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Hilbert-space Tikhonov amplifies data by at most $1 / (2 \sqrt{λ})$

Statement

Assume Countable Choice. Let $X, Y$ be real or complex Hilbert spaces and let $A : X \to Y$ be bounded, with Hilbert adjoint $A^{*}$ . For $λ > 0$ , put $R_{λ} = (A^{*} A + λ I_{X})^{- 1} A^{*}$ , which is well-defined by The shifted Hilbert-space normal operator is invertible with inverse norm at most $1 / λ$ . Then for every $g \in Y$ ,

$∥ R_{λ} g ∥ \leq \frac{∥ g ∥}{2 \sqrt{λ}}, ∥ R_{λ} ∥ \leq \frac{1}{2 \sqrt{λ}} .$

The proof uses the normal equation and the Hilbert-space inner product; it does not use a compact-operator convergence rate or a source condition.

Facts & Assumptions

Given: Countable Choice, Hilbert spaces $X, Y$ , a bounded linear map $A : X \to Y$ , its Hilbert adjoint $A^{*}$ , $λ > 0$ , and $g \in Y$ .

[F1]

The shifted Hilbert-space normal operator is invertible with inverse norm at most $1 / λ$ : For $B_{λ} = A^{*} A + λ I_{X}$ , the inverse exists and $∥ B_{λ}^{- 1} ∥ \leq 1 / λ$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is the choice assumption used to define the Hilbert adjoint in this argument; full Choice is not used.

[F2]

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

[F3]

Real and complex inner product spaces, with the inner product linear in the first argument: The inner product is linear in its first argument and conjugate symmetric.

[F4]

The induced length is a norm: The induced norm satisfies $∥ v ∥^{2} = ⟨ v, v ⟩$ .

[F5]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ .

[F6]

The operator norm as the least bound and as the unit-sphere or unit-ball supremum: For bounded $T$ , its operator norm is the supremum of $∥ T z ∥$ over $∥ z ∥ \leq 1$ .

Proof

1.1givenA1F1F2F3F4algebra

Let $x = R_{λ} g$ . Under [A1], [F2] gives the Hilbert adjoint used in the definition of $R_{λ}$ . By [F1], $(A^{*} A + λ I_{X}) x = A^{*} g$ . Pairing with $x$ and using [F2] gives $⟨ A^{*} A x, x ⟩_{X} = ⟨ A x, A x ⟩_{Y} = ∥ A x ∥^{2}$ and $⟨ A^{*} g, x ⟩_{X} = ⟨ g, A x ⟩_{Y}$ by [F2]–[F4]. Hence $∥ A x ∥^{2} + λ ∥ x ∥^{2} = ⟨ g, A x ⟩_{Y}$ .

2.1F5step 1.1algebra

The left side of step 1.1 is real and nonnegative. By [F5], with $s = ∥ A x ∥$ and $t = ∥ g ∥$ , step 1.1 gives $λ ∥ x ∥^{2} = ⟨ g, A x ⟩ - s^{2} \leq ∣ ⟨ g, A x ⟩ ∣ - s^{2} \leq t s - s^{2}$ . Completing the square gives $t s - s^{2} = \frac{t^{2}}{4} - (s - t / 2)^{2} \leq t^{2} / 4$ . Since $λ > 0$ , taking nonnegative square roots yields $∥ R_{λ} g ∥ = ∥ x ∥ \leq ∥ g ∥ / (2 \sqrt{λ})$ .

3.1F6step 2.1algebra∎

The pointwise inequality of step 2.1 holds for every $g \in Y$ . Taking the supremum over the unit ball in [F6] gives $∥ R_{λ} ∥ \leq 1 / (2 \sqrt{λ})$ .

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Sources

Guido Clason, Regularization of Inverse Problems, full lecture notes (standard reference, not scraped)
Per Christian Hansen, Introduction to Inverse Problems, Chapter 5: Regularization Parameter Choice (standard reference, not scraped)
Per Christian Hansen, Introduction to Inverse Problems, Chapter 4: Regularization Methods (standard reference, not scraped)

Hilbert-space Tikhonov amplifies data by at most 1/(2λ)1/(2λ​)