AlexKontorovich/PrimeNumberTheoremAnd
Source indexedlemma · leanprover/lean4:v4.32.0
ZeroFactorization
PrimeNumberTheoremAnd.StrongPNT · PrimeNumberTheoremAnd/StrongPNT.lean:382 to 427
Mathematical statement
Exact Lean statement
@[blueprint "ZeroFactorization"
(title := "ZeroFactorization")
(statement := /--
Let $f:\mathbb{C}\to\mathbb{C}$ be analytic on $\overline{\mathbb{D}_1}$ with $f(0)\neq 0$.
For all $\rho\in\mathcal{K}_f(R)$ with $R<1$ there exists $h_\rho(z)$ such that
$h_\rho(z)$ is analytic at $\rho$, $h_\rho(\rho)\neq 0$, and
$f(z)=(z-\rho)^{m_f(\rho)}\,h_\rho(z)$.
-/)
(proof := /--
Since $f$ is analytic on neighborhoods of points in $\overline{\mathbb{D}_1}$ we know
that there exists a series expansion about $\rho$:
$$f(z)=\sum_{0\leq n}a_n\,(z-\rho)^n.$$
Now if we let $m$ be the smallest number such that $a_m\neq 0$, then
$$f(z)=\sum_{0\leq n}a_n\,(z-\rho)^n=\sum_{m\leq n}a_n\,(z-\rho)^n
=(z-\rho)^m\sum_{m\leq n}a_n\,(z-\rho)^{n-m}=(z-\rho)^m\,h_\rho(z).$$
Trivially, $h_\rho(z)$ is analytic at $\rho$ (we have written down the series
expansion); now note that
$$h_\rho(\rho)=\sum_{m\leq n}a_n(\rho-\rho)^{n-m}=\sum_{m\leq n}a_n0^{n-m}=a_m\neq 0.$$
-/)
(latexEnv := "lemma")]
lemma ZeroFactorization {R : ℝ} {f : ℂ → ℂ} {ρ : ℂ}
(RleOne : R < 1)
(hfAnalytic : AnalyticOnNhd ℂ f (Metric.closedBall (0 : ℂ) 1))
(hf_neq_zero_at_zero : f 0 ≠ 0)
(hρ : ρ ∈ SetOfZeros R f) :
∃ h_ρ : ℂ → ℂ, AnalyticAt ℂ h_ρ ρ ∧ h_ρ ρ ≠ 0 ∧ ZeroFactor f ρ = h_ρ ρ ∧
f =ᶠ[nhds ρ] fun z ↦ (z - ρ) ^ analyticOrderNatAt f ρ * h_ρ zComplete declaration
Lean source
Full Lean sourceLean 4
@[blueprint "ZeroFactorization" (title := "ZeroFactorization") (statement := /-- Let $f:\mathbb{C}\to\mathbb{C}$ be analytic on $\overline{\mathbb{D}_1}$ with $f(0)\neq 0$. For all $\rho\in\mathcal{K}_f(R)$ with $R<1$ there exists $h_\rho(z)$ such that $h_\rho(z)$ is analytic at $\rho$, $h_\rho(\rho)\neq 0$, and $f(z)=(z-\rho)^{m_f(\rho)}\,h_\rho(z)$. -/) (proof := /-- Since $f$ is analytic on neighborhoods of points in $\overline{\mathbb{D}_1}$ we know that there exists a series expansion about $\rho$: $$f(z)=\sum_{0\leq n}a_n\,(z-\rho)^n.$$ Now if we let $m$ be the smallest number such that $a_m\neq 0$, then $$f(z)=\sum_{0\leq n}a_n\,(z-\rho)^n=\sum_{m\leq n}a_n\,(z-\rho)^n =(z-\rho)^m\sum_{m\leq n}a_n\,(z-\rho)^{n-m}=(z-\rho)^m\,h_\rho(z).$$ Trivially, $h_\rho(z)$ is analytic at $\rho$ (we have written down the series expansion); now note that $$h_\rho(\rho)=\sum_{m\leq n}a_n(\rho-\rho)^{n-m}=\sum_{m\leq n}a_n0^{n-m}=a_m\neq 0.$$ -/) (latexEnv := "lemma")]lemma ZeroFactorization {R : ℝ} {f : ℂ → ℂ} {ρ : ℂ} (RleOne : R < 1) (hfAnalytic : AnalyticOnNhd ℂ f (Metric.closedBall (0 : ℂ) 1)) (hf_neq_zero_at_zero : f 0 ≠ 0) (hρ : ρ ∈ SetOfZeros R f) : ∃ h_ρ : ℂ → ℂ, AnalyticAt ℂ h_ρ ρ ∧ h_ρ ρ ≠ 0 ∧ ZeroFactor f ρ = h_ρ ρ ∧ f =ᶠ[nhds ρ] fun z ↦ (z - ρ) ^ analyticOrderNatAt f ρ * h_ρ z := by have zero_mem_closedBall : 0 ∈ Metric.closedBall (0 : ℂ) 1 := by rw[mem_closedBall_iff_norm, sub_zero, norm_zero] exact zero_le_one have ρ_mem_closedBall : ρ ∈ Metric.closedBall (0 : ℂ) 1 := by rw[mem_closedBall_iff_norm, sub_zero] linarith[hρ.1] have orderAtZeroIsZero : analyticOrderAt f 0 = 0 := by rw[analyticOrderAt_eq_zero] exact Or.symm (Decidable.not_or_of_imp fun a a_1 ↦ hf_neq_zero_at_zero a) have finiteOrder : analyticOrderAt f ρ ≠ ⊤ := by refine AnalyticOnNhd.analyticOrderAt_ne_top_of_isPreconnected hfAnalytic (Metric.isPreconnected_closedBall) zero_mem_closedBall ρ_mem_closedBall (lt_top_iff_ne_top.mp ?_) rw[orderAtZeroIsZero] exact ENat.top_pos have AnalyticAt_ρ : AnalyticAt ℂ f ρ := by exact (hfAnalytic ρ ρ_mem_closedBall) obtain ⟨h_ρ, h_ρ_neq_zero_at_zero, f_eq⟩ := (AnalyticAt_ρ.analyticOrderAt_ne_top.mp finiteOrder).choose_spec set g := (AnalyticAt_ρ.analyticOrderAt_ne_top.mp finiteOrder).choose refine ⟨g, h_ρ, h_ρ_neq_zero_at_zero, ?_, f_eq⟩ simp only [ZeroFactor, AnalyticAt_ρ, ↓reduceDIte, ne_eq, finiteOrder, not_false_eq_true, smul_eq_mul, g]