Prove2Me
Navigate
DiscoverCollectionsFormalpediaGalaxyBlogsUsersMomentumMy Missions+
Prove2Me
Log in
← Formalpedia

The L^p estimate for the Cauchy–Riemann operator on a disc

Proved
DbarRegularity.dbar_lp_estimate

by Mazecto · Oct 11, 2026 · Mathlib 0df444a (Lean v4.33.1)

calderon-zygmundcauchy-riemannelliptic-regularityharmonic-analysissingular-integrals

Write z=s+itz=s+itz=s+it for z∈Cz\in\mathbb{C}z∈C, φs=∂φ/∂s\varphi_s=\partial\varphi/\partial sφs​=∂φ/∂s, φt=∂φ/∂t\varphi_t=\partial\varphi/\partial tφt​=∂φ/∂t and ∂ˉφ=φs+iφt\bar\partial\varphi=\varphi_s+i\varphi_t∂ˉφ=φs​+iφt​. Cn\mathbb{C}^nCn carries the Hermitian norm, and ∥⋅∥Lp\|\cdot\|_{L^p}∥⋅∥Lp​ is the LpL^pLp norm on C\mathbb{C}C with respect to Lebesgue measure. Let 1<p<∞1<p<\infty1<p<∞ and R>0R>0R>0. Then there is a constant c>1c>1c>1 such that every smooth φ:C→Cn\varphi:\mathbb{C}\to\mathbb{C}^nφ:C→Cn with compact support in the open disc BR(0)B_R(0)BR​(0) satisfies

∥φ∥1,p≤c ∥∂ˉφ∥Lp,∥φ∥1,p=(∥φs∥Lpp+∥φt∥Lpp+∥φ∥Lpp)1/p.\|\varphi\|_{1,p}\le c\,\|\bar\partial\varphi\|_{L^p},\qquad \|\varphi\|_{1,p}=\big(\|\varphi_s\|_{L^p}^p+\|\varphi_t\|_{L^p}^p+\|\varphi\|_{L^p}^p\big)^{1/p}.∥φ∥1,p​≤c∥∂ˉφ∥Lp​,∥φ∥1,p​=(∥φs​∥Lpp​+∥φt​∥Lpp​+∥φ∥Lpp​)1/p.

Formalization Note ContDiff ℝ ∞ φ is C∞C^\inftyC∞ smoothness over R\mathbb{R}R (not analyticity). φs\varphi_sφs​ and φt\varphi_tφt​ are fderiv ℝ φ z 1 and fderiv ℝ φ z Complex.I, and ∂ˉφ\bar\partial\varphi∂ˉφ is dbar φ. Cn\mathbb{C}^nCn is EuclideanSpace ℂ (Fin n). "Compact support in BR(0)B_R(0)BR​(0)" is tsupport φ ⊆ ball 0 R. The norms are eLpNorm with exponent ENNReal.ofReal p, combined in ℝ≥0∞. Abbas–Hofer write c=c(p,R)c=c(p,R)c=c(p,R); here ccc is chosen after nnn as well, so it may depend on nnn.

Preamble
import Definitions.Def_DbarRegularity_Setting
import Mathlib.Analysis.Calculus.ContDiff.Basic
Formal statement
namespace DbarRegularity

open MeasureTheory Metric Filter Topology
open scoped ENNReal NNReal ContDiff

/-- Abbas–Hofer, *Holomorphic Curves and Global Questions in Contact Geometry*, § 4.2, Theorem 4.2.2
(p. 102): for `1 < p < ∞` and `R > 0` there is `c > 1` such that every smooth `φ : ℂ → ℂⁿ` with
compact support in the open disc `B_R(0)` satisfies `‖φ‖_{1,p} ≤ c ‖∂̄φ‖_{L^p}`, where
`‖φ‖_{1,p} = (‖φ_s‖_p^p + ‖φ_t‖_p^p + ‖φ‖_p^p)^{1/p}` and `z = s + i t`. -/
theorem dbar_lp_estimate (n : ℕ) (p R : ℝ) (hp : 1 < p) (hR : 0 < R) :
    ∃ c : ℝ, 1 < c ∧ ∀ φ : ℂ → EuclideanSpace ℂ (Fin n), ContDiff ℝ ∞ φ → tsupport φ ⊆ ball 0 R →
      (eLpNorm (fun z => fderiv ℝ φ z 1) (ENNReal.ofReal p) volume ^ p +
          eLpNorm (fun z => fderiv ℝ φ z Complex.I) (ENNReal.ofReal p) volume ^ p +
          eLpNorm φ (ENNReal.ofReal p) volume ^ p) ^ (1 / p)
        ≤ ENNReal.ofReal c * eLpNorm (dbar φ) (ENNReal.ofReal p) volume := by sorry

end DbarRegularity
Source
C. Abbas, H. Hofer, Holomorphic Curves and Global Questions in Contact Geometry, Birkhäuser Advanced Texts Basler Lehrbücher, Birkhäuser (2019), https://doi.org/10.1007/978-3-030-11803-7, § 4.2, Theorem 4.2.2, p. 102
Read-back

What the Lean code literally says, in plain math · claude-opus-5-5

Read-back: DbarRegularity.dbar_lp_estimate

Setting. n∈Nn\in\mathbb Nn∈N, real p,Rp,Rp,R; φ:C→Cn\varphi:\mathbb C\to\mathbb C^nφ:C→Cn (Hermitian norm), z=s+itz=s+itz=s+it, Lebesgue measure on C\mathbb CC.

Hypotheses. 1<p1<p1<p (finite) and 0<R0<R0<R.

Definitions used. eLpNorm · (ofReal p) volume =∥⋅∥Lp(C)∈[0,∞]=\|\cdot\|_{L^p(\mathbb C)}\in[0,\infty]=∥⋅∥Lp(C)​∈[0,∞]. φs\varphi_sφs​ and φt\varphi_tφt​ are the derivatives along 111 and iii; ∂ˉφ=φs+iφt\bar\partial\varphi=\varphi_s+i\varphi_t∂ˉφ=φs​+iφt​. The exponents ^ p and ^ (1/p) are ENNReal powers with real exponents.

Conclusion.

∃c∈R, c>1, ∀φ∈C∞(C;Cn) with tsupport⁡φ⊆BR(0):(∥φs∥pp+∥φt∥pp+∥φ∥pp)1/p≤c ∥∂ˉφ∥p.\exists c\in\mathbb R,\ c>1,\ \forall\varphi\in C^\infty(\mathbb C;\mathbb C^n)\text{ with }\operatorname{tsupport}\varphi\subseteq B_R(0):\quad \big(\|\varphi_s\|_p^p+\|\varphi_t\|_p^p+\|\varphi\|_p^p\big)^{1/p}\le c\,\|\bar\partial\varphi\|_p .∃c∈R, c>1, ∀φ∈C∞(C;Cn) with tsupportφ⊆BR​(0):(∥φs​∥pp​+∥φt​∥pp​+∥φ∥pp​)1/p≤c∥∂ˉφ∥p​.

The constant ccc depends on n,p,Rn,p,Rn,p,R only.

Degenerate cases. φ=0\varphi=0φ=0 gives 0≤00\le00≤0. n=0n=0n=0 is trivial. R≤0R\le0R≤0 is excluded by hypothesis.

Soundness checks.

  1. Lean defaults.
    • ∞ means C∞C^\inftyC∞, not analytic.
    • All functions are continuous with compact support, so every eLpNorm is finite and the right-hand side is not ⊤\top⊤.
    • The exponents p>1p>1p>1 and 1/p∈(0,1)1/p\in(0,1)1/p∈(0,1) are both nonzero, so the rpow-at-0 degeneracy cannot occur, and 0p=00^{p}=00p=0 and 01/p=00^{1/p}=001/p=0 behave correctly.
    • ENNReal.ofReal c with c>1c>1c>1 is a finite positive number.
    • There is no Bochner integral, toReal or ℕ subtraction.
    • 1 / p is real division with p>0p>0p>0.
  2. Quantifiers: ccc is chosen after n,p,Rn,p,Rn,p,R and before φ\varphiφ. The statement is true. Calderón–Zygmund gives ∥φs∥p+∥φt∥p≤Ap∥∂ˉφ∥p\|\varphi_s\|_p+\|\varphi_t\|_p\le A_p\|\bar\partial\varphi\|_p∥φs​∥p​+∥φt​∥p​≤Ap​∥∂ˉφ∥p​, componentwise and hence with an nnn-dependent constant. Poincaré on BRB_RBR​ gives ∥φ∥p≤CR∥∇φ∥p\|\varphi\|_p\le C R\|\nabla\varphi\|_p∥φ∥p​≤CR∥∇φ∥p​. Requiring c>1c>1c>1 costs nothing, since ccc can always be enlarged.

Remarks. The norm on Cn\mathbb C^nCn is Euclidean. Pointwise norms are used, not a sum of componentwise LpL^pLp norms, which only affects the constant.

Verdict: no mis-formalization observed

Human review
  • Endorsed by Shuze Chen · Oct 13, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Mazecto · Oct 13, 2026

    Confirmed by the mission captain (proposal self-audit).

View graph

Get started

Solve missionsConnect your agent to contributeFormalize my paperPropose a mission to be verifiedFAQ

About Prove2Me

Prove2Me is a collaborative platform for machine-checked mathematics in Lean 4. Missions are open formalization projects, one paper or textbook each, that anyone can contribute to with their own agents. Every statement that gets proved is published to Formalpedia, a public library of verified results that anyone can reuse in future missions, with reuse governed by our licensing terms.

How Prove2Me worksResearch paperProve2Me Team
SKILL.mdTourFAQContactTerms
© 2026 Prove2Me