LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A frozen Newton step yields a controlled error inequality

Statement

Let $1 \leq p < \infty$ . Let $X, Y, Z$ be real Banach spaces, let $X_{0}$ be a closed linear subspace of $X$ , and let $D, U \subseteq X$ . Let $x_{0}, x^{†} \in D \cap U \cap X_{0}$ , let $F : D \to Y$ , $K \in L (Z, Y)$ , and $r : D \to Z$ satisfy $F (x) - F (x_{0}) = K r (x)$ for all $x \in D$ . Assume $r ∣_{U} : U \to \tilde{U}$ is a bijection, $r$ is Lipschitz on $D$ with constant $L_{r}$ , and $r$ has bounded linear Gâteaux derivatives on $U$ . Let $P : X \to X$ be a bounded linear projection with $P^{2} = P$ , $P ∣_{X_{0}} = 0$ , and $(I - P) (X) \subseteq X_{0}$ . Assume $P (x^{†}) = 0$ and $(I - P) r^{- 1} (\hat{r}) \in D \cap X_{0}$ for each admitted $\hat{r} \in \tilde{U}$ .

Let $\overset{ˇ}{R} : X_{0} \to [0, + \infty]$ be proper and convex, choose $ξ^{†} \in \partial \overset{ˇ}{R} (x^{†})$ , and write

$D_{\overset{ˇ}{R}}^{ξ^{†}} (z, x^{†}) : = \overset{ˇ}{R} (z) - \overset{ˇ}{R} (x^{†}) - ⟨ ξ^{†}, z - x^{†} ⟩_{X_{0}^{*}, X_{0}} .$

Suppose that, on the stated sublevel $\overset{ˇ}{R} (z) \leq R_{0}$ , the variational source inequality

$- ⟨ ξ^{†}, z - x^{†} ⟩_{X_{0}^{*}, X_{0}} \leq b D_{\overset{ˇ}{R}}^{ξ^{†}} (z, x^{†}) + ψ ⁣ (∥ F (z) - F (x^{†}) ∥_{Y}^{p})$

holds for some $R_{0} \geq \overset{ˇ}{R} (x^{†})$ and $0 \leq b < 1$ . Assume that $ψ : [0, \infty) \to [0, \infty)$ is continuous, nondecreasing and concave, with $ψ (0) = 0$ , and that for a constant $C_{ψ} > 0$ ,

$ψ (a + b) \leq C_{ψ} (ψ (a) + ψ (b)) (a, b \geq 0) .$

Let $H (q) : = \sup_{s \geq 0} (ψ (s) - q s)$ , and assume the values of $H$ used below are finite. Let $γ > C_{ψ}$ , $C_{P} \geq 2^{p - 1} ∥ K ∥^{p} L_{r}^{p}$ , and ${\tilde{C}}_{Q} > 0$ . Let $R_{P} : P (X) \to [0, + \infty]$ satisfy $R_{P} (0) = 0$ and

$R_{P} (z) \geq γ ψ (C_{P} ∥ z ∥_{X}^{p}) (z \in P (X)) .$

For each admitted $\hat{r} \in \tilde{U}$ , put $u (\hat{r}) : = r^{- 1} (\hat{r})$ and $\overset{ˇ}{x} (\hat{r}) : = (I - P) u (\hat{r})$ . Assume the compatibility bound

$∥ \hat{r} - r (x^{†}) ∥_{Z}^{p} \leq {\tilde{C}}_{Q} (D_{\overset{ˇ}{R}}^{ξ^{†}} (\overset{ˇ}{x} (\hat{r}), x^{†}) + ψ (C_{P} ∥ P u (\hat{r}) ∥_{X}^{p})) .$

Choose $c_{β α} > 0$ with $c_{β α} \leq \min {1 - b, γ - C_{ψ}} / (4 {\tilde{C}}_{Q})$ , and set

$m : = \min {1 - b, γ - C_{ψ}}, a : = 2^{1 - p}, λ : = 2^{1 - p}, t : = 2^{2 - 2 p} - 2^{p - 1} c_{t c r}^{p}, v : = (1 + 2^{p - 1}) c_{t c r}^{p}, \overline{C} : = C_{ψ} 2^{2 p - 2} .$

Assume $0 < c_{t c r} < 1$ and $t > 0$ , and for all $x_{1}, x_{2} \in U$ assume

$∥ r (x_{1}) - r (x_{2}) - r^{'} (x_{2}) (x_{1} - x_{2}) ∥_{Z} \leq c_{t c r} ∥ r (x_{1}) - r (x_{2}) ∥_{Z} .$

Assume the candidate projected point satisfies $\overset{ˇ}{R} (\overset{ˇ}{x} ({\hat{r}}_{n + 1})) \leq R_{0}$ .

Let $Q (a, b) : = ∥ a - b ∥_{Z}$ . Fix $δ > 0$ , $y^{δ} \in Y$ with $∥ y^{δ} - F (x^{†}) ∥_{Y} \leq δ$ , and $α_{n} > 0$ , with $β_{n} : = c_{β α} α_{n}$ . Suppose $({\hat{r}}_{n + 1}, x_{n + 1}) \in \tilde{U} \times U$ satisfies, for some $η_{n} \geq 0$ ,

$J_{n}^{δ} ({\hat{r}}_{n + 1}, x_{n + 1}) \leq J_{n}^{δ} (\hat{r}, x) + η_{n} ((\hat{r}, x) \in \tilde{U} \times U),$

where $J_{n}^{δ}$ is the frozen objective in the Joint and frozen-Newton objectives definition, with regularizer $R (\hat{r}) = \overset{ˇ}{R} (\overset{ˇ}{x} (\hat{r})) + R_{P} (P u (\hat{r}))$ and distance $Q$ .

Define

$E_{n + 1} : = \frac{c_{β α}}{C_{ψ}} (λ ∥ {\hat{r}}_{n + 1} - r (x^{†}) ∥_{Z}^{p} + t ∥ r (x_{n + 1}) - r (x^{†}) ∥_{Z}^{p}),$

$S_{n + 1} : = \frac{1}{C_{ψ}} (2^{1 - p} ∥ K ({\hat{r}}_{n + 1} - r (x^{†})) ∥_{Y}^{p} + ∥ P x_{n + 1} ∥_{X}^{p}),$

and

$d_{n} : = \frac{c_{β α}}{C_{ψ}} α_{n} v ∥ r (x_{n}) - r (x^{†}) ∥_{Z}^{p} + \frac{2 δ^{p} + η_{n}}{C_{ψ}} .$

Then

$S_{n + 1} + α_{n} E_{n + 1} \leq d_{n} + α_{n} ψ (\overline{C} S_{n + 1}) .$

Consequently,

$E_{n + 1} \leq \frac{d_{n}}{α_{n}} + H ⁣ (\frac{1}{\overline{C} α_{n}}), S_{n + 1} \leq 2 (d_{n} + α_{n} H ⁣ (\frac{1}{2 \overline{C} α_{n}})) .$

Statement references

Range-invariance factorization and domains

Joint and frozen-Newton objective definitions

Barbara Kaltenbacher, Convergence rates under a range invariance condition with application to electrical impedance tomography, equations (2.31)–(2.41)

Frederic Weidling, Variational Source Conditions and Conditional Stability Estimates for Inverse Problems in PDEs, Theorem 2.20

Facts & Assumptions

Given: All spaces, maps, penalties, parameters, source inequalities, compatibility inequalities, and the approximate frozen step satisfy the hypotheses in the Statement.

[A1]

The projected point $\overset{ˇ}{x} ({\hat{r}}_{n + 1})$ belongs to the regularizer sublevel where the VSC holds.

[F1]

Differential range invariance for Banach-space maps: $F (x) - F (x_{0}) = K r (x)$ on the specified domain, and $r ∣_{U}$ is invertible onto $\tilde{U}$ .

[F2]

Joint and frozen-Newton objectives for range-invariant relaxations: the frozen objective has data term $∥ K \hat{r} + F (x_{0}) - y^{δ} ∥^{p}$ , regularizer $α_{n} R (\hat{r})$ , coupling $β_{n} ∥ ℓ_{n} (x) - \hat{r} ∥^{p}$ , and penalty $∥ P x ∥^{p}$ , where $ℓ_{n} (x) = r (x_{n}) + r^{'} (x_{n}) (x - x_{n})$ .

[F3]

Banach-space convex subgradients and dual pairings: $ξ^{†} \in \partial \overset{ˇ}{R} (x^{†})$ means $\overset{ˇ}{R} (z) \geq \overset{ˇ}{R} (x^{†}) + ⟨ ξ^{†}, z - x^{†} ⟩$ for every $z \in X_{0}$ .

[F4]

Bregman distances for convex regularizers: $D_{\overset{ˇ}{R}}^{ξ^{†}} (z, x^{†}) = \overset{ˇ}{R} (z) - \overset{ˇ}{R} (x^{†}) - ⟨ ξ^{†}, z - x^{†} ⟩$ .

[F5]

Variational source conditions on a regularizer sublevel: $- ⟨ ξ^{†}, z - x^{†} ⟩ \leq b D_{\overset{ˇ}{R}}^{ξ^{†}} (z, x^{†}) + ψ (∥ F (z) - F (x^{†}) ∥^{p})$ on the stated sublevel.

[F6]

A conjugate bound for residual and Bregman errors: if $r + α e \leq d + α ψ (C r)$ , then $e \leq d / α + H (1 / (C α))$ and $r \leq 2 (d + α H (1 / (2 C α)))$ .

Proof

1.1A1F1F2givenalgebra

Write $r^{†} : = r (x^{†})$ , $u : = r^{- 1} ({\hat{r}}_{n + 1})$ , $\overset{ˇ}{x} : = (I - P) u$ , and $ℓ_{n} (x) : = r (x_{n}) + r^{'} (x_{n}) (x - x_{n})$ . Compare the approximate-minimum inequality with $(\hat{r}, x) = (r^{†}, x^{†})$ . The competitor has data residual at most $δ^{p}$ , penalty $R_{P} (0) = 0$ , and $P x^{†} = 0$ . If $g : = K {\hat{r}}_{n + 1} + F (x_{0}) - y^{δ}$ , then $K ({\hat{r}}_{n + 1} - r^{†}) = g + (y^{δ} - y)$ ; convexity of $s \mapsto s^{p}$ gives $∥ K ({\hat{r}}_{n + 1} - r^{†}) ∥^{p} \leq 2^{p - 1} (∥ g ∥^{p} + δ^{p})$ . Thus the comparison gives $2^{1 - p} ∥ K ({\hat{r}}_{n + 1} - r^{†}) ∥^{p} + α_{n} (R ({\hat{r}}_{n + 1}) - \overset{ˇ}{R} (x^{†})) + β_{n} ∥ ℓ_{n} (x_{n + 1}) - {\hat{r}}_{n + 1} ∥^{p} + ∥ P x_{n + 1} ∥^{p} \leq 2 δ^{p} + η_{n} + β_{n} ∥ ℓ_{n} (x^{†}) - r^{†} ∥^{p}$ . The remainder hypothesis at $(x^{†}, x_{n})$ and $c_{t c r} < 1$ imply $∥ ℓ_{n} (x^{†}) - r^{†} ∥^{p} \leq c_{t c r}^{p} ∥ r (x_{n}) - r^{†} ∥^{p}$ .

1.2givenalgebra

Let $A : = ∥ r (x_{n + 1}) - {\hat{r}}_{n + 1} ∥$ , $B : = ∥ ℓ_{n} (x_{n + 1}) - {\hat{r}}_{n + 1} ∥$ , and $D_{n} : = ∥ r (x_{n + 1}) - ℓ_{n} (x_{n + 1}) ∥$ . The triangle inequality gives $A \leq B + D_{n}$ , hence $B^{p} \geq 2^{1 - p} A^{p} - D_{n}^{p}$ . Applying the same power inequality to $∥ r (x_{n + 1}) - r^{†} ∥ \leq A + ∥ {\hat{r}}_{n + 1} - r^{†} ∥$ gives $A^{p} \geq 2^{1 - p} ∥ r (x_{n + 1}) - r^{†} ∥^{p} - ∥ {\hat{r}}_{n + 1} - r^{†} ∥^{p}$ . The Taylor hypothesis and another triangle-power estimate give $D_{n}^{p} \leq 2^{p - 1} c_{t c r}^{p} (∥ r (x_{n + 1}) - r^{†} ∥^{p} + ∥ r (x_{n}) - r^{†} ∥^{p})$ . Therefore $∥ ℓ_{n} (x_{n + 1}) - {\hat{r}}_{n + 1} ∥^{p} \geq t ∥ r (x_{n + 1}) - r^{†} ∥^{p} - λ ∥ {\hat{r}}_{n + 1} - r^{†} ∥^{p} - 2^{p - 1} c_{t c r}^{p} ∥ r (x_{n}) - r^{†} ∥^{p}$ , with $t = 2^{2 - 2 p} - 2^{p - 1} c_{t c r}^{p}$ and $λ = 2^{1 - p}$ .

1.3A1F1F4F5givenalgebra

Set $z : = P u$ . Since $∥ r (\overset{ˇ}{x}) - r (u) ∥ \leq L_{r} ∥ z ∥$ , the factorization and power inequality give $∥ F (\overset{ˇ}{x}) - F (x^{†}) ∥^{p} \leq 2^{p - 1} ∥ K ({\hat{r}}_{n + 1} - r^{†}) ∥^{p} + 2^{p - 1} ∥ K ∥^{p} L_{r}^{p} ∥ z ∥^{p}$ . By the VSC [F5], the Bregman identity [F4], the penalty lower bound, and $ψ (a + b) \leq C_{ψ} (ψ (a) + ψ (b))$ , we obtain $R ({\hat{r}}_{n + 1}) - \overset{ˇ}{R} (x^{†}) \geq m (D_{\overset{ˇ}{R}}^{ξ^{†}} (\overset{ˇ}{x}, x^{†}) + ψ (C_{P} ∥ z ∥^{p})) - C_{ψ} ψ (2^{p - 1} ∥ K ({\hat{r}}_{n + 1} - r^{†}) ∥^{p})$ . Indeed, the VSC yields coefficient $1 - b$ on the Bregman term, and the penalty leaves coefficient $γ - C_{ψ}$ on the penalty-index term; their minimum is $m$ .

2.1step 1.1step 1.2step 1.3A1F1F3F4F5givenalgebra

Substituting steps 1.1–1.3 into the objective comparison and applying the compatibility inequality gives $C_{ψ} S_{n + 1} + α_{n} m T + β_{n} t ∥ r (x_{n + 1}) - r^{†} ∥^{p} \leq β_{n} v ∥ r (x_{n}) - r^{†} ∥^{p} + β_{n} λ ∥ {\hat{r}}_{n + 1} - r^{†} ∥^{p} + α_{n} C_{ψ} ψ (2^{p - 1} ∥ K ({\hat{r}}_{n + 1} - r^{†}) ∥^{p}) + 2 δ^{p} + η_{n}$ , where $T : = D_{\overset{ˇ}{R}}^{ξ^{†}} (\overset{ˇ}{x}, x^{†}) + ψ (C_{P} ∥ P u ∥^{p})$ . By [F3], $D_{\overset{ˇ}{R}}^{ξ^{†}} (\overset{ˇ}{x}, x^{†}) \geq 0$ , so $T \geq 0$ . Since $∥ {\hat{r}}_{n + 1} - r^{†} ∥^{p} \leq {\tilde{C}}_{Q} T$ and $β_{n} λ {\tilde{C}}_{Q} \leq α_{n} m / 4$ , the left regularizer term absorbs the displayed compatibility term and a second copy of it, leaving $α_{n} m T / 2$ ; in particular it bounds another $β_{n} λ ∥ {\hat{r}}_{n + 1} - r^{†} ∥^{p}$ . The resulting inequality is $C_{ψ} S_{n + 1} + β_{n} λ ∥ {\hat{r}}_{n + 1} - r^{†} ∥^{p} + β_{n} t ∥ r (x_{n + 1}) - r^{†} ∥^{p} \leq β_{n} v ∥ r (x_{n}) - r^{†} ∥^{p} + α_{n} C_{ψ} ψ (2^{p - 1} ∥ K ({\hat{r}}_{n + 1} - r^{†}) ∥^{p}) + 2 δ^{p} + η_{n}$ .

3.1step 2.1F6givenalgebra∎

Divide step 2.1 by $C_{ψ}$ and use $β_{n} = c_{β α} α_{n}$ . Since $C_{ψ} S_{n + 1} \geq 2^{1 - p} ∥ K ({\hat{r}}_{n + 1} - r^{†}) ∥^{p}$ , we have $2^{p - 1} ∥ K ({\hat{r}}_{n + 1} - r^{†}) ∥^{p} \leq C_{ψ} 2^{2 p - 2} S_{n + 1} = \overline{C} S_{n + 1}$ . Monotonicity of $ψ$ gives $S_{n + 1} + α_{n} E_{n + 1} \leq d_{n} + α_{n} ψ (\overline{C} S_{n + 1})$ . Applying [F6] with $c_{1} = c_{2} = 1$ , $C_{3} = \overline{C}$ , $r = S_{n + 1}$ , and $e = E_{n + 1}$ proves both conjugate bounds.

Depends on

Used by

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Barbara Kaltenbacher, Convergence rates under a range invariance condition with application to electrical impedance tomography, equations (2.31)–(2.41) (standard reference, not scraped)
Frederic Weidling, Variational Source Conditions and Conditional Stability Estimates for Inverse Problems in PDEs, Theorem 2.20 (standard reference, not scraped)