Equation eqn:phi0-F is clear. Equation eqn:psiS-G is already shown in Lemma Lemma 8.19.

By eqn:psiS-define from Definition 8.18, \psi_I(it) = (\psi_S|_{-2}S)(it) = (it)^{2}\psi_S\left(-\frac{1}{it}\right) = -t^2 \psi_S\left(\frac{i}{t}\right). Combined with Lemma Lemma 9.4 we can rewrite eqn:ineqA as A(t) = -t^2 \phi_0\left(\frac{i}{t}\right) + \frac{36}{\pi^2} \psi_S\left(\frac{i}{t}\right) < 0 \Leftrightarrow \frac{F(it)}{\Delta(it)} + \frac{18}{\pi^2} \frac{G(it)}{\Delta(it)} > 0 for t > 0, which is equivalent to eqn:ineqAnew by Corollary Corollary 7.25. Equivalences of eqn:ineqB and eqn:ineqBnew follows similarly; just change the sign.

By Ramanujan's identity Theorem 7.48, we have F(z) = 9 E_4'(z)^2, and hence F(it) = 9E_4'(it)^2 = 9 \left(240\sum_{n \geq 1} n \sigma_3(n) e^{-2 \pi n t} \right)^{2} > 0. The inequality G(it) > 0 follows from positivity of H_2(it) and H_4(it) in Corollary 7.42.

This directly follows from Lemma Lemma 9.6.

Both identities can be shown by direct computations. By Ramanujan's identities (Theorem Theorem 7.48) and the product rule for Serre derivatives (Theorem Theorem 7.51), we have \partial_{5} (E_2 E_4 - E_6) = (E_2 E_4 - E_6)' - \frac{5}{12} E_2 (E_2 E_4 - E_6) = \frac{E_2^2 - E_4}{12} \cdot E_4 + E_2 \cdot \frac{E_2 E_4 - E_6}{3} - \frac{E_2 E_6 - E_4^2}{2} - \frac{5}{12}E_2 (E_2 E_4 - E_6) = -\frac{5}{12} (E_2 E_6 - E_4^2) and \partial_{7} (E_2 E_6 - E_4^2) = (E_2 E_6 - E_4^2)' - \frac{7}{12} E_2 (E_2 E_6 - E_4^2) = \frac{E_2^2 - E_4}{12} \cdot E_6 + E_2 \cdot \frac{E_2 E_6 - E_4^2}{2} - 2 E_4 \cdot \frac{E_2 E_4 - E_6}{3} - \frac{7}{12} E_2 (E_2 E_6 - E_4^2) = -\frac{7}{12} E_4 (E_2 E_4 - E_6). Using these, we compute \partial_{10} F = \partial_{10} (E_2 E_4 - E_6)^2 = 2 (E_2 E_4 - E_6) \partial_{5} (E_2 E_4 - E_6) = -\frac{6}{5} (E_2 E_4 - E_6) (E_2 E_6 - E_4^2), and then \partial_{12}\partial_{10} F = -\frac{5}{6} \partial_{12} ((E_2 E_4 - E_6) (E_2 E_6 - E_4)) = -\frac{5}{6} (\partial_{5}(E_2 E_4 - E_6)) (E_2 E_6 - E_4^2) - \frac{5}{6} (E_2 E_4 - E_6) (\partial_{7} (E_2 E_6 - E_4)) = \frac{25}{72} (E_2 E_6 - E_4^2)^2 + \frac{35}{72} E_4 (E_2 E_4 - E_6)^2. Hence \partial_{12}\partial_{10}F - \frac{5}{6} E_4 F = \frac{25}{72} ((E_2 E_6 - E_4^2)^2 - E_4 (E_2 E_4 - E_6)^2) = \frac{25}{72} (- E_2^2 E_4^3 + E_2^2 E_6^2 + E_4^4 - E_4 E_6^3) = -\frac{25}{72} (E_4^3 - E_6^2) (E_2^2 - E_4) = 7200 \Delta (-E_2'). This proves equation eqn:ddf. The second is proved similarly, using Proposition Lemma 7.50 and Lemma Lemma 7.41.

From equation eqn:E2 in Definition 7.17 and Lemma Corollary 7.25, 7200 (-E_2'(it)) \Delta(it) = 7200 \cdot 24 \left(\sum_{n \ge 1} n \sigma_1(n) e^{-2 \pi n t}\right) \cdot \Delta(it) > 0. Negativity of equation eqn:ddg, namely -640 \Delta(it) H_2(it) < 0, follows from Corollary 7.42 and Corollary 7.25.

We have \lim_{t \to 0^+} Q(t) = \lim_{t \to 0^+} \frac{F(it)}{G(it)} = \lim_{t \to \infty} \frac{F(i/t)}{G(i/t)}. By using the transformation laws of Eisenstein series eqn:E2-S-transform, eqn:Ek-trans-S (for k = 4, 6) and the thetanull functions, eqn:H2-transform-S, eqn:H4-transform-S, we get F\left(\frac{i}{t}\right) = t^{12} F(it) - \frac{12t^{11}}{\pi} (E_2(it)E_4(it) - E_6(it))E_4(it) + \frac{36t^{10}}{\pi^2}E_4(it)^2 and G\left(\frac{i}{t}\right) = t^{10} H_{4}(it)^{3}(2H_{4}(it)^{2} + 5 H_{4}(it)H_{2}(it) + 5 H_{2}(it)^{2}). Since F, E_2 E_4 - E_6 and H_2 are cusp forms, we have \lim_{t \to \infty}t^k A(it) = 0 when A(z) is one of these forms and k \geq 0. From \lim_{t \to \infty} E_4(it) = 1 = \lim_{t \to \infty}H_{4}(it), we get \begin{aligned} \lim_{t \to \infty} \frac{F(i/t)}{G(i/t)} &= \lim_{t \to \infty} \frac{t^{12} F(it) - \frac{12t^{11}}{\pi} (E_2(it)E_4(it) - E_6(it))E_4(it) + \frac{36t^{10}}{\pi^2}E_4(it)^2}{t^{10} H_{4}(it)^{3}(2H_{4}(it)^{2} + 5 H_{4}(it)H_{2}(it) + 5 H_{2}(it)^{2})} \\ &= \lim_{t \to \infty} \frac{t^{2} F(it) - \frac{12t}{\pi} (E_2(it)E_4(it) - E_6(it))E_4(it) + \frac{36}{\pi^2}E_4(it)^2}{H_{4}(it)^{3}(2H_{4}(it)^{2} + 5 H_{4}(it)H_{2}(it) + 5 H_{2}(it)^{2})} \\ &= \frac{18}{\pi^2}. \end{aligned}

By Lemma Lemma 7.44, \lim_{t \to \infty} \frac{F'(it)}{F(it)} = \lim_{t \to \infty} \frac{\sum_{n \ge n_0} n a_n e^{-2 \pi n t}}{\sum_{n \ge n_0} a_n e^{-2 \pi n t}} = \lim_{t \to \infty} \frac{n_0 a_{n_0} e^{-2 \pi n_0 t} + O(e^{-2 \pi (n_0 + 1) t})}{a_{n_0} e^{-2 \pi n_0 t} + O(e^{-2 \pi (n_0 + 1) t})} = n_0.

It is enough to show that \frac{\dd}{\dd t} \left(\frac{F(it)}{G(it)}\right) < 0 \Leftrightarrow (- 2\pi) \frac{F'(it)G(it) - F(it) G'(it)}{G(it)^{2}} < 0 \Leftrightarrow F'(it) G(it) - F(it) G'(it) > 0 \Leftrightarrow (\partial_{10}F)(it) G(it) - F(it) (\partial_{10}G)(it) > 0. Let \mathcal{L}_{1, 0} := (\partial_{10}F) G - F (\partial_{10} G) = F'G - FG'. Then \frac{\mathcal{L}_{1, 0}}{FG} = \frac{F'G - FG'}{FG} = \frac{F'}{F} - \frac{G'}{G}. The vanishing orders of F and G at infinity are 2 and \frac{3}{2} respectively, so by Lemma 9.11 we have \lim_{t \to \infty} \frac{\mathcal{L}_{1, 0}(it)}{F(it) G(it)} = \lim_{t \to \infty} \left(\frac{F'(it)}{F(it)} - \frac{G'(it)}{G(it)}\right) = 2 - \frac{3}{2} = \frac{1}{2} > 0, so \mathcal{L}_{1, 0}(it) > 0 for sufficiently large t. Its Serre derivative is positive by Corollary 9.9: \partial_{22} \mathcal{L}_{1, 0} = (\partial_{12} \partial_{10} F) G - F (\partial_{12}\partial_{10} G) = \Delta (7200 (-E_{2}') G + 640 H_2 F) > 0. Hence \mathcal{L}_{1, 0}(it) > 0 by Theorem 7.52, and the monotonicity follows.

We have \frac{F(it)}{G(it)} = Q(t) < \lim_{u \to 0^+} Q(u) = \frac{18}{\pi^2}, and by Lemma Lemma 9.6 the desired inequality follows.

First, we prove the first sign condition. By Propositions Lemma 8.14 and Lemma 8.28 we know that for r>\sqrt{2}, g(r)=\frac{\pi}{2160}\,\sin(\pi r^2/2)^2\,\int\limits_0^\infty A(t)\,e^{-\pi r^2 t}\,dt where A(t)=-t^2\phi_0(i/t)-\frac{36}{\pi^2}\,\psi_I(it). From Proposition Lemma 9.1 we know that A(t)<0 for t\in(0,\infty), so this identity implies the first sign condition. /- source paragraph break -/ Next, we prove the Fourier-side sign condition. By Propositions Lemma 8.15 and Lemma 8.29 we know that for r>0, \widehat{g}(r)=\frac{\pi}{2160}\,\sin(\pi r^2/2)^2\,\int\limits_0^\infty B(t)\,e^{-\pi r^2 t}\,dt where B(t)=-t^2\phi_0(i/t)+\frac{36}{\pi^2}\,\psi_I(it). /- source paragraph break -/ Finally, the normalization g(0)=\widehat g(0)=1 follows readily from Propositions Lemma 8.16 and Lemma 8.30. This finishes the proof.