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A compact diagonal model displays the Tikhonov bias-noise balance

Example

Assume Countable Choice and write $N_{+} = {1, 2, \dots}$ . Let $A$ be the compact diagonal map on $ℓ^{2} (N_{+}, R)$ given by $A e_{n} = n^{- 1} e_{n}$ . Fix $N \geq 2$ and $0 < ν \leq 1$ , take $w = e_{1} + e_{N}$ , source target $x^{†} = (A^{*} A)^{ν} w$ , noise $e = - N^{- (2 ν + 1)} e_{N}$ , and $λ = N^{- 2}$ . Then $∥ w ∥ = \sqrt{2}$ , the noise level is $δ = N^{- (2 ν + 1)}$ , and the exact Tikhonov bias and filtered noise each contribute $\frac{1}{2} N^{- 2 ν}$ in the $e_{N}$ direction. The total error is of order $N^{- 2 ν}$ , while the balancing parameter is comparable to $(δ / \sqrt{2})^{2 / (2 ν + 1)}$ . The same operator also exhibits saturation: for any $μ > 1$ , the unit source $w = e_{1}$ gives $x^{†} = e_{1}$ and exact-data bias $λ / (1 + λ)$ , which is not $O (λ^{μ})$ .

Facts & Assumptions

Given: Countable Choice, the real sequence Hilbert space $ℓ^{2} (N_{+}, R)$ , its standard basis $(e_{n})$ , the diagonal map $A e_{n} = n^{- 1} e_{n}$ , an integer $N \geq 2$ , an exponent $0 < ν \leq 1$ , and the data and parameter in the Statement; for the separate saturation calculation, also fix any exponent $μ > 1$ .

[F1]

Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ : The canonical sequence norm is $∥ x ∥_{2}^{2} = \sum_{n} ∣ x_{n} ∣^{2}$ , and $(e_{n})$ is orthonormal.

[F2]

The real square-summable sequence space is complete: The real $ℓ^{2}$ sequence space is complete.

[F3]

Hilbert space: A complete real inner-product space is a real Hilbert space.

[F4]

Bounded finite rank operators are compact: Bounded finite-rank maps are compact.

[F5]

Norm limit of compact operators is compact: A norm limit of compact operators into a Banach space is compact.

[F6]

Compact spectral filters, source conditions, and qualification: The source condition $S^{ν} w$ and Tikhonov filter use $S = A^{*} A$ and $g_{λ} (t) = 1 / (t + λ)$ .

[F7]

Hilbert-space Tikhonov amplifies data by at most $1 / (2 \sqrt{λ})$ : The filtered noise norm is bounded by $∥ e ∥ / (2 \sqrt{λ})$ .

[F8]

Tikhonov source-condition rates balance noise at qualification one: For $0 < ν \leq 1$ , source radius $R$ gives error bound $R λ^{ν} + δ / (2 \sqrt{λ})$ .

Verification

1.1givenF1F2F3F4F5algebra

Relabelling coordinates by $k \mapsto k + 1$ is an isometry from the complete space $ℓ^{2} (N, R)$ in [F2] to $ℓ^{2} (N_{+}, R)$ , so the latter is Hilbert. For $M \geq 1$ , let $A_{M} e_{n} = n^{- 1} e_{n}$ for $n \leq M$ and $A_{M} e_{n} = 0$ for $n > M$ . Since $n^{- 1} \leq 1$ , both $A$ and $A_{M}$ are bounded by one, and $A_{M}$ has finite rank. Also $∥ (A - A_{M}) x ∥_{2}^{2} = \sum_{n > M} n^{- 2} ∣ x_{n} ∣^{2} \leq (M + 1)^{- 2} ∥ x ∥_{2}^{2}$ , with equality at $x = e_{M + 1}$ . Thus $∥ A - A_{M} ∥ = 1 / (M + 1) \to 0$ . By [F3]–[F5], $A$ is a bounded compact operator on a Hilbert space.

1.2givenF1F6algebra

For every $x \in ℓ^{2}$ , the coordinate inner product gives $⟨ A x, e_{n} ⟩ = n^{- 1} x_{n} = ⟨ x, n^{- 1} e_{n} ⟩$ , so the adjoint identity yields $A^{*} e_{n} = n^{- 1} e_{n}$ and $S e_{n} = n^{- 2} e_{n}$ . Thus the source target is $x^{†} = e_{1} + N^{- 2 ν} e_{N}$ . Its forward image is $y = e_{1} + N^{- (2 ν + 1)} e_{N}$ . The chosen noise cancels the $e_{N}$ datum, so $y_{δ} = y + e = e_{1}$ and $∥ e ∥ = δ = N^{- (2 ν + 1)}$ .

2.1F6step 1.2algebra

The Tikhonov filter applied to $y_{δ} = e_{1}$ gives $x_{λ}^{δ} = (1 + λ)^{- 1} e_{1}$ . Hence $x^{†} - x_{λ}^{δ} = \frac{λ}{1 + λ} e_{1} + N^{- 2 ν} e_{N},$ and, at $λ = N^{- 2}$ , $∥ x^{†} - x_{λ}^{δ} ∥_{2} = \sqrt{(N^{2} + 1)^{- 2} + N^{- 4 ν}} .$ The exact-data bias in the $e_{N}$ coordinate is $λ N^{- 2 ν} / (N^{- 2} + λ) = \frac{1}{2} N^{- 2 ν}$ ; the noise output in that coordinate is $(1 / N) (N^{- 2} + λ)^{- 1} (- N^{- (2 ν + 1)}) = - \frac{1}{2} N^{- 2 ν}$ . Their contributions to solution error add.

3.1F7F8step 1.2step 2.1algebra

For $0 < ν \leq 1$ , $N^{- 2} \leq N^{- 2 ν}$ , so the displayed error lies between $N^{- 2 ν}$ and $\sqrt{2} N^{- 2 ν}$ . Also $δ / (2 \sqrt{λ}) = \frac{1}{2} N^{- 2 ν}$ and ${(\frac{δ}{\sqrt{2}})}^{2 / (2 ν + 1)} = N^{- 2} 2^{- 1 / (2 ν + 1)},$ so the selected $λ = N^{- 2}$ differs from the balancing value by a factor between $1$ and $2$ . These exact coordinates exhibit the stated bias-noise balance.

4.1F1F6step 1.1step 1.2algebra∎

For the separate exponent $μ > 1$ , take the unit source $w = e_{1}$ . Since $S e_{1} = e_{1}$ by step 1.2, the source target is $x^{†} = S^{μ} w = e_{1}$ and the exact datum is $y = A e_{1} = e_{1}$ . For every $x$ , $⟨ A x, e_{1} ⟩ = x_{1} = ⟨ x, e_{1} ⟩$ , so the adjoint identity gives $A^{*} e_{1} = e_{1}$ . The Tikhonov solution is therefore $x_{λ} = (1 + λ)^{- 1} e_{1}$ . Hence $\frac{∥ x^{†} - x_{λ} ∥}{λ^{μ}} = \frac{λ^{1 - μ}}{1 + λ} ⟶ + \infty (λ ↓ 0) .$ This unit source therefore has bias not $O (λ^{μ})$ .

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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 4: Regularization Methods (standard reference, not scraped)