Research proof index

Rudin's inequality, usual form.

Let $1<q\le2$, let $q'$ be its Hölder conjugate, and suppose the linearized nontangential operators have the required uniform $L^2$ bound. For bounded measurable $F,G$ and measurable $f$ with $\lVert f(x)\rVert\le\mathbf 1_F(x)$, the supremum over all truncated scale intervals $-B\le s_1\le s_2\le B$ satisfies

$\int_G^+\sup_{s_1\le s_2}\lVert T_{s_1,s_2}f(x)\rVert\,dx\le C(a,q)\,\mu(G)^{1/q'}\mu(F)^{1/q}.$

The Serre derivative of a bounded holomorphic function is bounded at infinity. serre_D k f = D f - (k/12)·E₂·f. Both terms are bounded: - D f is bounded by D_isBoundedAtImInfty_of_bounded - (k/12)·E₂·f is bounded since E₂ and f are bounded

Serre derivative is equivariant under the slash action. More precisely, if F is invariant under the slash action of weight k, then serre_D k F is invariant under the slash action of weight k + 2.

General limit: if f → c at i∞ and f is holomorphic and bounded, then serre_D k f → -k*c/12. This is the continuous mapping theorem applied to serre_D k f = D f - (k/12) * E₂ * f: - D f → 0 (Cauchy estimate from boundedness) - E₂ → 1 - f → c Therefore serre_D k f → 0 - (k/12) * 1 * c = -k*c/12.

The Serre derivative of E₂ is weight-4 slash-invariant. This requires explicit computation since E₂ is not modular. Proof strategy: Write serre_D 1 E₂ = serre_D 2 E₂ + (1/12) E₂². Then: - (serre_D 2 E₂) ∣[4] γ = serre_D 2 (E₂ ∣[2] γ) by serre_D_slash_equivariant - E₂ ∣[2] γ = E₂ - α D₂ γ where α = 1/(2ζ(2)) = 3/π² - (E₂²) ∣[4] γ = (E₂ ∣[2] γ)² After e...