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

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Convexity, lower semicontinuity, and BV agreement of L² TV

Statement

Let $Ω \subseteq R^{2}$ be open and let $T V_{2} (\cdot; Ω)$ be the extended distributional total variation on $L^{2} (Ω; R)$ . It is proper, convex, lower semicontinuous in the $L^{2}$ norm, and positively one-homogeneous for nonnegative scalars. For every $u \in B V (Ω) \cap L^{2} (Ω)$ , it agrees with spatial TV: $T V_{2} (u; Ω) = T V (u; Ω)$ .

Statement references

The extended L² distributional-TV functional is a supremum over smooth compactly supported vector fields.

For globally integrable BV functions, that test-field supremum equals the measure total variation.

Facts & Assumptions

Given: An open $Ω \subseteq R^{2}$ , the functional $T V_{2}$ on $L^{2} (Ω; R)$ , and the spatial BV total variation.

[F1]

Distributional TV as an extended functional on L²: $T V_{2} (u; Ω)$ is the supremum of $\int_{Ω} u div φ$ over smooth compactly supported vector fields with Euclidean norm at most one.

[F2]

Test-field formula and L1 lower semicontinuity of spatial TV: for $u \in B V (Ω)$ , $T V (u; Ω)$ equals the same test-field supremum.

[F3]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: each test functional $u \mapsto \int u div φ$ is continuous on $L^{2}$ , with absolute value at most $∥ u ∥_{2} ∥ div φ ∥_{2}$ .

Proof

1.1F1givenalgebra

The zero function has $T V_{2} (0; Ω) = 0$ , so the functional is proper; its defining test class is symmetric under $φ \mapsto - φ$ , hence its supremum is nonnegative. For $λ > 0$ , linearity of every test integral gives $T V_{2} (λ u; Ω) = λ T V_{2} (u; Ω)$ , and the equality also holds for $λ = 0$ . For $θ \in [0, 1]$ , every admissible field gives $\int (θ u + (1 - θ) v) div φ \leq θ T V_{2} (u; Ω) + (1 - θ) T V_{2} (v; Ω)$ ; taking the supremum proves convexity, with the extended-real cases automatic.

2.1F1F2F3step 1.1∎

By [F3], every fixed test integral is a continuous linear functional on $L^{2}$ . For each real $a$ , the sublevel set ${u : T V_{2} (u; Ω) \leq a}$ is the intersection, over all admissible $φ$ , of the closed half-spaces ${u : \int u div φ \leq a}$ ; hence it is closed and $T V_{2}$ is lower semicontinuous. If $u \in B V (Ω) \cap L^{2} (Ω)$ , [F1] and [F2] give $T V_{2} (u; Ω) = T V (u; Ω)$ .

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Sources

José A. Iglesias, Gwenael Mercier, and Otmar Scherzer, A note on convergence of solutions of total variation regularized linear inverse problems, §1.1 (standard reference, not scraped)
Gláucio Terra, Introduction to Geometric Measure Theory — Lecture Notes, version 1.6, §7.4 (standard reference, not scraped)