Bounding the derivative of H t

We continue using the notation from Effective bounds on H_t - second approach. We also assume [math]\displaystyle{ T \geq 10 }[/math] (so [math]\displaystyle{ x \geq 20 }[/math]).

Since

[math]\displaystyle{ H_t(z) = \frac{1}{8} \xi_t( s) }[/math]

with [math]\displaystyle{ s := \frac{1+iz}{2} }[/math], we have

[math]\displaystyle{ \frac{d}{dz} H_t(z) = \frac{i}{16} \frac{d}{ds} \xi_t(s). }[/math]

Next, we have

[math]\displaystyle{ \xi_t(s) = \sum_{n=1}^N F_{t,n}(s) + F_{t,n}(1-s) + G_{t,N}(s) + G_{t,N}(1-s) }[/math]

(using the convention [math]\displaystyle{ F(\bar{s}) = \bar{F(s)} }[/math] for [math]\displaystyle{ s }[/math] in the lower half-plane). Thus (assuming that we are not at a discontinuity for $latex N$) we have

[math]\displaystyle{ \frac{d}{ds} \xi_t(s) = \sum_{n=1}^N F'_{t,n}(s) - F'_{t,n}(1-s) + G'_{t,N}(s) - G'_{t,N}(1-s). }[/math]

Now we have for any [math]\displaystyle{ \alpha_n }[/math] that

[math]\displaystyle{ F_{t,n}(s) = \exp( - \frac{t}{4} \alpha_n^2 ) \int_{-\infty}^\infty \exp( - \sqrt{t} \alpha_n u) F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha_n ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du, }[/math]

hence in differentiation under the integral sign (justifiable for instance using the Cauchy integral formula and Fubini's theorem)

[math]\displaystyle{ F'_{t,n}(s) = \exp( - \frac{t}{4} \alpha_n^2 ) \int_{-\infty}^\infty \exp( - \sqrt{t} \alpha_n u) \frac{\partial}{\partial s} F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha_n ) \frac{1}{\sqrt{\pi}} e^{-u^2}\ du. }[/math]

This identity is true for any [math]\displaystyle{ \alpha_n }[/math]; we now set [math]\displaystyle{ \alpha_n = \alpha_n(s) }[/math] as in the above wiki page. One can replace [math]\displaystyle{ \frac{\partial}{\partial s} }[/math] on the RHS by [math]\displaystyle{ \frac{1}{\sqrt{t}} \frac{\partial}{\partial u} }[/math] and integrate by parts to conclude that

[math]\displaystyle{ F'_{t,n}(s) = \exp( - \frac{t}{4} \alpha_n(s)^2 ) \int_{-\infty}^\infty \exp( - \sqrt{t} \alpha_n(s) u) F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha_n ) \frac{1}{\sqrt{\pi}} (\alpha_n(s) + \frac{2u}{\sqrt{t}}) e^{-u^2}\ du. }[/math]

We have

[math]\displaystyle{ F_{0,n}( s + \sqrt{t} u + \frac{t}{2} \alpha_n(s)) = H_{0,n}(s) \exp( (\sqrt{t} u + \frac{t}{2} \alpha_n(s)) \alpha_n(s) + O_{\leq}( \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) ) ) }[/math]

and hence

[math]\displaystyle{ |F'_{t,n}(s)| \leq \exp( \frac{t}{4} \mathrm{Re}(\alpha_n(s)^2) ) |H_{0,n}(s)| \int_{-\infty}^\infty \exp( \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) ) ) \frac{1}{\sqrt{\pi}} (|\alpha_n(s)| + |\frac{2u}{\sqrt{t}}|) e^{-u^2}\ du. }[/math]

We can bound

[math]\displaystyle{ \frac{1}{4 (T - 3.08)} ( |\sqrt{t} u + \frac{t}{2} \alpha_n(s)|^2 + \frac{2}{3} ) \leq \frac{1}{2(T-3.08)} ( tu^2 + \frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3} ) }[/math]

to obtain

[math]\displaystyle{ |F'_{t,n}(s)| \leq \frac{\exp( \frac{t}{4} \mathrm{Re}(\alpha_n(s)^2) )}{n^\sigma} |H_{0,1}(s)| \frac{\exp( \frac{1}{2(T-3.08)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3}) ) }{\sqrt{\pi}} \int_{-\infty}^\infty (|\alpha_n(s)| + |\frac{2u}{\sqrt{t}}|) \exp( -(1-\frac{t}{2(T-3.08)}) u^2 )\ du. }[/math]

Noting that

[math]\displaystyle{ \int_{-\infty}^\infty e^{-a u^2}\ du = a^{-1/2} \sqrt{\pi} }[/math]

and

[math]\displaystyle{ \int_{-\infty}^\infty e^{-a u^2} |u|\ du = a^{-1} }[/math]

for any [math]\displaystyle{ a\gt 0 }[/math], we thus have

[math]\displaystyle{ |F'_{t,n}(s)| \leq \frac{\exp( \frac{t}{4} \mathrm{Re}(\alpha_n(s)^2) )}{n^\sigma} |H_{0,1}(s)| \exp( \frac{1}{2(T-3.08)} (\frac{t^2}{4} |\alpha_n(s)|^2 + \frac{1}{3}) ) ( |\alpha_n(s)| (1-\frac{t}{2(T-3.08)})^{-1/2} + \frac{2}{\sqrt{\pi t}} (1-\frac{t}{2(T-3.08)})^{-1}). }[/math]

We can bound [math]\displaystyle{ |\alpha_n(s)| \leq |\alpha_1(s)| }[/math] and [math]\displaystyle{ T \geq T_N }[/math] and conclude that

[math]\displaystyle{ \sum_{n=1}^N |F'_{t,n}(s)| \leq |H_{0,1}(s)| \exp( \frac{1}{2(T_N-3.08)} (\frac{t^2}{4} |\alpha_1(s)|^2 + \frac{1}{3}) ) ( |\alpha_1(s)| (1-\frac{t}{2(T_N-3.08)})^{-1/2} + \frac{2}{\sqrt{\pi t}} (1-\frac{t}{2(T_N-3.08)})^{-1}) \sum_{n=1}^N \frac{\exp( \frac{t}{4} \mathrm{Re}(\alpha_n(s)^2) )}{n^\sigma}. }[/math]