$\tilde{E}(t_{k},s)=\frac{\alpha_{t_{k}}\gamma(t_{k}y_{t_{k}})}{\sqrt{\tilde{w}(s)}}\cos[t_{k}(s-(\tilde{x}_{t_{k}}-iy_{t_{k}}))].$

By definition of the transfer matrix $M_{t_{1}\rightarrow t_{2}}$ and
Corollary 2 we find

$M_{t_{1}\rightarrow t_{2}}(z)=M(t_{2},1)M^{-1}(t_{1},z)$

$=\frac{1}{2}\begin{pmatrix}1&&1\\\
i&&-i\end{pmatrix}\begin{pmatrix}E(t_{2},z)&&\tilde{E}(t_{2},z)\\\
E^{\sharp}(t_{2},z)&&\tilde{E}^{\sharp}(t_{2},z)\end{pmatrix}\begin{pmatrix}E(t_{1},z)&&\tilde{E}(t_{1},z)\\\
E^{\sharp}(t_{1},z)&&\tilde{E}^{\sharp}(t_{1},z)\end{pmatrix}^{-1}2\begin{pmatrix}1&&-i\\\
1&&i\end{pmatrix}^{-1}.$

Again by Corollary 2

$\begin{pmatrix}E_{t_{1}\rightarrow t_{2}}&&\tilde{E}_{t_{1}\rightarrow
t_{2}}\\\ E_{t_{1}\rightarrow t_{2}}^{\sharp}&&\tilde{E}_{t_{1}\rightarrow
t_{2}}^{\sharp}\end{pmatrix}=2\begin{pmatrix}1&&1\\\
i&&-i\end{pmatrix}^{-1}M_{t_{1}\rightarrow t_{2}}(z).$

$=\begin{pmatrix}E(t_{2},z)&&\tilde{E}(t_{2},z)\\\
E^{\sharp}(t_{2},z)&&\tilde{E}^{\sharp}(t_{2},z)\end{pmatrix}\frac{1}{2i}\begin{pmatrix}\tilde{E}^{\sharp}(t_{1},z)&&-\tilde{E}(t_{1},z)\\\
-E^{\sharp}(t_{1},z)&&E(t_{1},z)\end{pmatrix}\begin{pmatrix}1&&-i\\\
1&&i\end{pmatrix}.$

By definition of $a_{t_{1}\rightarrow t_{2}}$ we obtain

$a_{t_{1}\rightarrow
t_{2}}(s)=\frac{e^{i(t_{2}-t_{1})z}}{2}(E_{t_{1}\rightarrow
t_{2}}(z)+i\tilde{E}_{t_{1}\rightarrow t_{2}}(z))$
$=\frac{e^{i(t_{2}-t_{1})z}}{4i}\bigg{(}\big{[}(E(t_{2},z)\tilde{E}^{\sharp}(t_{1},z)-\tilde{E}(t_{2},z)E^{\sharp}(t_{1},z))+(-E(t_{2},z)\tilde{E}(t_{1},z)+\tilde{E}(t_{2},z)E(t_{1},z))\big{]}$
$+i\big{[}-i(E(t_{2},z)\tilde{E}^{\sharp}(t_{1},z)-\tilde{E}(t_{2},z)E^{\sharp}(t_{1},z))+i(-E(t_{2},z)\tilde{E}(t_{1},z)+\tilde{E}(t_{2},z)E(t_{1},z))\big{]}\bigg{)}$
$=\frac{e^{i(t_{2}-t_{1})z}}{2i}(E(t_{2},z)\tilde{E}^{\sharp}(t_{1},z)-\tilde{E}(t_{2},z)E^{\sharp}(t_{1},z)).$

We will simplify notation by putting
$x_{k}=x_{t_{k}}-s,\tilde{x}_{k}=\tilde{x}_{t_{k}}-s$ and $y_{k}=y_{t_{k}}$
for $k=1,2.$ Thus, $t_{k}(s-(x_{t_{k}}-iy_{t_{k}}))=-t_{k}(x_{k}-iy_{k})$ and
$t_{k}(s-(\tilde{x}_{t_{k}}-iy_{t_{k}}))=-t_{k}(\tilde{x}_{k}-iy_{k})$.

Hence, by using Theorem 4

$a_{t_{1}\rightarrow
t_{2}}(s)=-\frac{e^{i(t_{2}-t_{1})s}}{2i}\bigg{(}\frac{\alpha_{t_{2}}\gamma(t_{2}y_{2})}{\sqrt{w(s)}}\sin[t_{2}(x_{2}-iy_{2})]\frac{\overline{\alpha}_{t_{1}}\gamma(t_{1}y_{1})}{\sqrt{\tilde{w}(s)}}\cos[t_{1}(\tilde{x}_{1}+iy_{1})]$
$-\frac{\alpha_{t_{2}}\gamma(t_{2}y_{2})}{\sqrt{\tilde{w}(s)}}\cos[t_{2}(\tilde{x}_{2}-iy_{2})]\frac{\overline{\alpha}_{t_{1}}\gamma(t_{1}y_{1})}{\sqrt{w(s)}}\sin[t_{1}(x_{1}+iy_{1})]\bigg{)}$

(34)
$=-\frac{e^{i(t_{2}-t_{1})s}}{2i}\frac{\gamma(t_{1}y_{1})\gamma(t_{2}y_{2})}{\sqrt{w(s)\tilde{w}(s)}}\big{|}\sin[t_{2}(x_{2}-iy_{2})]\cos[t_{1}(\tilde{x}_{1}+iy_{1})]-\cos[t_{2}(\tilde{x}_{2}-iy_{2})]\sin[t_{1}(x_{1}+iy_{1})]\big{|}.$

After using that

$\cos z=\frac{e^{iz}+e^{-iz}}{2},\quad\sin z=\frac{e^{iz}-e^{-iz}}{2i}$

and rearranging the terms, the expression inside the bracket in (34) is equal
to

$\frac{1}{2}\bigg{(}\sin[t_{2}(x_{2}-iy_{2})+t_{1}(\tilde{x}_{1}+iy_{1})]+\sin[t_{2}(x_{2}-iy_{2})-t_{1}(\tilde{x}_{1}+iy_{1})]$
$-\sin[t_{1}(x_{1}+iy_{1})+t_{2}(\tilde{x}_{2}-iy_{2})]-\sin[t_{1}(x_{1}+iy_{1})-t_{2}(\tilde{x}_{2}-iy_{2})]\bigg{)}$

Recall that for any two complex numbers $z,w\in\mathbb{C},$

$\sin z-\sin
w=2\sin\bigg{(}\frac{z-w}{2}\bigg{)}\cos\bigg{(}\frac{z+w}{2}\bigg{)}.$

Hence, the last expression can be written as

$=\cos\bigg{[}\frac{1}{2}\big{(}[t_{2}(x_{2}-iy_{2})+t_{1}(\tilde{x}_{1}+iy_{1})]+[t_{1}(x_{1}+iy_{1})+t_{2}(\tilde{x}_{2}-iy_{2})]\big{)}\bigg{]}$
$\times\sin\bigg{[}\frac{1}{2}\big{(}[t_{2}(x_{2}-iy_{2})+t_{1}(\tilde{x}_{1}+iy_{1})]-[t_{1}(x_{1}+iy_{1})+t_{2}(\tilde{x}_{2}-iy_{2})]\big{)}\bigg{]}$
$+\cos\bigg{[}\frac{1}{2}\big{(}[t_{2}(x_{2}-iy_{2})-t_{1}(\tilde{x}_{1}+iy_{1})]+[t_{1}(x_{1}+iy_{1})-t_{2}(\tilde{x}_{2}-iy_{2})]\big{)}\bigg{]}$
$\times\sin\bigg{[}\frac{1}{2}\big{(}[t_{2}(x_{2}-iy_{2})-t_{1}(\tilde{x}_{1}+iy_{1})]-[t_{1}(x_{1}+iy_{1})-t_{2}(\tilde{x}_{2}-iy_{2})]\big{)}\bigg{]}$
$=\cos\bigg{[}\frac{1}{2}(t_{2}(x_{2}+\tilde{x}_{2})+t_{1}(x_{1}+\tilde{x}_{1}))-i(t_{2}y_{2}-t_{1}y_{1})\bigg{]}\times\sin\bigg{[}\frac{1}{2}(t_{2}(x_{2}-\tilde{x}_{2})-t_{1}(x_{1}-\tilde{x}_{1}))\bigg{]}$
$+\cos\bigg{[}\frac{1}{2}(t_{2}(x_{2}-\tilde{x}_{2})+t_{1}(x_{1}-\tilde{x}_{1}))\bigg{]}\times\sin\bigg{[}\frac{1}{2}(t_{2}(x_{2}+\tilde{x}_{2})-t_{1}(x_{1}+\tilde{x}_{1}))-i(t_{2}y_{2}+t_{1}y_{1})\bigg{]}.$

By Theorem 4 (ii) we have
$\cos[t(\tilde{x}_{t}-x_{t})]=\sqrt{w(s)\tilde{w}(s)}$ for $t\in[t_{1},t_{2}]$
and moreover since $x_{t_{1}}$ moves continuously to $x_{t_{2}}$ we find
$t_{2}(x_{2}-\tilde{x}_{2})=t_{1}(x_{1}-\tilde{x}_{1}).$ Thus, the last
expression simplifies further to

$\sqrt{w(s)\tilde{w}(s)}\sin\bigg{[}\frac{1}{2}(t_{2}(x_{2}+\tilde{x}_{2})-t_{1}(x_{1}+\tilde{x}_{1}))-i(t_{2}y_{2}+t_{1}y_{1})\bigg{]}.$

Plugging this back into (34) and using $\sin(ix)=i\sinh(x)$ we obtain

$a_{t_{1}\rightarrow
t_{2}}(s)=-\frac{e^{i(t_{2}-t_{1})s}\alpha_{t_{2}}\overline{\alpha}_{t_{1}}}{i\sqrt{\sin[2it_{1}y_{1}]\sin[2it_{2}y_{2}]}}\big{(}\sin[(t_{2}x_{2}-t_{1}x_{1})-i(t_{2}y_{2}+t_{1}y_{1})]\big{)}.$

We now substitute
$t_{1}x_{1}=u,t_{2}x_{2}=u+\varepsilon_{1},t_{1}y_{1}=v,t_{2}y_{2}=v+\varepsilon_{2}.$
Our last equation can then be written as

$a_{t_{1}\rightarrow
t_{2}}(s)=-\frac{e^{i(t_{2}-t_{1})s}\alpha_{t_{2}}\overline{\alpha}_{t_{1}}}{i\sqrt{\sin[2iv]\sin[2i(v+\varepsilon_{2})]}}\sin[\varepsilon_{1}-2iv\color[rgb]{1,0,0}-\color[rgb]{0,0,0}i\varepsilon_{2}].$

We are going to simply this further to

$a_{t_{1}\rightarrow
t_{2}}(s)=e^{i(t_{2}-t_{1})s}\alpha_{t_{2}}\overline{\alpha}_{t_{1}}\bigg{[}1+i\varepsilon_{1}\coth[2v]+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2})\bigg{]},$

where the implicit constant depends only on $s$ and $A.$ Recall first that
$v=t_{1}y_{1}$ and $t_{2}y_{2}$ satisfy the bounds $2<v<A,$ so the terms with
$\sin[2iv]$ and $\sin[2i(v+\varepsilon_{2})]$ are bounded away from zero.
Further recall the identities

(35) $\sin(ix)=i\sinh(x),\quad\cos(ix)=\cosh(x),\text{ for all
}x\in\mathbb{R},$

(36) $\sin(z\pm w)=\sin z\cos w\pm\cos z\sin w,\text{ for all
}z,w\in\mathbb{C},$

(37) $\sinh(z\pm w)=\sinh z\cosh w\pm\sinh w\cosh z,\text{ for all
}z,w\in\mathbb{C},$

(38) $\cosh(z\pm w)=\cosh z\cosh w\pm\sinh z\sinh w,\text{ for all
}z,w\in\mathbb{C}.$

We decompose using (36)

$a_{t_{1}\rightarrow
t_{2}}(s)=\frac{1}{i\sqrt{\sin[2iv]\sin[2i(v+\varepsilon_{2})]}}\bigg{[}\sin\varepsilon_{1}\cos(2iv+i\varepsilon_{2})-\cos\varepsilon_{1}\sin(2iv+i\varepsilon_{2})\bigg{]}.$

We are going to approximate both summands seperately. For the first one we
decompose further using (35) and (38) to arrive at

$\sin\varepsilon_{1}\cosh(2v+\varepsilon_{2})=\sin\varepsilon_{1}\cosh
2v\cosh\varepsilon_{2}+\sin\varepsilon_{1}\sinh 2v\sinh\varepsilon_{2}.$

We thus have to approximate the following terms for the first summand.

(39) $\sqrt{\sinh
2v}\coth[2v]\sin\varepsilon_{1}\frac{\cosh\varepsilon_{2}}{\sinh[2(v+\varepsilon_{2})]},$

(40) $\sqrt{\sinh
2v}\sin\varepsilon_{1}\frac{\sinh\varepsilon_{2}}{\sqrt{\sinh[2(v+\varepsilon_{2})]}}.$

By Taylor’s Theorem we have that (39) and (40) are respectively

$\sqrt{\sinh
2v}\coth[2v]\sin\varepsilon_{1}\frac{\cosh\varepsilon_{2}}{\sinh[2(v+\varepsilon_{2})]}$
$=\sqrt{\sinh
2v}\coth[2v](\varepsilon_{1}+O(\varepsilon_{1}^{2}))\bigg{(}\frac{1}{\sqrt{\sinh
2v}}+O(\varepsilon_{2})\bigg{)}=\varepsilon_{1}\coth[2v]+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}),$

$\sqrt{\sinh
2v}\sin\varepsilon_{1}\frac{\sinh\varepsilon_{2}}{\sqrt{\sinh[2(v+\varepsilon_{2})]}}$
$=\sqrt{\sinh
2v}(\varepsilon_{1}+O(\varepsilon_{1}^{2}))\bigg{(}\frac{\varepsilon_{2}}{\sqrt{\sinh
2v}}+O(\varepsilon_{2}^{2})\bigg{)}=O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}).$

Thus, the first summand equals

(41) $\varepsilon_{1}\coth[2v]+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}).$

For the second summand we obtain using (35) and (37)

$\cos\varepsilon_{1}\sin(2iv+i\varepsilon_{2})=i(\cos\varepsilon_{1}\sinh
2v\cosh\varepsilon_{2}+\cos\varepsilon_{1}\sinh\varepsilon_{2}\cosh 2v).$

The first term equals using Taylor’s Theorem

$\sqrt{\sinh
2v}\cos\varepsilon_{1}\frac{\cosh\varepsilon_{2}}{\sqrt{\sinh[2(v+\varepsilon_{2})]}}$

$=\sqrt{\sinh 2v}(1+O(\varepsilon_{1}^{2}))\bigg{(}\frac{1}{\sqrt{\sinh
2v}}-\varepsilon_{2}\frac{\cosh 2v}{(\sinh
2v)^{\frac{3}{2}}}+O(\varepsilon_{2}^{2})\bigg{)}=1-\varepsilon_{2}\coth[2v]+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}).$

By another application of Taylor’s Theorem the second term equals

$\sqrt{\sinh
2v}\cos\varepsilon_{1}\coth[2v]\frac{\sin\varepsilon_{2}}{\sqrt{\sinh[2(v+\varepsilon_{2})]}}=$
$\sqrt{\sinh
2v}\coth[2v](1+O(\varepsilon_{2}^{2}))\bigg{(}\frac{\varepsilon_{2}}{\sqrt{\sinh
2v}}+O(\varepsilon_{2}^{2})\bigg{)}=\varepsilon_{2}\coth[2v]+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}).$

Hence, the second summand is

(42) $i+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}).$

Thus, by combining (41) and (42) we find as claimed

$a_{t_{1}\rightarrow
t_{2}}(s)=-\frac{1}{i}e^{i(t_{2}-t_{1})s}\alpha_{t_{2}}\overline{\alpha}_{t_{1}}\bigg{[}\varepsilon_{1}\coth[2v]-i+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2})\bigg{]}$
$=e^{i(t_{2}-t_{1})s}\alpha_{t_{2}}\overline{\alpha}_{t_{1}}\bigg{[}1+i\varepsilon_{1}\coth[2v]+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2})\bigg{]}.$

∎

One can show that on $H_{s}-$ and $\sigma-$ intervals we have the bound

$\int_{t_{1}}^{t_{2}}|f(t)|dt\ll\varepsilon_{1}.$

It was thus desirable that $\log a_{t_{1}\rightarrow t_{2}}$ was linear in
$\varepsilon_{1}$ to otbain by linearity of the integral that

$||\log a_{t_{1}\rightarrow
t_{2}}||_{L^{1}(\mathbb{R})}\gg\varepsilon_{1}\gg\int_{t_{1}}^{t_{2}}|f(t)|dt$

and deduce our result by a standard covering argument. However, this does not
quite work as the asymptotics from the previous Lemma show. Indeed, because we
have a non-trivial imaginary part the norm is giving us a quadratic
contribution in $\varepsilon_{2}$ and thus we can not exploit linearity of the
integral anymore. This was a mistake in older versions of [10] which was fixed
by using the following idea. While it turns out that $|\log
a_{t_{1}\rightarrow t_{2}}(s)|$ is quadratic in $\varepsilon_{1}$, we will
show that $\arg a_{t_{1}\rightarrow t_{2}}$ is barely enough linear on some
subset in the base of the box, so we can show on the interval $(t_{1},t_{2})$
an estimate of the form

$|\arg a_{t_{1}\rightarrow t_{2}}(s)|\gg\int_{t_{1}}^{t_{2}}|f(t)|dt.$

We now note that $\arg a_{t_{1}\rightarrow t_{2}}$ is a harmonic conjugate
$\log|a_{t_{1}\rightarrow t_{2}}|$ and thus may differ only by a constant from
the Hilbert transform of $\log|a_{t_{1}\rightarrow t_{2}}|.$ Since $\arg
a_{t_{1}\rightarrow t_{2}}(0)=0$ the constant is zero. Let $R$ denote the
subset where the above inequality holds. Thus, we can estimate the weak
$L^{1}-$norm of $\arg a_{t_{1}\rightarrow t_{2}}(s)$ from below by

$||\arg a_{t_{1}\rightarrow
t_{2}}(s)||_{L^{1,\infty}(\mathbb{R})}>|R|\int_{t_{1}}^{t_{2}}|f(t)|dt.$

Since the Hilbert transform is of weak type $(1,1)$ we find that

(43) $||\log|a_{t_{1}\rightarrow t_{2}}||_{L^{1}(\mathbb{R})}\gg||\arg
a_{t_{1}\rightarrow
t_{2}}(s)||_{L^{1,\infty}(\mathbb{R})}>|R|\int_{t_{1}}^{t_{2}}|f(t)|dt.$

This saves the idea of our argument to exploit the linearity of the integral
and the non-linear Parseval identity via a covering type argument. We are now
going to make these ideas rigorous.

######  Lemma 40.

Suppose that $t_{k}y_{k}>A/2$ for $k=1,2$ and that $(t_{1},t_{2})$ is a
$H_{s}-$ and $\sigma-$interval. Then, on some subset $R$ of the interval
$I_{t_{2}}=(s-\frac{10}{t_{2}},s+\frac{10}{t_{2}})$ of measure $|R|>D/t_{2},$
we have the lower bound

$|\arg a_{t_{1}\rightarrow t_{2}}|\geq
D|\varepsilon_{1}|+o(|\varepsilon_{1}|+|\varepsilon_{2}|),$

where $D>0$ is an absolute constant depending only on $A.$

###### Proof.

By the last lemma we have the asymptotics

$a_{t_{1}\rightarrow
t_{2}}(s)=e^{i(t_{2}-t_{1})s}\alpha_{t_{2}}\overline{\alpha}_{t_{1}}\bigg{(}1+i\varepsilon_{1}\coth[2t_{1}y_{1}]+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2})\bigg{)}.$

To estimate $\arg a_{t_{1}\rightarrow t_{2}}$ it is enough to work out the
argument of each term seprerately by additivity. We have trivially that

$\arg e^{i(t_{2}-t_{1})s}=(t_{2}-t_{1})s.$

By Theorem 4 we have that

(44)
$E(t_{k},s)=\alpha_{t_{k}}\sin[t_{k}(iy_{k}-x_{k})]=\alpha_{t_{k}}\sin[t_{k}(s-(x_{t_{k}}-iy_{t_{k}}))].$

We set $\phi(t,s)=\arg E(t,s).$ We know that $E(t,z)$ satisfies the
differential equation

$\frac{d}{dt}E(t,z)=-izE(t,z)+f(t)E^{\sharp}(t,z).$

Thus, we can use the same strategy as in the proof of Lemma 33 to calculate

$\phi(t,s)=-ts-\int_{0}^{t}f(t)\sin(2\phi(t,s))dt.$

Hence, by (44)

$\arg\alpha_{t_{k}}=\phi(t_{k},s)-\arg\sin[t_{k}(iy_{k}-x_{k})]$

By direct computations one shows that

$\arg\sin[t_{k}(iy_{k}-x_{k})]=\arg\sin[t_{k}(s-(x_{t_{k}}-iy_{t_{k}}))]=t_{k}(s-x_{k})-\frac{\pi}{2}+\eta(t_{k},s),$

where $\eta(t_{k},\cdot)$ is a $\pi/t_{k}-$periodic function whose integral
over each period vanishes. Since $(t_{1},t_{2})$ is an $H_{s}-$interval we
have the bound following bound on $\mathbb{R}$ for the difference function
$\eta(s)=\eta(t_{2},s)-\eta(t_{1},s)$

$|\eta(s)|\ll|\varepsilon_{1}|+|\varepsilon_{2}|\ll|\varepsilon_{1},$

where the implicit constant depends only on $A.$ Combining everything, we
obtain

$\arg a_{t_{1}\rightarrow
t_{2}}(s)=(t_{2}-t_{1})s+(\phi(t_{2},s)-\phi(t_{1},s))-\eta(s)-\varepsilon_{1}+\varepsilon_{1}\coth[2v]+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2})$
$=(\coth[2v]-1)\varepsilon_{1}-\int_{t_{1}}^{t_{2}}f(t)\sin(2\phi(t,s))dt-\eta(s)+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}).$

We now need to make sure that the oscillating term does not cancel the linear
term in $\varepsilon_{1}$ completely. Since by assumption $2v>A>0,$ we have
$\coth[2v]-1>0.$ Suppose that $\varepsilon_{1},\varepsilon_{2}$ and
$\int_{t_{1}}^{t_{2}}f(t)dt$ are all positive. We can treat other cases
similarly. Now as in the upcoming proof of Lemma 41, using that
$(t_{1},t_{2})$ is an $H_{s}-$interval and a $\sigma-$interval we can show
that

(45)
$\int_{t_{1}}^{t_{2}}|f(t)|dt\ll\int_{t_{1}}^{t_{2}}f(t)dt\ll\varepsilon_{1}.$

Together with the previously established bound $|\eta(s)|\ll\varepsilon_{1}$
we find that

(46)
$\bigg{|}\int_{t_{1}}^{t_{2}}f(t)\sin(2\phi(t,s))dt+\eta(s)\bigg{|}\ll\varepsilon_{1}.$

We note that Theorem 4 can be applied to any point $u$ in $I_{t_{2}}$ with the
same zeros $\xi_{k}$ and $\tilde{\xi}_{k}$ of $E(t,z)$ and $\tilde{E}(t,z),$
respectively. We denote by $x_{t_{k}}^{u},\tilde{x}_{t_{k}}^{u}$ and
$y_{t_{k}}^{u}$ the corresponding approximating zeros from the Theorem at the
point $u.$ Thus, by the very same calculations with the point $u$ instead of
$s$ we find again that

$\arg a_{t_{1}\rightarrow
t_{2}}(u)=(\coth[2v]-1)\varepsilon_{1}^{u}-\int_{t_{1}}^{t_{2}}f(t)\sin(2\phi(t,u))dt-\eta(u)+O((\varepsilon_{1}^{u})^{2}+(\varepsilon_{2}^{u})^{2}).$

Our goal is now to show that the left hand side of (46) is less than
$(\coth[2v]-1)\varepsilon_{1}^{u}/2$ on a large subset of $I_{t_{2}}(s).$ Due
to Corollary 10 (i) we have for each fixed $t\in[t_{1},t_{2}]$ that

$\sin(2\phi(t,u))=\sin(2\arg E(t,u))$

differs from a $\pi/t_{k}-$periodic function of $u$ with vanishing integral
over each period by at most $o(1).$ Recall that the integral of
$\eta(t_{k},\cdot)$ vanishes as well over each period. Hence,

$\bigg{|}\int_{s-\frac{\pi}{t_{2}}}^{s+\frac{\pi}{t_{2}}}\int_{x-\frac{\pi}{t_{1}}}^{x+\frac{\pi}{t_{1}}}\bigg{[}\int_{t_{1}}^{t_{2}}f(t)\sin(2\phi(t,u))dt+\eta(u)\bigg{]}dudx\bigg{|}$
$=\bigg{|}\int_{s-\frac{\pi}{t_{2}}}^{s+\frac{\pi}{t_{2}}}\int_{x-\frac{\pi}{t_{1}}}^{x+\frac{\pi}{t_{1}}}\bigg{[}\int_{t_{1}}^{t_{2}}f(t)\sin(2\phi(t,u))dt+\eta(t_{2},u)-\eta(t_{1},u)\bigg{]}dudx\bigg{|}$
$=\bigg{|}\int_{s-\frac{\pi}{t_{2}}}^{s+\frac{\pi}{t_{2}}}\int_{x-\frac{\pi}{t_{1}}}^{x+\frac{\pi}{t_{1}}}\bigg{[}\int_{t_{1}}^{t_{2}}f(t)\sin(2\phi(t,u))dt\bigg{]}dudx\bigg{|}$

By Fubini’s Theorem we can change the order of integration to arrive at

$=\bigg{|}\int_{s-\frac{\pi}{t_{2}}}^{s+\frac{\pi}{t_{2}}}\int_{t_{1}}^{t_{2}}f(t)\bigg{[}\int_{(x-\frac{\pi}{t},x+\frac{\pi}{t})}\sin(2\phi(t,u))du+\int_{(x-\frac{\pi}{t_{1}},x+\frac{\pi}{t_{1}})\setminus(x-\frac{\pi}{t},x+\frac{\pi}{t})}\sin(2\phi(t,u))du\bigg{]}dtdx\bigg{|}$

Since $\sin(2\phi(t,u))=\sin(2\arg E(t,u))$differs from a $\pi/t_{k}-$periodic
function of $u$ with vanishing integral over each period by at most $o(1)$ and
due to (45) we arrive at

$\ll\varepsilon_{1}\frac{o(1)}{t_{1}t_{2}}+\varepsilon_{1}\frac{t_{2}-t_{1}}{t_{1}t_{2}^{2}}.$

Thus, the first integral is for large enough $t_{1}$ less than
$\varepsilon_{1}(\coth[2v]-1)/t_{2}^{2}.$ Since by (46), which is true for any
point $u$ in the interval $I_{t_{2}}(s)$ we get

$\bigg{|}\int_{t_{1}}^{t_{2}}f(t)\sin(2\phi(t,u))dt+\eta(u)\bigg{|}\ll\varepsilon_{1}^{u},$

combining both estimates shows that the left hand side must be less than
$\varepsilon_{1}^{u}(\coth[2v]-1)/2$ on a set

$R\subset\bigg{(}s-\frac{\pi(t_{1}+t_{2})}{t_{1}t_{2}},s+\frac{\pi(t_{1}+t_{2})}{t_{1}t_{2}}\bigg{)},\quad|R|\asymp\frac{1}{t_{2}}.$

Therefore, on $R$ we have the lower bound

(47) $\arg a_{t_{1}\rightarrow
t_{2}}(u)\geq\frac{\coth[2v]-1}{2}\varepsilon_{1}^{u}+O((\varepsilon_{1}^{u})^{2}+(\varepsilon_{2}^{u})^{2}).$

In a last step we will have to make this lower bound uniform over $R,$ so the
constant does only depend on $A$ as desired. To that end, note that for large
enough $A$ the box $Q(s,A/t)$ contains multiple zeros of the approximating
functions and, switching to the next zero if necessary we can assume with out
loss of generality that $x_{t_{k}}^{u}-u>x_{t_{k}}-s.$ Indeed, this just boils
down to the fact that $\sin(tz)$ whose zeros differ by exactly $k\pi/t,$ for
$k\in\mathbb{Z}.$ If now $A$ is large, a lot of these zeros are in the box and
you can switch to any zeros with out changing the approximating function. By
Theorem 4

$(x_{t_{2}}^{u}-iy_{t_{2}}^{u})-(x_{t_{1}}^{u}-iy_{t_{1}}^{u})=\xi_{2}-\xi_{1}+o(\xi_{2}-\xi_{1}),$
$(\tilde{x}_{t_{2}}^{u}-iy_{t_{2}}^{u})-(\tilde{x}_{t_{1}}^{u}-iy_{t_{1}}^{u})=\tilde{\xi}_{2}-\tilde{\xi}_{1}+o(\xi_{2}-\xi_{1}),$

Again these relations do hold for all choices of the zeros of the
approximating functions in the box $Q(s,A/t).$ This follows from the proof of
Theorem 4, more precisely (74). Therefore, there exists $c>0$ such that

$\varepsilon_{1}^{u}=t_{2}(x_{t_{2}}^{u}-u)-t_{1}(x_{t_{1}}^{u}-u)=$
$t_{2}(x_{t_{2}}-s+c)-t_{1}{x_{t_{1}}-s+c}+o(\xi_{2}-\xi_{1})>\varepsilon_{1}+o(|\varepsilon_{1}|+|\varepsilon_{2}|).$

Combining this estimate with (47) gives uniform on $R$

$|\arg a_{t_{1}\rightarrow
t_{2}}(u)|>D\varepsilon_{1}+o(|\varepsilon_{1}|+|\varepsilon_{2}|),$

with a constant $D>0,$ depending only on $A.$ ∎

We can further refine this bound by using Theorem 4 which tells us that the
zeros of the approximating functions move like the zeros of $E(t,z).$ Indeed,
if $\xi_{t_{2}}$ and $\xi_{t_{1}}$ are the zeros like before, then by (25) we
have

$\xi_{2}-\xi_{1}=\int_{t_{1}}^{t_{2}}\frac{f(t)}{\theta_{z}(t,\overline{\xi}(t))}dt.$

Moreover, by Lemma 32 (i) we know that on $(t_{1},t_{2}),$ we have
$|\theta_{z}(t,\overline{\xi}_{t})|\asymp t.$ Hence, we expect a lower bound
in terms of $||f||_{L^{1}(t_{1},t_{2})}.$

######  Lemma 41.

In the situation of Lemma 40 set $J=(t_{1},t_{2})$ and suppose that $J$ is an
$H_{s}-$interval. In particular, by assumption of Lemma 40 $J$ is a
$\sigma-$interval. Then there exists a constant $D>0,$ depending only on $A,$
such that for large enough $t_{1},$ we have on $R$ the estimate

$|\arg a_{t_{1}\rightarrow t_{2}}(s)|\geq D\int_{t_{1}}^{t_{2}}|f(t)|dt$

###### Proof.

Let $\xi_{t}$ be the zero $E(t,z)$ from Theorem 4. By Lemma 32 we have
$|\theta_{z}(t,\overline{\xi}_{t})|\asymp t$ for all $t\in J.$ Furthermore, by
Lemma 33

$\frac{\pi}{2}+2\arg\alpha(s,t)=-\arg\theta_{z}(t,\overline{\xi}(t)).$

Moreover, since $J$ is a $H_{s}-$interval,
$|\Im\alpha^{2}(s,t)|>\frac{1}{200}.$ Since $|\alpha(s,t)|=1$ for all $t\in
J,$ we have $|\arg\alpha^{2}(s,t)|\in[\frac{1}{200},\pi-\frac{1}{200}].$
Hence, $|\Im e^{-i\arg\alpha^{2}(s,t)}|\geq\frac{1}{300}.$ Thus, for large
enough $t_{1},$

$|\Re\theta_{z}(t,\overline{\xi}_{t})|=|\theta_{z}(t,\overline{\xi}_{t})||\Re
e^{i\arg\theta_{z}(t,\overline{\xi}_{t})}|=|\theta_{z}(t,\overline{\xi}_{t})||\Re(ie^{i\arg\alpha^{2}(s,t)})|$

$=|\theta_{z}(t,\overline{\xi}_{t})||\Im
e^{i\arg\alpha^{2}(s,t)}|\geq\frac{|\theta_{z}(t,\overline{\xi}_{t})|}{300}.$

By continuity of $\theta_{z}$ this implies that for all $t\in J$ either

(48)
$\Re\theta_{z}(t,\overline{\xi}_{t})\geq\frac{|\theta_{z}(t,\overline{\xi}_{t})|}{300}\text{
or
}\Re\theta_{z}(t,\overline{\xi}_{t})\leq-\frac{|\theta_{z}(t,\overline{\xi}_{t})|}{300}.$

Assume that

(49)
$\Re(\xi_{2}-\xi_{1})=\Re\int_{J}\frac{f(t)}{\theta_{z}(t,\overline{\xi}_{t})}dt\geq
0.$

The case where $\Re(\xi_{t_{2}}-\xi_{t_{1}})<0$ can be proved similarly. Since
$J$ is a $H_{s}-$interval and $\sigma-$interval we find that

(50) $|\xi_{2}-\xi_{1}|\ll|\Re(\xi_{2}-\xi_{1})|.$

Indeed,

$|\xi_{2}-\xi_{1}|=\bigg{|}\int_{J}\frac{f(t)}{\theta_{z}(t,\overline{\xi}_{t})}dt\bigg{|}=\bigg{|}\int_{J}\frac{f(t)\overline{\theta_{z}(t,\overline{\xi}_{t})}}{|\theta_{z}(t,\overline{\xi}_{t})|^{2}}dt\bigg{|}=\bigg{|}\int_{J}\frac{f(t)\Re\theta_{z}(t,\overline{\xi}_{t})}{|\theta_{z}(t,\overline{\xi}_{t})|}dt-i\int_{J}\frac{f(t)\Im\theta_{z}(t,\overline{\xi}_{t})}{|\theta_{z}(t,\overline{\xi}_{t})|^{2}}dt\bigg{|}$

$=|\Re(\xi_{2}-\xi_{1})-i\Im(\xi_{2}-\xi_{1})|\leq|\Re(\xi_{2}-\xi_{1})|+|\Im(\xi_{2}-\xi_{1})|.$

We thus have to show that

$|\Im(\xi_{2}-\xi_{1})|\ll|\Re(\xi_{2}-\xi_{1})|.$

To see this, recall the following facts. By (48) we have that
$\Re\theta_{z}(t,\overline{\xi}_{t})$ has constant sign and
$|\Im\theta_{z}(t,\overline{\xi}_{t})|\leq|\theta_{z}(t,\overline{\xi}_{t})|\leq
300|\Re\theta_{z}(t,\overline{\xi}_{t})|$. Furthermore, by Lemma 32 we have
that $|\theta_{z}(t,\overline{\xi}_{t})|\asymp t$ on $J$ and from the
assumption $t_{2}-t_{1}\leq 1,$ we see immediately that
$\frac{t_{2}}{t_{1}}\leq 2.$ Thus,

$|\Im(\xi_{2}-\xi_{1})|=\bigg{|}\int_{J}\frac{f(t)\Im\theta_{z}(t,\overline{\xi}_{t})}{|\theta_{z}(t,\overline{\xi}_{t})|^{2}}dt\bigg{|}\leq\int_{J}\frac{|f(t)||\Im\theta_{z}(t,\overline{\xi}_{t})|}{|\theta_{z}(t,\overline{\xi}_{t})|^{2}}dt\leq\int_{J}\frac{|f(t)|}{|\theta_{z}(t,\overline{\xi}_{t})|}dt$

$\ll\frac{1}{t_{1}}\int_{J}|f(t)|dt\ll\frac{2(1-\sigma)}{t_{2}}\bigg{|}\int_{J}f(t)dt\bigg{|}\ll
2(1-\sigma)\bigg{|}\int_{J}\frac{f(t)}{|\theta_{z}(t,\overline{\xi}_{t})|}dt\bigg{|}$

$=2(1-\sigma)\bigg{|}\int_{J}\frac{f(t)|\theta_{z}(t,\overline{\xi}_{t})|}{|\theta_{z}(t,\overline{\xi}_{t})|^{2}}dt\bigg{|}=600(1-\sigma)\bigg{|}\int_{J}\frac{f(t)\Re\theta_{z}(t,\overline{\xi}_{t})}{|\theta_{z}(t,\overline{\xi}_{t})|}dt\bigg{|}\ll|\Re(\xi_{2}-\xi_{1})|.$

Thus, as claimed

$|\xi_{2}-\xi_{1}|\ll|\Re(\xi_{2}-\xi_{1})|.$

By Theorem 4 (iv), (49) and (50)

$x_{t_{2}}-x_{t_{1}}=\Re(\xi_{2}-\xi_{1})+(x_{t_{2}}-x_{t_{1}}-\Re(\xi_{2}-\xi_{1}))\leq|\Re(\xi_{2}-\xi_{1})|-|(x_{t_{2}}-x_{t_{1}}-\Re(\xi_{2}-\xi_{1}))|$
(51)
$\geq|\Re(\xi_{2}-\xi_{1})|-\psi(t_{1})|\xi_{2}-\xi_{1}|\geq|\Re(\xi_{2}-\xi_{1})|-D\psi(t_{1})|\Re(\xi_{2}-\xi_{1})|\geq
D_{1}|\Re(\xi_{2}-\xi_{1})|$

for some positive constants $D$ and $D_{1}.$

Since in Theorem 4 we have chosen $A>8\pi$ the box $Q(s,A/t_{1})$ contains at
least one zero $x_{t_{2}}$ of the approximating function with
$x_{t_{2}}>s+\frac{1}{t_{1}}.$ Recall aswell that
$t_{2}-t_{1}\leq\frac{1}{|s|+1}\leq 1.$Thus, using that
$\Re\theta_{z}(t,\overline{\xi}_{t})$ has constant sign we find

$(t_{2}(x_{t_{2}}-s)-t_{1}(x_{t_{1}}-s))=(t_{2}-t_{1})(x_{t_{2}}-s)+t_{1}(x_{t_{2}}-x_{t_{1}})$
$\geq\frac{t_{2}-t_{1}}{t_{1}}+t_{1}(x_{t_{2}}-x_{t_{1}})\overset{\eqref{misc19}}{\geq}\frac{t_{2}-t_{1}}{t_{1}}D_{1}|\Re
t_{1}(\xi_{2}-\xi_{1})|$
$=\frac{t_{2}-t_{1}}{t_{1}}+D_{1}t_{1}\bigg{|}\Re\int_{J}\frac{f(t)}{\theta_{z}(t,\overline{\xi}_{t})}dt\bigg{|}\geq\frac{t_{2}-t_{1}}{t_{1}}+\frac{D_{1}t_{1}}{300}\bigg{|}\int_{J}\frac{f(t)}{|\theta_{z}(t,\overline{\xi}_{t})|}dt\bigg{|}$
$\geq\frac{t_{2}-t_{1}}{t_{1}}+\frac{D_{1}D_{2}t_{1}}{300t_{2}}\bigg{|}\int_{J}f(t)dt\bigg{|}\geq\frac{t_{2}-t_{1}}{t_{1}}+\frac{D_{1}D_{2}(1-\sigma)}{600}\int_{J}|f(t)|dt.$

Furthermore, we now show that there is an absolute constant $D_{3}>0$ such
that

(52)
$|t_{2}y_{t_{2}}-t_{1}y_{t_{1}}|<D_{3}\bigg{(}\frac{t_{2}-t_{1}}{t_{1}}\int_{J}|f(t)|dt\bigg{)}.$

We separate two cases. First assume $t_{2}y_{t_{2}}\leq t_{1}y_{t_{1}}.$ Then,
since $t_{2}>t_{1}$

$|t_{2}y_{t_{2}}-t_{1}y_{t_{1}}|=t_{1}y_{t_{1}}-t_{2}y_{t_{2}}\leq
t_{2}y_{t_{1}}-t_{2}y_{t_{2}}=t_{2}\int_{t_{1}}^{t_{2}}\frac{f(t)}{\theta_{z}(t,\xi_{t})}dt\leq
D_{3}\int_{J}|f(t)|dt.$

Now assume that $t_{2}y_{t_{2}}\geq t_{1}y_{t_{1}}.$ Then,

$|t_{2}y_{t_{2}}-t_{1}y_{t_{1}}|=t_{2}y_{t_{2}}-t_{1}y_{t_{1}}=t_{2}y_{t_{2}}-{t_{2}-(t_{2}-t_{1})}y_{t_{1}}\leq
D_{3}\bigg{(}\frac{t_{2}-t_{1}}{t_{1}}+\int_{J}|f(t)|dt\bigg{)}.$

Combining everything, gives as claimed

$|\arg a_{t_{1}\rightarrow t_{2}}(s)|\geq D\int_{t_{1}}^{t_{2}}|f(t)|dt$

on $R$ for an absolute constant $D>0,$ depending only on $A.$

∎

Since $|S_{H}^{n}|\leq|S|<\infty$ we can consider a finite collection of
intverals $I_{1},\dots,I_{N}$ of size $|I_{k}|=4C2^{-n},$ centered at
$s_{1},\dots,s_{N}\in S_{H}^{n}$ such that each point is covered by at most
two intervals and $I_{1},\dots,I_{N}$ cover at least one half of $S_{H}^{n}.$
We consider the intervals $L_{s_{k}}^{n}$ for $k=1,\dots,N$ as defined before.

######  Lemma 42.

There exists a finite collection of pairwise disjoint intervals
$\mathcal{T}_{1},\dots\mathcal{T}_{M},\mathcal{T}_{l}=(\tau_{1}^{l},\tau_{2}^{l})$
satisfying the following properties.

  * (i)

Each $\mathcal{T}_{l}$ belongs to $HL_{s_{k}}$ for at least one $k=1,\dots,N.$

  * (ii)

All $\mathcal{T}_{l}$ are $\sigma-$intervals.

  * (iii)

For each $k=1,\dots,N$

$\int_{\bigcup_{\mathcal{T}_{l}\in
HL_{s_{k}}}\mathcal{T}_{l}}|f(t)|dt>\frac{\Delta}{500}.$

  * (iv)

For each $\mathcal{T}_{l}$ we have

$\int_{\mathcal{T}_{l}}|f(t)|dt<\frac{1}{100}.$

###### Proof.

We can assume without loss of generality that $s_{1},\dots,s_{n}$ are Lebesgue
points of $f.$ Then, the claim follows immediately from the fact that all
small enough intervals centered around $s_{1},\dots,s_{n}$ are
$\sigma-$intervals by Lemma 38 and the fact that subintervals of
$H_{s}-$intervals are $H_{s}-$intervals by definition. ∎

Let
$\mathcal{T}_{1},\dots\mathcal{T}_{M},\mathcal{T}_{l}=(\tau_{1}^{l},\tau_{2}^{l})$
be the finite collection of pairwise disjoint intervals from Lemma 42. Then,

$\sum_{l=1}^{M}||\log|a_{\tau_{1}^{l}\rightarrow\tau_{2}^{l}}|||_{L^{1}(\mathbb{R})}\overset{(1)}{\geq}\sum_{l=1}^{M}\frac{1}{2}\sum_{\begin{subarray}{c}1\leq
k\leq N:\\\ \mathcal{T}_{l}\in
HL_{s_{k}}\end{subarray}}||\log|a_{\tau_{1}^{l}\rightarrow\tau_{2}^{l}}|||_{L^{1}(I_{t_{2}}(s_{k}))}\overset{(2)}{\geq}\sum_{l=1}^{M}\frac{1}{2}\sum_{\begin{subarray}{c}1\leq
k\leq N:\\\ \mathcal{T}_{l}\in
HL_{s_{k}}\end{subarray}}\frac{D^{2}}{\tau_{2}^{l}}\int_{\tau_{1}^{l}}^{\tau_{2}^{l}}|f(t)|dt.$

$\overset{(3)}{\geq}\sum_{l=1}^{M}\sum_{\begin{subarray}{c}1\leq k\leq N:\\\
\mathcal{T}_{l}\in
HL_{s_{k}}\end{subarray}}\frac{D^{2}|I_{k}|}{32C}\int_{\mathcal{T}_{l}}|f(t)|dt\overset{(4)}{=}\sum_{k=1}^{N}\sum_{\begin{subarray}{c}1\leq
l\leq M:\\\ \mathcal{T}_{l}\in
HL_{s_{k}}\end{subarray}}\frac{D^{2}|I_{k}|}{32C}\int_{\mathcal{T}_{l}}|f(t)|dt$
$\overset{(5)}{=}D^{2}\sum_{k=1}^{N}\frac{|I_{k}|}{32C}\int_{\bigcup_{\mathcal{T}_{l}\in
HL_{s_{k}}}\mathcal{T}_{l}}|f(t)|dt\overset{(6)}{>}\frac{D^{2}\Delta}{16000C}\sum_{k=1}^{N}|I_{k}|\overset{(7)}{\geq}D^{2}\frac{\Delta}{32000C}|S_{H}^{n}|,$

where we used the following facts

  * (1)

By property $(i)$ of Lemma 42 each $\mathcal{T}_{l}$ is a $H_{s_{k}}-$interval
for at least one $1\leq k\leq N.$ Since $\mathcal{T}_{l}\subset L_{s}^{n},$
and $|I_{k}|=4C2^{-n},$ we have $I_{t_{2}}(s_{k})\subset I_{k}$ and since each
point is covered by at most two of those intervals, this gives the factor
$\frac{1}{2}.$

  * (2)

By property $(ii)$ of Lemma 42 all $\mathcal{T}_{l}$ are $\sigma-$intervals.
Hence, the conditions for Lemma 41 are satisfied and the inequality follows by
applying that Lemma. Moreover, we used (74) and that $|R|\geq
D^{\prime}/\tau_{2}$ with $D^{\prime}$ depending only on $A.$ Additionally, we
assume without loss of generality that $D^{\prime}>D,$ so we can write
$DD^{\prime}>D^{2}.$

  * (3)

Since by property $(i)$ of Lemma 42 each $\mathcal{T}_{l}$ belongs to
$HL_{s_{k}}$ for at least one $1\leq k\leq N.$ Thus, by definition
$\mathcal{T}_{l}\subset L_{s_{k}}^{n}\subset[2^{n},2^{n+2}].$ In particular,
$\tau_{2}^{l}\leq 2^{n+2}$ and thus

$\frac{D}{\tau_{2}^{l}}\geq\frac{D}{2^{n+2}}=\frac{D|I_{k}|}{16C}.$

  * (4)

By changing the order of summation.

  * (5)

Since the intervals $\mathcal{T}_{1}\dots,\mathcal{T}_{M}$ are pairwise
disjoint.

  * (6)

By property (iii) of Lemma 42.

  * (7)

Since the intervals $I_{1},\dots I_{N}$ cover at least one half of
$S_{H}^{n}.$

Applying now the non-linear Parseval identity (33) to
$a_{\tau_{1}^{l}\rightarrow\tau_{2}^{l}}$ for all $l=1,\dots,M$ and using the
property that the intervals $\mathcal{T}_{l}$ are pairwise disjoint and all
contained in $[2^{n},2^{n+2}]$ gives

$||f||_{L^{2}([2^{n},2^{n+2}])}^{2}\geq\sum_{l=1}^{M}||f||_{L^{2}(\mathcal{T}_{l})}^{2}=\sum_{l=1}^{M}||\log|a_{\tau_{1}^{l}\rightarrow\tau_{2}^{l}}|||_{L^{1}(\mathbb{R})}.$

Together with the last computation, we obtain the desired

######  Corollary 9.

There exists a constant $D>0$ such that

$D|S_{H}^{n}|<||f||_{L^{2}([2^{n},2^{n+2}])}^{2}.$

###  3.7. Stronger approximations of the Hermite-Biehler functions

In this section, we are going to prove Theorem 4. Since the statement is quite
rich, we are going to construct the stronger approximation in several steps.
First, we will prove that we can approximate $E(t,z)$ and $\tilde{E}(t,z)$
with the same imaginary part $y(t)$ and have the real parts satisfy
$\cos[t(\tilde{x}(t)-x(t))]=\sqrt{w(s)\tilde{w}(s)}.$ Since we now approximate
$E(t,z)$ and $\tilde{E}(t,z)$ at the same time, we will have to introduce
analogous notation as in the case of $E(t,z).$ We define for a positive
function $C=C(t)$ the sets

$\tilde{T}_{0}(s,C)=\\{t>0\mid\text{ there is a zero }z(t)\text{ of
}\tilde{E}(t,z)\text{ inside }Q(s,C/t)\\},$

$\tilde{T}_{1}(s,C)=\\{t>0\mid\text{ all zeros }z(t)\text{ of
}\tilde{E}(t,z)\text{ inside }Q(s,C/t),\text{ satisfy }\Im z(t)<-1/t\\}.$

Let us once again mention that all results out of the previous section that
hold for $E(t,z)$ hold aswell for $\tilde{E}(t,z)$ and recall that by
definition if there is no zero of $E(t,z),$ respectively $\tilde{E}(t,z),$
inside $Q(s,C/t),$ then $t\in T_{1}(s,C),$ respectively
$t\in\tilde{T}_{1}(s,C).$ In particular it will be convenient for us to state
Lemma 23 for $\tilde{E}(t,z)$ with $\cos(z)$ as the approximating function
instead of $\sin(z)$ as in the case of $E(t,z).$ This just corresponds to the
translation $\tilde{z}(t)\mapsto\tilde{z}(t)+\frac{\pi}{2t}.$

######  Lemma 43.

For almost all $s\in\mathbb{R}$ there exists a increasing function
$C(t)>0,C(t)\rightarrow\infty$ as $t\rightarrow\infty$ with the following
properties.

For all $t\in T_{1}(s,C)\cap\tilde{T}_{1}(s,C)$ there exist
$z(t)=u(t)-ip(t),\tilde{z}(t)=\tilde{u}(t)-ip(t)$ and $\alpha(t)=\alpha(s,t)$
such that $p(t)>0,$ $|\alpha(t)|=1$ and

$\sup_{z\in
Q(s,C(t)/t)}\bigg{|}E(t,z)-\frac{\alpha(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-z(t))]\bigg{|}=o(1),$

$\sup_{z\in
Q(s,C(t)/t)}\bigg{|}\tilde{E}(t,z)-\frac{\alpha(t)\gamma(ty(t))}{\sqrt{w(s)}}\cos[t(z-\tilde{z}(t))]\bigg{|}=o(1),$

as $t\rightarrow\infty.$ Furthermore, $p(t)>C(t)$ for $t\notin T_{0}(s,C)$ and
$p(t)\leq C(t)$ for $t\in T_{0}(s,C).$ The functions $u(t)$ and $\tilde{u}(t)$
satisfy

$\cos[t(\tilde{u}(t)-u(t))]=\sqrt{w(s)\tilde{w}(s)}.$

###### Proof.

We apply Corollary 6 to both $E$ and $\tilde{E}$ to obtain for any $D>4\pi$

(53) $\sup_{z\in
Q(s,D/t)}\bigg{|}E(t,z)-\frac{\beta(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-\xi(t))]\bigg{|}=o(\gamma(ty(t))),$

(54) $\sup_{z\in
Q(s,D/t)}\bigg{|}E(t,z)-\frac{\delta(t)\gamma(t\tilde{y}(t))}{\sqrt{\tilde{w}(s)}}\sin[t(z-\tilde{\xi}(t))]\bigg{|}=o(\gamma(ty(t))),$

for some $\xi(t)=x(t)-iy(t),\tilde{\xi}(t)=\tilde{x}(t)-i\tilde{y}(t)$ and
unimodular functions $\beta(t),\delta(t).$ Hence,

$\det\begin{pmatrix}E(t,z)&&\tilde{E}(t,z)\\\
E^{\sharp}(t,z)&&\tilde{E}^{\sharp}(t,z)\end{pmatrix}=\frac{\gamma(ty(t))\gamma(t\tilde{y}(t))}{\sqrt{w(s)\tilde{w}(s)}}\big{(}\beta(t)\overline{\delta}(t)\sin[t(z-\xi(t))]\cos[t(z-\overline{\tilde{\xi}}(t))]$

(55)
$-\overline{\beta}(t)\delta(t)\sin[t(z-\overline{\xi}(t))]\cos[t(z-\tilde{\xi}(t))]\big{)}+o(\gamma(ty(t))\gamma(t\tilde{y}(t))).$

If we plug in $z=x\in\mathbb{R},$ (55) simplifies to

$=2i\Im\bigg{(}\frac{\gamma(ty(t))\gamma(t\tilde{y}(t))}{\sqrt{w(s)\tilde{w}(s)}}\beta(t)\overline{\delta(t)}\sin[t(x-\xi(t))]\cos[t(x-\overline{\tilde{\xi}}(t))]\bigg{)}$

(56)
$=2i\Im\bigg{(}\frac{\gamma(ty(t))\gamma(t\tilde{y}(t))}{\sqrt{w(s)\tilde{w}(s)}}\beta(t)\overline{\delta}(t)\frac{1}{2}\bigg{[}\sin[t(\overline{\tilde{\xi}}(t)-\xi(t))]+\sin[t(2x-(\xi(t)+\overline{\tilde{\xi}}(t)))]\bigg{]}\bigg{)}.$

We know that by Corollary 2 for all $z\in\mathbb{C}$ and any $t\geq 0$

$\det\begin{pmatrix}E(t,z)&&\tilde{E}(t,z)\\\
E^{\sharp}(t,z)&&\tilde{E}^{\sharp}(t,z)\end{pmatrix}=2i.$

Hence, by (55) the expression in (56) must be within
$o(\gamma(ty(t))\gamma(t\tilde{y}(t))$ from $2i$ on the interval
$I_{t}=Q(s,D/t)\cap\mathbb{R}.$ Suppose that for $t\in
T_{1}(s,C(t))\cap\tilde{T}_{1}(s,C(t))$ we had the lower bound
$\limsup_{t\rightarrow\infty}t|\tilde{y}(t)-y(t)|>\Delta>0.$ In that case for
any fixed $t$ the first sine in (56) is constant, while the second has
absolute value $\geq\sinh\Delta$ and its argument grows by more than $2\pi$ on
$I_{t}.$ This immediately contradicts that the expression in (56) is within
$o(\gamma(ty(t))\gamma(t\tilde{y}(t))$ from $2i$ on $I_{t}.$

Thus, $y(t)=\tilde{y}(t)+o(1/t)$ and we can change either of $y(t)$ and
$\tilde{y}(t)$ into the other with (53) and (54) remaining still true, just
with a different error term of the same order. So we assume from now on that
$\tilde{y}(t)=y(t)$ and set $p(t)=y(t)=\tilde{y}(t).$ By choice of the
approximation from Corollary 6 we have $p(t)>C(t)$ for $t\notin T_{0}(s,C(t))$
and $p(t)\leq C(t)$ for $t\in T_{0}(s,C(t)).$

Suppose now that $\xi(t)=x(t)-ip(t)\in Q(s,D/t)$ and $tp(t)>1.$ By plugging
$z=\xi(t)$ and $z=\overline{\xi}(t)$ into (55) we obtain the equations

$\frac{2i\overline{\beta}(t)\delta(t)}{\sqrt{w(s)\tilde{w}(s)}}\cos[t(\tilde{x}(t)-x(t))]=2i+o(1),$

$\frac{2i\beta(t)\overline{\delta}(t)}{\sqrt{w(s)\tilde{w}(s)}}\cos[t(\tilde{x}(t)-x(t))]=2i+o(1).$

We immediately conclude from both equations that
$\arg\beta(t)=\arg\delta(t)=o(1)$ and since $|\beta(t)|=|\delta(t)|=1$ we find
that $\beta(t)=\delta+o(1)$ and thus set $\alpha(t)=\beta(t)=\delta+o(1)$ for
$t\in T_{0}(s,D)\subset T_{0}(s,C(t)).$

Furthermore, we conclude from these equations that

$\cos[t(\tilde{x}(t)-x(t))]=\sqrt{w(s)\tilde{w}(s)}+o(1).$

Hence, $\tilde{x}(t)=\tilde{u}(t)+o(1/t)$ and $x(t)=u(t)+o(1/t),$ where $u(t)$
and $\tilde{u}(t)$ are chosen continuously to satisfy the equation

$\cos[t(\tilde{u}(t)-u(t))]=\sqrt{w(s)\tilde{w}(s)}.$

Thus, by replacing the $x(t)$ with $u(t)$ and $\tilde{x}(t)$ with
$\tilde{u}(t)$ the approximations (53) and (54) remain valid with a different
error term of the same order.

Finally, by our standard argument, we can improve from fixed $D>4\pi$ to
$C(t)$ by making $C(t)$ grow a bit slower if necessary. ∎

In the next step, we are going to change the approximation, such that the
approximating function and $E(t,z)$ have the same value at the point
$s\in\mathbb{R}$ instead of sharing a common zero. The following simple lemma
will play an important role in this process.

######  Lemma 44.

Let $a,b,c,d\in\mathbb{C}\setminus\\{0\\}$ be such that
$\frac{a}{b}=\frac{c}{d}$ and

$\det\begin{pmatrix}a&&b\\\
\overline{a}&&\overline{b}\end{pmatrix}=\det\begin{pmatrix}c&&d\\\
\overline{c}&&\overline{d}\end{pmatrix}\neq 0.$

Then, $|a|=|c|$ and $|b|=|d|.$

###### Proof.

We write $b=\frac{ad}{c}$ and thus obtain

$a\overline{b}-\overline{a}b=a\frac{\overline{ad}}{\overline{c}}-\overline{a}\frac{ad}{c}=|a|^{2}\bigg{(}\frac{c\overline{d}-d\overline{c}}{|c|^{2}}\bigg{)}=c\overline{d}-d\overline{c}.$

We immediately see $|a|^{2}=|c|^{2}.$ We get $|b|^{2}=|d|^{2}$ similarly by
writing $a=\frac{bc}{d}.$ ∎

Thus, our strategy will be to change the approximating functions slightly such
that they satisfy the assumptions of the previous Lemma.

######  Lemma 45.

For almost all $s\in\mathbb{R}$ and for all $C>1$ there exists
$\varepsilon_{0}>0$ such that for all $\varepsilon<\varepsilon_{0}$ the
following holds. If

$\sup_{z\in
Q(s,3C/t)}\bigg{|}E(t,z)-\frac{\alpha\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x-iy))]\bigg{|}<\varepsilon,$

$\sup_{z\in
Q(s,3C/t)}\bigg{|}\tilde{E}(t,z)-\frac{\alpha\gamma(ty)}{\sqrt{\tilde{w}(s)}}\cos[t(z-(\tilde{x}-i\tilde{y}))]\bigg{|}<\varepsilon,$

for some $t>0,x,\tilde{x},y\in\mathbb{R},\alpha\in\mathbb{C},$ satisfying
$1<ty<2C,|\alpha|=1$ and

(57)
$-\frac{\pi}{2}<t(\tilde{x}-x)<\frac{\pi}{2},\quad\cos[t(\tilde{x}-x)]=\sqrt{w(s)\tilde{w}(s)},$

then there exist $y^{\prime},x^{\prime},\tilde{x}^{\prime}\in\mathbb{R}$ and
$\alpha^{\prime}\in\mathbb{C},|\alpha^{\prime}|=1,$ such that

$|tx^{\prime}-tx|+|t\tilde{x}^{\prime}-t\tilde{x}|+|ty^{\prime}-ty|+|\alpha^{\prime}-\alpha|<D\varepsilon,\quad\tilde{x}-x=\tilde{x}^{\prime}-x^{\prime}$

for some constant $D=D(C,s)$ and furthermore

(58)
$E(t,s)=\alpha^{\prime}\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(s-(x^{\prime}-iy^{\prime}))],\quad\tilde{E}(t,s)=\alpha^{\prime}\frac{\gamma(ty)}{\sqrt{\tilde{w}(s)}}\cos[t(s-(\tilde{x}^{\prime}-iy^{\prime}))].$

###### Proof.

Recall that for almost all $s\in\mathbb{R}$ we have that $w(s)\neq 0$ and
$\tilde{w}(s)\neq 0.$ So assume without loss of generality that
$w(s)\tilde{w}(s)\neq 0.$ We set for all $z\in\mathbb{C}$ where it is well-
defined

$f(z)=\sqrt{\frac{w(s)}{\tilde{w}(s)}}\cdot\frac{\sin[t(z-(x-iy))]}{\cos[t(z-(\tilde{x}-iy))]}.$

Denote by $J$ the middle-third of $I=Q(s,3C/t)\cap\mathbb{R}.$ and note that
by the same calculation as in the proof of Lemma 32 $(i)$ we find a constant
$D_{1}$ such that for all $s\in J,$ we have if $\varepsilon_{0}$ is small
enough

$\bigg{|}\frac{E(t,s)}{\tilde{E}(t,s)}-f(s)\bigg{|}<D_{1}\varepsilon.$

Since $1<ty<2C,$ $f$ is holomorphic in a $\frac{1}{2t}-$neighborhood of $J.$
Hence, for small enough $\varepsilon$ there exists by Lemma 30 a $D_{2}>0$
such that in the disk $B(s,D_{2}\varepsilon/t),$ f takes all values from
$B(f(s),D_{1}\varepsilon).$ We choose $a\in B(s,D_{2}\varepsilon/t)$ such that

$f(a)=\frac{E(t,s)}{\tilde{E}(t,s)}.$

If we now set
$\tilde{x}^{\prime}=\tilde{x}+\Re(s-a),x^{\prime}=x+\Re(s-a),y^{\prime}=y+\Im
a,$ then

$\frac{E(t,s)}{\tilde{E}(t,s)}=\sqrt{\frac{\tilde{w}(s)}{w(s)}}\cdot\frac{\sin[t(s-(x^{\prime}-iy^{\prime}))]}{\cos[t(s-(\tilde{x}^{\prime}-iy^{\prime}))]}.$

Using trigonometric identities and (57) one can check that

$\det\begin{pmatrix}\frac{\gamma(ty)}{w(s)}\sin[t(s-(x^{\prime}-iy^{\prime}))]&&\frac{\gamma(ty)}{\tilde{w}(s)}\cos[t(s-(\tilde{x}^{\prime}-iy^{\prime}))]\\\
\frac{\gamma(ty)}{w(s)}\sin[t(s-(x^{\prime}+iy^{\prime}))]&&\frac{\gamma(ty)}{\tilde{w}(s)}\cos[t(s-(\tilde{x}^{\prime}+iy^{\prime}))]\end{pmatrix}=2i.$

Together with Corollary 2 and Lemma 44 this implies

$|E(t,s)|=\bigg{|}\frac{\gamma(ty)}{w(s)}\sin[t(s-(x^{\prime}-iy^{\prime}))]\bigg{|},\quad|\tilde{E}(t,s)|=\bigg{|}\frac{\gamma(ty)}{\tilde{w}(s)}\cos[t(s-(\tilde{x}^{\prime}-iy^{\prime}))]\bigg{|}.$

Since we have additionally

$\arg\frac{E(t,s)}{\tilde{E}(t,s)}=\arg\frac{\sin[t(s-(x^{\prime}-iy^{\prime}))]}{\cos[t(s-(\tilde{x}^{\prime}-iy^{\prime}))]},$

we find $\alpha^{\prime}\in\mathbb{C},|\alpha^{\prime}|=1$ such that
$\eqref{misc25}$ holds. By the inequalities from the statement we have
$|\alpha-\alpha‘|\ll\varepsilon.$ Indeed,

$|\alpha^{\prime}-\alpha|\cdot\bigg{|}\frac{\alpha\gamma(ty)}{\sqrt{w(s)}}\sin[t(s-(x^{\prime}-iy^{\prime}))]\bigg{|}=\bigg{|}\alpha^{\prime}\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(s-(x^{\prime}-iy^{\prime}))]-\alpha\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x^{\prime}-iy^{\prime}))]\bigg{|}$

$=\bigg{|}E(t,s)-\frac{\alpha\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x^{\prime}-iy^{\prime}))]\bigg{|}\leq\bigg{|}E(t,s)-\alpha\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x-iy))]\bigg{|}$
$+\bigg{|}\alpha\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x-iy))]-\alpha\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x^{\prime}-iy^{\prime}))]\bigg{|}$

By assumption we have

$\bigg{|}E(t,s)-\alpha\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x-iy))]\bigg{|}<\varepsilon$

and by the mean value Theorem for complex functions, using that the box is
convex and compact, as well as the fact that $\gamma(ty)$ is bounded due to
$1<ty<2C$ we have

$\bigg{|}\alpha\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x-iy))]-\alpha\frac{\gamma(ty)}{\sqrt{w(s)}}\sin[t(z-(x^{\prime}-iy^{\prime}))]\bigg{|}\leq
D_{3}|(tx^{\prime}-tx)+(ty^{\prime}-ty)|<D_{3}\varepsilon,$

where $D_{3}$ is a constant depending only on $s$ and $C.$ Thus,
$|\alpha^{\prime}-\alpha|\ll\varepsilon$ by using again that $\gamma(ty)$ is
bounded from above and below since $1<ty<2C.$

∎

Let us briefly summarize what we have shown so far and restate Theorem 4 to
see what is still left to prove.

######  Corollary 10.

For almost all $s\in\mathbb{R}$ there exist positive functions
$C(t),C(t)\rightarrow\infty,\psi(t),\psi(t)\rightarrow 0$ as
$t\rightarrow\infty,t\in T_{1}(s,C(t))\cap\tilde{T}_{1}(s,C(t)),$ real
functions $x(t),\tilde{x}(t),y(t),$ with $ty(t)>1$ and a complex function
$\alpha(t)\in\mathbb{C},|\alpha(t)|=1$ such that for all $t\in
T_{1}(s,C(t))\cap\tilde{T}_{1}(s,C(t))$ we have

  * (i)

$\sup_{z\in
Q(s,3C(t)/t)}\bigg{|}E(t,z)-\frac{\alpha(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-(x(t)-iy(t)))]\bigg{|}<\psi(t),$

$\sup_{z\in
Q(s,3C(t)/t)}\bigg{|}\tilde{E}(t,z)-\frac{\alpha(t)\gamma(ty(t))}{\sqrt{\tilde{w}(s)}}\cos[t(z-(\tilde{x}(t)-iy(t)))]\bigg{|}<\psi(t).$

  * (ii)

$-\frac{\pi}{2}<t(\tilde{x}(t)-x(t))<\frac{\pi}{2},\quad\cos[t(\tilde{x}(t)-x(t))]=\sqrt{w(s)\tilde{w}(s)}.$

  * (iii)

For large enough $t$ for which $1<ty(t)<2C(t),$

$E(t,s)=\frac{\alpha(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(s-(x(t)-iy(t)))],$

$\tilde{E}(t,s)=\frac{\alpha(t)\gamma(ty(t))}{\sqrt{\tilde{w}(s)}}\cos[t(s-(\tilde{x}(t)-iy(t)))].$

Note in the following that we established all claims about the approximation
itself. We are left to establish the claims about the movement of the zeros of
$E,\tilde{E}$ and the approximating functions.

######  Theorem (Strong approximation).

For almost all $s\in\mathbb{R}$ there exists a increasing functions
$C(t)>0,C(t)\rightarrow\infty$ and a function $\psi(t)>0,\psi(t)\rightarrow 0$
as $t\rightarrow\infty$ such that the following holds.

Let $(t_{1},t_{2})\subset
T_{0}(s,C(t))\cap\tilde{T}_{0}(s,C(t)),t_{2}-t_{1}\leq\frac{1}{|s|+1}$ be a
$\sigma-$interval and $H_{s}-$interval such that

$\int_{t_{1}}^{t_{2}}|f(t)|dt<\frac{1}{100\cosh[2A]},$

and $C(t)>10\pi$ for $t>t_{1},$ where $A$ is a constant satisfying
$10\pi<A<C(t)$ for all $t>t_{1}.$ Let $\xi_{1}$ a zero of $E(t_{1},z)$ in
$Q(s,A/t_{1})$ which moves continuously inside $Q(s,A/t)$ to a zero $\xi_{2}$
of $E(t_{2},z)$ as $t$ changes from $t_{1}$ to $t_{2}.$ Let
$\tilde{\xi_{1}},\tilde{\xi_{2}}$ be similar zeros of $\tilde{E}$ inside
$Q(s,A/t).$ Suppose that $t_{k}\Im\xi_{k}>2,t_{k}\Im\tilde{\xi_{2}}>2$ for
$k=1,2.$

Then, the zeros of $E(t,z)$ and $\tilde{E}(t,z)$ change in similar ways as $t$
changes from $t_{1}$ to $t_{2},$

$|(\xi_{2}-\xi_{1})-(\tilde{\xi_{2}}-\tilde{\xi_{1}})|\leq\psi(t_{1})|\xi_{2}-\xi_{1}|,$

and there exist real continuous functions $y_{t},x_{t},\tilde{x}_{t}$ and a
unimodular continuous function $\alpha_{t}$ on the interval $[t_{1},t_{2}]$
such that $ty_{t}>1$ for all $t\in[t_{1},t_{2}]$ and

  * (i)

At $t=t_{1}$ we have

$\sup_{z\in
Q(s,3C(t_{1})/t_{1})}\bigg{|}E(t_{1},z)-\frac{\alpha_{t_{1}}\gamma(t_{1}y_{t_{1}})}{\sqrt{w(s)}}\sin[t_{1}(z-(x_{t_{1}}-iy_{t_{1}}))]\bigg{|}<\psi(t_{1}),$

$\sup_{z\in
Q(s,3C(t_{1})/t_{1})}\bigg{|}\tilde{E}(t_{1},z)-\frac{\alpha_{t_{1}}\gamma(t_{1}y_{t_{1}})}{\sqrt{\tilde{w}(s)}}\cos[t_{1}(z-(\tilde{x}_{t_{1}}-iy_{t_{1}}))]\bigg{|}<\psi(t_{1}).$

  * (ii)

For all $t\in[t_{1},t_{2}]$ we have

$-\frac{\pi}{2}<t(\tilde{x}_{t}-x_{t})<\frac{\pi}{2},\quad\cos[t(\tilde{x}_{t}-x_{t})]=\sqrt{w(s)\tilde{w}(s)}.$

  * (iii)

For $k=1,2$ and large enough $t_{1}$ we have

$E(t_{k},s)=\frac{\alpha_{t_{k}}\gamma(t_{k}y_{t_{k}})}{\sqrt{w(s)}}\sin[t_{k}(s-(x_{t_{k}}-iy_{t_{k}}))],$

$\tilde{E}(t_{k},s)=\frac{\alpha_{t_{k}}\gamma(t_{k}y_{t_{k}})}{\sqrt{\tilde{w}(s)}}\cos[t_{k}(s-(\tilde{x}_{t_{k}}-iy_{t_{k}}))].$

  * (iv)

As $t$ changes from $t_{1}$ to $t_{2}$ the zeros of the approximating
functions change similarly to the zeros of $E(t,z)$ and $\tilde{E}(t,z),$

$|[(x_{t_{2}}-iy_{t_{2}})-(x_{t_{1}}-iy_{t_{1}})]-[\xi_{2}-\xi_{1}]|\leq\psi(t_{1})|\xi_{2}-\xi_{1}|,$

$|[(\tilde{x}_{t_{2}}-iy_{t_{2}})-(\tilde{x}_{t_{1}}-iy_{t_{1}})]-[\tilde{\xi_{2}}-\tilde{\xi_{1}}]|\leq\psi(t_{1})|\xi_{2}-\xi_{1}|.$

###### Proof.

Assume $s\in\mathbb{R}$ is like in Corollary 10. Then,
$y_{t_{1}},x_{t_{1}},\tilde{x}_{t_{1}}$ and $\alpha_{t_{1}}$ satisfying
$(i)-(iii)$ do exist. We will have to find
$y_{t_{2}},x_{t_{2}},\tilde{x}_{t_{2}}$ and $\alpha_{t_{2}}$ such that $(iv)$
holds and $(i)$ and $(iii)$ remain true as well, for $t=t_{2}.$

Recall that rewriting the real Dirac system (3) gave us the differential
equation

(59) $\frac{\partial}{\partial t}E(t,z)=-izE(t,z)+f(t)E^{\sharp}(t,z)$

for the Hermite-Biehler function $E(t,z)=A(t,z)-iC(t,z).$ Let $S(t,z)$ be the
solution to (59) for $t\in[t_{1},t_{2}]$ satisfying the initial condition

$S(t_{1},z)=\alpha_{t_{1}}\frac{\gamma(t_{1}y_{t_{1}})}{\sqrt{w(s)}}\sin[t_{1}(z-(x_{t_{1}}-iy_{t_{1}}))].$

Note that Lemma 15 and Lemma 16 can be formualted aswell for the Hermite-
Biehler function $S(t,z)$ and hence the associated scattering function
$\mathfrak{S}(t,z)=e^{itz}S(t,z)$ satisfies the differential equation

(60) $\frac{\partial}{\partial
t}\mathfrak{S}(t,z)=iz\mathfrak{S}(t,z)+e^{izt}\frac{\partial}{\partial
t}E(t,z)=f(t)e^{2izt}\mathfrak{S}^{\sharp}(t,z).$

Hence, we have for all $x\in\mathbb{R}$

$\frac{\partial}{\partial t}|S(t,x)|\leq|S(t,x)||f(t)|.$

Indeed a simple calculation recalling that $\mathfrak{S}(t,z)=e^{itz}S(t,z)$
yields

$\frac{d}{dt}|S(t,x)|=\frac{d}{dt}|\mathfrak{S}(t,x)|=\Re\bigg{(}\frac{S(t,x)\overline{\frac{d}{dt}S(t,x)}}{|S(t,x)|}\bigg{)}\overset{\eqref{misc35}}{=}\frac{\Re(e^{ixt}S(t,x)f(t)e^{-ixt}S(t,x))}{|S(t,x)|}$

$=\frac{\Re(S^{2}(t,x)f(t))}{|S(t,x)|}\leq|S(t,x)||f(t)|.$

Integrating this inequality yields the pointwise bound

(61) $|S(t,x)|\leq|S(t_{1},x)|e^{\int_{t_{1}}^{t}|f(u)|du}.$

By assumption we have that

$\int_{t_{1}}^{t_{2}}|f(t)|dt<\frac{1}{100}.$

Using this together with the inequality $e^{x}\leq 4(1+x)$ for $0\leq
x\leq\frac{1}{100}$ gives

$e^{\int_{t_{1}}^{t}|f(u)|du}\leq
4\bigg{(}1+\int_{t_{1}}^{t}|f(u)|du\bigg{)}\leq 8.$

Moreover, we have chosen the initial condition such that

$|S(t_{1},x)|=\bigg{|}\alpha_{t_{1}}\frac{\gamma(t_{1}y_{t_{1}})}{\sqrt{w(s)}}\sin[t_{1}(x-(x_{t_{1}}-iy_{t_{1}}))]\bigg{|}\leq\bigg{|}\frac{1}{\sqrt{w(s)}}\frac{\sqrt{2}}{\sqrt{\sinh[2t_{1}y_{t_{1}}]}}\sin[t_{1}(x-(x_{t_{1}}-iy_{t_{1}}))]\bigg{|}$

$\leq\bigg{|}\frac{\sqrt{2}}{\sqrt{w(s)}}\frac{\sin[t_{1}(x-x_{t_{1}})]\cosh(t_{1}y_{t_{1}})+i\cos[t_{1}(x-x_{t_{1}})]\sinh(t_{1}y_{t_{1}})}{\sqrt{\sinh(2t_{1}y_{t_{1}})}}\bigg{|}$

Recall the following limits

$\lim_{t\rightarrow\infty}\frac{\cosh t}{\sqrt{\sinh
t}}=\frac{1}{\sqrt{2}},\quad\lim_{t\rightarrow\infty}\frac{\sinh
t}{\sqrt{\sinh t}}=\frac{1}{\sqrt{2}}.$

Inserting these limits in the last formula together with the fact that
$t_{1}y_{t_{1}}^{>}1$ gives us for large $t_{1}$ that

$|S(t_{1},x)|\leq\frac{\sqrt{2}}{\sqrt{w(s)}}\sqrt{2\bigg{(}\frac{1}{\sinh
2}\bigg{)}^{2}}<\frac{1}{\sqrt{w(s)}}.$

Combining these estimates with (61) yields the bound

$|S(t,x)|\leq|S(t_{1},x)|e^{\int_{t_{1}}^{t}|f(u)|du}\leq\frac{8}{\sqrt{w(s)}}.$

By Lemma 16 we have for all $t\in(t_{1},t_{2})$ and $x\in\mathbb{R}$ that

$|\mathfrak{S}(t_{1},x)-\mathfrak{S}(t,x)|=\bigg{|}e^{it_{1}x}S(t_{1},x)-e^{itx}S(t,x)\bigg{|}=\int_{t_{1}}^{t}\frac{d}{dt}|\mathfrak{S}(t,x)|=\int_{t_{1}}^{t}\frac{d}{dt}|S(t,x)|$
$\leq\int_{t_{1}}^{t}|S(t,x)||f(t)|\leq\frac{8}{\sqrt{w(s)}}\int_{t_{1}}^{t}|f(u)|du.$

Since $\mathfrak{S}(t,z)$ is outer and together with the maximum principle for
the upper half plane this implies that for all $z\in\mathbb{C}_{+}$ we have

$|\mathfrak{S}(t_{1},z)-\mathfrak{S}(t,z)|\leq\frac{8}{\sqrt{w(s)}}\int_{t_{1}}^{t}|f(u)|du.$

We are going to use this bound, to bound the distance of two zeros of
$S(t,z).$ We denote by $\zeta_{t_{1}}=\zeta_{1}$ the zero of $S(t_{1},z)$
inside $Q(s,A/t_{1})$ that is closest to $\xi_{1}.$ If there are more than one
at the same distance, we pick one of them. Denote by $\xi_{t}$ and $\zeta_{t}$
the zeros of $E(t,z)$ and $S(t,z)$ evolving from $\xi_{1}$ and $\zeta_{1},$
respectively.

From the previous bound we find

$|\mathfrak{S}(t_{1},\zeta_{t})-\mathfrak{S}(t,\zeta_{t})|=|\mathfrak{S}(t_{1},\zeta_{t})|\leq\frac{8}{\sqrt{w(s)}}\int_{t_{1}}^{t}|f(u)|du.$

Hence, we find a constant $C_{1},$ depending only on $s$ and $A$ such that

(62)
$|\zeta_{1}-\zeta_{t}|\leq\frac{C_{1}}{t_{1}}\int_{t_{1}}^{t_{2}}|f(u)|du.$

This inequality holds in fact for any zero $\nu_{1}$ of $S(t_{1},z)$ and its
evolution $\nu_{t}$ as the same calculation shows. Let now $z_{t_{1}}$ be any
zero of $E(t_{1},z)$ in $Q(s,A/t_{1})$ and $z_{t}$ its evolution. This
includes the choice $z_{t}=\xi_{t}.$ Using (25) and Lemma 32 we have that
$|\theta_{z}(t,z_{t})|\asymp t$ for $t\in[t_{1},t_{2}]$ and

(63)
$|z_{1}-z_{t}|=\bigg{|}\int_{t_{1}}^{t}\frac{f(t)}{\theta_{z}(t,z_{t})}\bigg{|}\leq\frac{C_{2}}{t_{1}}\int_{t_{1}}^{t_{2}}|f(u)|du$

for some constant $C_{2},$ depending only on $s$ and $A.$ Furthermore, we can
relate $\xi_{1}$ and $\zeta_{1}$ with the help of $(i).$ From

$\sup_{z\in
Q(s,3C(t_{1})/t_{1})}\bigg{|}E(t_{1},z)-\frac{\alpha_{t_{1}}\gamma(t_{1}y_{t_{1}})}{\sqrt{w(s)}}\sin[t_{1}(z-(x_{t_{1}}-iy_{t_{1}}))]\bigg{|}<\psi(t_{1}),$

we can first of all assume without loss of generality that
$\zeta_{1}=x_{t_{1}}-iy_{t_{1}},$ if not we can just change to $\zeta_{1},$
since all zeros of the approximating function have the same imaginary part and
the real parts differ by $2\pi.$ By periodicity of $\sin(z)$ this does not
change the approximation. Then, by plugging in $z=\xi_{1}$ we find

$\bigg{|}E(t_{1},\xi_{1})-\frac{\alpha_{t_{1}}\gamma(t_{1}y_{t_{1}})}{\sqrt{w(s)}}\sin[t_{1}(\xi_{1}-(x_{t_{1}}-iy_{t_{1}}))]\bigg{|}=\bigg{|}\frac{\gamma(t_{1}y_{t_{1}})}{\sqrt{w(s)}}\sin[t_{1}(\xi_{1}-\zeta_{1})]\bigg{|}<\psi(t_{1})$

Since $\psi(t_{1})\rightarrow 0,$ as $t_{1}\rightarrow\infty,$ we must have
because of $1<t_{1}y_{t_{1}}<A$ and the fact that $\zeta_{1}$ was chosen as
the zero of $S(t_{1},z)$ closest to $\xi_{1}$ that

(64) $\xi_{1}=\zeta_{1}+o(1/t_{1}).$

Now the equations (62), (63), and (64) imply immediately

(65)
$|\xi_{t}-\zeta_{t}|<\frac{C_{3}}{t_{1}}\int_{t_{1}}^{t_{2}}|f(u)|du+o(1/t_{1})$

for some constant $C_{3}$ depending only on $s$ and $A.$

Recall that we can associate to any Hermite-Biehler function $E$ a meromorphic
inner function by

$\theta(t,z)=\frac{E^{\sharp}(t,z)}{E(t,z)}.$

We will denote by $I(t,z)$ the meromorphic inner function associated to
$S(t,z).$ We have he estimate

$|\theta_{z}(t,\overline{\xi}_{1})-I_{z}(t_{1},\overline{\zeta}_{1})|=o(t_{1}).$

Indeed, looking at the initial condition of $S(t_{1},z)$ this is nothing, but
Lemma 32 $(i).$

Furthermore, since $E$ satisfies $(i)$ at $t=t_{1}$ we can apply Lemma 32 and
get

(66) $|\theta_{z}(t,\overline{\xi}_{t})|\asymp
t,\quad\bigg{|}\frac{\theta_{zz}(t,\overline{\xi}_{t})}{\theta_{z}(t,\overline{\xi}_{t})}\bigg{|}\ll
t.$

By (65) the zeros of $E(t,z)$ and $S(t,z)$ are close and thus one can show by
looking at the product representation of $\theta(t,z)$ and $I(t,z)$ that

(67) $|I_{z}(t,\overline{\zeta}_{t})|\asymp
t,\quad\bigg{|}\frac{I_{zz}(t,\overline{\zeta}_{t})}{I_{z}(t,\overline{\zeta}_{t})}\bigg{|}\ll
t.$

We now find by (26) that for all $t\in(t_{1},t_{2})$

(68) $|I_{z}(t,\overline{\zeta}_{t})-\theta_{z}(t,\overline{\xi}_{t})|\ll
t\bigg{(}\int_{t_{1}}^{t_{2}}|f(u)|du+o(1)\bigg{)}.$

Indeed, by equation (26)

$\frac{d}{dt}\theta_{z}(t,\overline{\xi}_{t})=g(t)\theta_{z}(t,\overline{\xi}_{t}),$

where

$g(t)=2i\overline{\xi}_{t}-f(t)\frac{\theta_{zz}(t,\overline{\xi}_{t})}{\theta_{z}^{2}(t,\overline{\xi}_{t})}.$

We then write

$\theta_{z}(t,\overline{\xi}_{t})=\theta_{z}(t_{1},\overline{\xi}_{1})e^{\int_{t_{1}}^{t_{2}}g(t)dt}.$

Similarly,

$I_{z}(t,\overline{\zeta}_{t})=I_{z}(t_{1},\overline{\zeta}_{1})e^{\int_{t_{1}}^{t_{2}}h(t)dt},\quad
h(t)=2i\overline{\zeta}_{t}+f(t)\frac{I_{zz}(t,\overline{\zeta}_{t})}{I_{z}^{2}(t,\overline{\zeta}_{t})}.$

Therefore, using the above established bounds (65), (66), and (67) we find

$|I_{z}(t,\overline{\zeta}_{t})-\theta_{z}(t,\overline{\xi}_{t})|\leq|\theta_{z}(t_{1},\overline{\xi}_{1})-I_{z}(t_{1},\overline{\zeta}_{1})|e^{\int_{t_{1}}^{t}|g|du}+|I_{z}(t_{1},\overline{\zeta}_{1})|\big{|}e^{\int_{t_{1}}^{t}g(u)du}-e^{\int_{t_{1}}^{t}h(u)du}\big{|}$

$\ll o(1)t_{1}+t\int_{t_{1}}^{t}|g-h|du\ll
t\bigg{(}\int_{t_{1}}^{t_{2}}|f(t)|dt+o(1)\bigg{)}.$

This estimate combined with (25) immediately gives a bound on the difference
of the velocities.

$|(\zeta_{t})^{\prime}(t)-(\xi_{t})^{\prime}(t)|=\bigg{|}\frac{f(t)}{I_{z}(t,\overline{\zeta}_{t})}-\frac{f(t)}{\theta_{z}(t,\overline{\xi}_{t})}\bigg{|}=|f(t)|\bigg{|}\frac{\theta_{z}(t,\overline{\xi}_{t})-I_{z}(t,\overline{\zeta}_{t})}{\theta_{z}(t,\overline{\xi}_{t})I_{z}(t,\overline{\zeta}_{t})}\bigg{|}$
(69) $\ll
t\bigg{(}\int_{t_{1}}^{t_{2}}|f(u)|du+o(1)\bigg{)}\frac{|f(t)|}{|\theta_{z}(t,\overline{\xi}_{t})||I_{z}(t,\overline{\zeta}_{t})|}\ll\bigg{(}\int_{t_{1}}^{t_{2}}|f(u)|du+o(1)\bigg{)}|(\xi_{t})^{\prime}(t)|.$

From (26) we infer

(70)
$|\arg\theta_{z}(t,\overline{\xi}_{t})-\arg\theta_{z}(t_{1},\overline{\xi}_{t_{1}})|<\frac{\pi}{2}.$

Indeed, by the same trick as in the proof of Lemma 33

$|\arg\theta_{z}(t,\overline{\xi}_{t})-\arg\theta_{z}(t_{1},\overline{\xi}_{t_{1}})|\leq\bigg{|}\int_{t_{1}}^{t_{2}}\frac{d}{dt}\arg\theta_{z}(t,\overline{\xi}_{t})dt\bigg{|}\leq\bigg{|}\int_{t_{1}}^{t_{2}}s+(\Re\xi_{t}-s)+f(t)A(t)dt\bigg{|}$

$\leq(t_{2}-t_{1})|s|+A/t_{1}+\int_{t_{1}}^{t_{2}}|f(t)||A(t)|dt\leq
1+o(1/t)+\frac{3\cosh[2A]}{100\cdot 3\cosh[2A]}<\frac{\pi}{2}.$

Since $(t_{1},t_{2})$ is a $\sigma-$interval we deduce

(71) $(\zeta_{2}-\zeta_{1})-(\xi_{2}-\xi_{1})=o(\xi_{2}-\xi_{2}).$

Let now $\nu_{t_{1}}$ be any zero of $S(t_{1},z)$ and $\nu_{t}$ its evolution.
Then, by (62) and the property that $(t_{1},t_{2})$ is a $\sigma-$interval and
$H_{s}-$interval, and $t_{2}-t_{1}\leq 1$ we find

(72)
$|\nu_{t_{2}}-\nu_{t_{1}}|\leq\frac{C_{1}}{t_{1}}\int_{t_{1}}^{t_{2}}|f(u)|du\leq\frac{2C_{1}}{t_{2}}\int_{t_{1}}^{t_{2}}|f(u)|du\ll\bigg{|}\int_{t_{1}}^{t_{2}}\frac{f(u)}{|\theta_{z}(u,\tilde{\xi}_{u})|}du\bigg{|}\leq|\xi_{2}-\xi_{1}|.$

Without loss of generality we assume $C(t)<\sqrt{t}.$ Then, all zeros of

$I(t_{1},z)=\overline{\alpha_{t_{1}}}^{2}\frac{\sin[t_{1}(z-(x_{t_{1}}+iy_{t_{1}}))]}{\sin[t_{1}(z-(x_{t_{1}}-iy_{t_{1}}))]}$

have the same imaginary part and their real parts differ by at most
$C(t_{1})/t_{1}<1/\sqrt{t_{1}}.$ Hence, we find that their velocities differ
at most by

(73) $\ll
t\bigg{(}\frac{1}{\sqrt{t}}+\int_{t_{1}}^{t_{2}}|f(u)|du\bigg{)}=o(|(\zeta_{t})^{\prime}(t)|).$

Again, we find that their arguments $\arg[(\nu_{t})^{\prime}(t)]$ change by
less than $\frac{\pi}{2}$ and thus we get

(74) $(\nu_{t_{2}}-\nu_{t_{1}})-(\xi_{2}-\xi_{1})=o(\xi_{2}-\xi_{1}).$

This estimate together with (72) implies that there is a $D_{1}\in\mathbb{C}$
such that for all $z\in Q(s,A/t)$ we have

(75)
$|S(t_{2},z)-D_{1}\sin[t_{2}(z-(\zeta_{1}+(\xi_{2}-\xi_{1})))]=o(t_{2}(\xi_{2}-\xi_{1})).$

Indeed, one can show this by a standard argument comparing the two product
representations and using that the zeros are close to each other. By the usual
trick we can improve this to

$\sup_{z\in
Q(s,C(t)/t)}|S(t_{2},z)-D_{1}\sin[t_{2}(z-(\zeta_{1}+(\xi_{2}-\xi_{1})))]=o(t_{2}(\xi_{2}-\xi_{1}))$

by making $C(t)$ grow slower if necessary. We repeat the same argument for
$\tilde{E}$ by defining $\tilde{S}(t,z)$ to be the solution of (59) with
initial condition

$\tilde{S}(t_{1},z)=\alpha_{1}\frac{\gamma(t_{1}y_{t_{1}})}{\tilde{w}(s)}\cos[t_{1}(z-(\tilde{x}_{t_{1}}-iy_{t_{1}}))].$

Once again, we define $\tilde{\zeta}_{1}$ to be the zero of
$\tilde{E}(t_{1},z)$ closest to $\tilde{x}_{1}.$ By the exact same argument we
obtain, making $C(t)$ grow slower if necessary that

$\sup_{z\in
Q(s,C(t)/t)}|\tilde{S}(t_{2},z)-D_{2}\cos[t_{2}(z-(\tilde{\zeta}_{1}+(\tilde{\xi}_{2}-\tilde{\xi}_{1})))]=o(t_{2}(\tilde{\xi}_{2}-\tilde{\xi}_{1})).$

Since $E$ and $\tilde{E}$ satisfy $(i)$ at $t=t_{1}$ we find as above with
$\theta_{z}(t_{1},\xi_{t})$ and $I_{z}(t_{1},\zeta_{1})$ that

$|\theta_{z}(t_{1},\overline{\xi}_{1})-\tilde{\theta}_{z}(t_{1},\overline{\tilde{\xi}}_{1})|=o(t).$

Since $|\xi_{1}-\tilde{\xi}_{1}|\ll\frac{1}{t}$ and
$\theta_{z}(t,\xi_{t}),\tilde{\theta}_{z}(t,\tilde{\xi}_{t})$ satisfy (26) we
find by using the same argument we used to show (68)that for all
$t\in[t_{1},t_{2}]$

$|\theta_{z}(t,\overline{\xi}_{1})-\tilde{\theta}_{z}(t,\overline{\tilde{\xi}}_{1})|=o(t).$

Hence, using (25) one more time we have by the same calculation we used to
prove (71) that

$(\xi_{2}-\xi_{1})-(\tilde{\xi}_{2}-\tilde{\xi}_{1})=o(\xi_{2}-\xi_{1}).$

Thus, we can replace $(\tilde{\xi}_{2}-\tilde{\xi}_{1})$ by
$(\xi_{2}-\xi_{1})$ in the approximation above to obtain

$\sup_{z\in
Q(s,C(t)/t)}|\tilde{S}(t_{2},z)-D_{2}\cos[t_{2}(z-(\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})))]=o(t_{2}(\xi_{2}-\xi_{1})).$

The solution to (59) is unique and thus

$S(t_{2},s)=E(t_{2},s),\quad\tilde{S}(t_{2},s)=\tilde{E}(t_{2},s).$

Combining this with earlier estimates gives by the same calculation as in the
proof of Lemma 32 $(i)$ that

$\frac{E(t_{2},s)}{\tilde{E}(t_{2},s)}=\frac{\sin[t_{2}(s-(\zeta_{1}+(\xi_{2}-\xi_{1})))]}{\cos[t_{2}(s-(\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})))]}+o(t_{2}(\xi_{2}-\xi_{1})).$

Note that in particular we have at least $o(t_{2}(\xi_{2}-\xi_{1}))=o(1).$
Hoewever, generally $o(t_{2}(\xi_{2}-\xi_{1}))$ might be much smaller than
$o(1).$ By the same way as in the proof of Lemma 45 we find a constant
$\Delta=o(1)(\xi_{2}-\xi_{1})$ such that

(76)
$\frac{E(t_{2},s)}{\tilde{E}(t_{2},s)}=\frac{\sin[t_{2}(s-(\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta))]}{\cos[t_{2}(s-(\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})+\Delta))]}.$

Note that we still have

$\sup_{z\in
Q(s,A/t)}|S(t_{2},z)-D_{1}\sin[t_{2}(z-(\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta))]=o(t_{2}(\xi_{2}-\xi_{1})),$

$\sup_{z\in
Q(s,A/t)}|\tilde{S}(t_{2},z)-D_{2}\cos[t_{2}(z-(\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})+\Delta))]=o(t_{2}(\xi_{2}-\xi_{1})).$

We set

$x_{t_{2}}=\Re[\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta],\quad\tilde{x}_{t_{2}}=\Re[\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})+\Delta]$

$y_{2}=-\Im[\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta]=-\Im[\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})+\Delta].$

We still need to show that $(ii)$ and $(iii)$ hold. Property $(ii)$ follows
from the construction. We will now prove $(iii).$ Note that since $S$ and
$\tilde{S}$ must satisfy a version of Corollary 2 we can choose $D_{1}$ and
$D_{2}$ so that

(77)
$D_{1}\overline{D}_{2}>0,\quad|D_{1}D_{2}|=\frac{\gamma^{2}(t_{2}y_{2})}{\sqrt{w(s)\tilde{w}(s)}}.$

Indeed, if we want to have that $E(t_{2},s)=S(t_{2},s)$ and
$\tilde{E}(t_{2},s)=\tilde{S}(t_{2},s)$ we must have

$\det\begin{pmatrix}D_{1}\sin[t_{2}(z-(\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta))]&&D_{2}\cos[t_{2}(z-(\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})+\Delta))]\\\
\overline{D}_{1}\sin[t_{2}(z-(\overline{\zeta}_{1}+(\overline{\xi}_{2}-\overline{\xi}_{1})+\Delta))]&&\overline{D}_{2}\cos[t_{2}(z-(\overline{\tilde{\zeta}}_{1}+(\overline{\xi}_{2}-\overline{\xi}_{1})+\Delta))]\end{pmatrix}=2i.$

Let $z,w\in\mathbb{C}.$ Then, we have

$\sin(z)\cos(\overline{w})-\sin(\overline{z})\cos(w)=\frac{\sin(z-\overline{w})}{2}+\frac{\sin(z+\overline{w})}{2}-\frac{\sin(\overline{z}+w)}{2}-\frac{\sin(\overline{z}-w)}{2}$

$=i(\Im\sin(z-\overline{w})+\Im\sin(z+\overline{w}))$

Now we set

$z=t_{2}(z-(\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta)),\quad
w=t_{2}(z-(\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})+\Delta)).$

Then,

$\Im\sin(z-\overline{w})=\Im\sin(t_{2}(\overline{\tilde{\zeta_{1}}}-\zeta_{1}-2i\Im(\xi_{2}-\xi_{1})+2i\Im\Delta))$
$=\cos(t_{2}\Re(\tilde{\zeta}_{1}-\zeta_{1}))\sinh(2y_{1}-2\Im(\xi_{2}-\xi_{1})+2\Im\Delta)$
$=\sqrt{w(s)\tilde{w}(s)}\sinh[2t_{2}y_{2}],$

where we used property $(ii)$ and the definition of $y_{2}$ in the last step.
Furthermore

$\Im(\sin(z+\overline{w}))=\Im\sin(t_{2}(2s-(\xi_{1}+\tilde{\xi}_{1})-2\Re(\xi_{2}-\xi_{1}))-2\Re\Delta)=0.$

Hence,

$\det\begin{pmatrix}D_{1}\sin[t_{2}(z-(\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta))]&&D_{2}\cos[t_{2}(z-(\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})+\Delta))]\\\
\overline{D}_{1}\sin[t_{2}(z-(\overline{\zeta}_{1}+(\overline{\xi}_{2}-\overline{\xi}_{1})+\Delta))]&&\overline{D}_{2}\cos[t_{2}(z-(\overline{\tilde{\zeta}}_{1}+(\overline{\xi}_{2}-\overline{\xi}_{1})+\Delta))]\end{pmatrix}$
$=i\sqrt{w(s)\tilde{w}(s)}\sinh[2t_{2}y_{2}].$

Thus, $D_{1}$ and $D_{2}$ have to satisfy as claimed

$D_{1}\overline{D}_{2}>0,\quad|D_{1}D_{2}|=\frac{\gamma^{2}(t_{2}y_{2})}{\sqrt{w(s)\tilde{w}(s)}}.$

If the point $s_{1}$ is satisifies
$t_{1}s_{1}=t_{1}(s+(\tilde{x}_{1}-x_{1}))-\frac{\pi}{2}$ we get that

(78) $\tilde{S}(t_{1},s_{1})=\sqrt{\frac{w(s)}{\tilde{w}(s)}}S(t_{1},s).$

By property $(ii)$ we have
$|s-s_{1}|\leq|\tilde{x}_{1}-x_{1}|+\frac{\pi}{2t_{1}}<\frac{\pi}{t_{1}}\leq\frac{2\pi}{t_{2}}$
and for all $t\in(t_{1},t_{2})$

$\arg\frac{\tilde{S}(t,s_{1})}{S(t,s)}<|s-s_{1}|(t_{2}-t_{1})+2\int_{t_{1}}^{t_{2}}|f(u)|du<\frac{2\pi}{t_{2}}+2\int_{t_{1}}^{t_{2}}|f(u)|du.$

Indeed, as in the proof of Lemma 33 we find that

$\frac{d}{dt}\arg\bigg{(}\frac{\tilde{S}(t,s_{1})}{S(t,s)}\bigg{)}$

can be calculated using $\operatorname{Proj}_{iS(t,s)}\frac{d}{dt}S(t,s).$ By
Lemma 15 we find that

$\operatorname{Proj}_{iS(t,s)}\frac{d}{dt}S(t,s)=-\operatorname{Proj}_{iS(t,s)}isS(t,s)+f(t)\operatorname{Proj}_{iS(t,s)}\overline{S(t,s)}$
$=\bigg{(}-s+\frac{2A(t,s)C(t,s)}{|S(t,s)|^{2}}\bigg{)}iS(t,s).$

Now by the same argument as in the proof of Lemma 33 we have

$\frac{d}{dt}\arg S(t,s)=-s+\frac{2F(t,s)G(t,s)}{|S(t,s)|^{2}},$

where we wrote $S(t,z)=F(t,z)-iG(t,z).$ Note as well that $2F(t,s)G(t,s)\leq
4(F(t,s)^{2}+G(t,s)^{2})=4|S(t,s)|^{2}.$

By (78) we find $\arg(\tilde{S}(t,_{1}s_{1}))=\arg(S(t,s))$ and thus, as
desired

$\bigg{|}\arg\frac{\tilde{S}(t,s_{1})}{S(t,s)}\bigg{|}=\bigg{|}\int_{t_{1}}^{t_{2}}\frac{d}{dt}\arg\bigg{(}\frac{\tilde{S}(t,s_{1})}{S(t,s)}\bigg{)}dt\bigg{|}\leq|s-s_{1}|(t_{2}-t_{1})+4\int_{t_{1}}^{t_{2}}|f(t)|dt.$

For the absolute values, taking into account the initial condition

$|\tilde{S}(t_{1},s_{1})|=\sqrt{\frac{\tilde{w}(s)}{w(s)}}|S(t_{1},s)|=D,$

we have by applying twice the mean value Theorem for all $t\in[t_{1},t_{2}]$

$\bigg{|}|\tilde{S}(t,s_{1})|-\sqrt{\frac{\tilde{w}(s)}{w(s)}}|S(t,s)|\bigg{|}=D\bigg{|}e^{\int_{t_{1}}^{t}f(u)\cos[2\arg\tilde{S}(u,s_{1})]du}-e^{\int_{t_{1}}^{t}f(u)\cos[2\arg
S(u,s)]du}\bigg{|}$

$\ll\int_{t_{1}}^{t}|f(u)|\cdot\big{|}\cos[2\arg\tilde{S}(u,s_{1})]-\cos[2\arg
S(u,s)]\big{|}du$
$\ll\int_{t_{1}}^{t}|f(u)|\cdot|\arg(\tilde{S}(u,s_{1}))-\arg(S(u,s))|du$
$\ll\bigg{(}\frac{1}{t_{2}}+\int_{t_{1}}^{t_{2}}|f(u)|du\bigg{)}\int_{t_{1}}^{t_{2}}|f(u)|du.$

Note that we used in the first application of the mean value Theorem that

$\int_{t_{1}}^{t_{2}}|f(t)|dt<\frac{1}{100\cosh[2A]},$

so that the bound is not dependent on the interval $(t_{1},t_{2}).$ In
particular we showed

(79)
$\bigg{|}|\tilde{S}(t_{2},s_{1})|-\sqrt{\frac{\tilde{w}(s)}{w(s)}}|S(t_{2},s)|\bigg{|}\ll\bigg{(}\frac{1}{t_{2}}+\int_{t_{1}}^{t_{2}}|f(u)|du\bigg{)}\int_{t_{1}}^{t_{2}}|f(u)|du.$

Combining this estimate with (77) now implies that the constants $D_{1},D_{2}$
can be chosen so that

(80) $D_{1}=\alpha\frac{\gamma(t_{2}y_{2})}{\sqrt{w(s)}},\quad
D_{2}=\alpha\frac{\gamma(t_{2}y_{2})}{\tilde{w}(s)}$

for some unimodular constant $\alpha.$ Using trigonometric identities we find
as above

$\det\begin{pmatrix}\alpha\frac{\gamma(t_{2}y_{2})}{\sqrt{w(s)}}\sin[t_{2}(z-(\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta))]&&\alpha\frac{\gamma(t_{2}y_{2})}{\tilde{w}(s)}\cos[t_{2}(z-(\tilde{\zeta}_{1}+(\xi_{2}-\xi_{1})+\Delta))]\\\
\alpha\frac{\gamma(t_{2}y_{2})}{\sqrt{w(s)}}\sin[t_{2}(z-(\overline{\zeta}_{1}+(\overline{\xi}_{2}-\overline{\xi}_{1})+\Delta))]&&\alpha\frac{\gamma(t_{2}y_{2})}{\tilde{w}(s)}\cos[t_{2}(z-(\overline{\tilde{\zeta}}_{1}+(\overline{\xi}_{2}-\overline{\xi}_{1})+\Delta))]\end{pmatrix}=2i.$

Combining Corollary 2 and Lemma 44 we find

$|E(t_{2},s)|=\bigg{|}\alpha\frac{\gamma(t_{2}y_{2})}{\sqrt{w(s)}}\sin[t_{2}(z-(\zeta_{1}+(\xi_{2}-\xi_{1})+\Delta))]\bigg{|},$

$|\tilde{E}(t_{2},s)|=\bigg{|}\alpha\frac{\gamma(t_{2}y_{2})}{\sqrt{w(s)}}\sin[t_{2}(z-(\overline{\zeta}_{1}+(\overline{\xi}_{2}-\overline{\xi}_{1})+\Delta))]\bigg{|}.$

We can now modify $\alpha$ such that $(iii)$ holds and define it to be
$\alpha_{t_{2}}.$ ∎

###  3.8. Pointwise convergence of the non-linear Fourier transform

In this section, we will finally prove Theorem 2. We begin with the following
result.

######  Theorem 5.

$|S|=0$

###### Proof.

By definition we have for all $k\geq 1$ that

$S\subset\bigcup_{n=k}^{\infty}S^{n}\subset\bigcup_{n=k}^{\infty}S_{V}^{n}\cup\bigcup_{n=k}^{\infty}S_{H}^{n}.$

Combining this with Corollary 8 and Corollary 9 we find

$|S|\leq\sum_{n=k}^{\infty}|S^{n}|=\sum_{n=k}^{\infty}|S_{V}^{n}|+\sum_{n=k}^{\infty}|S_{H}^{n}|\leq
2D\sum_{n=1}^{\infty}||f||_{L^{2}([2^{n},2^{n+2}])}^{2}=2D||f||_{L^{2}(\mathbb{R})}^{2}<\infty.$

Since each $s\in S$ belongs to infinitely many $|S^{n}|,$ we can estimate
$|S|$ against the tail of the first series, which tends to zero since it
converges. ∎

We state once again the non-linear version of Carleson’s Theorem.

######  Theorem (Non-linear Carleson’s theorem).

Let $f$ be in $L^{2}(\mathbb{R}_{+}).$ Then, for almost all $s\in\mathbb{R}$

$\lim_{T\rightarrow\infty}f_{T}^{\dagger}(s)=f^{\dagger}(s).$

###### Proof.

By Theorem 5 and Corollary 6 we find that for almost all $s\in\mathbb{R}$ and
all $D>0$ there is a unimodular function $\alpha(s,t)$ and a constant
$\beta\in\mathbb{C}$ such that

$\sup_{z\in
Q(s,D/t)}\bigg{|}E(t,z)-\frac{-i\alpha(s,t)}{\sqrt{w(s)}}e^{itz}\bigg{|}=o(1),$

$\sup_{z\in
Q(s,D/t)}\bigg{|}\tilde{E}(t,z)-\frac{-i\beta\alpha(s,t)}{\sqrt{\tilde{w}(s)}}e^{itz}\bigg{|}=o(1),$

as $t\rightarrow\infty.$ By Corollary 2 we can choose $\beta=e^{-i\varphi}$
with $\varphi=\arcsin\sqrt{w(s)\tilde{w}(s)}.$ Indeed, initially we have by
Corollary 6 that there exists a function $\alpha^{\prime}(s,t)$ such that

$\sup_{z\in
Q(s,D/t)}\bigg{|}\tilde{E}(t,z)-\frac{-i\alpha^{\prime}(s,t)}{\sqrt{\tilde{w}(s)}}e^{itz}\bigg{|}=o(1).$

We now define $\beta(t),|\beta(t)|=1$ such that
$\beta(t)\alpha(s,t)=\alpha^{\prime}(s,t).$ Plugging in this approximation in
the determinant equation from Corollary 2 we find

$\det\begin{pmatrix}\frac{-i\alpha(s,t)}{\sqrt{w(s)}}e^{its}&&\frac{-i\beta(t)\alpha(s,t)}{\sqrt{\tilde{w}(s)}}e^{its}\\\
\frac{i\overline{\alpha}(s,t)}{\sqrt{w(s)}}e^{-its}&&\frac{i\overline{\beta}(t)\overline{\alpha}(s,t)}{\sqrt{\tilde{w}(s)}}e^{-its}\end{pmatrix}=\frac{1}{\sqrt{w(s)\tilde{w}(s)}}\big{(}\overline{\beta}(t)-\beta(t)\big{)}=2i+o(1).$

Hence, in the limit

$\lim_{t\rightarrow\infty}\Im\beta(t)=\lim_{t\rightarrow\infty}\frac{\beta(t)-\overline{\beta}(t)}{2i}=-\sqrt{w(s)\tilde{w}(s)}.$

Together with the fact that $|\beta(t)|=1$ it follows that we can choose
$\beta=e^{-i\varphi}$ with $\varphi=\arcsin\sqrt{w(s)\tilde{w}(s)}$ in the
approximation above.

By definiton we have

$|a(t,s)|=\bigg{|}\frac{e^{its}E(t,s)+ie^{its}\tilde{E}(t,s)}{2}\bigg{|}=\frac{1}{2}|E(t,s)+i\tilde{E}(t,s)|.$

Combining the above estimates gives for almost all $s\in\mathbb{R}$ that

$\bigg{|}E(t,s)+i\tilde{E}(t,s)-\bigg{(}\frac{-i\alpha(s,t)}{\sqrt{w(s)}}+i\frac{-i\beta\alpha(s,t)}{\sqrt{\tilde{w}(s)}}\bigg{)}e^{its}\bigg{|}=o(1).$

Doing the same calculations for $b(t,s)$ and applying the inverse triangle
inequality now yields that

$|a(t,s)|=\frac{1}{2}|E(t,s)+i\tilde{E}(t,s)|\overset{t\rightarrow\infty}{\longrightarrow}\frac{1}{2}\bigg{|}\frac{1}{\sqrt{w(s)}}+\frac{i\beta}{\sqrt{\tilde{w}(s)}}\bigg{|}$

$|b(t,s)|=\frac{1}{2}|E(t,s)-i\tilde{E}(t,s)|\overset{t\rightarrow\infty}{\longrightarrow}\frac{1}{2}\bigg{|}\frac{1}{\sqrt{w(s)}}-\frac{i\beta}{\sqrt{\tilde{w}(s)}}\bigg{|}.$

Furthermore, we compute

$\arg\bigg{(}\frac{b}{a}\bigg{)}=\arg b(t,s)-\arg
a(t,s)=\arg\big{[}E(t,s)-i\tilde{E}(t,s)\big{]}-\arg\big{[}E(t,s)+i\tilde{E}(t,s)\big{]}$

$=\arg(-i\alpha(s,t))+\arg\bigg{(}\frac{1}{\sqrt{w(s)}}-\frac{i\beta}{\sqrt{\tilde{w}(s)}}\bigg{)}+tx-\arg(-i\alpha(s,t))-\arg\bigg{(}\frac{1}{\sqrt{w(s)}}+\frac{i\beta}{\sqrt{\tilde{w}(s)}}\bigg{)}-tx+o(1).$

$=\arg\bigg{(}\frac{\frac{1}{\sqrt{w(s)}}-\frac{i\beta}{\sqrt{\tilde{w}(s)}}}{\frac{1}{\sqrt{w(s)}}+\frac{i\beta}{\sqrt{\tilde{w}(s)}}}\bigg{)}=o(1).$

Thus, for almost all $s\in\mathbb{R}$ we have shown that

$\bigg{|}\frac{b}{a}\bigg{|}\overset{t\rightarrow\infty}{\longrightarrow}\bigg{|}\frac{\frac{1}{\sqrt{w(s)}}-\frac{i\beta}{\sqrt{\tilde{w}(s)}}}{\frac{1}{\sqrt{w(s)}}+\frac{i\beta}{\sqrt{\tilde{w}(s)}}}\bigg{|},\quad\arg\bigg{(}\frac{b}{a}\bigg{)}\overset{t\rightarrow\infty}{\longrightarrow}\arg\bigg{(}\frac{\frac{1}{\sqrt{w(s)}}-\frac{i\beta}{\sqrt{\tilde{w}(s)}}}{\frac{1}{\sqrt{w(s)}}+\frac{i\beta}{\sqrt{\tilde{w}(s)}}}\bigg{)}.$

Finally, by combining the convergence of the absolute value and the argument
we find that for almost all $s\in\mathbb{R}$

$\lim_{T\rightarrow\infty}f_{T}^{\dagger}(s)=\lim_{T\rightarrow\infty}\frac{b(T,s)}{a(T,s)}=\frac{\frac{1}{\sqrt{w(s)}}-\frac{i\beta}{\sqrt{\tilde{w}(s)}}}{\frac{1}{\sqrt{w(s)}}+\frac{i\beta}{\sqrt{\tilde{w}(s)}}}=f^{\dagger}(s).$

proving that the non-linear Fourier transform converges pointwise almost
everywhere.

∎

##  4\. Further results

In the last chapter we have shown that $|a(t,s)|$ and $|b(t,s)|$ converge for
almost all $s\in\mathbb{R}$ as $t\rightarrow\infty.$ It is natural to ask
whether $a(t,s)$ and $b(t,s)$ converge too. We are going to investigate this
problem by tracing it back to the question of convergence of an ordinary
Fourier integral. We will frequently abuse the notation by using the letter
$t$ in the integrand as well as in the domain of integration.

######  Lemma 46.

$\frac{d}{dt}|a(t,s)|=2f(t)\frac{|E(t,s)|^{2}\cos[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\cos[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}.$

###### Proof.

We have the equation

$\frac{d}{dt}|a(t,s)|=\frac{\Re(a(t,s)\overline{\frac{d}{dt}a(t,s)})}{|a(t,s)|^{2}}.$

To make the next calculations more clear, we will supress from time to time
the arguments of the appearing functions. Since $|e^{its}|=1$ we can ignore
the exponential term and it is sufficient to calculate with the same formula
as above the derivative

$\frac{d}{dt}|E(t,s)+i\tilde{E}(t,s)|.$

By Lemma 15 we have that

$\frac{d}{dt}E(t,s)+i\tilde{E}(t,s)=-isE(t,s)+f(t)E^{\sharp}(t,s)+i[-is\tilde{E}(t,s)+f(t)\tilde{E}^{\sharp}(t,s)]$
$=-isE(t,s)+s\tilde{E}(t,s)+f(t)E^{\sharp}(t,s)+if(t)\tilde{E}^{\sharp}(t,s).$

Thus, setting $\tilde{a}(t,s)=E(t,s)+iE(t,s)$ we find that

$\Re\bigg{(}\tilde{a}(t,s)\overline{\frac{d}{dt}\tilde{a}(t,s)}\bigg{)}=(E+i\tilde{E})(isE^{\sharp}+s\tilde{E}^{\sharp}+f(t)E-if(t)\tilde{E})$

$=s(AB+CD)+f(t)(A^{2}-C^{2})+f(t)(B^{2}-D^{2})-s(AB+CD)=f(t)[A^{2}-C^{2}]+f(t)[B^{2}-D^{2}].$

$=f(t)|E(t,s)|^{2}\cos[2\arg
E(t,s)]+f(t)|\tilde{E}(t,s)|\cos[2\arg\tilde{E}(t,s)].$

Hence,

$\frac{d}{dt}|a(t,s)|=2\frac{f(t)|E(t,s)|^{2}\cos[2\arg
E(t,s)]+f(t)|\tilde{E}(t,s)|^{2}\cos[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}$

∎

We proceed by estimating $\arg a.$ First note that

$\arg
a(t,s)=\arg\bigg{(}\frac{e^{its}}{2}[E(t,s)+i\tilde{E}(t,s)]\bigg{)}=ts+\arg\tilde{a}.$

Thus,

$|a(t,s)|=1+2\int_{0}^{t}f(t)\frac{|E(t,s)|^{2}\cos[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\cos[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}dt.$

######  Lemma 47.

$\frac{d}{dt}\tilde{a}(t,s)=-s-2f(t)\bigg{[}|E(t,s)|^{2}\sin[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\sin[2\arg\tilde{E}(t,s)]\bigg{]}.$

###### Proof.

$\frac{d}{dt}\arg\tilde{a}=\operatorname{Proj}_{i(E(t,s)+i\tilde{E}(t,s))}\frac{d}{dt}E(t,s)+i\tilde{E}(t,s)$

$=\operatorname{Proj}_{-\tilde{E}(t,s)+iE(t,s)}-isE(t,s)+s\tilde{E}(t,s)+f(t)E^{\sharp}(t,s)+if(t)\tilde{E}^{\sharp}(t,s)$

$=\operatorname{Proj}_{-\tilde{E}(t,s)+iE(t,s)}-isE(t,s)+s\tilde{E}(t,s)+f(t)\operatorname{Proj}_{-\tilde{E}(t,s)+iE(t,s)}E^{\sharp}(t,s)+i\tilde{E}^{\sharp}(t,s).$

$=-s+f(t)\frac{\bigg{\langle}\begin{pmatrix}A-D\\\
B+C\end{pmatrix},\begin{pmatrix}-B+C\\\
A+D\end{pmatrix}\bigg{\rangle}}{|a(t,s)|^{2}}=-s+\frac{2[AC+BD]}{|a(t,s)|^{2}}.$

$=-s-2\frac{|E(t,s)|^{2}\sin[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\sin[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}.$

∎

Thus,

$\arg a(t,s)=-2\int_{0}^{t}f(t)\frac{|E(t,s)|^{2}\sin[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\sin[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}dt.$

######  Lemma 48.

$\frac{d}{dt}|a(t,s)|=2f(t)\frac{|E(t,s)|^{2}\cos[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\cos[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}.$

###### Proof.

We have the equation

$\frac{d}{dt}|a(t,s)|=\frac{\Re(a(t,s)\overline{\frac{d}{dt}a(t,s)})}{|a(t,s)|^{2}}.$

To make the next calculations more clear, we will supress from time to time
the arguments of the appearing functions. Since $|e^{its}|=1$ we can supress
the exponential term and it is sufficient to calculate with the same formula
as above

$\frac{d}{dt}|E(t,s)+i\tilde{E}(t,s)|.$

By Lemma 15 we have that

$\frac{d}{dt}E(t,s)+i\tilde{E}(t,s)=-isE(t,s)+f(t)E^{\sharp}(t,s)+i[-is\tilde{E}(t,s)+f(t)\tilde{E}^{\sharp}(t,s)]$
$=-isE(t,s)+s\tilde{E}(t,s)+f(t)E^{\sharp}(t,s)+if(t)\tilde{E}^{\sharp}(t,s).$

Thus, setting $\tilde{a}(t,s)=E(t,s)+iE(t,s)$ we find that

$\Re\bigg{(}\tilde{a}(t,s)\overline{\frac{d}{dt}\tilde{a}(t,s)}\bigg{)}=(E+i\tilde{E})(isE^{\sharp}+s\tilde{E}^{\sharp}+f(t)E-if(t)\tilde{E})$

$=s(AB+CD)+f(t)(A^{2}-C^{2})+f(t)(B^{2}-D^{2})-s(AB+CD)=f(t)[A^{2}-C^{2}]+f(t)[B^{2}-D^{2}].$

$=f(t)|E(t,s)|^{2}\cos[2\arg
E(t,s)]+f(t)|\tilde{E}(t,s)|\cos[2\arg\tilde{E}(t,s)].$

Hence,

$\frac{d}{dt}|a(t,s)|=2\frac{f(t)|E(t,s)|^{2}\cos[2\arg
E(t,s)]+f(t)|\tilde{E}(t,s)|^{2}\cos[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}$

∎

Thus, we can write

$|a(t,s)|=1+2\int_{0}^{t}f(t)\frac{|E(t,s)|^{2}\cos[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\cos[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}dt.$

We proceed by estimating $\arg a.$ First note that

$\arg
a(t,s)=\arg\bigg{(}\frac{e^{its}}{2}[E(t,s)+i\tilde{E}(t,s)]\bigg{)}=ts+\arg\tilde{a}.$

######  Lemma 49.

$\frac{d}{dt}\tilde{a}(t,s)=-s-2f(t)\bigg{[}|E(t,s)|^{2}\sin[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\sin[2\arg\tilde{E}(t,s)]\bigg{]}.$

###### Proof.

$\frac{d}{dt}\arg\tilde{a}=\operatorname{Proj}_{i(E(t,s)+i\tilde{E}(t,s))}\frac{d}{dt}E(t,s)+i\tilde{E}(t,s)$

$=\operatorname{Proj}_{-\tilde{E}(t,s)+iE(t,s)}-isE(t,s)+s\tilde{E}(t,s)+f(t)E^{\sharp}(t,s)+if(t)\tilde{E}^{\sharp}(t,s)$

$=\operatorname{Proj}_{-\tilde{E}(t,s)+iE(t,s)}-isE(t,s)+s\tilde{E}(t,s)+f(t)\operatorname{Proj}_{-\tilde{E}(t,s)+iE(t,s)}E^{\sharp}(t,s)+i\tilde{E}^{\sharp}(t,s).$

$=-s+f(t)\frac{\bigg{\langle}\begin{pmatrix}A-D\\\
B+C\end{pmatrix},\begin{pmatrix}-B+C\\\
A+D\end{pmatrix}\bigg{\rangle}}{|a(t,s)|^{2}}=-s+\frac{2[AC+BD]}{|a(t,s)|^{2}}.$

$=-s-2\frac{|E(t,s)|^{2}\sin[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\sin[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}.$

∎

Thus, we have established

$\arg a(t,s)=-2\int_{0}^{t}f(t)\frac{|E(t,s)|^{2}\sin[2\arg
E(t,s)]+|\tilde{E}(t,s)|^{2}\sin[2\arg\tilde{E}(t,s)]}{|a(t,s)|^{2}}dt.$

######  Theorem 6.

Let $s\in\mathbb{R}.$ We have the following equivalence. As
$t\rightarrow\infty$

$a(t,s)\text{ converges if and only if the Fourier
integral}\int_{0}^{t}f(t)\frac{b(t,s)}{a^{\sharp}(t,s)}e^{-2its}dt\text{
converges}.$

###### Proof.

$|a(t,s)|-1-i\arg a(t,s)=2\int_{0}^{t}f(t)\frac{\bigg{[}|E(t,s)|^{2}e^{2i\arg
E(t,s)}+|\tilde{E}(t,s)|^{2}e^{2i\arg\tilde{E}(t,s)}\bigg{]}}{|a(t,s)|^{2}}dt$
$=2\int_{0}^{t}f(t)\frac{E(t,s)^{2}+\tilde{E}(t,s)^{2}}{|a(t,s)|^{2}}dt=2\int_{0}^{t}f(t)\frac{(E(t,s)+i\tilde{E}(t,s))(E(t,s)-i\tilde{E}(t,s))}{|a(t,s)|^{2}}dt$
$=8\int_{0}^{t}f(t)\frac{a(t,s)b(t,s)}{|a(t,s)|^{2}}e^{-2its}dt=8\int_{0}^{t}f(t)\frac{b(t,s)}{\overline{a(t,s)}}e^{-2its}dt.$

∎

Notice that $b(t,s)/a^{\sharp}(t,s)\leq 1$ for all $s\in\mathbb{R}$ and thus
we obtain that if $f(t)\in L^{p}(\mathbb{R})$ for $1\leq p<\infty,$ then as
well for all $s\in\mathbb{R}$

$f(t)\frac{b(t,s)}{a^{\sharp}(t,s)}\in L^{p}(\mathbb{R}).$

We obtain immediately the following

######  Corollary 11.

If $f(t)\in L^{1}(\mathbb{R}),$ then $a(t,s)$ converges for all
$s\in\mathbb{R}.$

Unfortunately we can not extend this with out furhter work to the case
$1<p<\infty.$ To that end we introduce for $(s,y)\in\mathbb{R}^{2}$ and $t>0$
the function

$I(t,s,y)=\int_{0}^{t}\int_{0}^{t}f(t)\frac{b(t,s)}{a^{\sharp}(t,s)}e^{-2ity}dt.$

Then, by the linear Carleson Theorem we find that for almost all
$(s,y)\in\mathbb{R}^{2}$ the following limit exists

$\lim_{t\rightarrow\infty}I(t,s,y).$

The statement that $a(t,s)$ converges almost everywhere on $\mathbb{R}$ is
equivalent to the statement that the above integral convergeges almost
everyhwere on the main diagonal in $\mathbb{R}^{2}.$ Therefore, we can think
about the question, whether $a(t,s)$ converges almost everywhere, being as
difficult as proving that the following Fourier integral exists

$\int_{0}^{\infty}f(t)f_{t}^{\dagger}(s)e^{-its}dt.$

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