operator with singular system (µn , ϕn , gn ) and let q : (0, ∞) × (0, ||A||] → R be a bounded function such that for every α > 0 there exists a positive constant c(α) such that |q(α, µ)| ≤ c(α)µ , 0 < µ ≤ ||A|| , and lim q(α, µ) = 1 α→0 , 0 < µ ≤ ||A|| .