Research proof index

H₂(it) > 0 for all t > 0. Blueprint: Lemma 6.43 - H₂ is positive on the imaginary axis. Proof strategy: Each term exp(-π(n+1/2)²t) > 0, so Θ₂(it) > 0, hence H₂ = Θ₂^4 > 0.

Use jacobiTheta₂_functional_equation

These three transformation laws follow directly from tsum definition.

H₄(it) > 0 for all t > 0. Blueprint: Corollary 6.43 - H₄ is positive on the imaginary axis. Proof strategy: Use the modular S-transformation relating H₄ and H₂. From H₄_S_action: (H₄ ∣[2] S) = -H₂ From ResToImagAxis.SlashActionS: relates values at t and 1/t. This gives H₂(i/t) = t² * H₄(it), so H₄(it) > 0 follows from H₂(i/t) > 0.

If F has a Fourier expansion ∑_{m≥0} a_m exp(2πi(m+n₀)z) with n₀ > 0, and the coefficients are absolutely summable at height im z = c, then F = O(exp(-2π n₀ · im z)) at atImInfty. The key bound is: for im z ≥ c, ‖F(z)‖ ≤ (∑_m ‖a_m‖ · exp(-2π c m)) · exp(-2π n₀ · im z)

If F : ℍ → ℂ is O(exp(-c * im τ)) at atImInfty for some c > 0, then the restriction to the imaginary axis t ↦ F(it) is O(exp(-c * t)) at atTop.