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Landweber iterates and residuals are spectral-filter iterates

Statement

Assume Countable Choice. Let $A : X \to Y$ and $ω$ satisfy the hypotheses of Landweber iteration with discrepancy and budget statuses and put $T = A^{*} A$ . For each integer $k \geq 0$ , define $g_{0} (t) = 0$ and $g_{k} (t) = ω \sum_{j = 0}^{k - 1} (1 - ω t)^{j}$ for $k \geq 1$ , with $r_{k} (t) = (1 - ω t)^{k}$ . For the associated infinite mathematical path, $x_{k} = g_{k} (T) A^{*} y_{δ}, y_{δ} - A x_{k} = r_{k} (A A^{*}) y_{δ} .$ If $y_{δ} = A x^{†}$ , then $x^{†} - x_{k} = r_{k} (T) x^{†} .$ Here $g_{k} (T)$ and $r_{k} (T)$ are the compact spectral multipliers defined in Compact spectral filters, source conditions, and qualification; $r_{k} (A A^{*})$ is the multiplier with scalar function $t \mapsto (1 - ω t)^{k}$ on the compact self-adjoint operator $A A^{*}$ , as supplied by Bounded multipliers on compact self-adjoint eigenspaces define bounded operators.

Facts & Assumptions

Given: Countable Choice, compact nonzero $A : X \to Y$ , $0 < ω < 2 / ∥ A ∥^{2}$ , the algorithm's initial value $x_{0} = 0$ , and $T = A^{*} A$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is assumed by the Hilbert-adjoint and compact spectral multiplier suppliers used here.

[F1]

Landweber iteration with discrepancy and budget statuses: Its associated infinite path is defined for every $j \geq 0$ by $x_{j + 1} = x_{j} + ω A^{*} (y_{δ} - A x_{j})$ ; a finite run computes only a prefix.

[F2]

A compact Hilbert-space Gram operator is positive and has the forward kernel: Both $A^{*} A$ and $A A^{*}$ are compact, positive, and self-adjoint, with $\ker (A^{*} A) = \ker A$ .

[F3]

Bounded multipliers on compact self-adjoint eigenspaces define bounded operators: Every bounded scalar function on the nonzero eigenspaces of a compact self-adjoint operator defines a bounded spectral multiplier.

[F4]

Compact spectral filters, source conditions, and qualification: For $T = A^{*} A$ , compact spectral multipliers include the residual filter $r_{k} (t) = (1 - ω t)^{k}$ .

[F5]

The Hilbert-space adjoint of a bounded operator: $⟨ A x, y ⟩_{Y} = ⟨ x, A^{*} y ⟩_{X}$ .

Proof

Given: The hypotheses above and the exact recurrence [F1].

1.1A1givenF1F5inductionalgebra

The adjoint identity [F5] gives $A^{*} A = T$ . Thus [F1] rewrites each update before stopping as $x_{j + 1} = (I - ω T) x_{j} + ω A^{*} y_{δ}$ . Starting from $x_{0} = 0$ , induction gives $x_{k} = ω \sum_{j = 0}^{k - 1} (I - ω T)^{j} A^{*} y_{δ}$ : the sum is empty for $k = 0$ , and multiplying the $k$ -term sum by $I - ω T$ and adding $ω A^{*} y_{δ}$ produces the $(k + 1)$ -term sum.

2.1A1F1F2F3step 1.1inductionalgebra

Put $ρ_{k} = y_{δ} - A x_{k}$ . Substitution of [F1] gives $ρ_{k + 1} = ρ_{k} - ω A A^{*} ρ_{k} = (I - ω A A^{*}) ρ_{k}$ . Induction from $ρ_{0} = y_{δ}$ yields $ρ_{k} = (I - ω A A^{*})^{k} y_{δ}$ . By [F2], $A A^{*}$ is compact positive self-adjoint, and [F3] therefore interprets this polynomial as the stated compact spectral multiplier. The scalar geometric identity $1 - (1 - ω t)^{k} = ω t \sum_{j = 0}^{k - 1} (1 - ω t)^{j}$ also identifies the vector expression from step 1.1 with $g_{k} (T) A^{*} y_{δ}$ .

3.1A1F2F4step 1.1algebra∎

If $y_{δ} = A x^{†}$ , then [step 1.1] and $A^{*} y_{δ} = T x^{†}$ give $x_{k} = ω \sum_{j = 0}^{k - 1} (I - ω T)^{j} T x^{†} = [I - (I - ω T)^{k}] x^{†}$ by the same polynomial identity. Since [F2] makes $T$ compact positive self-adjoint, [F4] interprets $r_{k} (T)$ as its residual multiplier; hence $x^{†} - x_{k} = (I - ω T)^{k} x^{†} = r_{k} (T) x^{†}$ .

Depends on

Used by

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Sources

Per Christian Hansen, Introduction to Inverse Problems, Chapter 6: Iterative Methods (standard reference, not scraped)
Christian Clason, Regularization of Inverse Problems, full lecture notes (standard reference, not scraped)