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Data, operator, discretization, and solve errors — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Vanishing data errors with persistent small-singular-value reconstruction error

Example

For every integer $n \geq 1$ , let $X_{n} = Y_{n} = R^{n}$ with the Euclidean inner product and let $A_{n} = A_{h, n} = diag (1, 1 / 2, \dots, 1 / n) .$ Set $x_{n}^{†} = 0$ , exact data $y_{n} = A_{n} x_{n}^{†} = 0$ , noise $e_{n} = (0, \dots, 0, 1 / n)$ , noise bound $δ_{n} = 1 / n$ , and regularization coefficient $λ_{n} = n^{- 4}$ . The observed data are $y_{n, δ} = y_{n} + e_{n}$ , and the returned output is the exact Tikhonov output ${\tilde{x}}_{n} = (A_{n}^{*} A_{n} + λ_{n} I)^{- 1} A_{n}^{*} y_{n, δ} .$ The smallest singular value of $A_{n}$ is $1 / n$ , while $∥ e_{n} ∥ = δ_{n} \to 0$ . Nevertheless $∥ {\tilde{x}}_{n} - x_{n}^{†} ∥ = \frac{n^{2}}{n^{2} + 1} ⟶ 1.$ At each resolution the discrete operator equals the forward operator and the discretized datum is formed exactly before adding noise, so the operator, pre-inversion, and algebraic-solve channels vanish; only data perturbation remains. Thus vanishing noise alone does not force vanishing reconstruction error when the retained singular values shrink and $λ_{n}$ is chosen as above.

Example references

Five-channel error-budget convention

Clason, Regularization of Inverse Problems, Theorem 8.8 and its proof

Hansen, Introduction to Inverse Problems, Chapter 5, slides 4–6

Van Leeuwen and Brune, 10 Lectures on Inverse Problems and Imaging, §3.4.1

Facts & Assumptions

Given: For each $n \geq 1$ , the real finite-dimensional spaces $X_{n} = Y_{n} = R^{n}$ , the diagonal map $A_{n} = A_{h, n}$ , $x_{n}^{†} = 0$ , $y_{n} = 0$ , $e_{n} = (0, \dots, 0, 1 / n)$ , $δ_{n} = 1 / n$ , and $λ_{n} = n^{- 4}$ .

[F1]

Error budget for computed Tikhonov inversion: The budget uses $y = A x^{†}$ , $y_{δ} = y + e$ , $y_{h, δ} = A_{h} x^{†} + e$ , $d_{h} = (A - A_{h}) x^{†}$ , and the five bias, data, operator, pre-inversion, and solve vectors; its shifted normal operators are assumed bijective.

[F2]

Bounded finite rank operators are compact: Every bounded linear operator whose range has a finite ordered basis is compact.

Verification

Given: The finite matrices and data in the Example, with the standard coordinate vectors $q_{1}, \dots, q_{n}$ in $R^{n}$ .

1.1givenF1F2constructalgebra

The map $A_{n}$ is bounded and has range $R^{n}$ , which has the finite ordered basis $q_{1}, \dots, q_{n}$ . Hence $A_{n}$ is compact by [F2]. Its adjoint is $A_{n}^{*} = A_{n}$ , and $A_{n}^{*} A_{n} + λ_{n} I$ is diagonal with positive entries $j^{- 2} + n^{- 4}$ for $1 \leq j \leq n$ ; therefore the adjoints and bijective shifted normal operator required by [F1] are supplied in this finite model.

2.1givenF1step 1.1algebra

Let $q_{n}$ be the last coordinate vector. Then $e_{n} = q_{n} / n$ , $A_{n}^{*} y_{n, δ} = q_{n} / n^{2}$ , and $(A_{n}^{*} A_{n} + λ_{n} I) q_{n} = (n^{- 2} + n^{- 4}) q_{n}$ . Solving this one diagonal coordinate gives ${\tilde{x}}_{n} = \frac{n^{2}}{n^{2} + 1} q_{n}$ .

3.1givenF1step 2.1algebra

Since $x_{n}^{†} = y_{n} = 0$ and $A_{h, n} = A_{n}$ , the bias, pre-inversion, and operator channels are zero. Since ${\tilde{x}}_{n}$ is the exact discrete Tikhonov output, the solve channel is zero. The only nonzero channel is the data vector $R_{λ_{n}} y_{n} - R_{λ_{n}} y_{n, δ} = - {\tilde{x}}_{n}$ . Therefore $∥ {\tilde{x}}_{n} - x_{n}^{†} ∥ = n^{2} / (n^{2} + 1)$ , while $δ_{n} = 1 / n \to 0$ and the smallest singular value is $1 / n \to 0$ .

4.1givenstep 3.1algebra∎

The scalar data-noise allowance in the general budget is $δ_{n} / (2 \sqrt{λ_{n}}) = n / 2$ , so it does not force the exact errors in step 3.1 to tend to zero. The sequence has resolved forward data at every level and asserts no convergence theorem for a fixed infinite-dimensional discretization.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Early stopping leaves a nonzero Tikhonov normal residual

Example

Let $X = Y = R$ with the usual inner product, let $A = A_{h} = I$ , let $x^{†} = 0$ , $y = y_{h, δ} = 0$ , $e = 0$ , and set $λ = 1$ . The exact Tikhonov output is $R_{λ, h} y_{h, δ} = 0$ , and its normal equation is $2 x = 0$ . Apply one Richardson step with step size $1 / 4$ from $x_{0} = 1 / 2$ : ${\tilde{x}}_{h} = x_{1} = x_{0} - \frac{1}{4} (2 x_{0}) = \frac{1}{4} .$ The returned output has exact normal-equation residual $r_{h, h} = 2 {\tilde{x}}_{h} = \frac{1}{2} \neq 0,$ and reconstruction error $∣ x^{†} - {\tilde{x}}_{h} ∣ = 1 / 4$ . The bias, data, operator, and pre-inversion channels are zero; only the algebraic solve channel is nonzero.

Example references

Five-channel Tikhonov error budget

Facts & Assumptions

Given: The one-dimensional real Hilbert spaces $X = Y = R$ , the identity maps $A = A_{h} = I$ , $x^{†} = 0$ , $y = y_{h, δ} = 0$ , $e = 0$ , $δ = 0$ , $λ = 1$ , initial iterate $x_{0} = 1 / 2$ , and Richardson step size $1 / 4$ .

[F1]

Error budget for computed Tikhonov inversion: The solve vector is $b_{s o l v e} = R_{λ, h} y_{h, δ} - {\tilde{x}}_{h}$ , and the exact normal residual is $r_{h, h} = (A_{h}^{*} A_{h} + λ I_{X}) {\tilde{x}}_{h} - A_{h}^{*} y_{h, δ}$ .

[F2]

Bounded finite rank operators are compact: Every bounded linear operator whose range has a finite ordered basis is compact.

Verification

Given: The scalar data and iteration in the Example.

1.1givenF1F2constructalgebra

The identity is bounded and has the finite range basis ${1}$ , so it is compact by [F2]. The shifted normal operator is $A_{h}^{*} A_{h} + λ I = 2 I$ , which is bijective with inverse $(1 / 2) I$ ; hence the finite scalar instance satisfies [F1].

1.2givenF1algebra

One Richardson step for $2 x = 0$ from $x_{0} = 1 / 2$ gives ${\tilde{x}}_{h} = 1 / 2 - (1 / 4) (2 \cdot 1 / 2) = 1 / 4$ . Substitution into the residual definition gives $r_{h, h} = 2 (1 / 4) - 0 = 1 / 2$ , while the exact regularized output is $R_{λ, h} 0 = 0$ .

2.1givenF1step 1.2algebra∎

The truth bias and data perturbation vanish because $x^{†} = y = e = 0$ . Since $A_{h} = A$ , $d_{h} = 0$ and the regularized-operator difference vanishes; the pre-inversion channel is therefore also zero. The solve channel is $0 - {\tilde{x}}_{h} = - 1 / 4$ , so its norm equals the reconstruction error $1 / 4$ and is at most the residual allowance $∣ r_{h, h} ∣ / λ = 1 / 2$ .

Data, operator, discretization, and solve errors — Examples and Counterexamples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

Vanishing data errors with persistent small-singular-value reconstruction error

Example

Example references

Facts & Assumptions

Verification

Early stopping leaves a nonzero Tikhonov normal residual

Example

Example references

Facts & Assumptions

Verification

Sources