# On pointwise convergence of the non-linear Fourier transform

Lukas Mauth  Mathematisches Institut, Endenicher Allee 60, 53115 Bonn
<EMAIL_ADDRESS>
###### Abstract.

In [10] A. Poltotaski proved an analog of Carleson’s Theorem on almost
everywhere convergence of Fourier series for a version of the non-linear
Fourier transform. We aim to present his proof in full detail and elaborate on
the ideas behind each step.

###### Contents

  1. 1 Introduction
    1. 1.1 Carleson’s Theorem
    2. 1.2 The non-linear Fourier transform
    3. 1.3 Idea of the proof for non-linear Carleson
  2. 2 Preliminaries
    1. 2.1 Bounded analytic functions
    2. 2.2 Hilbert spaces of entire functions
    3. 2.3 Properties of the non-linear Fourier transform
  3. 3 Proof of non-linear Carleson
    1. 3.1 Proximity of reproducing kernels
    2. 3.2 Approximation of the Hermite-Bieher functions
    3. 3.3 Resonances can not stay close to the real line forever
    4. 3.4 Meromorphic inner functions and dynamics of the zeros
    5. 3.5 Vertical intervals
    6. 3.6 Horizontal intervals
    7. 3.7 Stronger approximations of the Hermite-Biehler functions
    8. 3.8 Pointwise convergence of the non-linear Fourier transform
  4. 4 Further results

##  1\. Introduction

We aim to present a proof for an analog of Carleson’s Theorem for the non-
linear Fourier transform. This result has recently been established by Alexei
Poltoratski [10] and we will present his proof in full detail. We are going to
follow his proof closely by not modifying his argument, but changing the order
of steps and adding more intermediate results, making it easier to understand
the proof. In particular, we aim to present the proof in such a way that
anyone with a basic background in complex analysis can understand it. In this
section, we are going to introduce the non-linear Fourier transform and draw
the analogy to Carleson’s Theorem on the linear Fourier transform. At the end
of this section, we will describe the strategy of Poltoratski’s proof.

The next section aims to introduce the basic concepts needed from the theory
of bounded analytic functions in the upper half-plane and Krein-de Branges
theory. Besides these few preliminary results there is not much theory
involved in the proof. One could go even as far to say that this proof is
elementary in a sense that it almost purely relies upon well-known results.

In the third section, we present Poltoratski’s proof with all details
included. Poltoratski’s paper [10] is organized into $12$ sections. In our
treatment of his proof, we are following his structure closely but shifting
some sections around or dividing them into other sections. The biggest change
is probably moving section $8$ from [10], which establishes joint
approximations of the Hermite-Biehler functions to the very end of our
discussion to make the key elements of his argument stand out. In particular,
every Lemma, Theorem, and Corollary in Poltoratski’s paper has a corresponding
result with the same statement in our version. All of our other results do not
appear in Poltoratski’s paper explicitly and we added them to simplify the
understanding of his ideas. In the following table, there is a dictionary that
allows going back and forth between Poltoratski’s and our paper.

In the fourth and final chapter we give a brief outlook on the open question
whether $\arg a(t,s)$ and $\arg b(t,s)$ converge as well. In the present
version of this proof we only showed that $|a(t,s)|$ and $|b(t,s)|$ converge
for almost all $s\in\mathbb{R}.$ We will prove in some sense that with
probability $1$ this question can be answered affirmatively and we manage to
prove it completely for the special case $f\in L^{1}(\mathbb{R}).$ Another
interesting quesiton is whether we can extract bounds on the maximal operator.
This question was studied by G. Mnatsakanyan in [15].

Dictionary
---
Our Version |  Poltoratski’s Version [10]
Theorem 2 |  Theorem $1$
Lemma 29 |  Lemma $1$
Lemma 21 |  Lemma $2$
Corollary 3 |  Lemma $3$
Lemma 22 |  Lemma $4$
Corollary 4 |  Lemma $5$
Lemma 23 |  Lemma $6$
Lemma 33 |  Lemma $7$
Lemma 43 |  Lemma $8$
Lemma 45 |  Lemma $9$
Theorem 4 |  Lemma $10$
Lemma 40 |  Lemma $11$
Corollary 5 |  Corollary $1$
Lemma 32 |  Corollary $2$
Corollary 7 |  Corollary $3$
Corollary 6 |  Corollary $4$
Corollary 10 |  Corollary $5$
Corollary 8 |  Claim $1$
Lemma 41 |  Claim $2$
Corollary 9 |  Claim $3$

Acknowldegements I wish to thank Christoph Thiele for introducing me to non-
linear Fourier Analysis, giving me this extensive project to work on and being
a hospital and forecoming supervisor. Furthermore, I wish to thank Gevorg
Mnatsakanyan with whom I spent countless hours understanding and discussing
Poltoratski’s proof. I wish to thank Alexei Poltoratski for his hospitality
and patience when explaining us the remaining parts of the proof we did not
understand. Moreover, I wish to thank the Number Theory group of Bonn,
especially Valentin Blomer, Edgar Assing, Radu Toma and Gilles Felber for the
opportunity to learn more about, and discussing topics in Number Theory, when
working on this project was a little difficult from time to time.

###  1.1. Carleson’s Theorem

Let $f:[0,1]\rightarrow\mathbb{C}$ be a smooth function. We define for
$n\in\mathbb{Z}$ the Fourier coefficients of $f$ by

(1) $\hat{f}(n)=\int_{0}^{1}f(x)e^{-2\pi inx}dx.$

We can use the Fourier coefficients $\hat{f}$ to expand our initial function
$f$ by into a Fourier series

(2) $f(x)=\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{2\pi inx}.$

Indeed, since we assumed $f$ to be smooth it is easy to verify that this
series converges uniformly to $f.$ Looking at (1) we see that the Fourier
coefficients are well-defined for the much larger class of $L^{p}-$functions
with $1\leq p\leq\infty.$ It was desirable that $(\ref{fourier_series})$
remained true for $f\in L^{p}[0,1]$, i.e. that the Fourier series still
converges to $f$ in some sense. Indeed one shows easily that for $f\in
L^{p}[0,1]$ with $1<p<\infty$

$S_{N}f(x):=\sum_{|n|\leq N}\hat{f}(n)e^{2\pi
inx}\overset{L^{p}}{\longrightarrow}f(x),$

where the convergence is understood in the $L^{p}[0,1]-$norm. It is a much
deeper question to ask if $S_{N}f$ converges to $f$ pointwise almost
everywhere for $f\in L^{p}[0,1].$ This problem remained open for a long time.
The case $p=1$ was settled first in 1923, when Kolmogorov [8] constructed an
integrable function $f$, whose Fourier series diverges almost everywhere. The
first positive answer came from L. Carleson in 1966 [1] and is since
celebrated under the name Carleson Theorem.

######  Theorem 1 (Carleson Theorem).

Let $f$ be in $L^{2}[0,1].$ Then $S_{N}f$ converges pointwise to $f$ almost
everywhere.

By inclusion of $L^{p}$-spaces, the above result generalizes immediately to
$2\leq p\leq\infty.$ One year later R. Hunt [7] gave an affirmative answer for
the remaining cases $1<p<2.$ In conclusion, the Fourier series of $f$
converges pointwise almost everywhere for any $f$ in $L^{p}[0,1]$ with
$1<p\leq\infty.$

###  1.2. The non-linear Fourier transform

We are going to study scattering data of certain systems of differential
equations. The associated scattering functions behave similarly to the
classical Fourier transform, see below. In particular, there is a non-linear
version of Carleson’s Theorem which remained open for a long time and which we
are going to prove.

The scattering data we are going to consider is constructed from real Dirac
systems on $\mathbb{R}_{+}.$ Let $f\in L^{2}(\mathbb{R}_{+}).$ We define
matrices

$\Omega=\begin{pmatrix}0&&1\\\ -1&&0\end{pmatrix}\text{, and
}Q(t)=\begin{pmatrix}0&&f(t)\\\ f(t)&&0\end{pmatrix}.$

For a spectral parameter $z\in\mathbb{C}$ we consider the real Dirac system on
$\mathbb{R}_{+}$

(3) $\Omega\frac{d}{dt}X(t,z)=zX(t,z)-Q(t)X(t,z).$

A solution to the Dirac system (3) is a vector-valued function

$X(t,z)=\begin{pmatrix}u(t,z)\\\ v(t,z)\end{pmatrix}.$

We will consider two solutions of the Dirac system satisfying the following
initial conditions respectively

$X_{N}(t,z)=\begin{pmatrix}A(t,z)\\\ B(t,z)\end{pmatrix},\quad
X_{N}(0,z)=\begin{pmatrix}1\\\ 0\end{pmatrix}\quad\text{(Neumann boundary
condition)}$

$X_{D}(t,z)=\begin{pmatrix}C(t,z)\\\ D(t,z)\end{pmatrix},\quad
X_{D}(0,z)=\begin{pmatrix}0\\\ 1\end{pmatrix}\quad\text{(Dirichlet boundary
condition)}$

From the solutions to Dirichlet and Neumann conditions we construct two
functions

$E(t,z)=A(t,z)-iC(t,z)\text{ and }\tilde{E}(t,z)=B(t,z)-iD(t,z).$

Closely related to these functions are the scattering functions corresponding
to the Dirac system

$\mathcal{E}(t,z)=e^{itz}E(t,z)\text{ and
}\tilde{\mathcal{E}}(t,z)=e^{itz}\tilde{E}(t,z).$

Furthermore, define for each $t>0$

$a(t,z)=\frac{\mathcal{E}(t,z)+i\tilde{\mathcal{E}}(t,z)}{2}=\frac{e^{itz}}{2}(E(t,z)+i\tilde{E}(t,z)),$

$b(t,z)=\frac{\mathcal{E}(t,z)-i\tilde{\mathcal{E}}(t,z)}{2}=\frac{e^{itz}}{2}(E(t,z)-i\tilde{E}(t,z)).$

We are interested in the quotient $b(t,z)/a(t,z).$ We define the non-linear
Fourier transform of $f$ by the pointwise limit

$f^{\dagger}=\lim_{t\rightarrow\infty}\frac{b(t,z)}{a(t,z)},$

if it exists at $z.$ It is our main goal to show (see Theorem 2) that this
limit exists pointwise almost everywhere. Analogous to Carleson’s Theorem
where we investigated the limiting behavior of the partial sums, we consider a
partial non-linear Fourier transform. More precisely given $f\in
L^{2}(\mathbb{R}_{+})$ we look for any $T>0$ at the restriction of $f$ to the
interval $(0,T)$ given by $f_{T}:=f\chi_{(0,T)}.$ Then, non-linear Fourier
transform is

$f_{T}^{\dagger}(z)=\frac{b(T,z)}{a(T,z)}.$

It is a natural question to ask whether the classical results for the linear
Fourier-transform hold as well in the non-linear case. For instance there is
an analogue of Parseval’s identity. For all $0<t<\infty$ the functions
$a(t,\cdot)$ satisfy the non-linear Parseval identity

(4) $||\log|a(t,\cdot)|||_{L^{1}(\mathbb{R})}=||f||_{L^{2}(0,t)}^{2}.$

The non-linear Parseval identity in terms of $f^{\dagger}$ takes the form

$\frac{1}{2}||\log(1-|f^{\dagger})|||_{L^{1}(\mathbb{R})}=||f||_{L^{2}(\mathbb{R})}^{2}.$

This identity together with the fact that the linear term in the series
expansion of $f^{\dagger}$ is the Fourier transform $\hat{f}$ are mainly
responsible for calling $f^{\dagger}$ the non-linear Fourier transform of $f.$
For more information regarding the non-linear Fourier transform we refer to
the lecture notes by T. Tao and C. Thiele [14]. We want to prove the following
non-linear version of Carleson’s Theorem.

######  Theorem 2 (Non-linear Carleson theorem).

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

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

###  1.3. Idea of the proof for non-linear Carleson

The solutions $E(t,z)$ and $\tilde{E}(t,z)$ of (3) are Hermite-Biehler
functions, i.e. for all $z\in\mathbb{C}_{+}$ they satisfy the inequality

$|E(t,z)|>|E(t,\overline{z})|.$

It turns out that to each Hermite-Biehler function $E(t,z)$ one can associate
a Hilbert space of entire functions $B(E(t,z)).$ As sets, we have that the
spaces $B(E(t,z))$ are the same as the Paley-Wiener spaces
$\operatorname{PW}_{t},$ but with different norms. One key property that they
share with the Paley-Wiener spaces $\operatorname{PW}_{t}$ is that they are
reproducing kernel Hilbert spaces, which means that for all
$\lambda\in\mathbb{C}$ there exists a function $K(t,\lambda,\cdot)\in
B(E(t,z))$ such that for for all $f\in B(E(t,z))$ we can recover $f(\lambda)$
by the formula

$f(\lambda)=\langle f,K(t,\lambda,\cdot)\rangle.$

The reproducing kernel for the Paley-Wiener space $\operatorname{PW}_{t}$ is
by Fourier inversion the function

$\operatorname{Sinc}(t,\lambda,z)=\frac{1}{\pi}\frac{\sin[t(z-\overline{\lambda})]}{z-\overline{\lambda}}$

and the reproducing kernel of the space $B(E(t,z))$ is given by the expression

$K(t,\lambda,z)=\frac{1}{2\pi
i}\frac{E(z)E^{\sharp}(\overline{\lambda})-E^{\sharp}(\overline{z})E(\overline{\lambda})}{\overline{\lambda}-z}=\frac{1}{\pi}\frac{A(z)C(\overline{\lambda})-C(z)A(\overline{\lambda})}{\overline{\lambda}-z},$

where $E^{\sharp}(t,z)=\overline{E(t,\overline{z})}$ denotes the Schwarz
transform. From these two formulas, one could guess that if we showed that for
large times $t$ the two reproducing kernels are close, then $E(t,z)$ should be
approximated by linear combinations of trigonometric functions. This is
exactly what we are going to prove in sections $3.1$ and $3.2.$ Let
$s\in\mathbb{R}$ and $C>0.$ We define for $t>0$ the box

$Q(s,C/t):=\\{z\in\mathbb{C}\mid|\Re(s-z)|\leq C/t,|\Im z|\leq C/t\\}.$

We will show in Corollary 4 that for almost all $s\in\mathbb{R}$ and all $C>0$
we have

$\sup_{\lambda,z\in
Q(s,C/t)}\bigg{|}[A(t,z)C(t,\overline{\lambda})-C(t,z)A(t,\overline{\lambda})]-\frac{1}{w(s)}\sin[t(\overline{\lambda}-z)]\bigg{|}=o(1)$

as $t\rightarrow\infty.$ Based on this equation we find that there are real
functions $x_{1}(t),x_{2}(t),c_{1}(t),c_{2}(t)$ such that for all $z\in
Q(s,C/t)$ we have uniformly

$E(t,z)=-\frac{1}{c_{1}(t)}\cos[t(z-x_{1}(t))]-i\frac{1}{c_{2}(t)}\sin[t(z-x_{2}(t))]+o(1).$

If we now had that $c_{1}(t)=c_{2}(t)+o(1)$ and $t(x_{2}(t)-x_{1}(t))=o(1)$
mod $2\pi,$ then by Euler’s formula

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

for some complex constant $D$ and a complex function $\alpha(s,t)$ satisfying
$|\alpha(s,t)|=1$ for all $t>0.$ If we had this approximation unconditionally
we could derive the pointwise convergence of the non-linear Fourier transform
easily, see section 3.8.

Unfortunately we only have $c_{1}(t)=c_{2}(t)+o(1)$ and
$t(x_{2}(t)-x_{1}(t))=o(1)$ mod $2\pi$, when there is no zero of $E(t,z)$
inside the box $Q(s,C/t)$ for all large enough $t.$ If there is a zero $z(t)$
of $E(t,z)$ inside $Q(s,C/t)$ the approximation is pertubated to

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

compare to Lemma 23 and Corollary 6 for precise statements. This is the first
big milestone in the proof. We have shown that there is a good approximation
to $E(t,z)$ when there is no zero in the box $Q(s,C/t)$ for all large enough
$t$ and that the approximation to $E(t,z)$ is bad when there is a zero inside
$Q(s,C/t)$ for some time $t.$ Thus, for the remainder of the proof we will
show that the set

$T_{0}(s,C)=\\{t\in\mathbb{R}_{+}\mid Q(s,C/t)\text{ contains a zero of
}E(t,z)\\}$

is bounded for almost all $s\in\mathbb{R}$ and any $C>0.$ By the previous step
we can then immediately conclude Theorem 2. First, we will show that
$T_{0}(s,C)$ can not cover a half-line, i.e. the zeros of $E(t,z)$ can not
stay in the box $Q(s,C/t)$ forever. This is shown in section $3.3.$ We will
employ the fact the $|E(t,x)|$ converges in measure on $\mathbb{R}$ by Lemma
19. If we now had that the zeros of $E(t,z)$ remained in the box forever, then
$|E(t,x)|\approx|\sin tx|$ for points $x$ near $s.$ Since $|\sin tx|$ does not
converge in measure, neither should $|E(t,x)|,$ but this contradicts Lemma 19.
With the same argument, we could prove that zeros can not travel infinitely
often through the box $Q(s,C/t)$ for any fixed time scale. However, there
seems to be no minimal time scale, depending on $f,$ in which the zeros travel
through the box. Thus, with this approach we can not immediately conclude
Theorem 2. The next natural step is to study further the dynamics of the
zeros. This is done in section $3.4.$ From the study of the dynamics we get
stronger approximations for $E(t,z).$ Since the proof of the stronger
approximation (see Theorem 4) is quite technical and complicated we postpone
it to section $3.7.$

To show that $T_{0}(s,C)$ is bounded, we consider the boxes $Q(s,3C/t)$ and
$Q(s,C/t).$ Whenever a zero goes inside $Q(s,C/t)$ it has to travel through
the larger box $Q(s,3C/t).$ We have just seen that the zeros of $E(t,z)$ can
not stay inside $Q(s,C/t)$ forever. Thus, they would have to travel through
$Q(s,3C/t)$ infinitely often if $T_{0}(s,C)$ was unbounded. We will show that
for each time the zero travels through the box $Q(s,3C/t),$ this costs a chunk
of $||f||_{L^{2}(\mathbb{R})}.$ Since we assumed $f\in L^{2}(\mathbb{R})$ we
immediately obtain a contradiction if $T_{0}(s,C)$ was unbounded.

To prove such an estimate it is natural to relate the scattering data from
$E(t,z)$ to $f.$ In section 3.7 we will define scattering functions
$a_{t_{1}\rightarrow t_{2}}$ for any interval, where there is a zero of
$E(t,z)$ inside the box $Q(s,3C/t).$ Together with (4) we obtain the non-
linear Parseval identity

$||\log|a_{t_{1}\rightarrow
t_{2}}(s)|||_{L^{1}(\mathbb{R})}=||f||_{L^{2}((t_{1},t_{2}))}.$

When trying to estimate the quantity on the left hand side by its series
expansion we get a bound of the type

$||\log|a_{t_{1}\rightarrow t_{2}}(s)|||_{L^{1}(\mathbb{R})}\gg||\arg
a_{t_{1}\rightarrow t_{2}}||_{L^{1,\infty}(\mathbb{R})}\geq
D|\varepsilon_{1}|+O(\varepsilon_{1}^{2}+\varepsilon_{2}^{2}),$

where $\varepsilon_{1}$ measures the horizontal increment in the movement of
the zero in the time interval $(t_{1},t_{2})$ and $\varepsilon_{2}$ measures
the vertical component and $D>0$ is some constant depending only on $s$ and
the size of the box $C$. Since this estimate is linear in $\varepsilon_{1},$
but quadratic in $\varepsilon_{2}$ we will have to consider intervals where
the movement of the zero has a horizontal component comparable to the vertical
one and intervals where the zero moves almost verticaly separately. These two
cases will be covered in sections 3.6 and 3.7 respectively. Both arguments use
a version of Parseval’s identity. For the horizontal intervals we will use the
non-linear Parseval identity stated above and for vertical intervals, we are
going to use the linear Parseval identity. After both cases have been covered
Theorem 2 finally follows.

##  2\. Preliminaries

In this section we are going to introduce all the relevant concepts neeeded
for the proof of Theorem 2. We start with a short discussion on Hardy spaces,
which provide tools for the systematic study of bounded analytic functions in
the upper half-plane. We restrict ourselves to refering for all relevant
proofs to the literature and prove only the more special applications of the
theory here. We will most often refer to the book by J.Garnett [6]. The reason
for the rather short discussion on this interesting subject is, that they are
merely used as a tool in proving the approximations of the reproducing kernels
in section $3.1.$ After section $3.1,$ Hardy spaces are never used again and
the remainder of the proof is elementary but very technical complex analysis.

The more delicate input are the de Branges spaces, which are discussed in
section $3.2,$ which are reproducing kernel Hilbert spaces of entire functions
we can associate to the Hermite-Biehler functions $E(t,z).$ We will benefit a
lot from the existence of a Poisson finite spectral measure
$\mu=w(s)ds+d\mu_{s}$ which provides for all $t\geq 0$ an isometric embedding

$B(E(t,z))\hookrightarrow L^{2}(\mu).$

The key ingredient which tells us that $w(s)$ is reasonably well behaved is
the fact that

$\frac{\log w(t)}{1+t^{2}}\in L^{1}(\mathbb{R}).$

This fact is one of the key corner stones of the proof and is so important,
since on the one hand we see that $w(s)\neq 0$ for almost all $s\in\mathbb{R}$
and on the other hand it enables us to use the theory of Hardy spaces, see the
discussion on outer functions below or in [6].

Finally, we are going to establish basic properties of the functions $E(t,z)$
that we will need frequently throughout the proof. The most useful identity
which will be used throughout later when approximating $E(t,z)$ is Corollary 2
which states that

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

###  2.1. Bounded analytic functions

The upper half-plane will be denoted by

$\mathbb{C}_{+}:=\\{z=x+iy\in\mathbb{C}\mid\Im(z)=y>0\\}.$

We define the Hardy space on $\mathbb{C}_{+}$ as a certain subspace of the
analytic functions on the upper half-plane as follows. Let $1<p<\infty$ and
set

$\operatorname{H}^{p}(\mathbb{C}_{+}):=\bigg{\\{}f:\mathbb{C}_{+}\rightarrow\mathbb{C}\mid
f\text{ is analytic and
}\sup_{y>0}\int_{-\infty}^{\infty}|f(x+iy)|^{p}dx<\infty\bigg{\\}}.$

We turn $\operatorname{H}^{p}(\mathbb{C}_{+})$ into a normed space by defining
the $\operatorname{H}^{p}$ norm as

$||f||_{\operatorname{H}^{p}}:=\sup_{y>0}\bigg{(}\int_{-\infty}^{\infty}|f(x+iy)|^{p}dx\bigg{)}^{\frac{1}{p}}.$

######  Lemma 1.

Let $1<p<\infty$ and let $f$ be a function in
$\operatorname{H}^{p}(\mathbb{C}_{+}).$ Then, for almost all $x\in\mathbb{R}$
the non-tangential limit

$f(x):=\lim_{y\searrow 0}f(x+iy)$

exists and it satisfies the following properties

$||f||_{p}^{p}=||f||_{\operatorname{H}^{p}}^{p},$

$\lim_{y\searrow 0}||f(x+iy)-f(x)||_{p}=0.$

###### Proof.

[6] p.55 ∎

######  Lemma 2.

Let $1<p<\infty$ and let $f$ be a function in
$\operatorname{H}^{p}(\mathbb{C}_{+}).$ Then, for all $z\in\mathbb{C}_{+}$

$|f(x+iy)|\leq\bigg{(}\frac{2}{\pi
y}\bigg{)}^{\frac{1}{p}}||f||_{\operatorname{H}^{p}}.$

###### Proof.

[6] pp. 17-18. ∎

For any $z=x+iy\in\mathbb{C}_{+}$ we define the Poisson kernel in the upper
half-plane by

$P_{z}(t)=\frac{1}{\pi}\frac{y}{(x-t)^{2}+y^{2}}.$

Now let $d\mu(t)$ be a positive measure such that

$\int_{-\infty}^{\infty}\frac{1}{1+t^{2}}d\mu(t)<\infty.$

We will call such measures Poisson-finite and define the Poissson extension to
the upper half-plane by

$P\mu(x+iy):=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{y}{(x-t)^{2}+y^{2}}d\mu(t).$

######  Lemma 3.

Let $d\mu(t)$ be a positive Poisson-finite measure and let $P\mu(x+iy)$ be its
Poisson extension to the upper half plane. Then, $P\mu(x+iy)$ is a harmonic
function in $\mathbb{C}_{+}.$

###### Proof.

[6] pp.11-12. ∎

A family of functions $\\{\varphi_{y}(t)\\}_{y>0}$ which are integrable on
$\mathbb{R}$ is called an approximate identity, if the following conditions
are fulfilled

  * (i)

$\int_{-\infty}^{\infty}\phi_{y}(t)dt=1,$

  * (ii)

$||\phi_{y}||_{1}\leq M$ for some positive constant $M>0.$

  * (iii)

For all $\delta>0$ we have $\lim_{y\searrow
0}\sup_{|t|>\delta}|\phi_{y}(t)|=0,$

  * (iv)

For all $\delta>0$ we have $\lim_{y\searrow
0}\int_{|t|>\delta}|\phi_{y}(t)|dt=0.$

The Poisson kernel $P_{z}(t)$ is an example of an approximate identity.

######  Lemma 4.

Let $f(x)\in L^{p}(\mathbb{R})$ for any $1\leq p\leq\infty$ and assume that
$f$ is continous at some point $x_{0}\in\mathbb{R}.$ Furthermore, let
$\\{\varphi\\}_{y>0}$ an approximate identity and set $u(z)=P_{y}*f(x)$ for
$z=x+iy$ in $\mathbb{C}_{+}.$ Then,

$\lim_{(x,y)\rightarrow x_{0}}u(x+iy)=f(x_{0}).$

###### Proof.

[6] p.16 together with the remark on p.19. ∎

For $x\in\mathbb{R}$ define the cone

$\Gamma(x):=\\{z\in\mathbb{C}_{+}\mid|\Re(z-x)|<\Im z\\}.$

For any function $u:\mathbb{C}_{+}\rightarrow\mathbb{C}$ we introduce the
nontangential maximal function of $u$ on $\mathbb{R}$ as

$Mu(x):=\sup_{z\in\Gamma(x)}|u(z)|.$

Furthermore we can associate the maximal function to any bounded cone, i.e.
for $z\in\Gamma(x)$ with $\Im z<C$ for some $C>0.$ We will use the same
notation $Mu(x).$ It will unambigious in the text, whether we use the bounded
non-tangential maximal operator or the unbounded one. More precisely, Lemma 21
uses the bounded, Lemma 22 the unbounded operator.

######  Lemma 5.

Let $P\mu$ denote the Poisson extension of a Poisson finite measure on
$\mathbb{R}.$ Then, $MP\mu$ is weak-$L^{1},$ where $MP\mu$ is the maximal
function with respect to a bounded cone.

###### Proof.

This follows immediately from the correspong result in the disk, see [6]
pp.55-56, and the fact that Poisson-integrable functions on the real line
transfer into integrable functions on the disk under the Cayley transform. ∎

We obtain immediately

######  Corollary 1.

Let $P\mu$ be the Poisson extension of a Poisson-finite measure on
$\mathbb{R}.$ Then, $\sqrt{MP\mu}\in L_{loc}^{1}(\mathbb{R}).$

######  Lemma 6.

Let $f\in\operatorname{H}^{p}(\mathbb{C}_{+})$ for $1<p<\infty.$ Then, the
non-tangential maximal function $Mf(x)$ is in $L^{p}(\mathbb{R})$ and there is
an absolute constant $A_{p}>0$ such that

$||Mf||_{L^{p}(\mathbb{R})}\leq
A_{p}||f||_{\operatorname{H}^{p}(\mathbb{R})}.$

###### Proof.

[6] p.27 ∎

######  Lemma 7.

Let $\\{\varphi_{y}(t)\\}_{y>0}$ be an approximate identity and let $\nu$ be a
finite singular measure on $\mathbb{R}.$ Then, $(\varphi_{y}*\nu)(x)$
converges non-tangentially to zero almost everywhere.

###### Proof.

[6] p.30 ∎

Let $f\in L_{loc}^{1}(\mathbb{R}^{n}).$ We say that $x\in\mathbb{R}^{n}$ is a
Lebesgue point of $f,$ if

$\lim_{r\searrow 0}\frac{1}{|B_{r}(x)|}\int_{B_{r}(x)}f(y)dy=f(x).$

It is well known, that almost all $x\in\mathbb{R}^{n}$ are Lebesgue points.

Let $f:\mathbb{C}_{+}\rightarrow\mathbb{C}$ be an analytic function on the
upper half plane. We say that $f$ is an inner function on $\mathbb{C}_{+},$ if
the following two conditions are fulfilled

  * (i)

$f$ is bounded,

  * (ii)

the non-tangential limit exists almost everywhere and is unimodular, i.e. for
almost all $x\in\mathbb{R}$ we have

$\lim_{y\searrow 0}|f(x+iy)|=1.$

We say that $f$ is an outer function in $\mathbb{C}_{+}$ if it is of the form

$f(z)=\alpha e^{u(z)+iv(z)},$

where $\alpha$ is a unimodular constant, i.e. $|\alpha|=1,$ $u(z)=(P_{y}*\log
w)(x)$ for a non-negative real function $w(t)$ satisfying

$\int_{-\infty}^{\infty}\frac{\log w(t)}{1+t^{2}}dt<\infty$

and $v(z)$ is the harmonic conjugate of $u(z).$ We say that $w(t)$ is the
boundary function of $f.$

######  Lemma 8.

Let $f:\mathbb{C}_{+}\rightarrow\mathbb{C}$ be an outer function withboundary
function $w(t).$ Then, for almost all $x\in\mathbb{R}$ the non-tangential
limit exists and

$\lim_{z\in\Gamma(x)}|f(z)|=w(x).$

###### Proof.

[6] p.64 ∎

######  Lemma 9.

Let $f\in L^{2}(\mathbb{R}).$ Then, $f$ is the non-tangential limit of an
$\operatorname{H}^{2}(\mathbb{C}_{+})$ function if $\hat{f}(s)=0$ for almost
all $s<0.$

###### Proof.

[6] p.84 ∎

######  Lemma 10.

Let $g\in\operatorname{H}^{2}(\mathbb{C}_{+})$ and
$g^{\prime}\in\operatorname{H}^{2}(\mathbb{C}_{-}).$ Suppose
$H:\mathbb{C}\rightarrow\mathbb{C}$ is an entire function equal to $e^{-iz}g$
in $\mathbb{C}_{+}$ and equal to $e^{iz}g^{\prime}$ in $\mathbb{C}_{-},$ then
$H\in\operatorname{PW}_{1}.$

###### Proof.

This is an immediate consequence of the previous Lemma, which asserts that

$\operatorname{supp}\widehat{g}\subset[0,\infty),\quad\operatorname{supp}\widehat{g^{\prime}}\subset(-\infty,0].$

However, multipliying by $e^{-iz}$ and $e^{iz}$ respectively, shifts the
support of the Fourier transform such that

$\operatorname{supp}\widehat{e^{-iz}g}\subset[-1,\infty),\quad\operatorname{supp}\widehat{e^{iz}g^{\prime}}\subset(-\infty,1].$

By the assumption on $H$ we have that

$\operatorname{supp}\widehat{H}\subset(-\infty,1]\cap[1,\infty)=[-1,1].$

Therefore, $H$ is an entire function whose Fourier transform is supported on
$[-1,1]$ and thus $H\in\operatorname{PW}_{1}.$ ∎

We say that a sequence of function $f_{n}:\mathbb{C}\rightarrow\mathbb{C}$
converges normally to $f,$ if $f_{n}$ converges uniformly on compact subsets
of $\mathbb{C}.$

######  Lemma 11.

Let $g_{n}\in\operatorname{H}^{2}(\mathbb{C}_{+})$ be a sequence converging
weakly to some function $g\in\operatorname{H}(\mathbb{C}_{+}).$ Then,
$g_{n}\rightarrow g$ normally in $\mathbb{C}_{+}.$

###### Proof.

[6] ∎

###  2.2. Hilbert spaces of entire functions

The results of this section can be found in [2]. We say that an entire
function $E(z)$ is a Hermite-Biehler function if it satisfies for all
$z\in\mathbb{C}_{+}$ the inequality

$|E(z)|>|E(\overline{z})|.$

To each Hermite-Biehler function $E(z)$ we can associate a de Branges space

$B(E):=\\{f:\mathbb{C}\rightarrow\mathbb{C}\mid f\text{ is entire},\quad
f/E\in H^{2}(\mathbb{C}_{+}),\quad f^{\sharp}/E\in H^{2}(\mathbb{C}_{+})\\}.$

We can turn $B(E)$ into a Hilbert space with inner product

$\langle
f,g\rangle_{B(E)}=\int_{-\infty}^{\infty}\frac{f(x)g(x)}{|E(x)|^{2}}dx.$

By Krein’s Theorem all functions in $B(E)$ have exponential type at most that
of $E.$ Furthermore, for each $\lambda\in\mathbb{C}$ the pointwise evaluation
functional

$F_{\lambda}:B(E)\rightarrow\mathbb{C}$ $\quad\quad\quad f\mapsto f(\lambda)$

is bounded. Hence, by the Riesz representation Theorem there is a function
$K(\lambda,\cdot)\in B(E)$ such that for all $f\in B(E)$ we can recover its
value at $\lambda$ by integrating against $K(\lambda,\cdot),$ i.e.

$f(\lambda)=\langle f,K(\lambda,\cdot)\rangle_{B(E)}.$

######  Lemma 12.

Let $K(\lambda,\cdot)$ be the reproducing kernel from above. Then,

$||K(\lambda,\cdot)||_{B(E)}=||K(\lambda,\cdot)/E||_{H^{2}(\mathbb{C}_{+})}=\sqrt{K(\lambda,\lambda)}=\sup_{f\in
B(E),||f||_{B(E)}\leq 1}|f(\lambda)|.$

###### Proof.

By Riesz representation Theorem there is an antilinear isometry

$\varphi:B(E)\rightarrow B(E)^{*}$ $\quad\quad\quad f\mapsto\langle
f,\cdot\rangle.$

Because $K(\lambda,\cdot)$ is a reproducing kernel we have
$\varphi(K(\lambda,\cdot))=F_{\lambda}.$ Since $\varphi$ is an isometry

$||K(\lambda,\cdot)||_{B(E)}=||F_{\lambda}||_{B(E)^{*}}=\sup_{f\in
B(E),||f||_{B(E)}\leq 1}|f(\lambda)|.$

The remaining claims follow from the equation $K(\lambda,\lambda)=\langle
K(\lambda,\cdot),K(\lambda,\cdot)\rangle_{B(E)}.$ ∎

###  2.3. Properties of the non-linear Fourier transform

Recall that our analysis of the non-linear Fourier transform is based on the
solutions $E(t,z)$ and $\tilde{E}(t,z)$ of the real Dirac system (3) with
Neumann and Dirichlet initial conditions respectively. Most of the results of
this section can be stated with arbitrary solutions of the real Dirac system
regardless of the initial condition. To avoid confusion we will denote a
solution of (3) with arbitrary initial condition by $S(t,z).$

######  Lemma 13.

For each fixed $t>0$ the function $S(t,z)$ is a Hermite-Biehler function, i.e.
for all $z\in\mathbb{C}_{+},$

$|S(t,z)|>|S(t,\overline{z})|.$

###### Proof.

A proof can be found in [12]. ∎

Hence, we can consider the de-Branges spaces $B(E(t,z))$ and
$B(\tilde{E}(t,z)).$ By Krein’s Theorem the function $E(t,z)$ has exponential
type $t.$ We thus, have an inclusion $B(E(t,z))\subset\operatorname{PW}_{t}$
as sets. It is natural to ask whether the reverse inclusion holds as well,
i.e. if $B(E(t,z))=\operatorname{PW}_{t}$ as sets. Indeed, this is true if the
potential function of the Dirac system (3) satisfies $f\in
L_{loc}^{1}(\mathbb{R}_{+}).$

To prove this result, we will effectively have to show that the norms on
$B(E(t,z))$ and $\operatorname{PW}_{t}$ are equivalent. More, precisely the
norm on $\operatorname{PW}_{t}$ is the ordinary $L^{2}$-norm, and $B(E(t,z))$
is equipped with the $L^{2}$-norm, weighted by the function $|E(t,z)|.$ Hence,
it suffices to show that $|E(t,z)|$ is bounded from above and from below. This
will follow from a couple observations on the solutions of the Dirac system
(3).

First, let us more conveniently keep track of our two solutions $E(t,z)$ and
$\tilde{E}(t,z)$ simultaneously by defining the matrix

(5) $M(t,z)=\begin{pmatrix}A(t,z)&&B(t,z)\\\ C(t,z)&&D(t,z)\end{pmatrix}.$

Hence, $M(t,z)$ satisfies the differential equation

(6) $\begin{pmatrix}0&&1\\\
-1&&0\end{pmatrix}\frac{d}{dt}M(t,z)=zM(t,z)-\begin{pmatrix}0&&f(t)\\\
f(t)&&0\end{pmatrix}M(t,z)$

The following result shows, that $E(t,z)$ and $\tilde{E}(t,z)$ are related.

######  Lemma 14.

The matrix $M(t,z)$ satisfies $\det M(t,z)=1$ for all $t\geq 0$ and
$z\in\mathbb{C}.$

###### Proof.

We calculate

$\frac{d}{dt}\det
M(t,z)=\frac{d}{dt}A(t,z)D(t,z)+\frac{d}{dt}D(t,z)A(t,z)-\frac{d}{dt}B(t,z)C(t,z)-\frac{d}{dt}C(t,z)B(t,z)$

From (6) we find

$\begin{pmatrix}\frac{d}{dt}C(t,z)&&\frac{d}{dt}D(t,z)\\\
-\frac{d}{dt}A(t,z)&&-\frac{d}{dt}B(t,z)\end{pmatrix}=\begin{pmatrix}zA(t,z)-f(t)C(t,z)&&zB(t,z)-f(t)D(t,z)\\\
zC(t,z)-f(t)A(t,z)&&zD(t,z)-f(t)B(t,z)\end{pmatrix}.$

Hence, we arrive at

$\frac{d}{dt}\det
M(t,z)=(-zC(t,z)+f(t)A(t,z))D(t,z)+(zB(t,z)-f(t)D(t,z))A(t,z)$
$+(zD(t,z)-f(t)B(t,z))C(t,z)+(-zA(t,z)+f(t)C(t,z))B(t,z)=0.$

This implies, that $\det M(t,z)$ is constant. The initial conditions of
$E(t,z)$ and $\tilde{E}(t,z)$ give

$\det M(0,z)=\det\begin{pmatrix}1&&0\\\ 0&&1\end{pmatrix}=1.$

∎

This relation can be rewritten in terms of $E(t,z)$ and $\tilde{E}(t,z)$ and
will play a central role in the proof of Theorem 2.

######  Corollary 2.

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

###### Proof.

This follows immediately from the equation

$\begin{pmatrix}A(t,z)&&B(t,z)\\\
C(t,z)&&D(t,z)\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1&&1\\\
i&&-i\end{pmatrix}\begin{pmatrix}E(t,z)&&\tilde{E}(t,z)\\\
E^{\sharp}(t,z)&&\tilde{E}^{\sharp}(t,z)\end{pmatrix}.$

∎

In the following we will talk for simplicity only about $E(t,z).$ All results
remain true as well for $\tilde{E}(t,z).$ The next step towards showing
$B(E(t,z))=\operatorname{PW}_{t}$ as sets, is the following

######  Lemma 15.

The Hermite-Biehler function $S(t,z)$ solves the differential equation

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

###### Proof.

From (6) we get

$\begin{pmatrix}\frac{d}{dt}A(t,z)\\\
\frac{d}{dt}C(t,z)\end{pmatrix}=\begin{pmatrix}-zC(t,z)+f(t)A(t,z)\\\
zA(t,z)-f(t)C(t,z)\end{pmatrix}.$

We compute

$\frac{d}{dt}E(t,z)=\frac{d}{dt}A(t,z)-iC(t,z)=-zC(t,z)+f(t)A(t,z)-i(zA(t,z)-f(t)C(t,z))$

$=-z(C(t,z)+iA(t,z))+f(t)(A(t,z)+iC(t,z))=-izE(t,z)+f(t)E^{\sharp}(t,z),$

where we used in the last step that $A^{\sharp}=A$ and $C^{\sharp}=C.$ ∎

To any Hermite-Biehler function $S(t,z)$ we can associate a scattering
function by

$\mathfrak{S}(t,z)=e^{itz}S(t,z).$

######  Lemma 16.

The scattering function $\mathfrak{S}(t,z)$ satisfies the differential
equation

$\frac{d}{dt}\mathfrak{S}(t,z)=iz\mathfrak{S}(t,z)+e^{itz}\frac{d}{dt}S(t,z)=f(t)e^{2itz}\mathfrak{S}^{\sharp}(t,z).$

###### Proof.

A short calculation involving Lemma 15 gives

$\frac{d}{dt}\mathfrak{S}(t,z)=ize^{itz}S(t,z)+e^{itz}\frac{d}{dt}S(t,z)=iz\mathfrak{S}(t,z)+e^{itz}\frac{d}{dt}S(t,z)$
$=iz\mathfrak{S}(t,z)+e^{itz}\big{(}-izS(t,z)+f(t)S^{\sharp}(t,z)\big{)}=f(t)e^{2itz}\mathfrak{S}^{\sharp}(t,z).$

∎

We denote by $\arg E$ the continuous branch of the argument of $E$ in the
closed upper half-plane satisfying $\arg E(t,0)=0.$ For $\tilde{E}$ we choose
the argument such that $\arg\tilde{E}(t,0)=-\frac{\pi}{2}.$ The last Lemma
gives rise to a differential equation for $|E(t,x)|$ where $x\in\mathbb{R}.$

######  Lemma 17.

For all $x\in\mathbb{R}$ the function $|S(t,x)|$ satisfies

$\frac{d}{dt}|S(t,x)|=f(t)|S(t,x)|\cos[2\arg S(t,x)].$

###### Proof.

Recall that for all $z\in\mathbb{C}$ $A(t,\overline{z})=\overline{A(t,z)}$ and
$C(t,\overline{z})=\overline{C(t,z)}.$ Hence, $A(t,x)$ and $C(t,x)$ are both
real on the real line. Thus,

(7)
$\frac{d}{dt}|E(t,x)|=\frac{\Re(E(t,x)\overline{E^{\prime}(t,x)})}{|E(t,x)|}.$

Note that this expression is well-defined by Lemma 14. By Lemma 15 we find

$\Re[E(t,x)\overline{E^{\prime}(t,x)}]=\Re[(A(t,x)-iC(t,x))(ix(A(t,x)+iC(t,x))+f(t)(A(t,x)-iC(t,x)))]$

$=f(t)(A(t,x)^{2}-C(t,x)^{2})=f(t)\Re[(A(t,x)-iC(t,x))^{2}]=f(t)\Re[E(t,x)^{2}]$
$=f(t)|E(t,x)|^{2}\cos[2\arg E(t,x)].$

We conclude by plugging this back into (7). ∎

Again, the same result holds for $\tilde{E}(t,z).$ We are finally ready to
prove the main result of this chapter.

######  Lemma 18.

The de Branges spaces $B(E(t,z))$ and $B(\tilde{E}(t,z))$ are equal to the
Paley-Wiener space $\operatorname{PW}_{t}$ as sets

###### Proof.

We will show that the norms on $B(E(t,z))$ and $\operatorname{PW}_{t}$ are
equivalent. Recall that the norm on the Paley-Wiener space
$\operatorname{PW}_{t}$ is just the ordinary $L^{2}$-norm on the real line.
Hence, it suffices to show that $E(t,z)$ is bounded from above and below.

Let us first show that $E(t,z)$ is bounded from above. By Lemma 14,
$|E(t,x)|\neq 0$ for all $x\in\mathbb{R}.$ From Lemma 17 we get

$\frac{\frac{d}{dt}{|E(t,x)|}}{|E(t,x)|}=f(t)\cos[2\arg E(t,x)].$

Integrating this equation and taking exponentials yields

$|E(t,x)|=|E(0,x)|\int_{0}^{t}f(s)\cos[2\arg E(s,x)]ds.$

Since $E(0,z)=1$ for all $z\in\mathbb{C},$ the last equation simplifies to

$|E(t,x)|\leq\int_{0}^{t}|f(s)|ds,$

which concludes the proof of the upper bound. The same bound holds for
$\tilde{E}(t,x).$ Let us now turn towards the lower bound. Lemma 14 tells us
that for all $z\in\mathbb{C}$

$\det\begin{pmatrix}A(t,x)&&B(t,x)\\\ C(t,x)&&D(t,x)\end{pmatrix}=1.$

If $E(t,x)=A(t,x)-iC(t,x)$ was not bounded away from zero, then
$\tilde{E}(t,x)=B(t,x)-iD(t,x)$ had to be arbtitrarily large, contradicting
the upper bound of $\tilde{E}(t,x).$ Switching the roles of $E(t,x)$ and
$\tilde{E}(t,x)$ in the previous argument gives the lower bound for
$\tilde{E}(t,x).$ ∎

Since we have shown that $B(E(t,z))=\operatorname{PW}_{t}$ as sets, we see
that if $t<t^{\prime},$ then $B(E(t,z))\subset B(E(t^{\prime},z)).$ Hence one
could guess that there is a measure $\mu$ on $\mathbb{R}$ such that we have an
isometric embedding $B(E(t,z))\hookrightarrow L^{2}(\mu)$ for all $t\geq 0.$
This is indeed true. We will call $\mu$ the spectral measure of $E(t,z).$
Moreover by Lebesgue’s decomposition Theorem we can decompose

$\mu=w(s)ds+d\mu_{s}$

into an absolute continuous and singular part. Again the same construction for
$\tilde{E}(t,z)$ gives rise to a spectral measure $\tilde{\mu}.$ The following
non-trivial Theorem is crucial for the proof of our main result and used to be
a difficult problem in its own right. Essentially this boils down to show that
the one-dimensional Schrödinger operator with $L^{2}$-potential has absolutely
continuous spectrum which covers the real line.

######  Theorem 3.

Let $w(s)$ and $\tilde{w}(s)$ be the densities of the absolute continuous
parts of the spectral measures $\mu$ and $\tilde{\mu}$ of $E(t,z)$ and
$\tilde{E}(t,z),$ respectively. Then,

$\frac{\log w(t)}{1+t^{2}}\in
L^{1}(\mathbb{R}),\quad\frac{\log\tilde{w}(t)}{1+t^{2}}\in L^{1}(\mathbb{R}).$

###### Proof.

A proof can be found in [3] and [4]. ∎

This result implies in particular that $w(s),\tilde{w}(s)\neq 0$ almost
everywhere on $\mathbb{R}.$ As a corollary we obtain the following

######  Lemma 19.

$|E(t,x)|$ converges in measure to $\frac{1}{\sqrt{w(s)}}$ on $\mathbb{R}$

###### Proof.

Since we have the isometric embedding for all $t>0$

$\int_{\mathbb{R}}\frac{|f(x)|^{2}}{|E(t,x)|^{2}}dx=\int_{\mathbb{R}}|f(x)|^{2}d\mu,$

one sees that as $t\rightarrow\infty$

$\frac{1}{|E(t,s)|^{2}}\rightarrow d\mu(s)=w(s)$

in measure on $\mathbb{R}.$ Since $w(s)\neq 0$ almost everywhere, we obtain
the desired result. ∎

Thus, combining Lemma 12, Lemma 18 and the isometric embeddings
$B(E(t,z))\hookrightarrow L^{2}(\mu)$ and $B(\tilde{E}(t,z))\hookrightarrow
L^{2}(\mu)$ gives

(8)
$||K(t,\lambda,z)||_{B(E(t,z)}=\sup_{f\in\operatorname{PW}_{t},||f||_{\mu}\leq
1}|f(\lambda)|,\quad||\tilde{K}(t,\lambda,z)||_{B(\tilde{E}(t,z)}=\sup_{f\in\operatorname{PW}_{t},||f||_{\tilde{\mu}}\leq
1}|f(\lambda)|,$

where $K(t,\lambda,\dot{)}$ and $\tilde{K}(t,\lambda,\dot{)}$ are the
reproducing kernels of $B(E(t,z))$ and $B(\tilde{E}(t,z)).$

Furthermore, we can even give an explicit expression for the reproducing
kernels.

######  Lemma 20.

For any $t>0$ and $\lambda,z\in\mathbb{C}$ we have

$K(t,\lambda,z)=\frac{1}{2\pi
i}\frac{E(z)E^{\sharp}(\overline{\lambda})-E^{\sharp}(\overline{z})E(\overline{\lambda})}{\overline{\lambda}-z}=\frac{1}{\pi}\frac{A(z)C(\overline{\lambda})-C(z)A(\overline{\lambda})}{\overline{\lambda}-z},$

$\tilde{K}(t,\lambda,z)=\frac{1}{2\pi
i}\frac{\tilde{E}(z)\tilde{E}^{\sharp}(\overline{\lambda})-\tilde{E}^{\sharp}(\overline{z})\tilde{E}(\overline{\lambda})}{\overline{\lambda}-z}=\frac{1}{\pi}\frac{B(z)D(\overline{\lambda})-D(z)B(\overline{\lambda})}{\overline{\lambda}-z}.$

###### Proof.

A proof can be found in [2]. ∎

We will now solve the Dirac System (6) for $f=0.$ This is referred to as the
free case. In this case the system of differential equations we have to solve
simplifies to

$\begin{pmatrix}0&&1\\\ -1&&0\end{pmatrix}\frac{d}{dt}M(t,z)=zM(t,z).$

One can check quite fast that the solutions are given by
$E(t,z)=e^{-itz}=\cos(tz)-i\sin(tz)$ and
$\tilde{E}(t,z)=-ie^{itz}=-\sin(tz)-i\cos(tz).$ Since
$|E(t,z)|=|\tilde{E}(t,z)|=1$ for all $z\in\mathbb{C}$ we find that
$B(E(t,z))=B(\tilde{E}(t,z))=\operatorname{PW}_{t}$ as normed vector spaces.
Recall that in general we only have equality as sets.

##  3\. Proof of non-linear Carleson

###  3.1. Proximity of reproducing kernels

The first key insight used to get an approximation for $E(t,z)$ is that the
reproducing kernel $K(t,\lambda,z)$ of the associated de Branges space
$B(E(t,z))$ allows reconstructing the function $E(t,z)$ since by Lemma 20

$K(t,\lambda,z)=\frac{1}{2\pi
i}\frac{E(z)E^{\sharp}(\overline{\lambda})-E^{\sharp}(\overline{z})E(\overline{\lambda})}{\overline{\lambda}-z}=\frac{1}{\pi}\frac{A(z)C(\overline{\lambda})-C(z)A(\overline{\lambda})}{\overline{\lambda}-z}.$

Since the potential function $f$ of the Dirac-system (6) is in
$L^{2}(\mathbb{R})$ we have at least

$\liminf_{t\rightarrow\infty}f(t)=0.$

For most well-behaved functions, we even have

$\lim_{t\rightarrow\infty}f(t)=0.$

Carrying this information toward the Dirac system

$\begin{pmatrix}0&&1\\\
-1&&0\end{pmatrix}\frac{d}{dt}M(t,z)=zM(t,z)-\begin{pmatrix}0&&f(t)\\\
f(t)&&0\end{pmatrix}M(t,z),$

suggests that the last term should play a less significant role for large
$t>0$ and hence gives rise to an approximation of the free system where $f=0.$
Following this logic, we would expect that the solution of the Dirac system
will be approximated by the solution of the free system. This is precisely
what we will show under the additional condition that there are no zeros of
$E(t,z)$ too close to the real line for large times $t.$ We will show this
estimate by proving first that the reproducing kernels of the associated de
Brange spaces are close, since we can then easily pass to the functions
$E(t,z)$ themselves as hinted above.

For $f=0$ and $t>0$ the associated de Brange space is the Paley-Wiener space
$\operatorname{PW}_{t}$ (as explained in the previous section) and its
reproducing kernel is the $\operatorname{Sinc}$ function

$\operatorname{Sinc}(t,\lambda,z)=\frac{1}{\pi}\frac{\sin(t(z-\overline{\lambda}))}{z-\overline{\lambda}}.$

Hence, the big goal of this section will be to show that approximately

$K(t,\lambda,z)=\frac{1}{\pi}\frac{A(z)C(\overline{\lambda})-C(z)A(\overline{\lambda})}{\overline{\lambda}-z}\approx\frac{1}{\pi}\frac{\sin(t(z-\overline{\lambda}))}{z-\overline{\lambda}}=\operatorname{Sinc}(t,\lambda,z).$

The strategy will be to first show that the norms of the reproducing kernels
are close and in the next step to pass from the norms to the functions
themselves. During this procedure Theorem 3 shows its importance as it allows
us to construct an outer function $G(s)$ with boundary function $\sqrt{w(s)}.$

######  Lemma 21.

For almost all $s\in\mathbb{R}$ and all fixed $C>0$ we have

$\sup_{z\in
Q(s,C/t)}\bigg{(}\frac{w(s)||K(t,z,\cdot)||_{\mu}^{2}}{||\operatorname{Sinc}(t,z,\cdot)||_{2}^{2}}-1\bigg{)}=o(1)$

as $t\rightarrow\infty.$

###### Proof.

Let us first prove the convergence pointwise. Choose for all $t>0$ a point
$z_{t}\in Q(s,C/t).$ We will show that

$\frac{w(s)||K(t,z_{t},\cdot)||_{\mu}^{2}}{||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}}-1=o(1).$

The first step will be to rewrite this equation in the more convenient form

$w(s)||K(t,z_{t},\cdot)||_{\mu}^{2}=(1+o(1))||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}.$

This equation is true if and only if and only if

$\sqrt{w(s)}||K(t,z_{t},\cdot)||_{\mu}=(1+o(1))||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}.$

The advantage of this formulation is that we can now plug in the relations

(9)
$||K(t,z_{t},\cdot)||_{\mu}=\sup_{f\in\operatorname{PW}_{t},||f||_{\mu}\leq
1}|f(z_{t})|,\quad||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}=\sup_{f\in\operatorname{PW}_{t},||f||_{2}\leq
1}|f(z_{t})|,$

to arrive at the equation we are going to show

(10) $\sqrt{w(s)}\sup_{f\in\operatorname{PW}_{t},||f||_{\mu}\leq
1}|f(z_{t})|=(1+o(1))\sup_{f\in\operatorname{PW}_{t},||f||_{2}\leq
1}|f(z_{t})|.$

We will show both inequalities separately. Let us first show that the left-
hand side is at most as large as the right-hand side. We define for all $t>0$
and $x\in\mathbb{R}$ the function

$s_{t}(x)=\frac{|\operatorname{Sinc}(t,z_{t},x)|^{2}}{||\operatorname{Sinc}(t,z_{t},x)||_{2}^{2}}.$

Note that $s_{t}(x)$ is an approximate identity at the point $s$ and it
follows from Lemma 4 and Lemma 7 that for almost all $s\in\mathbb{R}$

(11) $\int_{-\infty}^{\infty}s_{t}(x)d\mu=w(s)+o(1).$

Indeed, we can write

$\int_{-\infty}^{\infty}s_{t}(x)d\mu=\int_{-\infty}^{\infty}s_{t}(x)dw(x)+\int_{-\infty}^{\infty}s_{t}(x)d\mu_{s}.$

The first integral converges to $w(s)$ by Lemma 4 since $s_{t}(x)$ is an
approximate identity at $s.$ Since $\mu_{s}$ is Poisson-finite it follows by
Dominated Convergence Theorem that the second integral over the complement of
any finite interval is converging to zero. Restricting $\mu_{s}$ to the finite
interval, gives a finite measure, so we can conclude by Lemma 7 that the
second integral tends to zero.

Recall that $\operatorname{Sinc}(t,z_{t},\cdot)\in\operatorname{PW}_{t}$ and
it is thus rescaled by its $||\cdot||_{\mu}$ norm a valid test function for
the operator norm of $K(t,z_{t},\cdot).$ Moreover, since
$\operatorname{Sinc}(t,z_{t},\cdot)$ is the reproducing kernel of
$\operatorname{PW}_{t},$ we have by Lemma 12 that
$||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}=\sqrt{\operatorname{Sinc}(t,z_{t},z_{t})}.$
Hence,

$||K(t,z_{t},\cdot)||_{\mu}=\sup_{f\in\operatorname{PW}_{t},||f||_{\mu}\leq
1}||f(z)|\geq\frac{\operatorname{Sinc}(t,z_{t},z_{t})}{||\operatorname{Sinc}(t,z_{t},\cdot)||_{\mu}}=\frac{\operatorname{Sinc}(t,z_{t},z_{t})}{\sqrt{||\operatorname{Sinc}(t,z_{t},\cdot)||_{\mu}^{2}}}$

$=\frac{\operatorname{Sinc}(t,z_{t},z_{t})}{\sqrt{||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}\int_{-\infty}^{\infty}s_{t}(x)d\mu}}=\frac{\operatorname{Sinc}(t,z_{t},z_{t})}{\sqrt{||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}(w(s)+o(1))}}=\frac{||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}}{\sqrt{w(s)+o(1)}}$

Now, using (9) and rearranging the terms gives

$\sqrt{w(s)}\sup_{f\in\operatorname{PW}_{t},||f||_{\mu}\leq
1}||f(z_{t})|\leq(1+o(1))\sup_{f\in\operatorname{PW}_{t},||f||_{2}\leq
1}||f(z_{t})|.$

Since $||f||_{\mu}\geq||f||_{w}$ we can instead prove

(12) $\sqrt{w(s)}\sup_{f\in\operatorname{PW}_{t},||f||_{w}\leq
1}||f(z_{t})|\leq(1+o(1))\sup_{f\in\operatorname{PW}_{t},||f||_{2}\leq
1}||f(z_{t})|.$

We will now show the reverse inequality, proving (10). The first step is to
rescale the problem in the variable $t,$ so that we can work on the fixed
domain $Q(s,C).$ The hard part will be to control the function $w(s)$ during
this process. It is precisely for this reason, that we will use the theory of
Hardy spaces and Theorem 3, which told us that

$\frac{\log(w(t))}{1+t^{2}}\in L^{1}(\mathbb{R}).$

In particular, the last statement is still true if we replace $w(t)$ by
$\sqrt{w(t)}.$ Hence, we can pick an outer function
$G:\mathbb{C}_{+}\rightarrow\mathbb{C}$ with boundary function $\sqrt{w(t)}.$
By Lemma 8 $|G(x)|=\sqrt{w(x)}$ for almost all $x\in\mathbb{R}.$ We can assume
without loss of generality that $|G(s)|=\sqrt{w(s)}.$ By changing the
unimodular constant in the definition of $G,$ we can assume that
$G(s)=\sqrt{w(s)}.$ Without loss of generality we can assume $w(s)=1,$ since
$w(s)\neq 0$ for almost all $s\in\mathbb{R}.$

Let us now turn towards rescaling the problem. For any $t>0$ the map

$\varphi_{t}:Q(s,C)\rightarrow Q(s,C/t)$ $\quad\quad\quad\quad z\mapsto
s+\frac{z-s}{t}$

is bijective. Hence, we can lift our chosen points $z_{t}$ on $Q(s,C/t)$ to
$Q(s,C)$ by setting $\tilde{z_{t}}=\varphi^{-1}(z_{t}).$ By abuse of notation
we will just write $z_{t}$ instead of $\tilde{z_{t}}$ for the rest of the
proof.

We will now have to rescale the norms in (12). For each
$f\in\operatorname{PW}_{t}$ we set

$\tilde{f}(z_{t})=\frac{1}{\sqrt{t}}f\bigg{(}s+\frac{z_{t}-s}{t}\bigg{)}.\text{
and }G_{t}(z)=G\bigg{(}s+\frac{z_{t}-s}{t}\bigg{)}.$

Then, $f\in\operatorname{PW}_{1},$ because
$\operatorname{supp}\hat{\tilde{f}}\subset\frac{1}{t}\operatorname{supp}\hat{f}\subset\frac{1}{t}[-t,t]=[-1,1].$
Now recall that on $\mathbb{R}$ we have $|G(t)|^{2}=w(t)$ and thus,

$||\tilde{f}||_{w}=\frac{1}{\sqrt{t}}\bigg{(}\int_{-\infty}^{\infty}\bigg{|}f\bigg{(}s+\frac{x-s}{t}\bigg{)}G(x)\bigg{|}^{2}dx\bigg{)}^{\frac{1}{2}}=\frac{1}{\sqrt{t}}\bigg{(}\int_{-\infty}^{\infty}\sqrt{t}\bigg{|}f(x)G_{t}(x)\bigg{|}^{2}dx\bigg{)}^{\frac{1}{2}}=||fG_{t}||_{2}.$

Hence, we have the rescaled equation

$\sup_{f\in\operatorname{PW}_{t},||f||_{w}\leq
1}\bigg{|}f\bigg{(}s+\frac{z_{t}-s}{t}\bigg{)}\bigg{|}=\sqrt{t}\sup_{f\in\operatorname{PW}_{1},||fG_{t}||_{2}\leq
1}|f(z_{t})|.$

and similarly, we prove

$\sup_{f\in\operatorname{PW}_{t},||f||_{2}\leq
1}\bigg{|}f\bigg{(}s+\frac{z_{t}-s}{t}\bigg{)}\bigg{|}=\sqrt{t}\sup_{f\in\operatorname{PW}_{1},||f||_{2}\leq
1}|f(z_{t})|$

Thus, the inequality we will need to prove becomes

$\sup_{f\in\operatorname{PW}_{1},||fG_{t}||_{2}\leq
1}||f(z_{t})|\leq(1+o(1))D_{t},$

where we put to simplify notation

$D_{t}=\sup_{f\in\operatorname{PW}_{1},||f||_{2}\leq 1}|f(z_{t})|.$

We will prove this inequality by contradiction. Suppose that for some
$\varepsilon>0$ there is a sequence of functions
$f_{n}\in\operatorname{PW}_{1}$ such that for some subsequence
$k_{n}\rightarrow\infty$ and we have

$||f_{n}G_{k_{n}}||_{2}\leq 1,\text{ and
}|f_{n}(z_{k_{n}})|>D_{k_{n}}+\varepsilon.$

By construction all $z_{t}\in Q(s,C)$ and hence we can without loss of
generality assume that $z_{k_{n}}\rightarrow z_{0}\in Q(s,C),$ for the box is
compact. We set

$D=\sup_{f\in\operatorname{PW}_{1},||f||_{2}\leq 1}|f(z_{0})|.$

Notice that since $\operatorname{Sinc}(t,\lambda,z)\in\operatorname{PW}_{t}$
is the reproducing kernel, we have by continuity

$D_{t}=\sqrt{\operatorname{Sinc}(1,z_{t},z_{t})}\rightarrow\sqrt{\operatorname{Sinc}(1,z_{0},z_{0})}=D.$

Our strategy is to construct an admissible test function $H,$ i.e.
$H\in\operatorname{PW}_{1}$ and $||H||_{2}\leq 1$ satisfying $|H(z_{0})|\geq
D+\varepsilon.$ To construct the function $H,$ we will use the theory of Hardy
spaces. By Lemma 9 the function
$e^{iz}f_{n}\in\operatorname{H}^{2}(\mathbb{C}_{+})$. Indeed, we have that
$\operatorname{supp}\hat{f_{n}}\subset[-1,1].$ Now
$\operatorname{supp}\widehat{e^{iz}f_{n}}\subset[0,\infty)$ and thus
$e^{iz}f_{n}\in\operatorname{H}^{2}(\mathbb{C}_{+})$ by Lemma 9.

Since $G_{k_{n}}$ is outer and $||f_{n}G_{k_{n}}||_{2}\leq 1$ we find
$g_{n}\in\operatorname{H}^{2}(\mathbb{C}_{+})$ Moreover, by Lemma 1
$||g_{n}||_{H^{2}(\mathbb{C}_{+})}=||g_{n}||_{2}\leq 1,$ since
$||f_{n}G_{k_{n}}||_{2}\leq 1.$

Hence, we can without loss of generality assume, passing to a subsequence if
necessary, that $g_{n}\rightarrow g$ for some
$g\in\operatorname{H}^{2}(\mathbb{C}_{+}).$ By Lemma 11 $g_{n}\rightarrow g$
normally in $\mathbb{C}_{+}.$ Moreover, since $g_{n}\rightarrow g$ weakly, we
have by Lemma 1

$||g||_{2}=||g||_{H^{2}(\mathbb{C}_{+})}\leq\liminf_{n\rightarrow\infty}||g_{n}||_{H^{2}(\mathbb{C}_{+})}\leq
1.$

Since by Lemma 8

$\lim_{z\in\Gamma(x)}|G(z)|=1,$

we find $G_{t}\overset{t\rightarrow\infty}{\longrightarrow}1$ normally in
$\mathbb{C}_{+}.$ We will conclude that $f_{n}$ converges normally to some
function in $\mathbb{C}_{+}.$ Let $K\subset\mathbb{C}_{+}$ be compact and
denote by $||\cdot||_{\infty,K}$ the supremum norm on $K.$ We will drop $K$
from the notation. Recall that by definition outer functions vanish nowhere in
$\mathbb{C}_{+}.$

$||e^{iz}f_{n}-g||_{\infty}\leq\bigg{|}\bigg{|}e^{iz}f_{n}-\frac{g}{G_{k_{n}}}\bigg{|}\bigg{|}_{\infty}+\bigg{|}\bigg{|}\frac{g}{G_{k_{n}}}-g\bigg{|}\bigg{|}_{\infty}.$

However,

$\bigg{|}\bigg{|}e^{iz}f_{n}-\frac{g}{G_{k_{n}}}\bigg{|}\bigg{|}_{\infty}\leq||G_{k_{n}}||_{\infty}||e^{iz}f_{n}G_{k_{n}}-g||_{\infty}\rightarrow
0,$

$\bigg{|}\bigg{|}\frac{g}{G_{k_{n}}}-g\bigg{|}\bigg{|}_{\infty}\leq||g||_{\infty}\bigg{|}\bigg{|}\frac{1}{G_{k_{n}}}\bigg{|}\bigg{|}_{\infty}||G_{k_{n}}-1||_{\infty}\rightarrow
0.$

Hence, $e^{iz}f_{n}\rightarrow g$ normally in $\mathbb{C}_{+}.$ From this, we
obtain quickly that $f_{n}\rightarrow e^{-iz}g$ normally in $\mathbb{C}_{+}.$
Indeed,

$||f_{n}-e^{-iz}g||_{\infty}\leq||e^{iz}||_{\infty}||e^{iz}f_{n}-g||_{\infty}\rightarrow
0.$

By repeating the same argument in the lower half-plane $\mathbb{C}_{-}$ we
obtain a function $g^{\prime}$ such that $f\rightarrow e^{iz}g^{\prime}$
normally in $\mathbb{C}_{-}.$ Since a normal limit of holomorphic functions is
holomorphic, we have a candidate function, which is analytic in the upper and
lower half-planes. However, we still do not know yet what happens around the
real line. To remedy this, we will try to find an integrable function that
dominates $f_{n}$ close to the real line and conclude with dominated
convergence. By Lemma 2 applied to $g_{n}=e^{iz}f_{n}G_{k_{n}}$ we have for
all $z\in\mathbb{C}$ the bound

$|f_{n}G_{k_{n}}(x+iy)|\leq\bigg{(}\frac{2}{\pi|y|}\bigg{)}^{\frac{1}{2}}||g_{n}||_{H^{2}(\mathbb{C}_{+})}\leq\frac{1}{\sqrt{|y|}}.$

Hence, we now have to work on bounding the outer function $G_{k_{n}}$ away
from zero. From the definition of the outer function $G$ we see that

(13) $|G(z)|=|e^{P\log\sqrt{w}(z)}|\geq e^{-MP\log\sqrt{w}(\Re z)},$

where $MP\log\sqrt{w}$ is the non-tangential maximal function of
$P\log\sqrt{w}$ with respect to a large bounded cone $\Gamma(x).$ By Corollary
1 the function $\sqrt{MP\log\sqrt{w}}$ is locally integrable. Hence, we can
assume that $s$ is a Lebesgue point of $\sqrt{MP\log\sqrt{w}}.$

Fix a constant $L>0.$ For large enough $n$ we can choose $c_{n}$ with
$L<c_{n}<2L$ such that $P\log\sqrt{w}$ is uniformly bounded on the union of
the lines $\Re(z-s)=\pm\frac{c_{n}}{k_{n}}.$ Indeed, suppose that for all
$x\in B_{\frac{2L}{k_{n}}}(s)\setminus B_{\frac{L}{k_{n}}}(s)$ we had
$\sqrt{MP\log\sqrt{w(x)}}>3\sqrt{MP\log\sqrt{w(s)}}.$ Then,

$\int_{B_{\frac{2L}{k_{n}}}(s)\setminus
B_{\frac{L}{k_{n}}}(s)}\sqrt{MP\log\sqrt{w(t)}}dt>3\sqrt{MP\log\sqrt{w(s)}}\big{|}B_{\frac{L}{k_{n}}}(s)\big{|},$

But this is a contradiction to the fact that $s$ is a Lebesgue point of
$MP\log\sqrt{w},$ since

$\frac{1}{\big{|}B_{\frac{L}{k_{n}}}(s)\big{|}}\int_{B_{\frac{2L}{k_{n}}}(s)\setminus
B_{\frac{L}{k_{n}}}(s)}\sqrt{MP\log\sqrt{w(t)}}dt\leq
2\frac{1}{\big{|}B_{\frac{2L}{k_{n}}}(s)\big{|}}\int_{B_{\frac{2L}{k_{n}}}(s)}\sqrt{MP\log\sqrt{w(t)}}dt\rightarrow
2f(s).$

Hence, for each large enough $n,$ without loss of generality for all $n\geq
1,$ there has to be a point $c_{n}\in B_{\frac{2L}{k_{n}}}(s)\setminus
B_{\frac{L}{k_{n}}}(s)$ such that
$\sqrt{MP\log\sqrt{w(\frac{c_{n}}{k_{n}})}}\leq 3\sqrt{MP\log\sqrt{w(s)}},$
which shows as claimed, that $P\log\sqrt{w}$ is uniformly bounded on the union
of the lines $\Re(z-s)=\pm\frac{c_{n}}{k_{n}}.$

Now consider for each $n\geq 1$ a sequence of squares $R_{n},$ centered at $s$
with side length $2c_{n}.$ Hence, on the vertical sides of $R_{n}$ we have by
construction and (13).

$|G_{k_{n}}(s\pm c_{n}+iy)|=\bigg{|}G\bigg{(}s+\frac{s\pm
cn+iy-s}{k_{n}}\bigg{)}\bigg{|}\geq
e^{MP\log\sqrt{w}(s\pm\frac{c_{n}}{k_{n}})}=\delta>0.$

Note that since $\frac{c_{n}}{k_{n}}\leq 2$ it is really sufficient to
consider only a bounded sector $\Gamma(x)$ to estimate $MP\log\sqrt{w}$ on the
square $R_{n}.$

Thus, on the vertical sides of $R_{n}$ we have

$|f_{n}(s\pm c_{n}+iy)|\leq\frac{1}{\delta\sqrt{|y|}}.$

We can now finally conclude, that $f_{n}$ converges to some entire function in
all of $\mathbb{C}.$ Denote by $R$ the space inscribed by the square centered
as $s$ with side length $L.$ Then, by using the Cauchy integral formula with
the paths traced out by $R_{n},$ and $R,$ together with the dominated
convergence Theorem we find that for all $a\in R$ that

$f_{n}(a)=\frac{1}{2\pi}\int_{R}\frac{f_{n}(z)}{z-a}dz\rightarrow\frac{1}{2\pi}\int_{R}\frac{f(z)}{z-a}dz=:H(z).$

The function $H$ is thus entire and by earlier calculations equal to
$e^{-iz}g$ in $\mathbb{C}_{+}$ and equal to $e^{iz}g^{\prime}$ in
$\mathbb{C}_{-}.$ Since $g\in\operatorname{H}^{2}(\mathbb{C}_{+})$ and
$g^{\prime}\in\operatorname{H}^{2}(\mathbb{C}_{-})$ we have by Lemma 10 that
$H\in\operatorname{PW}_{1}.$ Furthermore, by construction
$||H||_{2}=||g||_{2}\leq 1.$ Thus, $H$ is an admissible test function.
Finally, because $f_{n}$ converges normally to $H$ we have

$|H(z_{0})|=\lim_{n\rightarrow\infty}|f_{n}(z_{k_{n}})|\leq\lim_{n\rightarrow\infty}D_{k_{n}}+\varepsilon=D+\varepsilon,$

achieving our desired contradiction. Hence, we proved for our chosen points
$z_{t}\in Q(s,C/t)$ that

$\sqrt{w(s)}\sup_{f\in\operatorname{PW}_{t},||f||_{\mu}\leq
1}||f(z)|\geq(1+o(1))\sup_{f\in\operatorname{PW}_{t},||f||_{2}\leq 1}||f(z)|.$

and thus we have all together shown that

$\frac{w(s)||K(t,z_{t},\cdot)||_{\mu}^{2}}{||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}}-1=o(1).$

To upgrade this convergence to

$\sup_{z\in
Q(s,C/t)}\bigg{(}\frac{w(s)||K(t,z,\cdot)||_{\mu}^{2}}{||\operatorname{Sinc}(t,z,\cdot)||_{2}^{2}}-1\bigg{)}=o(1)$

we first note, that since
$||K(t,z,\cdot)||_{\mu}^{2}=K(t,z,z)\in\operatorname{PW}_{t}$ and
$||\operatorname{Sinc}(t,z,\cdot)||_{2}^{2}=\operatorname{Sinc}(t,z,z)\in\operatorname{PW}_{t}$
both expressions are continuous in $z.$ Hence, for each $t>0$ we choose by
compactness of $Q(s,C/t)$ the point $z_{t}^{*}$ which maximizes

$\bigg{|}\frac{w(s)||K(t,z_{t}^{*},\cdot)||_{\mu}^{2}}{||\operatorname{Sinc}(t,z_{t}^{*},\cdot)||_{2}^{2}}-1\bigg{|}.$

Since we have proven the convergence for all choices $z_{t}$ and $z_{t}^{*}$
is a legitimate choice, we achieve an error term $o(1)$ uniform over all
choices $z_{t},$ which finishes the proof. ∎

######  Corollary 3.

For almost all $s\in\mathbb{R}$ and for all $C>0$

$\sup_{z\in
Q(s,C/t)}\bigg{|}\bigg{|}K(t,z,\cdot)-\frac{1}{w(s)}\operatorname{Sinc}(t,z,\cdot)\bigg{|}\bigg{|}_{\mu}^{2}=o(t),$

as $t\rightarrow\infty.$

Note that since $w(s)\neq 0$ for almost all $s\in\mathbb{R},$ this expression
is well-defined almost everywhere.

###### Proof.

Let $z_{t}\in Q(s,C/t)$ for all $t>0$ and let $s_{t}$ be the function defined
in the proof of Lemma 21. Recall that

$\int_{-\infty}^{\infty}s_{t}(x)d\mu=w(s)+o(1).$

Using that $L^{2}(\mu)$ is a Hilbert space with inner product
$\langle\cdot,\cdot\rangle_{\mu}$ and the fact that $K(t,z_{t},\cdot)$ is a
reproducing kernel of the space $\operatorname{PW}_{t}$ equipped with the
$||\cdot||_{\mu}$ norm, we find

$\bigg{|}\bigg{|}K(t,z_{t},\cdot)-\frac{1}{w(s)}\operatorname{Sinc}(t,z_{t},\cdot)\bigg{|}\bigg{|}_{\mu}^{2}=\bigg{\langle}K(t,z_{t},\cdot)-\frac{1}{w(s)}\operatorname{Sinc}(t,z_{t},\cdot),K(t,z_{t},\cdot)-\frac{1}{w(s)}\operatorname{Sinc}(t,z_{t},\cdot)\bigg{\rangle}_{\mu}$

$=||K(t,z_{t},\cdot)||_{\mu}^{2}-\frac{2}{w(s)}\langle\operatorname{Sinc}(t,z_{t},\cdot),K(t,z_{t},\cdot)\rangle_{\mu}+\frac{1}{w^{2}(s)}||\operatorname{Sinc}(t,z_{t},\cdot)||_{\mu}^{2}$

$=||K(t,z_{t},\cdot)||_{\mu}^{2}-\frac{2}{w(s)}\operatorname{Sinc}(t,z_{t},z_{t})+\frac{||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}}{w^{2}(s)}\int_{-\infty}^{\infty}s_{t}(x)d\mu$

$=||K(t,z_{t},\cdot)||_{\mu}^{2}-\frac{2}{w(s)}||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}+\frac{||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}}{w^{2}(s)}(w(s)+o(1))$

$=||K(t,z_{t},\cdot)||_{\mu}^{2}-\frac{1}{w(s)}||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}+o(||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2})=o(||\operatorname{Sinc}(t,z_{t},\cdot)||_{2}^{2}),$

where we used 21 in the last step. Now the proof is finished by noting that

$\operatorname{Sinc}(t,z_{t},z_{t})=||\operatorname{Sinc}(t,z_{t},z_{t})||_{2}^{2}\asymp
t.$

∎

We are now ready to prove our main result of this section.

######  Lemma 22.

For almost all $s\in\mathbb{R}$ and for all $C>0$

$\sup_{\lambda,z\in
Q(s,C/t)}\bigg{|}K(t,\lambda,z)-\frac{1}{w(s)}\operatorname{Sinc}(t,\lambda,z)\bigg{|}=o(t),$

as $t\rightarrow\infty.$

###### Proof.

Let $z_{t}\in Q(s,C/t)$ for all $t>0$ and set

$\Delta(t,z)=K(t,z_{t},z)-\frac{1}{w(s)}\operatorname{Sinc}(t,z_{t},z).$

By an argument similar to the one in the proof of Lemma 21 we find that the
function
$F(t,z)=e^{itz}G(z)\Delta(t,z)\in\operatorname{H}^{2}(\mathbb{C}_{+}),$ where
$G$ is again an outer function with boundary function $\sqrt{w}.$ Since
neither $G,$ nor $e^{itz}$ has zeros in $\mathbb{C}_{+},$ we can write

$\Delta(t,z)=F(t,z)e^{itz}G(z).$

We will make use of Lemma 3 to get an upper bound on $F,$ while we can treat
the other terms with almost elementary methods. Let us first estimate $F.$ We
calculate

$||F(t,\cdot)||_{H^{2}(\mathbb{C}_{+})}^{2}=||\Delta(t,x)G(x)||_{2}^{2}=\int_{-\infty}^{\infty}|\Delta(t,x)G(x)|^{2}dx$

$=\int_{-\infty}^{\infty}|\Delta(t,x)|^{2}w(x)dx\leq\int_{-\infty}^{\infty}|\Delta(t,x)|^{2}d\mu(x)=o(t),$

where we used Lemma 3 in the last step. By Lemma 6

$||MF(t,\cdot)||_{2}^{2}=o(t).$

From this estimate of the norm of the maximal function, we will derive an
estimate for $MF.$ We set $I_{t}=Q(s,C/t)\cap\mathbb{R}$ and we define for any
positive scalar $\lambda>0$ the set $\lambda I_{t}$ as the interval, where
each point in $I_{t}$ is rescaled by $\lambda.$ Consider the set
$5I_{t}\setminus 3I_{t}$ which is a union of two intervals $J_{t}^{1}$ and
$J_{t}^{2}.$ We will prove that on two-thirds of each of $J_{t}^{1}$ and
$J_{t}^{2}$ we have

(14) $(MF(x))^{2}\leq\frac{3||MF||_{2}^{2}t}{C}=o(t^{2}).$

Indeed, let $A$ be the subset of $J_{t}^{1}$ such that (14) fails. Suppose
that $|A|>\frac{1}{3}\frac{2C}{t}.$ Then,

$\int_{J_{t}^{1}}(MF(x))^{2}dx\geq\int_{A}(MF(x))^{2}dx\geq\int_{A}\frac{3||MF||_{2}^{2}t}{C}dx=|A|\frac{3||MF||_{2}^{2}t}{C}>2||MF||_{2}^{2},$

which gives a contradiction. The proof for the interval $J_{t}^{2}$ is
completely analogous. We denote by $S_{t}^{1}$ and $S_{t}^{2}$ the subsets of
$J_{t}^{1}$ and $J_{t}^{2},$ respectively, where (14) holds.

The next step will be to estimate the outer function $G(z).$ We will do it
similarly to the way we did it in the proof of Lemma 21. We recall that the
function $R=MP\log\sqrt{w}$ is locally integrable by Corollary 1 and assume
that $s$ is a Lebesgue point for this function. Then a calculation like in the
proof of Lemma 21 shows that there are points $x_{1}\in S_{t}^{1}$ and
$x_{2}\in S_{t}^{2}$ such that $R(x_{1}),R(x_{2})<R(s).$ Thus, we have for
$k=1,2$

(15) $\sup_{z\in\Gamma(x_{k}),\Im
z\leq\frac{5C}{t}}|\Delta(t,z)|=\sup_{z\in\Gamma(x_{k}),\Im
z\leq\frac{5C}{t}}|e^{itz}G(z)F(t,z)|\leq e^{5C}e^{2R(s)}MF(x_{k})=o(t),$

where the error term $o(t)$ is uniform over all choices $z_{t}\in Q(s,C/t).$
The last inequality holds in particular on the boundary of the rhombus

$\mathbb{C}\setminus\bigcup_{x\notin(x_{1},x_{2})}\big{(}\Gamma(x)\cup\overline{\Gamma(x)}\big{)}$

which lies in the upper half-plane. Repeating the same construction with an
outer function $\tilde{G}$ on the lower half-plane, gives the same estimate
(15) in $\mathbb{C}_{-}.$ Hence, the inequality (15) holds on the complete
boundary of the rhombus, after possibly adjusting $x_{1}$ and $x_{2},$ and
hence it holds by the maximum principle in all of the interior. Our result
follows since the rhombus contains the box $Q(s,C/t).$ ∎

###  3.2. Approximation of the Hermite-Bieher functions

In this section, we are going to use the results from last section to get an
approximation for the Hermite-Biehler functions $E(t,z)$ and $\tilde{E}(t,s).$
First of all we notice that the functions

$K(t,\lambda,z)=\frac{1}{\pi}\frac{A(z)C(\overline{\lambda}-C(z)A(\overline{\lambda})}{\overline{\lambda}-z},\quad\operatorname{Sinc}(t,\lambda,z)=\frac{1}{\pi}\frac{\sin(t(\overline{\lambda}-z))}{\overline{\lambda}-z}$

have the same demoninators. Together with Lemma 22 we find that the numerators
are close. More precisely,

######  Corollary 4.

For almost all $s\in\mathbb{R}$ and all $C>0$ we have

$\sup_{\lambda,z\in
Q(s,C/t)}\bigg{|}[A(t,z)C(t,\overline{\lambda})-C(t,z)A(t,\overline{\lambda})]-\frac{1}{w(s)}\sin(t(\overline{\lambda}-z))\bigg{|}=o(1).$

###### Proof.

By Lemma 22

$\sup_{\lambda,z\in
Q(s,C/t)}\bigg{|}[A(t,z)C(t,\overline{\lambda})-C(t,z)A(t,\overline{\lambda})]-\frac{1}{w(s)}\sin(t(\overline{\lambda}-z))\bigg{|}$
$=\sup_{\lambda,z\in
Q(s,C/t)}|\pi(\overline{\lambda}-z)|\bigg{|}K(t,\lambda,z)-\frac{1}{w(s)}\operatorname{Sinc}(t,\lambda,z)\bigg{|}$

$=\frac{\pi\sqrt{2}C}{t}\sup_{\lambda,z\in
Q(s,C/t)}\bigg{|}K(t,\lambda,z)-\frac{1}{w(s)}\operatorname{Sinc}(t,\lambda,z)\bigg{|}=\frac{\pi\sqrt{2}C}{t}o(t)=o(1).$

∎

Since the last statement holds for all $C>0,$ we can formulate it equivalently
by considering a increasing function $C(t)>0,$ $C(t)\rightarrow\infty$ as
$t\rightarrow\infty,$ instead of a constant $C>0.$

######  Corollary 5.

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

$\sup_{\lambda,z\in
Q(s,C(t)/t)}\bigg{|}[A(t,z)C(t,\overline{\lambda})-C(t,z)A(t,\overline{\lambda})]-\frac{1}{w(s)}\sin[t(\overline{\lambda}-z)]\bigg{|}=o(1).$

###### Proof.

Choose some $C>0.$ We consider two ascending chains of real numbers

$C_{1}=C<C_{2}=C+1<C_{3}=C+2<\dots,$

$t_{1}<t_{2}<t_{3}<\dots$

such that for all $i\geq 1$ we have that for all $t>t_{i}$

$\sup_{\lambda,z\in
Q(s,C_{i}/t)}\bigg{|}[A(t,z)C(t,\overline{\lambda})-C(t,z)A(t,\overline{\lambda})]-\frac{1}{w(s)}\sin[t(\overline{\lambda}-z)]\bigg{|}<1/{2^{i}}.$

Define $C(t)=0$ for $t\in[0,t_{1}]$ and $C(t)=C_{i}$ for
$t\in[t_{i},t_{i+1}].$ Now let $\varepsilon>0.$ There is $i\geq 1$ such that
$2^{-i}<\varepsilon.$ Thus, for all $t>t_{i}$

$\sup_{\lambda,z\in
Q(s,C(t)/t)}\bigg{|}[A(t,z)C(t,\overline{\lambda})-C(t,z)A(t,\overline{\lambda})]-\frac{1}{w(s)}\sin[t(\overline{\lambda}-z)]\bigg{|}<\varepsilon.$

∎

From this it follows that the Hermite-Biehler function $E(t,z)$ is close to
$\sin z,$ whenever there is a zero in the box $Q(s,C/t).$ We will introduce
some notation to keep track of when a zero enters the box. Let $C$ be a
constant or slowly growing function and $s\in\mathbb{R}$. We define

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

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

Recall that all zeros of $E(t,z)$ are in the lower half plane, since for all
$z\in\mathbb{C}_{+}$

$\overline{|E(t,z)|}>|E(t,\overline{z})|.$

For $t>0$ we introduce the function

$\gamma(t)=\frac{\sqrt{2}}{\sqrt{\sinh[2t]}}.$

######  Lemma 23.

For almost all $s\in\mathbb{R}$ and for all $D>1$ such that $T_{0}(s,D)$ is
unbounded, there exists an increasing function $C(t)>D$ for all $t>0,$
$C(t)\rightarrow\infty$ as $t\rightarrow\infty$ such that the following holds.

Let $z(t)=x(t)-iy(t)$ be a continuous function in $Q(s,C/t)$ for $t\in
T_{0}(s,C(t))$ with $E(t,z(t))=0.$ Then, there exists a unimodular function
$\alpha(t),$ such that the following holds

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

as $t\rightarrow\infty,$ $t\in T_{0}(s,C(t))\cap T_{1}(s,C(t)).$ The
conclusion holds as well, if we only consider $t\in T_{0}(s,C(t)),$ however in
that case the error term in (16) becomes $o(\gamma(ty(t))).$

###### Proof.

Let $s\in\mathbb{R}$ and $C_{1}(t)$ be such that they satisfy Corollary
$\ref{corollary_2}.$ By assumption $E(t,z(t))=A(t,z(t))-iC(t,z(t))=0.$ Hence,

$A(t,z(t))=iC(t,z(t))=\beta(t).$

Note that by (14) we have $\beta(t)\neq 0.$ For all $\lambda\in\mathbb{C}$ we
thus have

$A(t,z(t))C(t,\overline{\lambda})-C(t,z(t))A(t,\overline{\lambda})=\beta(t)C(t,\overline{\lambda})+i\beta(t)A(t,\overline{\lambda})=i\beta
E(t,\overline{\lambda}).$

Hence, by Corollary 5 we have

(17) $\sup_{z\in
Q(s,C_{1}(t)/t)}\bigg{|}i\beta(t)E(t,z)-\frac{1}{w(s)}\sin[t(z-z(t))]\bigg{|}=o(1).$

We are left to work out $\beta(t).$ It will suffice to show (16) for any
constant $L>2D.$ By the standard argument we then obtain the statement for an
increasing function $C(t).$ The trick will be to cleverly calculate the
determinant of the matrix

$\begin{pmatrix}E(t,z)&&E(t,\overline{\lambda})\\\
E^{\sharp}(t,z)&&E^{\sharp}(t,\overline{\lambda})\end{pmatrix}$

for any $z,\lambda\in\mathbb{C}$ in two different ways. Once by using our
approximation of $E(t,z)$ from above and the other way by using Corollary 4 by
noting that

$A(t,z)C(t,\overline{\lambda})-C(t,z)A(t,\overline{\lambda})=\det\begin{pmatrix}A(t,z)&&A(t,\overline{\lambda})\\\
C(t,z)&&C(t,\overline{\lambda})\end{pmatrix}=\frac{1}{2i}\det\begin{pmatrix}E(t,z)&&E(t,\overline{\lambda})\\\
E^{\sharp}(t,z)&&E^{\sharp}(t,\overline{\lambda})\end{pmatrix}.$

For all $z,\lambda\in Q(s,L/t)$ we have by (17)

$\det\begin{pmatrix}i\beta(t)E(t,z)&&i\beta(t)E(t,\overline{\lambda})\\\
-i\overline{\beta}(t)E^{\sharp}(t,z)&&-i\overline{\beta}(t)E^{\sharp}(t,\overline{\lambda})\end{pmatrix}$
$=\frac{1}{w(s)^{2}}\det\begin{pmatrix}\sin[t(z-z(t))]&&\sin[t(\overline{\lambda}-z(t))]\\\
\sin[t(z-\overline{z}(t))]&&\sin[t(\overline{\lambda}-\overline{z}(t))]\end{pmatrix}+o(1)\psi_{1}(t,z,\lambda).$

Multiplying out the determinant and using the formula

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

the last line further simplifies to

$=\frac{1}{w(s)^{2}}\sin[2ity(t)]\sin[t(\overline{\lambda}-z)]+o(1)\psi_{1}(t,z,\lambda)=\frac{i}{w(s)^{2}}\sinh[2ty(t)]\sin[t(\overline{\lambda}-z)]+o(1)\psi_{1}(t,z,\lambda).$

as $t\rightarrow\infty,t\in T_{0}(s,L)$ for some bounded function $\psi_{1}.$
On the other hand by Corollary 4

$\frac{2i}{w(s)}\sin[t(\overline{\lambda}-z)]=\det\begin{pmatrix}E(t,z)&&E(t,\overline{\lambda})\\\
E^{\sharp}(t,z)&&E^{\sharp}(t,\overline{\lambda})\end{pmatrix}+o(1)\psi_{2}(t,z,\lambda)$

as $t\rightarrow\infty,t\in T_{0}(s,L)$ for some bounded function $\psi_{2}.$
Combining both equations we arrive at

$2|\beta(t)|^{2}w(s)=(1+o(1))\sinh[2ty(t)].$

Hence, we now find an unimodular continuous function $\alpha$ such that

$\beta(t)=(1+o(1))\frac{\alpha(t)\sinh[2ty(t)]}{2w(s)}.$

Substituting this back into (17) we get

$\sup_{z\in
Q(s,L/t)}\bigg{|}E(t,z)-\frac{\alpha(t)\sqrt{2}}{\sqrt{w(s)}\sqrt{\sinh[2ty(t)]}}\sin[t(z-z(t))]\bigg{|}=o\bigg{(}\frac{1}{\sqrt{\sinh[2ty(t)]}}\bigg{)}$

as $t\rightarrow\infty,t\in T_{0}(s,L).$ If additionally, $t\in T_{1}(s,L)$
the error term becomes $o(1),$ as desired. ∎

Let us now investigate what happens when there is no zero of $E(t,z)$ close to
the real line. Suppose $z(t)=x(t)-iy(t)$ is a continuous path traced out by a
zero in the lower half-plane. Assume that for some large $C>0$ the zero $z(t)$
is close to the bottom boundary of the box $Q(s,C/t)$ and (16) holds for all
$t>0$. Then, $y(t)$ is very large. We calculate

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

$=\frac{\gamma(ty(t))}{\sqrt{w(s)}}\bigg{(}\sin[t(z-x(t))]\cos[ity(t)]+\sin[ity(t)]\cos[t(z-x(t))]\bigg{)}.$

If $y(t)\rightarrow\infty$ we observe that

$\frac{\sqrt{2}}{\sinh[2ty(t)]}\cos[ity(t)]\rightarrow
1,\quad\frac{\sqrt{2}}{\sqrt{\sinh[2ty(t)]}}\sin[ity(t)]\rightarrow i.$

Hence,

$\bigg{|}\frac{\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-z(t))]-\frac{\beta(t)}{\sqrt{w(s)}}e^{itz}\bigg{|}=o(1)$

for $\beta(t)=e^{-itx(t)}.$ Thus, we expect that $E(t,z)$ is close to
$e^{itz},$ if the zeros are far away from the real line. The calculation above
is rigorous only if there is a zero near the boundary of the box and (16)
holds. Thus, let us check what happens if there is no zero inside the box.

Let $s\in\mathbb{R}$ and $C(t)$ be like in Corollary 5. We will consider only
$t>0$ such that the box $Q(s,C(t)/t)$ does not contain any zeros of $E(t,z),$
i.e. $t\notin T_{0}(s,C(t)).$ We set $I_{t}=\mathbb{R}\cap Q(s,C(t)/t).$ For
any $x,w\in I_{t}$ we can interpret Corollary 5 as an approximation of a
scalar product of two real vectors

(18) $\bigg{\langle}\begin{pmatrix}A(t,x)\\\
C(t,x)\end{pmatrix},\begin{pmatrix}C(t,w)\\\
-A(t,w)\end{pmatrix}\bigg{\rangle}=\frac{1}{w(s)}\sin[t(w-x)]+o(1)\psi(t,x,w)$

as $t\rightarrow\infty$ for some bounded function $\psi.$ We will assume that
$t$ is large enough, such that $o(1)\psi(t,x)\leq\frac{1}{w(s)}$ and $C(t)\geq
4\pi.$ Choose points $w_{1}(t),w_{2}(t)=w_{1}(t)+\frac{\pi}{2t}.$ Since (18)
holds for all $x\in I_{t},$ we see that the vector

$\begin{pmatrix}A(t,x)\\\ C(t,x)\end{pmatrix}$

has modulus bounded away from zero and rotates around the origin. In
particular, since $C(t)\geq 4\pi$ the vector makes at least one full rotation.
Hence, there are $x_{0}(t)$ and $x_{1}(t)$ such that $E(t,x_{0}(t))$ is
positive and $E(t,x_{1}(t))$ is positive imaginary, i.e. $C(t,x_{0}(t))=0,$
and $A(t,x_{1}(t))=0.$ If we set $x=x_{0}(t)$ and $w=x_{1}(t)$ in (18), we see
that $A(t,x_{0}(t))=c_{1}(t)$ and $C(t,x_{1}(t))=-c_{2}(t),$ where $c_{1}(t)$
and $c_{2}(t)$ are positive numbers satisfying
$-c_{1}(t)c_{2}(t)=\frac{1}{w(s)}\sin[t(x_{0}(t)-x_{1}(t)]+o(1).$

We set $x_{2}(t)=x_{1}(t)-\frac{\pi}{2t}.$ By Corollary 5 we have for $z\in
Q(s,C(t)/t)$

$C(t,z)=\frac{1}{c_{1}(t)w(s)}\sin[t(z-x_{0}(t))]+o(1),\quad
A(t,z)=-\frac{1}{c_{2}(t)w(s)}\cos[t(z-x_{2}(t))]+o(1).$

Hence, for all $z\in Q(s,C(t)/t)$ we find the approximation

$E(t,z)=-\frac{1}{c_{2}(t)w(s)}\cos[t(z-x_{2}(t))]-i\frac{1}{c_{1}(t)w(s)}\sin[t(z-x_{0}(t))]+o(1).$

This is almost $-e^{itz}.$ We have the following

######  Lemma 24.

$|t(x_{2}(t)-x_{0}(t))|=o(1)\mod 2\pi,\quad|c_{1}(t)-c_{2}(t)|=o(1).$

We will postpone the proof to section $3.4$ where we study the dynamics of the
zeros. Thus, $c_{1},c_{2}=\frac{1}{\sqrt{w(s)}}+o(1)$ and there is a
unimodular function $\beta(t)$ and an increasing function $C_{1}(t),$
$C_{1}(t)\rightarrow\infty$ as $t\rightarrow\infty,$ such that for all $z\in
Q(s,C_{1}(t)/t)$ we have

$\frac{\beta(t)}{\sqrt{w(s)}}e^{itz}.$

We summarize these results in the following

######  Corollary 6.

For almost all $s\in\mathbb{R}$ there exists a function
$C(t)>0,C(t)\rightarrow\infty$ as $t\rightarrow\infty$ and
$z(t)=x(t)-iy(t)\in\mathbb{C}_{-}$ such that

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

for some $\alpha=\alpha(s,t),|\alpha|=1$ as $t\rightarrow\infty,t\in
T_{1}(s,C(t)).$ If $t\in T_{0}(s,C(t)),$ $z(t)$ can be chosen as a zero of
$E(t,z).$ If additionally $t\notin T_{0}(s,C(t))$ for all sufficiently large
$t,$ then for all $D>0,$

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

One can drop the assumption $ty(t)>1,$ if one replaces the error term $o(1)$
by $o(\gamma(ty(t)))$ in (19).

In conclusion, if the zeros of $E(t,z)$ are bounded away by at least $1/t$
from the real line, then $E(t,z)$ behaves like $e^{itz}.$ In that case, we can
immediately conclude the convergence of the non-linear Fourier transform, see
section $3.8.$ However, if a zero is close to the real line, $E(t,z)$ is close
to $\sin z,$ which is not good enough to conclude convergence. Hence, from
this point on we will need to show that all zeros of $E(t,z)$ are bounded
locally at least by $1/t$ away from the real line.

If the zeros of $E(t,z)$ do not stay away from the box $Q(s,C(t)/t),$ i.e.
$T_{0}(s,C)$ is unbounded, there are precisely two situations in which this
can happen. Fix $s\in\mathbb{R}$ and $C>0.$

  * (i)

There is a time $T>0$ s.t. for all $t>T$ the box $Q(s,C/t)$ contains a zero of
$E(t,z),$ i.e. $T_{0}(s,C)$ covers a half-line, so after time $T$ at least one
zero stays in the box forever.

  * (ii)

For all times $T>0,$ such that $Q(s,C/T)$ does not contain a zero of $E(t,z),$
there exists $t>T$ such that the box $Q(s,C/t)$ contains a zero of $E(t,z),$
i.e. the zeros jump in and out of the box infinitely often.

While we can rule out (i) by the results proven already, we will have to put a
lot more effort into showing that case (ii) can not occur.

###  3.3. Resonances can not stay close to the real line forever

In this section, we are going to show that for almost all $s\in\mathbb{R}$ and
any $C>0$ the zeros of $E(t,z)$ can not stay in the box $Q(s,C/t)$ forever.
Our strategy will be as follows. If at least one zero of $E(t,z)$ were in the
box $Q(s,C/t)$ for all large enough $t,$ then by Lemma 23 $|E(t,x)|$ is close
to $|\sin(tx)|$ near $s.$ Since $|\sin(tx)|$ does not converge in measure on
$\mathbb{R},$ the same holds for the function $|E(t,x)|.$ However, this
contradicts Lemma 19. We set

$\Sigma=\\{s\in\mathbb{R}\mid\text{There exists }T>0\text{ such that for all
}t>T\text{ the box }Q(s,C/t)\text{ contains a zero}\\}.$

We can assume without loss of generality that $0<|\Sigma|<\infty.$ For two
sets $A$ and $B$ we denote the symmetric difference by

$A\Delta B=(A\setminus B)\cup(B\setminus A).$

The following well-known result allows us to assume that $\Sigma$ is a finite
union of intervals.

######  Lemma 25.

Let $\Sigma$ be a measurable set with $|\Sigma|<\infty.$ Then, for all
$\varepsilon>0$ there are finitely many intervals $I_{1},\dots I_{n},$ such
that

$\bigg{|}\Sigma\Delta\bigcup_{k=1}^{n}I_{k}\bigg{|}<\varepsilon.$

We will contradict convergence on each interval separately, so we might as
well assume from the very beginning that $\Sigma$ is an interval.

The following criterion allows us to show that a given sequence of functions
does not converge in measure.

######  Lemma 26.

Let $\Sigma$ be a measurable set with $|\Sigma|>0$ and let $f_{n}(x)$ be a
sequence of real functions. Suppose there exists $\delta>0$ and $N\geq 1$ such
that for all $x\in\Sigma$ and all $n,m\geq N$ we have

$||f_{n}(x)|-|f_{m}(x)||>\delta.$

Then, $|f_{n}|$ does not converge in measure on $\Sigma.$

###### Proof.

Suppose $|f_{n}|\rightarrow f$ as $n\rightarrow\infty$ for some function $f.$
Then, there exists $N\geq 0$ such that for all $n\geq N$

$\big{|}\\{x\in\Sigma:||f_{n}|-f|>\delta\\}\big{|}<\frac{|\Sigma|}{2}.$

Thus, we immediately obtain a contradiction

$|\Sigma|=\big{|}\\{x\in\Sigma:||f_{n}(x)|-|f_{m}(x)||>\delta\\}\big{|}$
$\leq\big{|}\\{x\in\Sigma:||f_{n}(x)|-f|+||f_{m}(x)|-f|>\delta\\}\big{|}$
$\leq\big{|}\\{x\in\Sigma:||f_{n}(x)|-f|>\delta\\}\big{|}+\big{|}\\{x\in\Sigma:||f_{m}(x)|-f|>\delta\\}\big{|}$
$<\frac{|\Sigma|}{2}+\frac{|\Sigma|}{2}=|\Sigma|.$

∎

Furthermore, we will need the following technical inequality.

######  Lemma 27.

Let $\Sigma$ be a non-empty interval. There exists $\delta>0$ independent of
$\Sigma$ and $N\geq 1$ dependent on $\Sigma,$ such that for all $a,b>0$ and
$z,w\in Q(s,C)\cap\mathbb{C}_{-}$ and all $n\geq N$ we have

(21)
$\bigg{|}a\big{|}\sin[2^{n}(x-z)]\big{|}-b\big{|}\sin[2^{n+1}(x-w)]\big{|}\bigg{|}>\max(a,b)\delta$

on a subset of $\Sigma$ of measure at least $\frac{\delta}{24}|\Sigma|.$

###### Proof.

Let us first prove the case where $a=b=1$ and $\Sigma=[0,\frac{\pi}{2^{n}}]$
for some $n\geq 1.$ The function

$f_{z,w,n}(x)=\big{|}\sin[2^{n}(x-z)]\big{|}-\big{|}\sin[2^{n+1}(x-w)]\big{|}$

is periodic with shortest period $\frac{\pi}{2^{n}}.$ The factor $2^{n}$ just
squeezes the function from the interval $\Sigma=[0,\frac{\pi}{2}]$ onto the
interval $[0,\frac{\pi}{2^{n}}]$ which covers exactly the shortest period of
$f_{z,w,n}.$ Hence, it suffices to show the inequality for $f_{z,w}=f_{z,w,1}$
and $\Sigma=[0,\frac{\pi}{2}].$

Observe that by continuity of $\sup$ and $\inf$ on compact sets, we have the
following min-max inequalities

$\delta_{1}=\sup_{z,w\in
Q(s,C)}\inf_{x\in[0,\frac{\pi}{2}]}\big{|}\sin[2(x-z)]\big{|}-\big{|}\sin[4(x-w)]\big{|}<0,$

$\delta_{2}=\inf_{z,w\in
Q(s,C)}\sup_{x\in[0,\frac{\pi}{2}]}\big{|}\sin[2(x-z)]\big{|}-\big{|}\sin[4(x-w)]\big{|}>0.$

Set $\delta=\min(|\delta_{1}|,|\delta_{2}|)>0.$ Furthermore, $f_{z,w}$ is
Lipschitz continuous, since $f_{z,w}^{\prime}$ is uniformly bounded for all
$z,w\in Q(s,C)\cap\mathbb{C}_{-}.$ Indeed, by the same method as in the proof
of Lemma 17 we find for $z=a+ib\in Q(s,C)\cap\mathbb{C}_{-}$ that

$\frac{d}{dx}\big{|}\sin[2(x-z)]\big{|}=\frac{d}{dx}\big{|}\sin[2(x-a)]\cosh(2b)+i\cos[2(x-a)]\sinh(2b)\big{|}$
$=\frac{2\cos[2(x-a)]\sin[2(x-a)](\cosh^{2}(2b)-\sinh^{2}(2b))}{\big{|}\sin[2(x-z)]\big{|}}$
$=2\frac{\cos[2(x-a)]\sin[2(x-a)]}{\big{|}\sin[2(x-z)]\big{|}}.$

Notice that for all $z\in Q(s,C)$ we have

$\frac{\cos[2(x-a)]\sin[2(x-a)]}{\big{|}\sin[2(x-z)]\big{|}}\leq 1.$

Hence we have for all $z,w\in Q(s,C)$ and for $x\in[0,\frac{\pi}{2}]$

$|f_{z,w}^{\prime}(x)|\leq 6.$

Choose a point $x_{0}\in[0,\frac{\pi}{2}]$ such that $f_{z,w}(x_{0})>\delta.$
Then, if $|x_{0}-y|<\frac{\delta}{12},$ we have

$|f_{z,w}(x_{0})-f_{z,w}(y)|<\frac{\delta}{2}.$

Thus, on an interval $I_{1}$ of length $\frac{\delta}{6}$ the inequality (21)
holds with $\frac{\delta}{2}.$ Similarly take a point
$x_{1}\in[0,\frac{\pi}{2}]$ such that $f_{z,w}(x_{1})<-\delta.$ By the same
argument we obtain an interval $I_{2}$ of length $\frac{\delta}{6}$ on which
the inequality (21) holds with $\frac{\delta}{2}.$

Let now $a,b>0$ and suppose that $a\geq b.$ Then for all $x\in I_{1}$

$a\big{|}\sin[2(x-z)]\big{|}-b\big{|}\sin[4(x-w)]\big{|}\geq
a\big{(}\big{|}\sin[2(x-z)]\big{|}-\big{|}\sin[4(x-w)]\big{|}\big{)}=\max(a,b)\frac{\delta}{2}.$

In the case $a\leq b$, a similar inequality holds on $I_{2}.$ By translation,
this concludes the proof for the case
$\Sigma=[k\frac{\pi}{2^{n}},(k+1)\frac{\pi}{2^{n}}]$ for any $n\geq 1$ and
$k\in\mathbb{Z}.$ Let now $\Sigma$ be any non-empty interval. Then, choose $N$
large enough such that there are finitely many scaled dyadic intervals
$I_{1}\dots I_{m}$ of the form $[k\frac{\pi}{2^{N}},(k+1)\frac{\pi}{2^{N}}],$
all contained in $\Sigma,$ satisfying

$\bigg{|}\Sigma\setminus\bigcup_{n=1}^{m}I_{n}\bigg{|}\leq\frac{|\Sigma|}{2}.$

Then, the inequality is satisfied for all $n\geq N$ on a subset of $\Sigma$ of
measure at least

$\frac{\delta}{12}\bigg{|}\bigcup_{n=1}^{m}I_{n}\bigg{|}=\frac{\delta}{12}\bigg{(}|\Sigma|-\bigg{|}\Sigma\setminus\bigcup_{n=1}^{m}I_{n}\bigg{|}\bigg{)}\geq\frac{\delta}{24}|\Sigma|.$

∎

######  Corollary 7.

$|\Sigma|=0.$

###### Proof.

Suppose $|\Sigma|>0.$ For all $s\in\Sigma$ there is $T(s)>0$ such that the box
$Q(s,C/t)$ contains a zero of $E(t,z)$ for all $t>T(s)$ and the left hand side
of (16) is smaller than $\varepsilon\gamma(ty(t)),$ where $\varepsilon>0$ is a
small number, which we specify later. By Lusin’s Theorem, we can assume that
$T(s)$ is the same for all $s\in\Sigma.$ Using Lusin’s Theorem again we can
assume that $w(s)<D<\infty$ for all $s\in\Sigma.$ Let $n\geq 1$ such that
$2^{n}>T.$ Observe that

$\bigg{|}E(2^{n},x)-\alpha(2^{n})\frac{\gamma(2^{n}y(2^{n}))}{\sqrt{w(s)}}\sin[2^{n}(x-z(2^{n}))]\bigg{|}<\varepsilon\gamma(2^{n}y(2^{n}))$

implies

$|E(2^{n},x)|\leq\varepsilon\gamma(2^{n}y(2^{n}))+\frac{\gamma(2^{n}y(2^{n}))}{\sqrt{w(s)}}\big{|}\sin[2^{n}(x-z(2^{n}))]\big{|}.$

Note that by inverse triangle inequality

$|E(2^{n+1},x)|\geq\frac{\gamma(2^{n+1}y(2^{n+1}))}{\sqrt{w(s)}}\big{|}\sin[2^{n+1}(x-z(2^{n+1}))]\big{|}$
$-\bigg{|}E(2^{n+1},x)-\alpha(2^{n+1})\frac{\gamma(2^{n+1}y(2^{n+1}))}{\sqrt{w(s)}}\sin[2^{n+1}(x-z(2^{n+1}))]\bigg{|}$
$\geq\bigg{|}\frac{\gamma(2^{n+1}y(2^{n+1}))}{\sqrt{w(s)}}\sin[2^{n+1}(x-z(2^{n+1}))]\bigg{|}-\varepsilon\gamma(2^{n+1}y(2^{n+1})).$

Thus,

$\bigg{|}|E(2^{n},x)|-|E(2^{n+1},x)|\bigg{|}\geq|E(2^{n},x)|-|E(2^{n+1},x)|$
$\geq\frac{\gamma(2^{n+1}y(2^{n+1}))}{\sqrt{w(s)}}\big{|}\sin[2^{n+1}(x-z(2^{n+1}))]\big{|}-\frac{\gamma(2^{n}y(2^{n}))}{\sqrt{w(s)}}\big{|}\sin[2^{n}(x-z(2^{n}))]\big{|}$
$-\varepsilon\gamma(2^{n+1}y(2^{n+1}))-\varepsilon\gamma(2^{n}y(2^{n})).$

We do the same calculation after reversing the roles of $E(2^{n},x)$ and
$E(2^{n+1},x).$ Combining both inequalities gives

$\bigg{|}|E(2^{n},x)|-|E(2^{n+1},x)|\bigg{|}\geq|E(2^{n},x)|-|E(2^{n+1},x)|$
$\geq\bigg{|}\frac{\gamma(2^{n+1}y(2^{n+1}))}{\sqrt{w(s)}}\big{|}\sin[2^{n+1}(x-z(2^{n+1}))]\big{|}-\frac{\gamma(2^{n}y(2^{n}))}{\sqrt{w(s)}}\big{|}\sin[2^{n}(x-z(2^{n}))]\big{|}\bigg{|}$
$-\varepsilon\gamma(2^{n+1}y(2^{n+1}))-\varepsilon\gamma(2^{n}y(2^{n})).$

Note that since $z(t)\in Q(s,C/t)$ we have for all $t>0$

$\gamma(ty(t))=\frac{\sqrt{2}}{\sqrt{\sinh[2ty(t)]}}\geq\frac{\sqrt{2}}{\sqrt{\sinh[2C]}}=C_{1}>0.$

Recall that we assumed that $w(s)<D<\infty.$ Hence, we can apply Lemma 27 with
$\Sigma$ to obtain $\delta>0$ such that on a subset of $\Sigma$ of measure at
least $\frac{\delta}{12}|\Sigma|$ we have

$\bigg{|}|E(2^{n},x)|-|E(2^{n+1},x)|\bigg{|}>\max\bigg{(}\gamma(2^{n+1}y(2^{n+1})),\gamma(2^{n+1}y(2^{n+1}))\bigg{)}\delta$
$-\varepsilon\gamma(2^{n+1}y(2^{n+1}))-\varepsilon\gamma(2^{n}y(2^{n})).$

We choose now $\varepsilon=\frac{\delta}{4}.$ Then,

$\bigg{|}|E(2^{n},x)|-|E(2^{n+1},x)|\bigg{|}>\max\bigg{(}\gamma(2^{n+1}y(2^{n+1})),\gamma(2^{n+1}y(2^{n+1}))\bigg{)}\frac{\delta}{2}\geq\frac{C_{1}}{2\sqrt{D}}\delta>0$

on a subset of $\Sigma$ of measure at least $\frac{\delta}{12}|\Sigma|,$
contradicting the convergence of $|E(t,x)|$ in measure on $\mathbb{R}.$ ∎

We now showed that the zeros of $E(t,z)$ can not stay inside the box
$Q(s,C/t)$ forever. For the remainder of this discussion, we will have to rule
out that the zeros travel infinitely many times in and out of the box
$Q(s,C/t).$ It is a natural question to ask, whether one can modify the
methods we used slightly to get this stronger result. The key inequality which
made everything work is Lemma 27. It is stated in particular such that we can
use the approximation from Lemma 23 for the times $t_{1}=2^{n}$ and
$t_{2}=2^{n+1}.$ This is only possible since we know by assumption that there
is a zero inside the box $Q(s,C/t)$ for all large enough times. In the
stronger case we want to prove, this does not hold anymore, i.e. one would
have to replace $t_{1}=2^{n},t_{2}=2^{n+1}$ by other values of $t.$ The
problem arises since we can not give a lower bound on the time interval a zero
has to stay inside the box $Q(s,C/t),$ i.e. the inequality (21) breaks down.
More precisely, $|t_{2}-t_{1}|\rightarrow 0$ gives $\delta\rightarrow 0,$ so
we can not hope anymore to contradict convergence in measure of $|E(t,x)|.$

Thus, we will need to further analyze the dynamics of the zeros to have extra
leverage for proving the stronger claim. It will turn out that the
contradiction arises through a different observation. Namely, we will observe
that each time the zero travels in and out of the box $Q(s,C/t),$ this will
require a chunk of $L^{2}-$norm of $f.$ Hence, if a zero of $E(t,z)$ traveled
infinitely often in and out of the box $Q(s,C/t),$ this would immediately
contradict the fact that $f\in L^{2}(\mathbb{R}).$

###  3.4. Meromorphic inner functions and dynamics of the zeros

We are going to study the dynamics of the zeros of $E(t,z).$ The appropriate
tool for doing so is the class of meromorphic inner functions, which can be
assigned to any Hermite-Biehler function.

Let $f:\mathbb{C}_{+}\rightarrow\mathbb{C}$ be an inner function. We say that
$f$ is an meromorphic inner function, if $f$ admits a meromorphic continuation
to the whole complex plane. Meromorphic inner functions enjoy the property,
that they have a product representation in terms of their zeros and poles.

######  Lemma 28.

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a meromorphic inner function. Then,
there exists a sequence of points $\lambda_{n}\in\mathbb{C}_{+}$ with
$|\lambda_{n}|\rightarrow\infty$ as $n\rightarrow\infty$ satisfying the
Blaschke condition

$\sum_{n=1}^{\infty}\frac{\Im\lambda_{n}}{1+|\lambda_{n}|^{2}}<\infty.$

Furthermore, there is $c\geq 0$ and a unimodular constant $\alpha$ such that

$f(z)=\alpha
e^{icz}\prod_{n=1}^{\infty}\frac{\overline{\lambda}_{n}}{\lambda_{n}}\frac{z-\lambda_{n}}{z-\overline{\lambda}_{n}}.$

###### Proof.

A proof and many more results on meromorphic inner functions, as well as
further applications can be found in [11]. ∎

From this we immediately conclude that every meromorphic inner function is
analytic in a neighborhood of the real line and hence all boundary values and
derivatives are well-defined everywhere on $\mathbb{R}.$ Furthermore, the
derivatives on $\mathbb{R}$ can be used to locate the zeros.

######  Lemma 29.

Let $\theta$ be a meromorphic inner function and let
$0<\varepsilon<\varepsilon_{0}$ for some small fixed constant
$\varepsilon_{0}.$ Suppose there are $x,y\in\mathbb{R}$ such that

(22) $\frac{|\theta^{\prime}(x)|}{|\theta^{\prime}(y)|}>1+\varepsilon.$

Then, the ball $\\{|z-x|<4|y-x|/\varepsilon\\}$ contains a zero of $\theta.$

###### Proof.

We use Lemma 28 to write

$\theta(z)=\alpha
e^{icz}\prod_{n=1}^{\infty}\frac{\overline{\lambda}_{n}}{\lambda_{n}}\frac{z-\lambda_{n}}{z-\overline{\lambda}_{n}}$

with a unimodular constant $\alpha,$ a non-negative constant $c$ and a
sequence of points $\lambda_{n}\in\mathbb{C}_{+}$ satisfying

$\sum_{n=1}^{\infty}\frac{\Im\lambda_{n}}{1+|\lambda_{n}|^{2}}<\infty.$

Assume without loss of generality that $\alpha=1.$ Since $|\theta(x)|=1$ for
almost all $x\in\mathbb{R},$ we can write $\theta(x)=e^{i\arg\theta(x)}.$
Together with the relation $\ln\theta(x)=i\arg\theta(x)$ we see that

$|\theta^{\prime}(x)|=\frac{d}{dx}\arg\theta(x)=\frac{1}{i}\frac{d}{dx}\ln\theta(x).$

Using the product representation of $\theta(x)$ we calculate

$\frac{d}{dx}\ln\theta(x)=\frac{d}{dx}\ln\bigg{(}e^{icx}\prod_{n=1}^{\infty}\frac{\overline{\lambda}_{n}}{\lambda_{n}}\frac{x-\lambda_{n}}{x-\overline{\lambda}_{n}}\bigg{)}=ic+\sum_{n=1}^{\infty}\frac{d}{dx}\ln\bigg{(}\frac{\overline{\lambda}_{n}}{\lambda_{n}}\frac{x-\lambda_{n}}{x-\overline{\lambda}_{n}}\bigg{)}.$

We will now calculate the derivative of a single Blaschke factor. Write
$\lambda_{n}=x_{n}+iy_{n}.$

$\frac{d}{dx}\ln\bigg{(}\frac{\overline{\lambda}_{n}}{\lambda_{n}}\frac{x-\lambda_{n}}{x-\overline{\lambda}_{n}}\bigg{)}=\frac{d}{dx}\ln(x-\lambda_{n})-\frac{d}{dx}\ln(x-\overline{\lambda}_{n})$
$=\frac{1}{(x-x_{n})-iy_{n}}-\frac{1}{(x-x_{n})+iy_{n}}=\frac{2iy_{n}}{(x-x_{n})^{2}+y_{n}^{2}}.$

Thus, combining everything we see

$|\theta^{\prime}(x)|=c+\sum_{n=1}^{\infty}\frac{2y_{n}}{(x-x_{n})^{2}+y_{n}^{2}}.$

By (22) there is an $n\geq 1$ such that

(23)
$\frac{y_{n}}{(x-x_{n})^{2}+y_{n}^{2}}\frac{(y-x_{n})^{2}+y_{n}^{2}}{y_{n}}=\frac{|y-\lambda_{n}|^{2}}{|x-\lambda_{n}|^{2}}>1+\varepsilon.$

Indeed, suppose the reverse inequality would hold for all $n\geq 1.$ Then,

$\frac{|\theta^{\prime}(x)|}{|\theta^{\prime}(y)|}=\frac{c+\sum_{n=1}^{\infty}\frac{2y_{n}}{(x-x_{n})^{2}+y_{n}^{2}}}{c+\sum_{n=1}^{\infty}\frac{2y_{n}}{(y-x_{n})^{2}+y_{n}^{2}}}$

$\leq\frac{c+(1+\varepsilon)\sum_{n=1}^{\infty}\frac{2y_{n}}{(y-x_{n})^{2}+y_{n}^{2}}}{c+\sum_{n=1}^{\infty}\frac{2y_{n}}{(y-x_{n})^{2}+y_{n}^{2}}}=1+\varepsilon\frac{\sum_{n=1}^{\infty}\frac{2y_{n}}{(y-x_{n})^{2}+y_{n}^{2}}}{c+\sum_{n=1}^{\infty}\frac{2y_{n}}{(y-x_{n})^{2}+y_{n}^{2}}}\leq
1+\varepsilon,$

contradicting (22). Hence, pick $n\geq 1$ such that (23) holds. Then,
$\lambda_{n}$ is contained in the ball
$\\{|z-x|<\frac{4}{3}|y-x|/\varepsilon\\}.$ Indeed, from

$\frac{|y-\lambda_{n}|}{|x-\lambda_{n}|}>\sqrt{1+\varepsilon}$

we find by triangle inequality that

$\sqrt{1+\varepsilon}|x-\lambda_{n}|<|x-y|+|x-\lambda_{n}|\implies|x-\lambda_{n}|<\frac{|x-y|}{\sqrt{1+\varepsilon}-1}.$

We use Taylor series expansion of $\sqrt{1+x}$

$\sqrt{1+x}=\sum_{k=0}^{\infty}\binom{1/2}{k}x^{k}$

and cut it off at the linear term to write
$\sqrt{1+\varepsilon}=1+\frac{\varepsilon}{2}+o(\varepsilon).$ For
$0<\varepsilon<\varepsilon_{0}$ small enough we have that
$|o(\varepsilon)|<\frac{1}{4}\varepsilon.$ Thus, as claimed

$|x-\lambda_{n}|<\frac{|x-y|}{\sqrt{1+\varepsilon}-1}<\frac{|x-y|}{1+\frac{\varepsilon}{2}-\frac{\varepsilon}{4}-1}=4\frac{|x-y|}{\varepsilon}.$

∎

If $E(z)$ is any Hermite-Biehler function we can associate a meromorphic inner
function by

$\theta_{E}(z)=\frac{E^{\sharp}(z)}{E(z)}.$

In particular, we will denote for the remainder of the paper the associated
meromorphic inner functions to the Hermite-Biehler functions $E(t,z)$ and
$\tilde{E}(t,z)$ respectively by

$\theta_{E}(t,z)=\frac{E^{\sharp}(t,z)}{E(t,z)},\quad\theta_{\tilde{E}}=\frac{\tilde{E}^{\sharp}(t,z)}{\tilde{E}(t,z)}.$

Note that the zeros of $E(t,z)$ are the precisely the poles of $\theta_{E}$
and the conjugates of the zeros of $E(t,z)$ are precisely the zeros of
$\theta_{E}.$

We now have the tools to give a proof for Lemma 24.

###### Proof of Lemma 24.

We set $\phi(x)=\arg E(t,x)$ and set
$J_{t}(s-\frac{4\pi}{t},s+\frac{4\pi}{t}).$ Note that

$|\theta_{E}^{\prime}(x)|=\bigg{|}\frac{d}{dx}e^{-2i\arg
E(t,x)}\bigg{|}=2|\phi^{\prime}(x)|.$

If there now was $\delta>0$ such that
$\limsup_{t\rightarrow\infty}|t(x_{2}(t)-x_{0}(t)|>\delta$ or
$\limsup_{t\rightarrow\infty}|c_{1}(t)-c_{2}(t)|>\delta,$ then

$\frac{\sup_{J_{t}}\phi^{\prime}}{\inf_{J_{t}}\phi^{\prime}}>1+\varepsilon,$

for some $\varepsilon>0$ depending on $\delta.$ However, since
$C(t)\rightarrow\infty,$ Lemma 29 gives that there is a zero of $E(t,z)$ in
$Q(s,C(t)/t)$ for large enough $t,$ contradicting our assumption $t\notin
T_{0}(s,C(t)).$ ∎

Let now $z(t)$ be a differentiable curve in $\mathbb{C}$ such that
$\theta=\theta_{E}(t,z(t))=0.$ We want to study further $z^{\prime}(t)$ and
$\theta_{z}(t,z(t)).$ Suppose in the following that $z\in\mathbb{C}$ is such
that $E(t,z)\neq 0.$ We have

$\frac{d}{dt}\theta(t,z)=\frac{(\frac{d}{dt}A+i\frac{d}{dt}C)(A-iC)-(A+iC)(\frac{d}{dt}A-i\frac{d}{dt}C)}{(A-iC)^{2}}=2i\frac{A\frac{d}{dt}C-C\frac{d}{dt}A}{(A-iC)^{2}}$

$=2i\frac{z(A^{2}+C^{2})-2fAC}{(A-iC)^{2}}=2iz\frac{A+iC}{A-iC}-4if\frac{AC}{(A-iC)^{2}}$

$=2iz\frac{A+iC}{A-iC}-f\frac{[(A+iC)-(A-iC)][(A+iC)+(A-iC)]}{(A-iC)^{2}}=2iz\theta(t,z)+f(1-\theta^{2}(t,z)).$

Thus, the meromorphic inner function $\theta$ satisfies the differential
equation

(24) $\frac{d}{dt}\theta=2iz\theta+f(1-\theta^{2}).$

Let now $t>0$ be such that $\theta_{z}(t,z(t))\neq 0.$ Thus,

$\frac{d}{dt}\theta(t,z(t))=0=\theta_{t}(t,z(t))+\theta_{z}(t,z(t))z^{\prime}(t),$

which immediately yields using (24)

(25) $z^{\prime}(t)=-\frac{f(t)}{\theta_{z}(t,z(t))}.$

Furthermore, it will be useful to know the change of the derivative of
$\theta_{z}$ at the zero $z(t).$ Using (24) and (25) we obtain

(26)
$\frac{d}{dt}\theta_{z}(t,z(t))=\theta_{zt}(t,z(t))+\theta_{zz}(t,z(t))z^{\prime}(t)=2iz(t)\theta_{z}(t,z(t))-f(t)\frac{\theta_{zz}(t,z(t))}{\theta_{z}(t,z(t))}$

The natural question to ask is when (25) is well-defined, i.e. when
$\theta_{z}(t,z(t))\neq 0.$ It turns out that when there is a zero in the box
$Q(s,C/t),$ then its is a simple zero for large enough $t.$ To see this we
will prove a more general result that will turn out useful in many situations
later in the proof. Essentially it is a quantitative version of the open
mapping Theorem. The proof is taken from [13].

######  Lemma 30.

Let $\Omega\subset\mathbb{C}$ be a domain and $f:\Omega\rightarrow\mathbb{C}$
a holomorphic function. Let $z_{0}\in\Omega$ and $w_{0}=f(z_{0})$ and choose
$\delta>0$ such that $\overline{B(z_{0},\delta)}\subset\Omega$ and
$f(z_{0})\neq w_{0}$ on the circle $|z-z_{0}|=\delta.$

If $\varepsilon>0$ is such that $|f(z)-w_{0}|\geq\varepsilon$ on the circle
$|z-z_{0}|=\delta,$ then $f$ attains every value in $B(w_{0},\varepsilon).$

###### Proof.

Define $g(z)=f(z)-w$ and write $g(z)=(f(z)-w_{0})+(w_{0}-w)=F(z)+G(z).$ If now
$|w-w_{0}|<\varepsilon$ we have $|F(z)|>|G(z)|$ on the circle
$|z-z_{0}|=\delta$ and thus by Rouché’s Theorem we find that $g=F+G$ has a
zero inside the circle since $F$ has one. ∎

######  Lemma 31.

For almost all $s\in\mathbb{R}$ and for all $C>4\pi$ such that the box
$Q(s,C/t)$ contains a zero of $E(t,z),$ the zeros of $E(t,z)$ in $Q(s,C/t)$
are simple for all large enough $t.$

###### Proof.

This follows immediately from Lemma 23 combined with the fact that the zeros
of $\sin(z)$ are simple and a double zero of $E(t,z)$ would imply a double
zero of $\sin(z)$ by Lemma 30. ∎

Furthermore, by Lemma 23 we can find an approximation for $\theta(t,z)$ and
$\theta_{z}(t,z(t)).$ We set $Q_{+}(s,C(t)/t)=Q(s,C(t)/t)\cap\mathbb{C}_{+}.$
Note that in the following Lemma $\overline{z}(t)$ will play the role of
$z(t),$ since the conjugates of the zeros of $E(t,z)$ are the zeros of
$\theta(t,z).$

######  Lemma 32.

Let $s,C(t),\alpha(t),D$ and $z(t)$ be like in Lemma 23.

  * (i)

(27) $\sup_{z\in
Q_{+}(s,C(t)/t)}\bigg{|}\theta(t,z)-\overline{\alpha}^{2}(t)\frac{\sin[t(z-\overline{z}(t))]}{\sin[t(z-z(t))]}\bigg{|}=o(1),$

as $t\rightarrow\infty,t\in T_{1}(s,C(t))=T_{0}(s,C(t))\cap\\{ty(t)>1\\}.$

  * (ii)

$\alpha^{2}(t)\theta_{z}(t,\overline{z}(t))=(1+o(1))t\frac{1}{\sin[2ity(t)]},$

  * (iii)

$\alpha^{2}(t)\theta_{zz}(t,\overline{z}(t))=-(1+o(1))t^{2}\frac{2\cos[2ity(t)]}{\sin^{2}[2ity(t)]},$

as $t\rightarrow\infty,t\in T_{1}(s,D)=T_{0}(s,D)\cap\\{ty(t)>1\\}.$

###### Proof.

Let $z\in Q_{+}(s,C(t)/t).$ We calculate

$\bigg{|}\theta(t,z)-\overline{\alpha}^{2}(t)\frac{\sin[t(z-\overline{z}(t))]}{\sin[t(z-z(t))]}\bigg{|}\leq\bigg{|}\frac{E^{\sharp}(t,z)-\frac{\overline{\alpha}(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-\overline{z}(t))]}{E(t,z)}\bigg{|}$
$+\bigg{|}\frac{\frac{\overline{\alpha}(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-\overline{z}(t))]}{\frac{\overline{\alpha}(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-z(t))]E(t,z)}\bigg{|}\cdot\bigg{|}E(t,z)-\frac{\overline{\alpha}(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-z(t))]\bigg{|}.$

Now by (16) there is a $T>0$ such that for all $t>T$ and $z\in
Q_{+}(s,C(t)/t)$

$|E(t,z)|>\bigg{|}\frac{\frac{\overline{\alpha}(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-z(t))]}{2}\bigg{|}.$

Thus,

$\bigg{|}\theta(t,z)-\overline{\alpha}^{2}(t)\frac{\sin[t(z-\overline{z}(t))]}{\sin[t(z-z(t))]}\bigg{|}\leq\bigg{|}\frac{2}{\frac{\overline{\alpha}(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-z(t))]}\bigg{|}\bigg{|}E^{\sharp}(t,z)-\sin[t(z-\overline{z}(t))]\bigg{|}$

$+2\bigg{|}\frac{\frac{\overline{\alpha}(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-\overline{z}(t))]}{\bigg{(}\frac{\overline{\alpha}(t)\gamma(ty(t))}{\sqrt{w(s)}}\sin[t(z-z(t))]\bigg{)}^{2}}\bigg{|}\bigg{|}E(t,z)-\sin[t(z-z(t))]\bigg{|}.$

Since $ty(t)>1$ there exists a constant $D>0$ such that

$\bigg{|}\theta(t,z)-\overline{\alpha}^{2}(t)\frac{\sin[t(z-\overline{z}(t))]}{\sin[t(z-z(t))]}\bigg{|}\leq
D\bigg{[}\big{|}E^{\sharp}(t,z)-\sin[t(z-\overline{z}(t))]\big{|}+\big{|}E(t,z)-\sin[t(z-z(t))]\big{|}\bigg{]}.$

Now again by Lemma 23, the last term tends to zero as $t\rightarrow\infty$
uniformly over all $z\in Q_{+}(s,C(t)/t).$ This proves $(i).$

The functions in (27) are holomorphic on $Q_{+}(s,C(t)/t).$ Suppose that
$C(t)>2D.$ Hence, by the Cauchy estimates for the derivative we find on the
smaller box $Q_{+}(s,D/t)$

$\sup_{z\in
Q_{+}(s,D/t)}\bigg{|}\theta_{z}(t,z)-\frac{d}{dz}\overline{\alpha}^{2}(t)\frac{\sin[t(z-\overline{z}(t))]}{\sin[t(z-z(t))]}\bigg{|}=o(t).$

On the other hand we compute

$\frac{d}{dz}\frac{\sin[t(z-\overline{z}(t))]}{\sin[t(z-z(t))]}\bigg{|}_{z=\overline{z}(t)}=\frac{t}{\sin[2ity(t)]}.$

Combining both formulas we obtain as desired

$\alpha^{2}(t)\theta_{z}(t,\overline{z}(t))=(1+o(1))t\frac{1}{\sin[2ity(t)]}.$

The proof for $(iii)$ is immediate by repeating the argument from the proof of
$(ii).$ ∎

The function $\theta_{z}(t,\overline{z}(t))$ recognizes the direction the zero
moves in. By our previous approximations, we will be able to carry over this
information to $\alpha(t).$

######  Lemma 33.

Let $s,C(t),\alpha(t),D$ and $z(t)$ be like in Lemma 23. Then there exists
$T>0$ such that for all intervals $[t_{1},t_{2}]\subset
T_{1}(s,D)\cap[T,\infty)$ we can change $\alpha(t)$ slightly so that it
satisfies Lemma 23 and we have the following relation

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

Furthermore, there is a real function $\psi$ on $[t_{1},t_{2}]$ satisfying

$\alpha(s,t)=e^{i[st+\psi(t)]}$

and for all $t_{1}\leq x_{1}\leq x_{2}\leq t_{2}$

(29)
$|\psi(x_{2})-\psi(x_{1})|\leq\int_{x_{1}}^{x_{2}}\bigg{(}3\cosh(2D)\cdot|f(t)|+\frac{D}{t}\bigg{)}dt.$

###### Proof.

By Lemma 32 $(ii),$

$\alpha^{2}(s,t)\theta_{z}(t,\overline{z}(t))=(1+o(1))t\frac{1}{\sin[2ity(t)]}.$

After rearranging the terms, using $\sin[2ity(t)]=i\sinh[2ty(t)]$ and using
the fact that $1<ty(t)<D$ we arrive at

$\frac{\sinh[2ty(t)]}{t}\alpha^{2}(s,t)\theta_{z}(t,\overline{z}(t))-1=o(1).$

This immediately gives us

$\arg i\alpha^{2}(s,t)\theta_{z}(t,\overline{z}(t))=o(1).$

After changing $\alpha$ slightly we can eliminate the error term. Using the
functional equation of the argument we arrive at (28)

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

We can thus recover $\alpha(s,t)$ by studying

$\frac{d}{dt}\arg\theta_{z}(t,\overline{z}(t)).$

For that, we will use orthogonal projections. In the following, we will
identify $\mathbb{C}\cong\mathbb{R}^{2}$ and denote for two vectors
$u,v\in\mathbb{R}^{2},v\neq 0$ the orthogonal projection of $u$ onto $v$ by

$\operatorname{Proj}_{v}u=\frac{\langle u,v\rangle}{||v||^{2}}v.$

We compute for all $z$ such that $\theta_{z}(t,z)\neq 0$

$\operatorname{Proj}_{i\theta_{z}(t,z)}\frac{d}{dt}\theta_{z}(t,z)=\operatorname{Proj}_{i\theta_{z}(t,z)}|\theta_{z}(t,z)|e^{i\arg\theta_{z}(t,z)}$

$=\operatorname{Proj}_{i\theta_{z}(t,z)}\bigg{(}\frac{d}{dt}|\theta_{z}(t,z)|e^{i\arg\theta_{z}(t,z)}+i\frac{d}{dt}\arg\theta_{z}(t,z)|\theta_{z}(t,z)|e^{i\arg\theta_{z}(t,z)}\bigg{)}=\frac{d}{dt}\arg\theta_{z}(t,z)i\theta_{z}(t,z).$

Recall that since the zeros $E(t,z)$ are simple,
$\theta_{z}(t,\overline{z}(t))\neq 0.$ Combining the above formula with (26)
gives

$\frac{d}{dt}\arg\theta_{z}(t,\overline{z}(t))i\theta_{z}(t,\overline{z}(t))=\operatorname{Proj}_{i\theta_{z}(t,\overline{z}(t))}\frac{d}{dt}\theta_{z}(t,\overline{z}(t))$

$=\operatorname{Proj}_{i\theta_{z}(t,\overline{z}(t))}2iz(t)\theta_{z}(t,\overline{z}(t))-f(t)\frac{\theta_{zz}(t,\overline{z}(t))}{\theta_{z}(t,\overline{z}(t))}$

$=i\Re\overline{z}(t)\theta_{z}(t,\overline{z}(t))-\operatorname{Proj}_{i\theta_{z}(t,\overline{z}(t))}f(t)\frac{\theta_{zz}(t,\overline{z}(t))}{\theta_{z}(t,\overline{z}(t))}$

$=(s+\Re(\overline{z}(t)-s)-f(t)A(t))i\theta_{z}(t,\overline{z}(t)),$

for some real valued function $A(t).$ Using the definition of
$\operatorname{Proj}_{v}u$ together with Lemma 32 $(ii)$ and $(iii)$ we find
for large enough $t$

$|A(t)|\leq\frac{\bigg{|}\frac{\theta_{zz}(t,\overline{z}(t))}{\theta_{z}(t,\overline{z}(t))}\bigg{|}}{|\theta_{z}(t,\overline{z}(t))|}\leq
2\cosh[2ty(t)]+o(1)\leq 3\cosh(2D).$

Moreover, since $\overline{z}(t)\in Q(s,D/t),$ we find
$|\Re\overline{z}(t)-s|\leq D/t.$ We now define for $t_{1}\leq x\leq t_{2}$

$\psi(x)=\int_{t_{1}}^{x}\Re(\overline{z}(t)-s)-f(t)A(t).$

Hence, we have for all $t_{1}\leq x_{1}\leq x_{2}\leq t_{2}$

$|\psi(x_{2})-\psi(x_{1})|\leq\frac{1}{2}\int_{x_{1}}^{x_{2}}\bigg{|}\frac{d}{dt}\arg\theta_{z}(t,\overline{z}(t))\bigg{|}dt\leq\int_{x_{1}}^{x_{2}}\bigg{(}3\cosh(2D)\cdot|f(t)|+D/t\bigg{)}dt.$

Finally, by using (28) we find

$\alpha(s,t)=e^{i[st+\psi(t)]}.$

∎

Hence by virtue of (28), $\alpha(t)$ carries information on the movement of
the zeros. If $|\Im\alpha(t)|$ is small, the zero moves almost vertically and
if $|\Im\alpha(t)|$ close to one, then the zero moves almost horizontally. We
now have the tools at hand to continue our proof that $T_{0}(s,C)$ is bounded
for almost all $s\in\mathbb{R}$ and for all $C>0.$

For the remainder of this section let $C>1$ be a constant. We define

$S=\\{s\in\mathbb{R}\mid T_{0}(s,C)\text{ is unbounded}\\}.$

We will assume without loss of generality that $|S|<\infty.$ By Corollary 7
the set $T_{0}(s,3C)$ can not cover a half-line for almost all
$s\in\mathbb{R}.$ After changing $S$ by a set of measure zero, we can assume
that for all $s\in S$ this is the case. Thus, the zeros will have to travel
infinitely often from outside of $Q(s,3C/t)$ into $Q(s,C/t).$ We will show
that every time this happens, this will require a chunk of the $L^{2}-$norm of
$f.$ More precisely, for all $s\in S$ there will exist infinitely many
intervals $\mathcal{L}_{s}(T_{1},T_{2})$ with the following properties
arbitrarily far in time.

  * (i)

There exists a zero $z(s,t)$ of $E(t,z)$ in $Q(s,3C/t)$ for all
$t\in\mathcal{L}_{s};t\mapsto z(s,t)$ is a differentiable curve for
$t\in\mathcal{L}_{s}$ and $\Im z(s,t)<-1/t$ for $t\in\mathcal{L}_{s},$

  * (ii)

$z(s,T_{1})\in\partial Q(s,3C/T_{1}),$

  * (iii)

If $2^{n}\leq T_{1}<2^{n+1},$ then $T_{1}<T_{2}\leq 2^{n+2}$ is such that
$z(s,T_{2})\in\partial Q(s,C/T_{2}).$

  * (iv)

For all $t\in(T_{1},T_{2})$ moves continuously inside $Q(s,3C/t)\setminus
Q(s,C/t).$

Indeed, we can get $(iii)$ by looking at the proof of Corollary7. That we can
achieve $(i)$ follows immediately by Lemma 30, since we can pick $z(t)$ as the
first zero to reenter the box $Q(s,C/t)$ at $t=t_{2}.$ If there now was a
point $t\in(t_{1},t_{2})$ such that $\Im z(s,t)\geq-\frac{1}{t},$ then
immediately by Lemma 23 and Lemma 30 we see that there were already other
zeros in the box $Q(s,C/t)$ at time $t,$ contradicting our choice of $z(t).$
For each $s\in S$ we now have infinitely many intervals
$\mathcal{L}_{s}=(T_{1},T_{2})$ with the above properties. If $2^{n}\leq
T_{1}\leq 2^{n+1},$ we will call that interval $\mathcal{L}_{s}^{n}.$ If there
is more than one interval satisfying $(i)-(iv)$ and $2^{n}\leq T_{1}<2^{n+1},$
we will choose one of them. We will denote by $S^{n}$ the set of all $s\in S,$
such that there exists an interval $\mathcal{L}_{s}^{n}$ with the above
properties. Thus, for any $k\geq 1,$

$S\subset\bigcup_{n=k}^{\infty}S^{n}.$

######  Lemma 34.

Let $s\in S$ and let $\mathcal{L}_{s}^{n}=(T_{1},T_{2})$ be one of the
considered intervals. There exists a constant $\Delta>0,$ depending only on
$C,$ such that

$\int_{T_{1}}^{T_{2}}|f(t)|dt>\Delta.$

###### Proof.

By Lusin’s Theorem, we can assume that all error terms in Lemma 32 are uniform
over all $s\in S.$ Then, by Lemma 32 $(ii)$ and property $(i)$ of
$\mathcal{L}_{s}^{n}$ we have $|\theta_{z}(t,z(s,t))|\asymp t.$ Furthermore,
by properties $(ii)$ and $(iii)$ of $\mathcal{L}_{s}^{n}$ we get
$1/t\ll|z(s,T_{1})-z(s,T_{2})|.$ Both implicit constants depend only on $C.$
Thus, by (25)

$\frac{1}{T_{2}}\leq\frac{1}{t}\ll|z(s,T_{1})-z(s,T_{2})|\leq\int_{T_{1}}^{T_{2}}\frac{|f(t)|}{|\theta_{z}(t,\overline{z}(s,t))|}dt\ll\int_{T_{1}}^{T_{2}}\frac{|f(t)|}{t}dt\ll\frac{1}{T_{1}}\int_{T_{1}}^{T_{2}}|f(t)|dt.$

∎

By making $\Delta>0$ smaller if necessary, we will from now on assume that

(30) $0<\Delta<\frac{1}{100\cosh 6C}.$

We will also need upper bounds. The following Lemma tells us that this is
possible by passing to a subinterval $L_{s}^{n}\subset\mathcal{L}_{s}^{n}.$

######  Lemma 35.

Let $s\in S^{n}$ and $\mathcal{L}_{s}^{n}$ the associated interval satisfying
properties $(i)-(iv)$ from above. Then, there exists a subinterval
$L_{s}^{n}=(\tau_{1},\tau_{2})\subset\mathcal{L}_{s}^{n}$ satisfying
properties $(i)$ and $(iv)$ from above and we have the following upper bounds
by making $\Delta$ a bit smaller if necessary

$0<\Delta<\int_{\tau_{1}}^{\tau_{2}}|f(t)|dt<2\Delta,$

$\int_{\tau_{1}}^{\tau_{2}}\frac{3C}{t}dt<\frac{1}{100}.$

###### Proof.

Since $T_{2}-T_{1}\leq 3\cdot 2^{n},$ we find

$\int_{T_{1}}^{T_{2}}\frac{3C}{t}dt\leq\int_{T_{1}}^{T_{2}}\frac{3C}{T_{1}}dt\leq\frac{3\cdot
2^{n}\cdot 3C}{2^{n}}=9C.$

Dividing $\mathcal{L}_{s}^{n}$ into $k$ subintervals of equal size, where
$k>900C,$ gives by the pigeon hole principle a subinterval
$I=(\tau_{1},\tau_{2}),$ where

$0<\frac{\Delta}{k}<\int_{\tau_{1}}^{\tau_{2}}|f(t)|dt,$

$\int_{\tau_{1}}^{\tau_{2}}\frac{3C}{t}dt<\frac{3\cdot 2^{n}\cdot 3C}{k\cdot
2^{n}}<\frac{1}{100}.$

Redefining $\Delta$ as $\Delta/k$ and decreasing $\tau_{2},$ we can achieve
further

$0<\Delta<\int_{\tau_{1}}^{\tau_{2}}|f(t)|dt<2\Delta.$

The desired interval is then $L_{s}^{n}=(\tau_{1},\tau_{2}).$

∎

In the following, we will only consider the intervals $L_{s}^{n}.$ We will
have to further subdivide $S^{n}$ into two parts, depending on whether the
zero moves almost vertically or has a horizontal component, comparable to the
total increment. Each case will require a different argument. To rule out a
criterion for when the zero moves horizontally or vertically, we will use
Lemma 33. Let $\alpha(s,t)$ be the function from Lemma 33. By (29),
$\alpha(s,t)$ is continuous. For fixed $s\in S$ let $I_{s}\subset
T_{0}(s,3C).$ We say that

  * (i)

$I_{s}$ is a $V_{s}-$interval if $|\Im\alpha^{2}(s,t)|<\frac{1}{100}$ for all
$t\in I_{s},$

  * (ii)

$I_{s}$ is a $H_{s}-$interval if $|\Im\alpha^{2}(s,t)|\geq\frac{1}{200}$ for
all $t\in I_{s}.$

Note that an interval can be both an $V_{s}-$and $H_{s}-$interval. By
continuity of $\alpha(s,t)$ we can partition $L_{s}^{n},$ up to countably many
points, into disjoint intervals, which are either $V_{s}-$ or $H_{s}-$
intervals or both.

We can now split up the set $S^{n}$ into disjoint subsets

$S_{V}^{n}=\bigg{\\{}s\in S^{n}\bigg{|}\int_{\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}|f(t)|dt\geq\frac{99}{100}\int_{L_{s}^{n}}|f(t)|dt\bigg{\\}},$

$S_{H}^{n}=\bigg{\\{}s\in S^{n}\bigg{|}\int_{\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}|f(t)|dt<\frac{1}{100}\int_{L_{s}^{n}}|f(t)|dt\bigg{\\}}.$

Note that since $L_{s}^{n}$ is partitioned, up to countably many points, into
$V_{s}-$ and $H_{s}-$intervals, we find for all $s\in S_{H}^{n}$

$\int_{\bigcup_{I_{s}\in
HL_{s}^{n}}I_{s}}|f(t)|>\frac{1}{100}\int_{L_{s}^{n}}|f(t)|dt.$

Hence, we now have for all $k\geq 1$

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

In the next two sections we are going to prove that there exists a constant
$D>0$ such that for all $n\geq 1$

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

From this, we find quickly that $|S|>0$ contradicts $f\in L^{2}(\mathbb{R}),$
compare Theorem 5.

###  3.5. Vertical intervals

In this section, we are going to show that there exists a constant $D>0$ such
that

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

Consider $s\in S_{V}^{n}\subset S^{n}.$ Then there exists an interval
$(\tau_{1},\tau_{2})$ with $2^{n}\leq\tau_{1}<2^{n+1}.$ We will split
$L_{s}^{n}$ into three disjoint intervals

$L_{s}^{n}=T_{-}^{n}\cup T_{s}^{n}\cup T_{+}^{n}$

such that $T_{-}^{n}$ is to the left of $T_{s}^{n}$ and $T_{+}^{n}$ is to the
right of $T_{s}^{n}$ and

$\int_{T_{-}^{n}}|f(t)|dt>\frac{\Delta}{3},\quad\int_{T_{s}^{n}}|f(t)|dt=\frac{\Delta}{3},\quad\int_{T_{+}^{n}}|f(t)|dt>\frac{\Delta}{3}.$

Furthermore, we choose $T_{s}^{n}$ to be open. We split $f$ into its positive
and negative part $f=f_{+}-f_{-}.$ We have

$\int_{T_{s}^{n}}f_{+}(t)dt\geq\frac{\Delta}{6},\quad\text{ or
}\int_{T_{s}^{n}}f_{-}(t)dt\geq\frac{\Delta}{6},$

and we will assume that the inequality with $f_{+}$ holds for all $T_{s}^{n}.$
In the following we consider the collection

$W=\bigcup_{s\in S_{V}^{n}}\\{T_{s}^{n}\\}.$

Our next result tells us that we can pick a finite subcollection and proceed
with that subcollection.

######  Lemma 36.

There exists a finite subcollection

$\mathcal{T}=\bigcup_{k=1}^{N}\\{T_{s_{k}}^{n}\\}$

with the following properties (abusing slightly the notation by identifying
the collection with a union of intervals)

  * (i)

$\int_{W\setminus\mathcal{T}}|f(t)|dt<\frac{\Delta}{100},$

  * (ii)

Each $T_{s_{k}}^{n}\in\mathcal{T}$ intersects at most two other intervals from
$\mathcal{T}$ and each point in $\mathcal{T}$ is contained at most in two
intervals from $\mathcal{T}.$

###### Proof.

First, we show that there exists a countable subcollection $W^{\prime}$ of $W$
that already covers $W.$ Indeed enumerate by $\mathbb{N}$ all open balls
$B_{r}(x),$ where $x,r\in\mathbb{Q}.$ For any $B_{r}(x)$ we choose if possible
an interval $\mathcal{T}$ from $W$ such that $B_{i}\subset\mathcal{T}$ and add
this interval to the collection $W^{\prime}.$ By construction $W^{\prime}$ is
countable since its indexed by the balls $B_{r}(x),$ which are countably many.
We now show that $W^{\prime}$ covers $W.$ Indeed, assume that $x\in W.$ Then,
there exists an interval $T_{s}^{n}$ with $x\in T_{s}^{n}.$ Since $T_{s}^{n}$
is open there is a ball $B_{r}(x)\subset T_{s}^{n}.$ But then by definiton of
$W^{\prime}$ there is an interval $\mathcal{T}$ (possibly different from
$T_{s}^{n}$) with $x\in\mathcal{T}$ and $\mathcal{T}\in W^{\prime}.$

Enumerate $W^{\prime}=\\{\mathcal{T}^{1},\mathcal{T}^{2},\dots\\}$ by
$\mathbb{N}.$ We now choose an interval $\mathcal{T}^{1}$ from $W^{\prime}$
such that

$\int_{\mathcal{T}^{1}}|f(t)|dt>\frac{1}{2}\sup_{n\in
N}\int_{\mathcal{T}^{n}}|f(t)|dt,\quad\mathcal{I}^{1}=\mathcal{T}^{1}$

In the $n$-th step choose an interval $\mathcal{T}^{n}\in W^{\prime}$ such
that

$\int_{\mathcal{T}^{n}}|f(t)|dt>\frac{1}{2}\sup_{m\in\mathbb{N}}\int_{\big{(}W\setminus\bigcup_{k=1}^{n-1}\mathcal{T}^{k}\big{)}\cap\mathcal{T}^{m}}|f(t)|dt,\quad\mathcal{I}^{n}=\bigg{(}W\setminus\bigcup_{k=1}^{n-1}\mathcal{T}^{k}\bigg{)}\cap\mathcal{T}^{n}.$

By construction the sets $I^{n}$ are disjoint and cover $W^{\prime}=W.$ Hence,
we have

$\int_{W}|f(t)|dt=\int_{W^{\prime}}|f(t)|dt=\int_{\bigcup_{n=1}^{\infty}\mathcal{I}^{n}}|f(t)|dt=\sum_{n=1}^{\infty}\int_{\mathcal{I}^{n}}|f(t)|dt.$

Since the series converges there exists $N\in\mathbb{N}$ such that

$\sum_{n=N+1}^{\infty}\int_{\mathcal{I}^{n}}|f(t)|dt<\frac{\Delta}{100}.$

Set $\mathcal{T}=\\{\mathcal{T}^{1},\dots\mathcal{T}^{N}\\}.$ Since
$\mathcal{I}^{n}\subset\mathcal{T}^{n}$ property $(i)$ holds. Indeed,

$\int_{W\setminus\mathcal{T}}|f(t)|dt=\int_{W\setminus\bigcup_{n=1}^{N}\mathcal{I}^{n}}|f(t)|dt=\sum_{n=N+1}^{\infty}\int_{\mathcal{I}^{n}}|f(t)|dt<\frac{\Delta}{100}.$

We now need to reduce the collection. Note that if three intervals
$I_{1},I_{2},I_{3}\subset\mathbb{R}$ intersect in one point, then one of them
is contained in the union of the others. We remove all those intervals which
are contained in the union of two others intervals from the collection. But
then at most two intervals intersect and each point in $\mathcal{T}$ is
contained in at most two intervals from $\mathcal{T},$ which is precisely
condition $(ii).$ ∎

Then by property $(i)$, for all intervals $T_{s}^{n},s\in S_{V}^{n},$ there is
an interval $\mathcal{T}^{m}\in\mathcal{T}$ such that

(31) $\int_{T_{s}^{n}\cap\mathcal{T}^{m}}f_{+}(t)dt>0.$

######  Lemma 37.

Let $s\in S_{V}^{n}$ and $\mathcal{T}^{m}\in\mathcal{T}$ such that (31) holds.
Then,

$\bigg{|}\int_{\mathcal{T}^{m}}f_{+}(t)e^{its}dt\bigg{|}>\frac{\Delta}{25}.$

###### Proof.

Since $T_{s}^{n}\cap\mathcal{T}^{m}\neq\emptyset,$ we have
$\mathcal{T}^{m}\subset L_{s}^{n}.$ By definition of $S_{V}^{n}$ and Lemma 35

$\int_{\mathcal{T}^{m}\setminus\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)dt\leq\int_{L_{s}^{n}\setminus\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}|f(t)|dt=\int_{L_{s}^{n}}|f(t)|dt-\int_{\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}|f(t)|dt.$

$\leq\int_{L_{s}^{n}}|f(t)|dt-\frac{99}{100}\int_{L_{s}^{n}}|f(t)|dt=\frac{1}{100}\int_{L_{s}^{n}}|f(t)|dt<\frac{\Delta}{50}$

Hence,

(32) $\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)dt=\int_{\mathcal{T}^{m}}f_{+}(t)dt-\int_{\mathcal{T}^{m}\setminus\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)dt\geq\frac{\Delta}{6}-\frac{\Delta}{50}>\frac{\Delta}{10}.$

The interval was partitioned, up to countably many points, into the
collections of intervals $VL_{s}^{n}$ and $HL_{s}^{n}.$ By definiton an
interval $I\in VL_{s}^{n},$ if and only if
$|\Im\alpha^{2}(s,t)|<\frac{1}{100}$ for all $t\in I.$ Since
$|\alpha(s,t)|=1,$ this implies $|\Re\alpha(s,t)|\geq\frac{1}{2}.$ Thus,

$\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in VL_{s}^{n}}I_{s}}f_{+}(t)dt\leq
2\bigg{|}\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)\alpha(s,t)dt\bigg{|}.$

Since $\alpha(s,t)$ is like in Lemma 33 and using Lemma 35, we find for any
$\tau_{1}\leq t\leq\tau_{2},$

$|\alpha(s,t)\overline{\alpha}(s,\tau_{1})-e^{is(t-\tau_{1})}|=|e^{i[\psi(t)-\psi(\tau_{1})]}-1|=2\bigg{|}\sin\bigg{(}\frac{\psi(t)-\psi(\tau_{1})}{2}\bigg{)}\bigg{|}\leq|\psi(t)-\psi(\tau_{1})|$

$\leq\int_{\tau_{1}}^{t}\bigg{(}3\cosh(6C)\cdot|f(t)|+D/t\bigg{)}dt\leq
6\cosh(6C)+\frac{1}{100}<\frac{1}{10}.$

Thus, we estimate further

$2\bigg{|}\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)\alpha(s,t)dt\bigg{|}$
$=2\bigg{|}\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)(\alpha(s,t)-\alpha(s,\tau_{1})e^{is(t-\tau_{1})}+\alpha(s,\tau_{1})e^{is(t-\tau_{1})})dt\bigg{|}$

$\leq 2\bigg{|}\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)e^{ist}dt\bigg{|}+2\bigg{|}\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)(\alpha(s,t)-\alpha(s,\tau_{1})e^{is(t-\tau_{1})})dt\bigg{|}$

$<2\bigg{|}\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)e^{ist}dt\bigg{|}+\frac{1}{5}\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)dt.$

Combining this estimate with (32) gives

$\bigg{|}\int_{\mathcal{T}^{m}}f_{+}(t)e^{ist}dt\bigg{|}\geq\bigg{|}\int_{\mathcal{T}^{m}\cap\bigcup_{I_{s}\in
VL_{s}^{n}}I_{s}}f_{+}(t)e^{ist}dt\bigg{|}>\frac{\Delta}{25}.$

∎

Define $\mathcal{S}^{m}$ to be the set of all $s\in S_{V}^{n}$ such that

$\int_{T_{s}^{n}\cap\mathcal{T}^{m}}f_{+}(t)dt\geq\frac{\Delta}{12}.$

Then, as we have shown above $\cup_{k=1}^{N}\mathcal{S}^{k}=S_{V}^{n}$ and by
the previous Lemma we have for all $s\in\mathcal{S}^{m},$

$\bigg{|}\int_{\mathcal{T}^{m}}f_{+}(t)e^{its}dt\bigg{|}>\frac{\Delta}{25}.$

Since $f\in L^{2}(\mathbb{R})$ we have Plancherel’s Theorem

$\int_{-\infty}^{\infty}|f(t)|^{2}dt=\int_{-\infty}^{\infty}|\hat{f}(t)|^{2}dt.$

This gives

$||f_{+}||_{L^{2}(\mathcal{T}^{m})}^{2}=\int_{-\infty}^{\infty}|\widehat{f_{+}\chi_{\mathcal{T}^{m}}}(s)|^{2}ds=\int_{-\infty}^{\infty}\bigg{|}\int_{\mathcal{T}^{m}}f_{+}(t)e^{-2\pi
ist}dt\bigg{|}^{2}ds$

$=\frac{1}{4\pi^{2}}\int_{-\infty}^{\infty}\bigg{|}\int_{\mathcal{T}^{m}}f_{+}(t)e^{ist}dt\bigg{|}^{2}ds>\frac{1}{4\pi^{2}}\int_{S^{m}}\bigg{|}\int_{\mathcal{T}^{m}}f_{+}(t)e^{ist}dt\bigg{|}^{2}ds>\frac{\Delta^{2}}{2500\pi^{2}}|S^{m}|.$

If we set $D=\frac{\Delta^{2}}{2500\pi^{2}}>0,$ we obtain

$\sum_{k=1}^{N}||f_{+}||_{L^{2}(\mathcal{T}^{m})}^{2}\geq
D\sum_{k=1}^{N}|S^{m}|\geq D|S_{V}^{n}|,$

Thus, since each point in $\cup_{k=1}^{N}\mathcal{T}^{k}$ is covered by at
most two intervals, we have as desired the following result.

######  Corollary 8.

There exists a constant $D>0$ such that

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

###  3.6. Horizontal intervals

In this section, we are going to prove that there exists a constant $D>0$ such
that

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

In contrast to the previous section, we will now need to work with the
scattering data and conclude with the non-linear Parseval identity. More
precisely, for an interval $(t_{1},t_{2})\subset\mathbb{R}$ we associate the
scattering function $a_{t_{1}\rightarrow t_{2}}$ constructed from the solution
of the Dirac system (3) with function $f\chi_{(t_{1},t_{2})}.$

We can construct $a_{t_{1}\rightarrow t_{2}}$ from the solution of (3) with
potential function $f$ as follows. If $M(t,z)$ is the matrix from (5), then we
define the transfer matrix from $t=t_{1}$ to $t=t_{2}$ as

$M_{t_{1}\rightarrow t_{2}}(z)=\begin{pmatrix}A_{t_{1}\rightarrow
t_{2}}(z)&&B_{t_{1}\rightarrow t_{2}}(z)\\\ C_{t_{1}\rightarrow
t_{2}}(z)&&D_{t_{1}\rightarrow
t_{2}}(z)\end{pmatrix}=M(t_{2},z)M^{-1}(t_{1},z).$

The associated Hermite-Biehler functions $E_{t_{1}\rightarrow t_{2}}(z)$ and
$\tilde{E}_{t_{1}\rightarrow t_{2}}(z)$ are then defined as

$E_{t_{1}\rightarrow t_{2}}=A_{t_{1}\rightarrow t_{2}}-iC_{t_{1}\rightarrow
t_{2}},\quad\tilde{E}_{t_{1}\rightarrow t_{2}}=B_{t_{1}\rightarrow
t_{2}}-iD_{t_{1}\rightarrow t_{2}}.$

Hence, the scattering function $a_{t_{1}\rightarrow t_{2}}$ can be written as

$a_{t_{1}\rightarrow
t_{2}}(z)=\frac{1}{2}e^{i(t_{2}-t_{1})z}(E_{t_{1}\rightarrow
t_{2}}(z)+i\tilde{E}_{t_{1}\rightarrow t_{2}}(z)).$

In this setting the non-linear Parseval identity (4) becomes

(33) $||\log|a_{t_{1}\rightarrow
t_{2}}|||_{L^{1}(\mathbb{R})}=||f||_{L^{2}((t_{1},t_{2}))}^{2}.$

Therefore, the strategy will be to derive lower bounds on

$||\log|a_{t_{1}\rightarrow t_{2}}|||_{L^{1}(\mathbb{R})}$

for special intervals $(t_{1},t_{2})\subset\mathbb{R}.$ This program will
require a stronger approximation of the functions $E(t,z)$ and
$\tilde{E}(t,z),$ than the approximations, we have used so far. In particular,
we wish to approximate both of the functions simultaneously in the imaginary
component and show that the zeros of $E(t,z)$ and $\tilde{E}(t,z)$ move
similar to the zeros of the approximating functions. We are going to formulate
the precise result below and postpone the proof to the next section.

First, we will need the notion of $\sigma-$interval for $f$ which is an
interval $I\subset\mathbb{R}$ with the property that

$\bigg{|}\int_{I}f(t)dt\bigg{|}\geq(1-\sigma)\int_{I}|f(t)|dt,$

where $0<\sigma<1$ is a small number. On a $\sigma-$interval, the function $f$
has almost the same sign. We fix once and for all a number
$0<\sigma<\frac{1}{100},$ whose exact value does not matter.

######  Lemma 38.

Let $f\in L_{loc}(\mathbb{R}).$ Then, for almost all $s\in\mathbb{R}$ with
$f(s)\neq 0$ all sufficiently small intervals $I,$ which are centered around
$s,$ are $\sigma-$intervals.

###### Proof.

Let $I_{n}$ be any sequence of intervals $I$ centered at $s$ with $|I|=2^{n}.$
Almost all $s\in\mathbb{R}$ are Lebesgue-points of $f$ and $|f|.$ Thus, we can
assume without loss of generality that $s$ is indeed a Lebesgue-point for both
functions. Hence,

$\lim_{n\rightarrow\infty}\frac{1}{|I_{n}|}\bigg{|}\int_{I_{n}}f(t)dt\bigg{|}\rightarrow|f(s)|=\lim_{n\rightarrow\infty}\frac{1}{|I_{n}|}\int_{I_{n}}|f(t)|dt.$

Then, for large enough $n$ we find

$\frac{1}{|I_{n}|}\bigg{|}\int_{I_{n}}f(t)dt\bigg{|}\geq\bigg{(}1+\frac{\sigma}{2}\bigg{)}(1-\sigma)|f(s)|\geq\frac{(1+\frac{\sigma}{2})(1-\sigma)}{(1+\frac{\sigma}{2})}\frac{1}{|I_{n}|}\int_{I_{n}}|f(t)|dt.$

∎

We are now ready to formulate the stronger approximation result.

######  Theorem 4 (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 $(t_{1},t_{2})\subset T_{1}(s,C(t)),t_{2}-t_{1}\leq\frac{1}{|s|+1}$ be a
$\sigma-$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
$Q(s,A/t).$ Since $(t_{1},t_{2})\subset T_{1}(s,C(t)/t),$ we have
$t_{k}\Im\xi_{k}>2,t_{k}\Im\tilde{\xi_{2}}>2$ 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
unimodular continuous functions $\alpha_{t}$ 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

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

  * (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 $\tilde{E}(t,z),$

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

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

As already mentioned we will prove Theorem 4 in the next section. We are now
going to use the stronger approximation to deduce asymptotics for
$a_{t_{1}\rightarrow t_{2}}.$ For the remainder of this section we will assume
that $s$ is such that Theroem 4 holds and we fix the notation for the time
interval $(t_{1},t_{2}),$ the size of the box $A,$ the zeros $\xi_{1}$ and
$\tilde{\xi}_{t}$ of $E(t,z)$ and $\tilde{E}(t,z),$ respectively, as well as
the approximating zeros $x_{t}-iy_{t}$ and $\tilde{x}_{t}-iy_{t}$. We
introduce two quantities $\varepsilon_{1}$ and $\varepsilon_{2}$ which measure
the horizontal increment and vertical increment of the blown up approximating
zeros as they travel from time $t_{1}$ to $t_{2},$ respectively. More
precisely, we set

$\varepsilon_{1}=t_{2}(x_{t_{2}}-s)-t_{1}(x_{t_{1}}-s),\quad\varepsilon_{2}=t_{2}y_{2}-t_{1}y_{1}.$

Directly from the definition one sees that $E(t,0)=1$ and $\tilde{E}(t,0)=-i$
for all $t>0.$ Hence, $a_{t_{1}\rightarrow t_{2}}(0)=1$ for all intervals
$(t_{1},t_{2})$ and we denote by $\arg E$ and $\arg a_{t_{1}\rightarrow
t_{2}}$ continuous branches of the argument on $\mathbb{R}$ and $\arg
E(t,0)=\arg a_{t_{1}\rightarrow t_{2}}(0)=0.$ We now have the following
asymptotics.

######  Lemma 39.

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

where the implicit constant depends only on $A.$

In the proof we are going to apply Theorem 4 to both functions $E(t,z)$ and
$\tilde{E}(t,z)$ at the same time. This requires that both functions have a
zero in the box $Q(s,A/t).$ In this section we supposed only that $E(t,z)$ has
a zero, so Theorem4 seems not applicable at a first glance. However, by Lemma
43, which we will prove in the next section and is essential for the proof of
Theorem 4, $E(t,z)$ is close to $\sin z$ and $\tilde{E}(t,z)$ is close to
$\cos z.$ Therefore, by Lemma 30 if one of the functions has a zero in the
box, the other one has a zero in the box $Q(s,(A+\varepsilon)/t)$ (we may have
to consider a larger box, because the zero could be located at the boundary.
If the zero is in the centre, we do not have to consider a bigger box). For
larger $t$ we can choose $\varepsilon$ smaller. Hence, one should think of the
zeros of $E(t,z)$ and $\tilde{E}(t,z)$ to be similar to the zeros of $\sin z$
and $\cos z,$ i.e. they arrange in a straight line the more $t$ increases.
Since we consider a large box with $A>8\pi$ it is really enough to look at
times $t,$ where the function $E(t,z)$ has a zero in the box $Q(s,C/t)$

###### Proof.

For $k=1,2$ we have by Theorem 4 for large enough $t_{1}$

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