text:
\begin{document}

\newcommand{\norm}[1]{\ensuremath{\left\|#1\right\|}}
\newcommand{\abs}[1]{\ensuremath{\left\vert#1\right\vert}}
\newcommand{\ip}[2]{\ensuremath{\left\langle#1,#2\right\rangle}}
\newcommand{\p}{\ensuremath{\partial}}
\newcommand{\pr}{\mathcal{P}}

\newcommand{\pbar}{\ensuremath{\bar{\partial}}}
\newcommand{\db}{\overline\partial}
\newcommand{\D}{\mathbb{D}}
\newcommand{\B}{\mathbb{B}}
\newcommand{\Sp}{\mathbb{S}}
\newcommand{\T}{\mathbb{T}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\N}{\mathbb{N}}

\newcommand{\supp}{\operatorname{supp}}
\newcommand{\spa}{\operatorname{span}}

\def\to{\rightarrow}
\def\La{{\Leftarrow}}
\def\Ra{{\Rightarrow}}

\def\no{\noindent}

\def\PWa{\mathcal{P}\mathcal{W}_a}

\def\AA{{\mathcal{A}}}
\def\BB{{\mathcal B}}
\def\CC{{\mathcal C}}
\def\DD{{\mathcal{D}}}
\def\EE{{\mathcal E}}
\def\FF{{\mathcal F}}
\def\GG{{\mathcal{G}}}
\def\HH{{\mathcal H}}
\def\II{{\mathcal{I}}}
\def\JJ{{\mathcal{J}}}
\def\LL{{\mathcal{L}}}
\def\MM{{\mathcal M}}
\def\NN{{\mathcal N}}
\def\PP{{\mathcal{P}}}
\def\QQ{{\mathcal{Q}}}
\def\RR{\mathcal{R}}
\def\SS{{\mathcal{S}}}
\def\TT{{\mathcal{T}}}
\def\UU{{\mathcal{U}}}
\def\XX{{\mathcal{X}}}
\def\YY{{\mathcal{Y}}}
\def\ZZ{{\mathcal{Z}}}

\def\WW{{\mathcal W}}
\def\KK{{\mathcal K}}
\def\BM{{\mathcal{BM}}}

\def\a{{\alpha}}
\def\e{{\epsilon}}
\def\d{{\delta}}
\def\D{{\Delta}}
\def\l{{\lambda}}

\def\o{{\omega}}
\def\sL{{\sigma_{\Lambda}}}
\def\sG{{\sigma_{\Gamma}}}

\def\lan{{\lambda_{n}}}
\def\lam{{\lambda_{m}}}
\def\lak{{\lambda_{k}}}
\def\lmn{{\lambda_{mn}}}
\def\gmn{{\gamma_{mn}}}

\def\k0{{\xi_{0}}}
\def\ms{{\mu^*}}

\def\Th{{\Theta}}
\def\L{{\Lambda}}
\def\G{{\Gamma}}

\def\lfl{{\left \lfloor}}
\def\rfl{{\right \rfloor}}
\def\mh{{\hat{\mu}}}
\def\nh{{\hat{\nu}}}
\def\sh{{\hat{\sigma}}}
\def\mup{{\sigma^+}}
\def\mum{{\sigma^-}}
\def\nup{{\nu^+}}
\def\num{{\nu^-}}

\def\const{\text{\rm const}}
\def\ti{\tilde}

\title{Determinacy for measures}
\subjclass[2010]{}
\keywords{}

\begin{abstract} We study the general moment problem for measures on the real line, with polynomials replaced by more general spaces
of entire functions. As a particular case, we describe measures that are uniquely determined by a restriction of their Fourier transform
to a finite interval. We apply our results to prove an extension of a theorem by Eremenko and Novikov on the frequency of oscillations of
measures with a spectral gap (high-pass signals) near infinity.
\end{abstract}

\author[M. Mitkovski]{Mishko Mitkovski$^\dagger$}
\address{Mishko Mitkovski, Department of Mathematical Sciences\\ Clemson University\\ Clemson, SC USA }
\email{mmitkov@clemson.edu}
\urladdr{http://people.clemson.edu/$\sim$mmitkov}
\thanks{$\dagger$ Research supported in part by National Science Foundation DMS grant \# 1101251.}

\author[A. Poltoratski]{Alexei Poltoratski$^\ddagger$}
\address{Alexei Poltoratski, Department of Mathematics\\ Texas A\&M University\\ College Station, TX USA}
\email{alexeip@math.tamu.edu}
\urladdr{www.math.tamu.edu/$\sim$alexeip}
\thanks{$\ddagger$ Research supported in part by National Science Foundation DMS grant \# 1101278. }

\subjclass[2000]{}
\keywords{Determinacy, spectral gap, Fourier transform, moment problem}

\maketitle

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{example}[theorem]{Example}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{remark}[theorem]{Remark}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem*{remark*}{Remark}

\newenvironment{proofA}
{{\noindent{\bf \textit{Proof of Theorem 1.2:}}}}{\hfill$\Box$}

\numberwithin{equation}{section}

\section{Introduction}

For a finite measure $\mu$ on the real line we define its Fourier transform as
$$\mh(x)=\int e^{ixt}d\mu(t).$$
We say that a positive finite measure $\mu$ is $a$-determinate if there exists no other positive finite measure $\nu$ such that  Fourier transforms of $\mu$ and $\nu$ coincide on $[-a,a]$.

One of the  problems that we consider here can be formulated as follows.

Suppose that we are given a finite positive measure $\mu$. For a given $a>0$, how can we tell whether $\mu$ is $a$-determinate?
This question can be viewed as the analog of the determinacy part of the classical moment problem, which received a lot of attention throughout the years.  The study of the trigonometric version of the  problem stated above originates from  a classical paper of Krein \cite{Krein3}.
Despite a number of deep results, it appears that no explicit solution is known, even in the classical polynomial case.

In this paper we consider the general case of determinacy in the moment problem, with polynomials (exponentials) replaced with broader classes of entire functions.
The correct space for such purposes seems to be the de Branges space, which can become a space of polynomials or a Paley-Wiener space with a proper choice of a generating Hermite-Biehler function $E$.

We apply our methods to generalize classical theorems by Beurling, de Branges, and Levinson on the absence of gaps in the support of the Fourier transform (spectral gaps) of a finite measure on $\R$. Another application is a refinement of a theorem by Eremenko and Novikov \cite{EN, EN1}
on the oscillations of high-pass signals.

Let $\mu$ be a finite real measure on $\R$ with the property that its Fourier transform $\hat\mu$ vanishes on the interval $[-a,a]$ for some $a>0$.
Such measures are called high-pass signals in electrical engineering. One of the classical problems  is to estimate
how fast a high-pass signal should oscillate near infinity. Significant progress in this direction was provided by the following result.

\begin{restatable}{theorem}{ErNov}\cite{EN, EN1}
 If $\sigma$ is a nonzero signed measure with spectral gap $(-a/2, a/2)$  then the number of sign changes $s(r,\sigma)$ of $\sigma$ on the interval $(0,r)$ satisfies
$$\liminf_{r\to\infty}\frac{s(r,\sigma)}{r}\geq \frac{a}{2\pi}.$$
\end{restatable}

Here the number of sign changes $s(r,\sigma)$ can be understood in the usual sense for absolutely continuous measures with continuous densities and in any reasonable sense for more general
measures, see section \ref{last}. This theorem solves an old problem by Grinevich from 1964 included in V. Arnold's list of problems (2000 \cite{Arn}). A problem of achieving a better understanding of the asymptotic interplay between the positive and negative parts of a high-pass signal, including a possibility of replacing the inequality in the last theorem with an equality of some sort, served as an initial motivation for the present paper.

In this article we obtain a short and self-contained proof of the last statement, that does not rely on any of the  deep tools of Harmonic Analysis
often applied in this area, such as the Beurling-Malliavin theory or advanced estimates of singular integrals. At the same time, with the use of such advanced tools, we are able to prove the following stronger statement.

 For a finite real measure $\mu$ on $\R$ we denote by $\mu^+$ and $\mu^-$ its mutually singular positive and negative parts, $\mu=\mu^+-\mu^-,\ \mu^+\perp\mu^-$.
 If $A$ and $B$ are two closed subsets of $\R$,  let
$\MM_a(A,B)$ be the class of all finite real measures $\sigma$ with a spectral gap $[-a,a]$ such that $\supp \sigma^+\subset A$ and $\supp \sigma^-\subset B$. We define the gap characteristic of a pair of closed subsets $A$ and $B$ of $\R$ as
$$G(A,B)=\sup\{a>0: \MM_a(A,B)\neq\{0\}\}.$$

\begin{restatable}{theorem}{oscillation} \label{oscillation} For any closed sets $A, B\subset\R$,
$$ G(A,B)=\pi\sup\{d: \exists\ \   d\text{-uniform sequence }\{\lambda_n\},\ \{\lambda_{2n}\}\subset A, \{\lambda_{2n+1}\}\subset B\}.$$
\end{restatable}

The definition of a $d$-uniform sequence was given in \cite{PolGap} and is discussed in the next section.
One can say that, in a certain sense made precise in section \ref{prelim}, $d$-uniform sequences are similar to arithmetic progressions with a difference term equal to $1/d$.
Such sequences are discrete (i.e. have no accumulation points on $\R$) and are enumerated in the natural increasing order.

Theorem~\ref{oscillation} follows from a general discussion of $a$-determinacy of measures and properties of extreme indeterminate measures
that we undertake in sections \ref{Riesz} and \ref{extreme}. First, we prove a generalization of a classical result by M. Riesz for the
polynomial moment problem that characterizes indeterminate measures in terms of the norms of point-evaluation functionals.
A similar theorem was recently presented by M. Sodin~\cite{Sod}.
We then apply that result to extend classical gap theorems.

Following Beurling and Malliavin, we say that a union of disjoint open intervals $\bigcup (a_n,b_n)$ is long if $$\sum_n \left(\frac{b_n-a_n}{a_n}\right)^2=\infty.$$
The classical Beurling's gap theorem~\cite{Beu} states that if the complement of $\supp\mu$ is long than $\mu$ does not have
any spectral gaps, unless $\mu\equiv 0$. Like most results by Beurling, this statement is sharp in its scale:
the condition of longness of the complement of the support cannot be weakened, as was proved by M. Benedicks~\cite{Ben}.
However, as we show in section \ref{Riesz}, the result can be strengthened in the following way:

\begin{restatable}{corollary}{gap}\label{gap} If $\mu$ is a positive finite measure which is supported on a set whose complement is long then $\mu$ is $a$-determinate for every $a>0$.
\end{restatable}
\no   Equivalently, if $\mu$ is a real finite non-zero measure and the complement of $\mu_+$ is long, then $\mu$ has no spectral gaps. Notice that we do not put any conditions on the support of the negative part of the measure.  Other classical gap theorems by de Branges and Levison can be similarly
strengthened, see section \ref{Riesz}.

The last statement implies that if $\mu>0$ is $a$-indeterminate for some $a>0$ then $\mu$ must contain a sequence of positive interior Beurling-Malliavin density in its support. It is natural to expect further relations of this type between the  determinacy of  $\mu$ and density of its support. Our next result makes this observation precise.

\begin{restatable}{theorem}{indeterminate} \label{indeterminate} Let $\mu$ be a positive finite measure. If $\mu$ is $a$-indeterminate then the support of $\mu$ contains an $a/2\pi$-uniform sequence.
\end{restatable}

Extending the last statement, we prove that for a general regular de~Branges space $\BB_E$ a positive measure $\mu$ whose support is sufficiently sparse (in a certain precise sense) is uniquely determined by its action on the space, meaning that there is no other positive measure $\nu$ such that
$$\int F(t)d\mu(t)=\int F(t) d\nu(t), $$ for all $F\in\BB_E$.

Clearly, every $a$-indeterminate measure is a positive part of a signed measure $\sigma$ with a spectral gap $(-a,a)$, i. e.,  such that $\hat{\sigma}(x)=0,$ for $x\in [-a,a]$. Conversely, every non-zero measure $\sigma$ with a spectral gap $(-a,a)$ gives a rise to two $a$-indeterminate measures $\sigma^+$ and $\sigma^-$. Therefore, the determinacy problem is closely related to the gap problem which was recently studied by the second author in~\cite{PolGap}. We prove the following result that gives a precise quantitive relation between these two problems.

For a closed set $X\subset \R$ its gap characteristic is defined by $$G(X)=\sup\{a: \exists\text{ a real non-zero measure } \mu  \text{ with spectral gap } [-a, a]\text{ supported on } X\}.$$ Define the determinacy characteristic of $X$ by
$$Det(X)=\inf\{a : \text{all } \mu \text{ supported on } X \text{ are } a\text{-determinate}\}.$$

\begin{restatable}{theorem}{DetGap}\label{DetGap} For any closed set $X\subset\R$, $$2Det(X)=G(X).$$\end{restatable}

\noindent\textbf{Acknowledgement.} We are grateful to A. Eremenko and M. Sodin who brought the problem on oscillations of Fourier integrals to our attention.

\section{Preliminaries}\label{prelim}

We  denote by $\Pi$ the Poisson measure on $\R$, $d\Pi(x)=dx/(1+x^2)$.
The notation $\CC_a$ stands for the Cartwright class, the class of entire functions $F$ of exponential type no greater than $a$  such that $\log|F|\in L^1(\Pi)$. We will also use the standard notation $\PWa$ for the Paley-Wiener class, and $\BB_a$ for the Bernstein class consisting of entire functions of type at most $a$ which are bounded on $\R$. Finally, we will denote by $\KK_a$ the Krein class  consisting of entire functions $F(z)$ of exponential type no greater than $a$ which have only real simple zeros $\{\lambda_n\}$, such that $\sum_n1/|F'(\lambda_n)|<\infty,$ and satisfy $$\frac{1}{F(z)}=\sum_n\frac{1}{(z-\lambda_n)F'(\lambda_n)}.$$  By a well known theorem of M. Krein each function from $\KK_a$ belongs in $\CC_a$

If $E$ is an entire function we denote $E^\#(z)=\bar E(\bar z)$. An entire function $E$ belongs to the Hermite-Biehler class if
it satisfies
$$|E(z)|>|E^\#(z)|$$
for all $z\in \C_+=\{\Im z>0\}$. Hermite-Biehler functions are often called de Branges functions in the literature.
For any Hermite-Biehler (de Branges) function $E$, the de Branges space $\BB_E$ is defined as a space of all
entire functions $F$ such that both $F/E$ and $F^\#/E$ belong to the Hardy space $H^2(\C_+)$.
For more on de Branges spaces see \cite{dBr}.

A de~Branges space $\BB_E$ is called regular if
\begin{itemize}
\item[(i)] $\BB_E\subset\CC_a$ and
\item[(ii)] If $F\in \EE$ , then $(F(z)-F(w))/(z-w)\in \EE$ for any $w\in\C$.
\end{itemize}

The main example of a regular de Branges space is the Payley-Wiener space $\PWa$. In that case $E$ is the standard
exponential function $S^a=e^{iaz}$.

If $\Th$ is an inner function in the upper half-plane, we denote by $\KK_\Th$ the model space $H^2(\C_+)\ominus\Th H^2(\C_+)$.
Such spaces play a central role in the Nagy-Foias functional model theory, see \cite{Nik}.

Every de~Branges function $E(z)$ gives rise to an inner function $\Th(z):=E^\#(z)/E(z)$ and a model space $\KK_{\Th}$ that this inner function generates. There exists a well known isometric isomorphism between $\BB_E$ and $\KK_{\Th}$ given by $F\to F/E$.

Each inner function $\Th(z)$ determines a family of positive measures $\mu_{\alpha}$ on $\R$ indexed by $\abs{\alpha}=1$ in the following way
\begin{equation}\label{Herglotz}
\Re{\frac{\alpha+\Th(z)}{\alpha-\Th(z)}}=p_{\alpha}\Im{z}+\frac{1}{\Im{z}}\int\frac{d\mu_{\alpha}(t)}{\abs{t-z}^2},
\end{equation}
for some $p_{\alpha}\geq 0$. The number $p_{\alpha}$ can be viewed as a point mass at infinity for $\mu_{\alpha}$. Each measure $\mu_{\alpha}$ is singular, supported on the set $\{\Th=\alpha\}$, and is Poisson-finite. They are usually called the Clark measures for $\Th(z)$. Throughout the paper when we say that $\Th(z)$ corresponds to a measure $\mu$ on $\R$ we will always mean that $\mu$ is the Clark measure $\mu_1$ for $\Th$ and that $p_1=0$.
In this case, each function $f\in\KK_{\Th}$ can be represented by the Clark formula
$$f(z)=\frac{1-\Th(z)}{2\pi i}K(f\mu)(z),$$ where $K(f\mu)$ stands for the Cauchy integral $$K(f\mu)(z)=\int \frac{f(t)}{t-z}d\mu(t).$$ This formula gives an isometric isomorphism between $\KK_{\Th}$ and $L^2(\mu)$. For more on Clark measures see for instance \cite{PolSar}.

For those inner functions $\Th$ that can be represented as  $\Th(z)=E^\#(z)/E(z)$ for some de Branges function $E$, all  Clark measures $\mu_{\alpha}$ are discrete and their point masses can be computed by $\mu_{\alpha}(\lambda)=2\pi/\abs{\Th'(\lambda)}$ for $\lambda\in\{\Th=\alpha\}$.

We will call the measures $|E|^2\mu_{\alpha}$, where $\mu_\alpha$ is a Clark measure for $\Th(z)=E^\#(z)/E(z)$,  spectral measures of the corresponding de~Branges space. It is well known that for any spectral measure $\nu$ of a de Branges space $\BB_E$ the natural embedding gives
and isometric isomorphism between $\BB_E$ and $L^2(\nu)$.

On the real line each inner $\Th(z)$ coming from a de~Branges function can be written as $\Th(t)=e^{i\theta(t)},\ t\in\R$, where $\theta(t)$ is real analytic strictly increasing function. The function $\theta(t)$ is a continuous branch of the argument of $\Th(z)$ on $\R$. The phase function of the corresponding de~Branges space is defined by $\phi(t)=\theta(t)/2$ and is equal to $-\arg E$.

Next we give the definitions of the densities that will be used in our statements. Recall that the family of disjoint intervals $\{I_n\}$ on the real line (or the union of the intervals from that family, when it is convenient) is called long if $$\sum_n \left(\frac{|I_n|}{1+\text{dist}(0, I_n)}\right)^2=\infty.$$ Otherwise, we will say that the family is short. In the case when $\{I_n\}$ is short, $\bigcup I_n=\R$, and $\abs{I_n}\to \infty$ as $n\to \pm\infty$ we will call the family $\{I_n\}$ a short partition.

In all our definitions and statements the intervals of a short partition, as well as discrete sequences of points, are assumed to be enumerated in
a natural increasing order.

\begin{definition} A sequence $\Lambda=\{\lambda_n\}$  with no finite accumulation point is said to be regular with density $a>0$ if there exists a short partition $\{I_n\}$ such that $$\#(\L\cap I_n)-a|I_n|=o(|I_n|) \text{ as } |n|\to \infty.$$
\end{definition}

\begin{remark} It can be shown that $\L$ is regular with density $a$ if and only if the following integral condition holds
$$\int \frac{\abs{n_\L(x)-ax}}{1+x^2}<\infty.$$ This definition of regularity was used in~\cite{Koo2} in the definition of Beurling-Malliavin densities.

\end{remark}

\begin{definition} The interior Beurling-Malliavin density $D_{BM}^-(X)$ of a closed set $X\subset \R$ is defined to be the supremum of all $a>0$ for which there exists a regular subsequence $\Lambda\subset X$ with density $a$. If no such subsequence exists $D_{BM}^-(X)=0$.

Similarly, for real sequences $\L$ one defines (the more famous) exterior density $D_{BM}^+(\L)$ as the infimum of $a$ such that
$\L$ is contained in a regular sequence with density $a$.
\end{definition}

The famous result of Beurling and Malliavin~\cite{BM} says that the zero set $\L$ of a function $F(z)\in\CC_{a\pi}$ must satisfy $D^+(\L)\leq a$. Conversely, for every sequence $\L$ with $D^+(\L)=a$ and every $k>a$, there exists non-zero $F(z)\in\CC_{k\pi}$ which vanishes at that sequence. However, $F(z)$ may have additional zeros.

In~\cite{PolGap} it was proved that a very similar statement holds for the zeros of functions in the Krein class. It turns out that in this case the regularity notion above needs to be refined to take into account a certain delicate separation condition. Namely, for any finite interval $I\subset \R$ and a Borel measure $\mu$ we define the energy of $\mu$, $E_I(\mu)$ by $$E_I(\mu)=\iint_{I, I} \log|x-y|d\mu(x)d\mu(y).$$ This is just the usual energy of the compactly supported measure $1_I(x)d\mu(x)$.

\begin{definition} We say that a real sequence $\Lambda=\{\lambda_n\}$  with no finite accumulation points is $a$-uniform if it is regular with density $a$ and satisfies the following energy condition: there exists a short partition $\{I_n\}$ such that

 $$ \sum_n\frac{\#(\L\cap I_n)^2\log_+|I_n|-E_{I_n}(dn_{\Lambda})}{1+\text{dist}(0,I_n)^2}<\infty.$$
\end{definition}

\begin{remark}

Notice that the series in the energy condition is positive.
The definition
given here is slightly different from the one given in \cite{PolGap, PolType}. Either one can be used in all of the results
and formulas from \cite{PolGap, PolType} or the present paper.

Note also that every separated regular sequence of density $a$ satisfies the energy condition automatically and therefore is $a$-uniform.

\end{remark}

The following technical lemma will be used several times in what follows.

\begin{lemma}\label{uff} If a discrete sequence $\L$ contains a $(d-\e)$-uniform subsequence for
every $\e>0$, then it also contains
a $d$-uniform subsequence. If $\L$ alternates between two sets $A$ and $B$, then this $d$-uniform
subsequence can be chosen to also
alternate between $A$ and $B$.
\end{lemma}

\begin{proof} Fix a small $\e>0$. Let $\L_1\subset\L$ be a $(d-\e)$-uniform sequence and let $\{I^{(1)}_n\}$ be the corresponding
short partition. Let $\L'_1\subset\L$ be a $(d-\e/2)$-uniform
sequence with corresponding short partition $\{J_n\}$.

We form a third short partition $\{K_n\}$ with the following properties:

\begin{itemize}

\item[(i)] $\{K_n\}$ is a superpartition of $\{J_n\}$, i.e. each interval $J_n$ is a subset of some interval $K_m$,

\item[(ii)] the intervals of $\{K_n\}$ are big enough, so that If two intervals $I^{(1)}_{n_k}$ and $I^{(1)}_{m_k}$
contain the endpoints of some interval $K_k$
then $|I^{(1)}_{n_k}|+|I^{(1)}_{m_k}|= o( |K_k|)$.
\end{itemize}

By lemma~4 in~\cite{PolGap} one can choose a subsequence $\Sigma$ of $\L'_1$
which is also $(d-\e/2)$-uniform with
a corresponding short partition $\{K_n\}$.

Let $\delta>0$. Choose $N$ big enough so that  the energy
sum for $\Sigma$ over $\{K_n\}_{|n|>N}$ is less than $\delta$, and so
that $I_{n_k}$ and $I_{m_k}$ for $k=\pm N$ are small in comparison
to $K_{\pm N}$.

We form a new sequence $\L_2$ in the following way. We keep $\L_1$ on all the
intervals $I^{(1)}_n$ that lie strictly between $K_{-N}$ and $K_N$.
On the rest of the line, keep $\Sigma$ on each $K_n$ with $|n|\geq N$.

We form a new short partition $\{I_n^{(2)}\}$  for $\L_2$ that consists of all intervals $I^{(1)}_n$ that lie between $K_{-N}$
and $K_N$, all $\{K_n\}_{|n|> N}$, instead of $K_N$ we include the interval $K_N\cup I^{(1)}_{n_N}$ and instead of
$K_{-N}$ we include $K_{-N}\cup I^{(1)}_{m_{-N}}$.

Then $\L_2$ is a $(d-\e/2)$-uniform sequence with a corresponding short partition $\{I_n^{(2)}\}$. The
corresponding energy sum  for $\L_2$ is
at most the energy sum for $\L_1$ plus $\delta$.

Reiterating this process we obtain sequences $\{\L_k\}$ that are $(d-\e/2^k)$-uniform with corresponding short partitions $\{I_n^{(k)}\}$ with energy sums at most the energy of $\L_1$ plus $1/2+\cdots +1/2^k$ (we choose $\delta = 2^{-i}$ in the $i$-th step). Using a diagonal process we
obtain a regular sequence of density $d$ whose energy sum converges, which is the desired sequence.

Finally, if $\L$ alternates between $A$ and $B$, in each step we can choose $\L_k$ to also alternate. The diagonal process will then also produce a sequence that alternates.

\end{proof}

\section{M.~Riesz-type criterion and its consequences}\label{Riesz}

For a given set of functions $\AA$ we will say that a positive measure $\mu$ is $\AA$-determinate if $\AA\subset L^1(\mu)$ and there is no other positive measure $\nu\neq \mu$, $\AA\subset L^1(\nu)$, such that
$$\int f(t)d\mu(t)=\int f(t) d\nu(t),$$ for all $f\in\AA$. Otherwise, we say that $\mu$ is $\AA$-indeterminate. We will study determinacy on linear spaces of entire functions $\EE$ satisfying the following properties:

\begin{itemize}
\item[(E1)] If $F(z)\in \EE$, then $F^\#(z):=\overline{F(\bar{z})}\in \EE$,
\item[(E2)] If $F(z)\in \EE$ and $w\in\C$, then $(F(z)-F(w))/(z-w)\in \EE$.
\item[(E3)] There exists $a\geq 0$ such that $\EE\subset \CC_a$.
\end{itemize}

There are many examples of sets $\EE$ that satisfy (E1)-(E3). The set of all polynomials $\PP$ is one such example. Other examples that we will use include the Bernstein class $\BB_a$, the Paley-Wiener space $\PP\WW_a$, and other regular de~Branges spaces (those de~Branges spaces which satisfy (E2) and (E3)).

It is not hard to see that a positive finite measure $\mu$ is $a$-determinate if and only if it is $\BB_a$-determinate or equivalently $\PP\WW_a$-determinate.

The next result gives a criterion for determinacy which generalizes the classical M. Riesz criterion for determinacy in the moment problem.
Our proof will also rely on the M. Riesz original idea. However, we will use  the de~Branges spaces as a substitute for the Gauss quadrature formula. We will use the following notation

$$\EE *\EE=\{F : F= GH, \text{ for some } G, H\in \EE\}.$$

\begin{theorem}\label{maj} Let $\EE$ be a set which satisfies (E1)-(E3) and let $\mu$ be a positive measure such that $\EE\subset L^1(\mu)\cap L^2(\mu)$. The following are equivalent:

\begin{itemize}
\item[(a)] $\mu$ is $\EE * \EE$-indeterminate
\item[(b)] The $L^2(\mu)$-closure of $\EE$ is a de~Branges space.
\item[(c)] The majorant  $m(x):=\sup\{\abs{F(x)} : F\in \EE, \norm{F}_{L^2(\mu)}\leq 1\}$ satisfies
$$\int \frac{\log{m(x)}}{1+x^2} dx < \infty,$$
\end{itemize}

\end{theorem}

As we already mentioned, a statement similar to the equivalence of (a) and (c) was recently presented in~\cite{Sod}. As a simple corollary we obtain the following result.

\begin{corollary}\label{Riesz1} A positive finite measure $\mu$ is $a$-determinate if and only if $$\int \frac{\log{m_{a/2}^{\mu}(x)}}{1+x^2} dx = \infty,$$ where $$m_{a/2}^{\mu}(w) := \sup \{ |F(w)| : F \in \spa\{e^{ixt}: t\in[-a/2,a/2]\}\text{ and } \|F\|_{L^2(\mu)} \leq 1 \}.$$
\end{corollary}

\begin{proof} First we prove that the integrability of  $\log{m(x)}/(1+x^2)$ implies that the $L^2(\mu)$-closure of $\EE$ is a de~Branges space. We will denote this closure by $\BB(\mu)$.

Each function $F$ in $\EE$ is in the Cartwright class $\CC_a$. Therefore $$\log|F(z)|\leq a|\Im z|+\frac{|\Im z|}{\pi}\int\frac{\log^+|F(t)|}{|t-z|^2}dt.$$ Taking a supremum over all $F(z)\in \EE$ with norm no greater than $1$ we first obtain
\begin{equation}\label{Cart}
\log|F(z)|\leq a|\Im z|+\frac{|\Im z|}{\pi}\int\frac{\log^+m(t)}{|t-z|^2}dt,
\end{equation}
for all non-real $z$. Using the simple estimate
$$\sup_{t\in\R}\abs{\frac{t-i}{t-z}}\leq \frac{1+\abs{z}}{\abs{\Im z}},$$ which is valid for all $z\notin \R$, we obtain
\begin{equation}
\log|F(z)|\leq a|\Im z|+C\frac{(1+\abs{z})^2}{\abs{\Im z}},
\end{equation}
where
$$C=\frac{1}{\pi}\int\frac{\log^+m(t)}{1+t^2}dt<\infty.$$
This immediately shows that $m(z)$ is bounded on compact sets that don't intersect the real axis. Standard application of the Levinson's log log theorem shows that $m(z)$ is also bounded on compact sets that intersect the real axis.

Let now $\{F_n\}$ be a sequence of functions in  $\EE$ which converges in $L^2(\mu)$ to some limit $f\in L^2(\mu)$. The goal is to show that $f$ is actually a restriction of some entire function $F\in  \BB(\mu)$. To see this notice that $\{F_n\}$ being Cauchy and the inequality  $$|F_k(z)-F_l(z)|\leq m(z)\|F_k-F_l\|_{L^2(\mu)},$$ together with the fact that $m(z)$ is locally bounded imply that $\{F_n\}$ converges uniformly on compact sets to some entire function $F$ which agrees with $f$ a.e. $\mu$. Since $F$ satisfies~\eqref{Cart}, $F(z)\in\CC_a$.

Let $F \in \BB(\mu)$ and let $F_n \in \EE$ be a sequence that converges to $F$ in $L^2(\mu)$.  Since $F_n(z)$ converges to $F(z)$ uniformly on compact sets, we have that for any $w \in \C$
\begin{equation}\label{exptype}
\frac{|F(w)|}{\|F\|} = \lim_{n \to \infty}\frac{|F_n(w)|}{\|F_n\|} \leq m(w).
\end{equation}
Therefore, non-real point evaluations are bounded in $\BB(\mu)$.  By considering $F_n^\#\in\EE$ we see that $F^{\#} \in \BB(\mu)$. By the axiomatic definition of de~Branges spaces, see \cite{dBr}, it remains to show that $F(z)(z-\bar{z}_0)/(z-z_0)\in \BB(\mu)$ for all $F\in \BB(\mu)$ and $z_0\notin\R$ such that $F(z_0)=0$. Since $$ F(z)\frac{z-\bar{z}_0}{z-z_0}=F(z)+(z_0-\bar{z}_0)\frac{F(z)}{z-z_0},$$ it is enough to show that $F(z)/(z-z_0)\in \BB(\mu)$, which again follows from the property that $(F_n(z)-F_n(z_0))/(z-z_0)\in\EE$. Therefore $\BB(\mu)$ is a de~Branges space.

If $\BB(\mu)$ is a de~Branges space pick any spectral measure $\nu$ of $\BB(\mu)$ which is different from $\mu$ (in case $\mu$ is one of them). Let $F\in\EE*\EE$ be arbitrary. Then by definition $F=GH$ for some $G, H\in \EE$. Since $\BB(\mu)$ is isomorphic to $L^2(\nu)$ and is isometrically contained in $L^2(\mu)$ we obtain that
$$\int F(t)d\mu(t)=\int G(t) \overline{H^\#(t)}d\mu(t)=\int F(t)d\nu(t).$$

Thus, $\mu$ is  $\EE*\EE$-indeterminate.

Next we show that if $\mu$ is $\EE*\EE$-indeterminate then $$\int \frac{\log{m(x)}}{1+x^2} dx < \infty.$$ If $\mu$ is $\EE$-indeterminate there exists another positive finite measure $\nu\neq\mu$ such that $\int F(t)d\mu(t)=\int F(t)d\nu(t)$ for all $F\in\EE*\EE$. This implies
\begin{equation}\label{eq1}
\int |F(t)|^2d\mu(t)=\int |F(t)|^2d\nu(t),
\end{equation} for all $F(z)\in \EE$.

Consider $\sigma=\mu-\nu$. We have that $$\int \frac{F(t)-F(z)}{t-z}d\sigma(t)=0,$$ for every  function $F(z)\in \EE$  and every $z\in \C$. This implies that $$F(z)=\frac{1}{G(z)}\int\frac{F(t)}{t-z}d\sigma(t),$$ where $G(z)=\int \frac{d\sigma(t)}{t-z}$. In particular, for $z=x+i$ with $x\in \R$ we have, $$|F(x+i)|\leq  \frac{1}{|G(x+i)|}\int |F(t)|d|\sigma|(t)\leq \frac{C}{|G(x+i)|}\left(\int|F(t)|^2d|\sigma |(t)\right)^{1/2}.$$ Now since $|\sigma| \leq \mu+\nu$ we obtain, $$|F(x+i)|\leq \frac{\sqrt{2}C}{|G(x+i)|}\left(\int|F(t)|^2d\mu(t)\right)^{1/2},$$ for every $F(z)\in \EE$. Consequently, $$m(x+i)\leq \frac{C'}{|G(x+i)|}.$$ The fact that $G(z+i)$, as a function of $z$, is analytic and bounded in $\C_+$ implies that $$\int \frac{\log{m(x+i)}}{1+x^2} dx \leq \int \frac{\log^-|G(x+i)|}{1+x^2}dx  + C''<\infty.$$

Now, let $F(z)$ be arbitrary element of $\EE$ with norm no greater than $1$. Then $F(z+i)$ as a function of $z$ is in the Cartwright class $\CC_{a}$. Therefore, for all $x\in \R$ we have $$\log|F(x)|\leq a+\frac{1}{\pi}\int\frac{\log^+|F(t+i)|}{(t-x)^2+1}dt.$$ Taking a supremum over all such $F(z)$ we first obtain $$\log|F(x)|\leq a+\frac{1}{\pi}\int\frac{\log^+|m(t+i)|}{(t-x)^2+1}dt,$$ and then again by taking a supremum on the left we deduce
$$\log m(x)\leq a+\frac{1}{\pi}\int\frac{\log^+|m(t+i)|}{(t-x)^2+1}dt.$$  Finally,
\begin{eqnarray*} \int \frac{\log{m(x)}}{1+x^2} dx &\leq& \frac{a\pi}{2}+\frac{1}{\pi}\int\int\frac{\log^+|m(t+i)|}{(t-x)^2+1}\frac{1}{1+x^2}dtdx \\ &=& a\pi+\int\frac{2\log m(x+i)}{4+x^2}dx<\infty.
\end{eqnarray*}
This completes the proof.

\end{proof}

Notice that the set of all polynomials $\PP$ satisfies $\PP*\PP=\PP$, so we derive the classical M.~Riesz criterion as a special case. In the case $\EE=\BB_a$ we obtain Corollary~\ref{Riesz1}.

As we mentioned in the introduction, the usefulness of the criterion above depends on the possibility to estimate the majorant $m(w)$. Below we consider some cases when this can be done.

\begin{corollary}\label{uniform} Let $\mu$ be a positive finite measure.  If there exists a non negative uniformly continuous function $w(t)$ on $\R$ satisfying $e^{w(t)}\in L^1(\mu)$ and $$\int\frac{w(t)}{1+t^2}dt=\infty,$$ then $\mu$ is $a$-determinate for every $a>0$.
\end{corollary}

\begin{remark} This lemma can be viewed as a stronger version of de Branges' gap theorem \cite{dBr}, which claims that if there is a function $w$ like in the lemma,
that satisfies $e^{w(t)}\in L^1(|\mu|)$ for a finite non-zero real measure $\mu$, then the support of $\hat\mu$ may not have any gaps. Our
statement can be equivalently reformulated as follows: if $e^{w(t)}\in L^1(\mu_+)$ then the support of $\hat\mu$ does not have any gaps.
\end{remark}

\begin{proof} First note that we may replace the uniform continuity assumption on $w(t)$ by the stronger one that $w(t)$ is uniformly Lipschitz (see~\cite[pg. 96]{Koo}). Next notice that every entire function $F(z)$ of exponential type at most $a$ for which $F(t)/e^{w(t)/2}\leq 1$ on $\R$ must also satisfy $\int |F(t)|^2d\mu(t)\leq 1$. Therefore, by combining the corollary in~\cite[pg. 236]{Koo} (for $W(t)=e^{w(t)/2}$) with Theorem~\ref{Riesz1}  we obtain $$\int \frac{\log{m_{a}^{\mu}(t)}}{1+t^2} dt = \infty,$$ for every $a>0$. The result now follows from Theorem~\ref{maj}. It should be noted that the idea used here is due to de~Branges.
\end{proof}

We next prove a strengthening of Levinson's gap result~\cite{Lev}.

\begin{corollary} Let $\mu$ be a positive finite measure.  If there exists a non negative function $w(t)$ on $\R$ which is increasing on $[0,\infty)$ and satisfies $e^{w(t)}\in L^1(\mu)$ and $$\int_1^{\infty}\frac{w(t)}{1+t^2}dt=\infty,$$ then $\mu$ is $a$-determinate for every $a>0$.
\end{corollary}

\begin{remark}
Equivalently, if $w$ is like in the last corollary and $\mu$ is a real non-zero finite measure such that $e^{w(t)}\in L^1(\mu_+)$ then $\supp\hat\mu$ has no gaps. The original theorem by Levinson has a stronger requirement that $e^{w(t)}\in L^1(|\mu|)$
\end{remark}

\begin{proof} Let $v(t)$ be the largest minorant of $w(t)$ which is uniformly Lipschitz with Lipschitz constant $1$. Then (see the lemma in~\cite[pg. 239]{Koo})  $$\int \frac{v(t)}{1+t^2}dt=\infty,$$ and the result follows from the previous corollary.
\end{proof}

Finally, we prove a strengthening of the Beurling's gap theorem as discussed in the introduction.

\gap*

\begin{proof} Define $w(t)$ as zero outside of the intervals $(a_n,b_n)$ and
  as $\textrm{dist} (t, \R\setminus (a_n,b_n))$ on $(a_n,b_n)$. This function is clearly non negative, uniformly continuous and satisfies $e^{w(t)}\in L^1(\mu)$. Moreover, the assumption that $\cup (a_n,b_n)$ is long yields that $$\int\frac{w(t)}{1+t^2}dt=\infty.$$ The result now follows from Corollary~\ref{uniform}.
\end{proof}

\section{Extreme measures in the indeterminate case}\label{extreme}
Let $\AA$ be a set of functions on $\R$.
For a given positive measure $\mu$ with $\AA\subset L^1(\mu)$ denote by $\MM_{\AA}(\mu)$ the set of all finite positive measures $\nu$ such that
$\AA\subset L^1(\nu)$ and $$\int F(t)d\mu(t)=\int F(t) d\nu(t)$$ for all $F\in \AA$. Clearly, this set is convex and therefore it is interesting to describe its extreme points (if they exist). The following result, which in essence goes back to M. Naimark, gives a description of the extreme points of $\MM_{\AA}(\mu)$.

\begin{theorem} A positive finite measure $\nu\in \MM_{\AA}(\mu)$ is an extreme point of $\MM_{\AA}(\mu)$ if and only if the span of the set $\AA$ is dense in  $L^1(\nu)$.
\end{theorem}

\begin{proof} Assume that  $\nu\in \MM_{\AA}(\mu)$ is not an extreme point of $\MM_{\AA}(\mu)$. Then $\nu=\alpha\nu_1+(1-\alpha)\nu_2$ for some $\nu_1, \nu_2\in \MM_{\AA}(\mu)$ and $0<\alpha<1$. It is easy to see that
$$\phi(f)=\int f(t)d\nu(t)-\int f(t)d\nu_1(t),$$ is a non trivial bounded linear functional on $L^1(\nu)$ which vanishes on the set $\AA$. Therefore, the span of $\AA$ cannot be dense in  $L^1(\nu)$.

Conversely, assume that the span of $\AA$ is not dense in  $L^1(\nu)$. Then there exists a nontrivial bounded linear functional $\phi$ on $L^1(\nu)$ which vanishes identically on $\AA$. We can assume that its norm is $1$. In this case both $\phi_1(f)=\int f(t)d\nu(t)-\phi(f)$ and $\phi_2(f)=\int f(t)d\nu(t)+\phi(f)$ are nonnegative linear functionals on $C_0(\R)$ in the sense that both $\phi_1(f)\geq 0$ and $\phi_2(f)\geq 0$ whenever $f(t)\geq 0$ on $\R$.  Therefore, there exist positive finite measures $\nu_1$ and $\nu_2$ such that $\phi_1(f)=\int f(t)d\nu_1(t)$ and  $\phi_2(f)=\int f(t)d\nu_2(t)$ for all $f\in L^1(\nu)$. Notice that $\phi\neq 0$ implies that $\phi_1\neq \phi \neq \phi_2$ and hence $\nu_1\neq \nu \neq \nu_2$.  Also for all $F\in\AA$ we have that $\int F(t)d\nu_1(t)=\int F(t)d\nu(t)=\int F(t)d\nu_2(t)$ and therefore $\nu_1, \nu_2 \in \MM_{\AA}(\mu)$. Finally, since clearly $$\nu=\frac{1}{2}\nu_1+\frac{1}{2}\nu_2$$ we obtain that $\nu$ cannot be an extreme point for $\MM_{\AA}(\mu)$.
\end{proof}

We will call a lower semi-continuous function $W:\R\to [1,\infty)$ a weight on $\R$. We will say that a measure $\mu$ on $\R$ is $W$-finite
if
$$||\mu||_W=\int Wd|\mu|<\infty.$$

Let  $\BB(E)$ be a de Branges space. We will call  a weight $W$ an $E$-weight if
 $F(x)/W(x)=o(1)$ as $x\to\pm\infty$ for any $F\in \BB(E)$.

Such weights exist for any de Branges space and can, in most cases, be chosen in a natural way.
For instance, for the generating Hermit-Biehler function $E$ one may consider entire functions $A=\frac 12(E^\#+E)$ and $B=\frac 1{2i}(E^\#-E)$ and
define  the weight $$W_E(x)=1+(B'(x)A(x)-B(x)A'(x))^{1/2}.$$ Since the reproducing kernels of $\BB(E)$ satisfy
$$||K^E_x||_{\BB(E)}=(B'(x)A(x)-B(x)A'(x))^{1/2}$$
for any $x\in\R$,
any weight $W$, such that $W_E=o(W)$ near infinity, is an $E$-weight.
 Note that $W_E$ itself is not always an $E$-weight: consider for instance the case when $E$ is
a polynomial. However, in the most important example for our purposes, $\BB(E)=\PWa$, $W_E=1+a/\pi$ is an $E$-weight.

If $W$ is an $E$-weight then $\BB(E)\subset L^1(|\mu|)$ for any $W$-finite measure $\mu$.
We say that a measure $\mu$ annihilates  $\BB(E)$ if it is $W$-finite for some $E$-weight $W$ and satisfies
$$\int Fd\mu=0$$
for all $F\in\BB(E)$.

Notice that positive and negative parts of a $\BB_E$-annihilating measure are $\BB_E$-indeterminate.

It will be crucial for us to examine the properties of the extreme points of the set of all  annihilating measures of a given de~Branges space $\BB_E$. Our approach is based on de~Branges' extreme point method~\cite{dBrSW, dBrA, dBrB, dBr}.

Let $A$ and $B$ be two disjoint closed subsets of $\R$ and let $W$ be an $E$-weight.
Denote by $\MM_{E}^W(A,B)$ the set of all real measures $\sigma$, $\|\sigma\|_W\leq 1$, which annihilate $\BB_E$, and such that $\supp\mup\subset A$, $\supp\mum\subset B$.  Here, as usual, $\mup$ and $\mum$ denote the positive and negative parts of $\sigma$ in the canonical Jordan decomposition $\sigma=\mup-\mum$. Although, in view of the problem of $a$-determinacy, the most important particular case for us is when
$E=\exp{(iaz)}$, $\BB_E=\PWa$ and $W=W_E=1+a/\pi$, in this section we study the general case.

In the rest of the statements of this section we assume that $\BB_E\neq \{0\}$ is a non-trivial de Branges space and $W$ is an $E$-weight.

\begin{lemma}\label{KrMil} The set $\MM_E^W(A, B)$ is convex and compact with respect to the  weak-* topology of the set of all $W$-finite measures on $\R$. In particular, if it contains a non-zero measure then it also contains non-zero extreme points.
\end{lemma}

\begin{proof} Convexity part is obvious. To show that  $\MM^W_E(A,B)$ is weak-* compact it is enough to show that it is weak-* closed. Let $\nu_n\in \MM^W_E(A,B)$ be a sequence of measures that converges to some measure $\nu$ in the weak-* sense. Notice that the predual space of the space of all $W$-finite measures is $C_0(W)$, the space of all
continuous functions $f$ on $\R$ satisfying $f=o(W)$ at infinity. Therefore, $\BB_E$ is a subspace of the predual space $C_0(W)$ and hence $\nu$ must also annihilate $\BB_E$.

It remains to show that $\supp\nup\subset A$ and $\supp\num\subset B$. Suppose that $\supp\nup\nsubseteq A$. Then there exists an open set $S$  disjoint from $A$ such that $\nup(S)>0$. Notice that if $S\cap \supp\num\neq \emptyset$ then $\num(S\cap \supp\num)>0$ in which case $\nup(S\cap \supp\num)=0$. Therefore, without loss of generality we can assume that $\num(S)=0$ because otherwise we can replace $S$ with $S\cap (\supp\num)^c$. Inner regularity of $\nup$ implies that there exists a compact set $K\subset S$ with $\nup(K)>\nup(S)/2$. Similarly, outer regularity of $\num$ implies existence of an open set $G$ such that  $S\subset G\subset A^c$ and $\num(G)<\num(S)+\nup(S)/4=\nup(S)/4$. By Urysohn's lemma there exists a continuous function $f$ which vanishes at infinity and such that $f=1$ on $K$, $f=0$ outside of $G$ and $0\leq f\leq 1$ everywhere. Clearly, $f\in C_0(W)$. Now, it is easy to see that $$\int fd\nu\geq \nup(K)-\num(G)\geq \nup(S)/4>0.$$ On the other hand, for all $n\in\N$ we have $\int f d\nu_n=\int f d\nu^-_n \leq 0$. We have a contradiction. The proof that $\supp\num\subset B$ is similar.

The Krein-Milman theorem now implies that $\MM^W_E(A,B)$ is a closed convex hull of its extreme points, and clearly some of them must be non-zero since $\MM^W_E(A,B)$ contains a non-zero element.

\end{proof}

\begin{lemma}\label{extremel} Let $\nu$ be an extreme point in $\MM^W_E(A,B)$. Then any function $f\in L^1(|\nu|)$ which is not in the $L^1(|\nu|)$-closure of $\BB_E$ must satisfy $$\int f(t)d\nu(t)\neq 0.$$
Moreover, the space of all bounded linear functionals on $L^1(|\nu|)$ which vanish on $\BB_E$ is one-dimensional.
\end{lemma}

\begin{proof} Let $f\in L^1(|\nu|)$ be a non-zero function which is not in the $L^1(|\nu|)$-closure of $\BB_E$. Then there exists a bounded linear functional which is zero on $\BB_E$ and non-zero at $f$. Each such functional $L$ on $L^1(|\nu|)$ can be represented as $L(f)=\int f(t)h(t)d|\nu|(t)$ for some $h\in L^{\infty}(|\nu|)$. Therefore, we have $\int F(t)h(t)d|\nu|(t)=0,$ for every $F\in \BB_E$ and  $\int f(t)h(t)d|\nu|(t)\neq 0$. Let $d\nu(t)=b(t)d|\nu|(t)$ be the polar decomposition of $\nu$. Using the fact that $F\in \BB_E$ implies $F^{\#}\in\BB_E$ we can assume with no loss in generality that $h/b$ is real valued. By adding a constant if necessary, we may assume that $h/b$ is a nonnegative function. Multiplying
 by a constant if necessary, we can assume that $ h/b$ is bounded by $1$ on the real line. Choose $\l\in (0,1)$ and put  $\nu_1= hd|\nu|$, $\nu_2=(\nu-\l hd|\nu|)/(1-\l)$. Both $\nu_1$ and $\nu_2$ are members of $\MM^W_E(A,B)$ and $\nu=\l \nu_1+(1-\l)\nu_2$. Since $\nu$ is an extreme point and clearly $\nu_1\neq 0$ we must have that $\nu_2=0$. Therefore, $\nu=\l h |\nu|$ and hence $$\int f(t)d\nu(t)\neq 0.$$
For the second part, notice that  $\nu=\l h |\nu|$ implies $\l h=b$ and therefore $L(f)=\l\int f(t)d\nu(t)$.
\end{proof}

\begin{lemma}\label{close} Let  $\BB_E$ be a regular de~Branges space such that $\BB_E\subset \CC_a$ and let $\mu$ be a finite positive measure
such that $\BB_E$ is not dense in $L^1(\mu)$. Then for every function $f\in L^1(\mu)$ which is in the $L^1(\mu)$-closure of $\BB_E$ there exists an entire function $F$ in the Cartwright class $\CC_a$ such that $F=f$, $\mu$-a.e.
\end{lemma}
\begin{proof} The proof is very similar to the proof of Theorem~\ref{maj}. Since  $\BB_E$ is not dense in $L^1(\mu)$, there exists
$h\in L^\infty(\mu)$ such that $\int hf\mu=0$ for all $f$ in the $L^1(\mu)$-closure of $\BB_E$. Put $\nu=h\mu$.
Let $$m(w) = \sup \{ |F(w)| : F \in \BB_E \text{ and } \|F\|_{L^1(|\nu|)} \leq 1 \}.$$ Using the fact that $\nu$ annihilates $\BB_E$ it follows that $$\int{\frac{F(t)-F(z)}{t-z}}d\nu(t)=0,$$ for all $F\in \BB_E$ and non-real $z$.  Then clearly $$m(z)\abs{\int \frac{d\nu(t)}{t-z}}\leq \frac{C}{|\Im{z}|}.$$ Exactly as in the proof of Theorem~\ref{maj} this implies that $m(t)$ is Poisson summable, that $m(w)$ is locally bounded, and that $$\log m(w)\leq a|\Im w|+\frac{|\Im w|}{\pi}\int\frac{\log^+m(t)}{|t-w|^2}dt,$$ for all non-real $w$.  Let now $\{F_n(z)\}$ be a sequence of functions in  $\BB_E$ which converges in $L^1(\mu)$ to some limit $f\in L^1(|\nu|)$. Then $\{F_n\}$ being Cauchy and the inequality  $$|F_k(z)-F_l(z)|\leq m(z)\|F_k-F_l\|_{L^1(|\nu|)},$$ together with the fact that $m(z)$ is locally bounded imply that $\{F_n\}$ converges uniformly on compact sets to some entire function $F$. On the other hand, $F_n$ converge to $f$ in $L^1(\mu)$. Therefore, $F$ agrees with $f$, $\mu$-a.e. Finally, it is then easy to see that $F\in \CC_a$.
\end{proof}

\begin{lemma}\label{Krein} Let $\BB_E\subset \CC_a$ be a regular de~Branges space. If $\nu\in\MM^W_E(A,B)$ is a non-zero extreme point then it is supported
on a discrete sequence and its Cauchy integral $K\nu:\C\to\hat\C$ is a non-vanishing function.
 Moreover, the supports of $\nup$ and $\num$ are interlaced, i.e., for any two points $a,b$ in the support of $\nup$ there exists a point $c$ from the support of $\num$ such that $a<c<b$ and vice versa.
\end{lemma}

\begin{proof}
Since $\BB_E$ is a regular space, it has no common zeros and the support of $\nu$ contains more than one point. Let $(a,b)$ be any finite interval that contains at least two points of the support of $\nu$. There exists a function $f\in L^1(|\nu|)$ which vanishes outside of $(a,b)$ such that $\int |f(t)|d|\nu|(t)=1$ and $\int f(t)d\nu(t)=0$. By Lemma~\ref{extremel} and Lemma~\ref{close}, there exists an entire function $F\in \CC_a$ such that $F=f$ a.e. $|\nu|$. It follows that the support of $\nu$ outside of $(a,b)$ must be contained in the zeros of $F$. Since $F$ is non zero, the support of $\nu$ outside of $(a,b)$ cannot have a finite accumulation point. Finally, choosing a new interval $(a,b)$ to be disjoint from the old one and repeating the argument we can show the whole support of $\nu$ must be a discrete set.

The statement that  $K\nu$ has no zeros is now equivalent to the statement that
$G=\frac{1}{K\nu}$ is an entire function, see the proof of Theorem~66 in~\cite{dBr}.
(See also Lemma 4 in \cite{PolType}.)
To prove the interlacing of $\supp\nu_\pm$ notice that the function $G(z)$ must vanish at every point $c$ in the support of $\nu$, and for all such $c$,  $G'(c)=1/\nu(c)$. Since $G$ is a real analytic function on $\R$ with simple zeros, for any two of its consecutive zeros $a$ and $b$ we must have $G'(a)G'(b)<0$. Therefore, $\nu(a)\nu(b)<0$ for any two consecutive points in the support of $\nu$.
\end{proof}

Let $\Th=E^\#/E$ be the inner function that corresponds to the de~Branges function $E$. If $\nu$ annihilates $\BB_E$, then $E\nu$ annihilates $K_\Th$ and hence the Cauchy integral of $\bar{E}\nu$  is divisible by $\Th$,
see \cite{Aleksandrov} or \cite{PolGap}.
In particular, if $\nu\in\MM^W_E(A,B)$ is an extreme point, then, by Lemma~\ref{extremel}, the only (up to a constant) bounded linear functional on $L^1(|\nu|)$ which annihilates the model space $\KK_{\Th}$ is given by  $L(f)=\int f(t)E(t)d\nu(t).$ Therefore, there is no non-constant bounded function $h$ for which the Cauchy integral $K(h\bar{E}\nu)$ is divisible by $\Th$. We will use this observation in the proof of our next result.

For a discrete sequence $\L=\{\lan\}$ we denote by $n_\L$ its counting function, that is defined to be $0$ at $0$, is constant on each
interval $(\lan,\l_{n+1}]$ and jumps up by one at each $\lan$.

\begin{theorem}\label{poissonsummable} Let $\BB_E$ be a regular de~Branges space and let $\phi(t)$ be the corresponding phase function.
Let $W$ be an $E$-weight and let $\mu$ be a $W$-finite measure. Suppose that there
 exists another $W$-finite measure $\nu$ such that
 $$\int Fd\mu=\int Fd\nu$$
 for all $F\in \BB(E)$. Then the support of $\mu$ contains a sequence $\L$ whose counting function satisfies $2\pi n_{\L}(t)-\phi(t)=\tilde{h}(t)$ for some $h\in L^1(\Pi)$.
\end{theorem}

\begin{proof}  Notice that $\sigma=\mu-\nu$ is a finite signed measure which annihilates $\BB_E$ and (after scaling) $\sigma\in\MM_E^W(A,B)$ for $A:=\supp\sigma^+, B=\supp\sigma^-$. Notice that clearly $A\subset \supp\mu$ and $B\subset \supp\nu$. By Lemma~\ref{KrMil} there exists an extreme measure $\eta\in \MM^W_E(A,B)$ which has to be discrete by Lemma~\ref{Krein}.

Let $\Th=E^\#/E$ be the inner function that corresponds to the de~Branges function $E$. Then $\Th(t)=e^{2i\phi(t)}$. Let $\Psi$ be the inner function corresponding to the Clark measure $\abs{E}|\eta|$. The goal is to show that there exists an outer function $h\in \ker T_{\bar{\Psi}\Th}$ such that $\Psi\bar{\Th}\bar{h}=h$ is also outer.

Define $g=(1-\Psi)K(\bar{E}\nu)$. Then clearly $\bar{E}\eta=g|E||\eta|$ and $g\in \KK_{\Psi}$. Moreover, the fact that $\eta$ is real and annihilates $\BB_E$ implies that the Cauchy integral $K(\bar{E}\eta)$ is divisible by $\Th$. Therefore, $g=\Th h$ for some $h\in\HH^2\cap\HH^\infty$. Assume that $h$ is not outer, i. e., there exists an inner function $\Phi$ such that $h=\Phi k$ for some $k\in\HH^2\cap \HH^\infty$. Then $\Th(1+\Phi) k\in \KK_{\Psi}$ and hence, by Clark's formula $$\Th(1+\Phi) k=(1-\Psi)K(\Th(1+\Phi)k|E||\eta|).$$ So, the Cauchy integral of the measure $\Th(1+\Phi)k|E||\eta|$ is also divisible by $\Th$. However,
$$\Th(1+\Phi)k|E||\eta|=(1+\Phi)\frac g\Phi|E||\eta|=\frac{1+\Phi}{\Phi}\bar{E}\eta.$$ Thus, we obtained a non-constant bounded function $(1+\Phi)\bar{\Phi}$ such that the Cauchy integral of $(1+\Phi)\bar{\Phi}\bar{E}\eta$ is also divisible by $\Th$. This is in contradiction with the fact that $\eta$ is an extreme point, see Lemma \ref{extremel} and the comment before the statement of the theorem.  Therefore, $h$ is outer.

To show that $\Psi\bar{\Th}\bar{h}=h$ is simple. Just notice that $$\Psi\bar{\Th}\bar{h}=\Psi\bar{g}=(1-\Psi)K(\bar{\Th}\bar{E}\eta)=(1-\Psi)K(\bar{\Th}g|E||\eta|)=(1-\Psi)K(h|E||\eta|)=h.$$ Therefore, $$e^{i(\psi(t)-2\phi(t))}=\bar{\Psi}(t)\Th(t)=\frac{\bar{h(t)}}{h(t)}=\frac{e^{h(t)-i\tilde{h}(t)}}{e^{h(t)+i\tilde{h}(t)}}=e^{-2i\tilde{h}(t)},$$ where $\psi(t)$ denotes the increasing argument of the inner function $\Psi$. Thus, $\psi(t)-2\phi(t)\in \tilde{L^1}(\Pi)$.

Let $\Lambda:=\supp \eta_+\subset A$. Recall that $\supp\eta=\{\Psi=1\}$. By Lemma \ref{Krein}, the supports of $\eta_\pm$ are interlaced and hence $\abs{4\pi n_{\L}(t)-\psi(t)}\leq 4\pi$. So, we also have that $4\pi n_{\L}(t)-\psi(t)\in \tilde{L^1}(\Pi)$ and we get the desired conclusion $$2\pi n_{\L}(t)-\phi(t)\in \tilde{L^1}\left(\Pi\right).$$

\end{proof}

The last statement can also be formulated  as follows:

\begin{corollary}\label{poissonsummablecor} Let $\BB_E$ be a regular de~Branges space and  let $\phi(t)$ be the corresponding phase function. If $\mu$ is a  non-zero measure that annihilates $\BB_E$ then the support of $\mu_+$ ($\mu_-$) contains a sequence $\L$ whose counting function satisfies $2\pi n_{\L}(t)-\phi(t)=\tilde{h}(t)$ for some $h\in L^1(\Pi)$.
\end{corollary}

In the case $\BB_E=\PWa$ we have a more precise result in the form of Theorem \ref{indeterminate} that will be proved below.

\section{Measures annihilating $\PWa$.}

Note that in the case $E=S^a$ we have $W_E=\const$.
Since $F(x)=o(1)$ as $x\to\pm\infty$ for all $F\in\PWa$, $W=1$ is an $E$-weight that can be used in all statements from the last section.
In what follows we write $\MM_a(A,B)$ instead of $\MM^1_{PW_a}(A,B)$, i.e. $\MM_a(A,B)$ is the set of real finite measures $\mu$ of norm at most 1
that annihilate $\PWa$ and satisfy $\supp\mu_+\subset A,\supp\mu_-\subset B$.

\indeterminate*

\begin{proof} If $\mu$ is $a$-indeterminate, then there exists a finite positive measure $\nu\neq\mu$ such that $\mu-\nu$
annihilates $\PWa$. Put $A=\supp\mu$, $B=\supp\nu$ and consider $\MM_a(A,B)$. Let $\eta$ be a non-trivial extreme point of
$\MM_a(A,B)$. Let $\L=\{\lan\}=\supp \eta$.
Since $\eta$ is extreme, it is a zero set of an entire function from the Krein class and therefore $\L$ must be
a regular sequence with density $a/\pi$.
By  lemma \ref{Krein}, $\L'=\{\l_{2n}\}\subset \supp\mu$.

The main theorem of \cite{PolGap} implies that $\L$ contains
a $(\frac{a}{\pi}-\e)$-uniform subsequence for any $\e>0$. It follows that $\L'$ contains a $(\frac{a}{2\pi}-\e)$-uniform subsequence for any $\e>0$.
Using lemma~\ref{uff}, one can select a  $\frac{a}{2\pi}$-uniform subsequence of $\L'$.
 \end{proof}

The following result represents a refinement of de~Branges Theorem~66 in~\cite {dBr}.

\begin{theorem} Let $A$ and $B$ be disjoint closed subsets of $\R$. The necessary and sufficient condition for $ \MM_a(A,B)$ to contain a non-zero measure is that there exists a sequence $\Lambda=\{\lambda_n\}$, satisfying  $ \{\lambda_{2n}\}\subset A, \{\lambda_{2n+1}\}\subset B$, which is the zero set of an entire function in the Krein class $\KK_a$.
\end{theorem}

\begin{proof}  Assume that $\MM_a(A,B)$ contains a non-zero measure. It follows from Lemma~\ref{Krein} that  there exists an entire function $G(z)$ in the Krein class $\KK_a$ with zero set $\Lambda=\{\lambda_n\}$ satisfying  $ \{\lambda_{2n}\}\subset A, \{\lambda_{2n+1}\}\subset B$. Conversely, if such function exists define a signed discrete measure $\sigma$ supported on $\L$ simply by  $\sigma(\{\lambda_n\})=1/G'(\lambda_n)$. Using the properties of $G(z)$ it is easy to show that this measure (or perhaps $-\sigma$) belongs in $\MM_a(A,B)$.
\end{proof}

To prove our oscillation result below we will need the following technical lemma. It basically says that any subsequence of real zeros of a given function $F\in\PWa$ can be replaced by double zeros without changing the exponential type and still keeping the boundedness on the real line.

\begin{lemma} Let $F\in\PWa$ with infinitely many real zeros and let $\Lambda=\{\lan\}$ be a subsequence of the real zeros of $F$ indexed so that
$$\dots<\l_{-2}<\l_{-1}<0<\l_1<\l_2<\dots $$
Assume that $\sup_n \l_{n+1} -\l_n<\infty$.
Put $\gamma_n=(\l_{2n-1}+\l_{2n})/2$ for $n$ positive and $\gamma_n = (\l_{2n}+\l_{2n+1})/2$ for $n$ negative.

Then the entire function $G(z)$ defined by
$$G(z)=\frac{\prod_{\gamma\in\Gamma} \left(z-\gamma\right)^2}{\prod_{\l\in\L} \left(z-\l\right)}F(z)$$
is bounded on $\R$.

\end{lemma}

\begin{proof} Let us assume for the time being that the products in the definition of $G$ converge (we address convergence at the end of the proof).
 Define $u(t)=2n_{\G}(t)-n_{\L}(t)$.
On the real line we have
$$\log\abs{G(t)}=-\ti u(t) +\log\abs{F(t)}+\const.$$
Denote $I_n=(\l_{2n-1},\l_{2n})$ for $n$ positive and $I_n = (\l_{2n},\l_{2n+1})$ for $n$ negative.
Let $x$ be a real point.
Since $u$ forms an atom on each $I_n$, its harmonic conjugate can be calculated as
$$\ti u(x)=\frac 2\pi \int\frac{u(t)}{x-t}dt+\const,$$
i.e. the integral on the right converges.
 Let $U$ be the union of $I_n$ that intersect the interval $(x-1,x+1)$.
 Then for any $I_n\not\in U$, $n>0$,
 $$\left|\int_{I_n}\frac{u(t)}{x-t}dt\right|=\left|\ln\frac{(\l_{2n-1}-x)(\l_{2n}-x)}{(\gamma_n-x)^2}\right|=$$
 $$\left|\ln\left(1-\frac{(\l_{2n-1}-\l_{2n})^2}{4(\gamma_n-x)^2}\right)\right|<D |I_n|^2 dist^{-2}(x,I_n)$$
 for some positive constant $D$ depending only on $C=\sup_n\abs{\l_{n+1} -\l_n}$.
 A similar estimate can be written for $n<0$. Hence
$$\left|\int_{\R\setminus U}\frac{u(t)}{x-t}dt\right|=\left|\sum_{I_n\not\in U}\int_{I_n}\frac{u(t)}{x-t}dt\right|
<C_1.$$
Suppose that $U=I_n\cup I_{n+1}\cup ...\cup I_{N}$ for some $0<n\leq N$. Let $x>0$, $x\in [\gamma_k,\gamma_{k+1}]$, $n\leq k\leq N$.
 Notice that $\L\cap [\gamma_k,\gamma_{k+1}]=\{\l_{2k},\l_{2k+1}\}$ and
$$F(x)/(x-\l_{2k})(x-\l_{2k+1})<C_2$$
because $F(x)$ is a $\PWa$ function and $\l_{2k},\l_{2k+1}$ are its zeros.
Also notice that
$$\abs{\frac{\prod_{n\leq m\leq N} \left(x-\gamma_m\right)^2}{\prod_{2n-1\leq m\leq 2N,\ m\neq 2k, 2k+1} \left(x-\l_m\right)}}<(1+C)^2.$$
Altogether we obtain
$$ \abs{G(x)}=\abs{\frac{\prod_{\gamma\in\Gamma} \left(x-\gamma\right)^2}{\prod_{\l\in\L} \left(x-\l\right)}F(z)}=
$$$$\abs{
\exp{\left[-\frac 2\pi \int_{\R\setminus U}\frac{u(t)}{x-t}dt\right]}\frac{\prod_{n\leq m\leq N} \left(x-\gamma_m\right)^2}{\prod_{2n-1\leq m\leq 2N} \left(x-\l_m\right)}F(z)}\leq e^{C_1}C_2(1+C)^2<\const.$$

Note that our estimates of $\ti u$ show that the products in the definition of $G$ converge for $z\in\R\setminus\L$. Via  standard argument,
convergence on the line implies convergence on $\C\setminus \L$.

\end{proof}

Now we can prove our main result stated in the introduction. Recall that for a pair of closed sets of real numbers $A$ and $B$ the
gap characteristic is defined as
$$G(A,B)=\sup\{a>0: \MM_a(A,B)\neq\{0\}\}.$$

\oscillation*

\begin{proof} Assume that $\MM_a(A,B)$ contains a non-zero measure. By the previous theorem, it then must contain a non-zero extreme measure $\nu$ with discrete support $\L=\{\lan\}$ such that $\{\lambda_{2n}\}\subset A, \{\lambda_{2n+1}\}\subset B$.
By the main theorem from \cite{PolGap}, $\L$ contains a $(a/\pi-\e)$-uniform subsequence $\L'$. One can then choose a $(a/\pi-2\e)$-uniform subsequence $\L''$ of $\L'$ that alternates properly between the sets $A$ and $B$.
Hence we are done with  one of the inclusions.

In the opposite direction, let $\Lambda=\{\lan\}$ be a $d$-uniform sequence with $\{\lambda_{2n}\}\subset A, \{\lambda_{2n+1}\}\subset B$. Let $\e>0$ be arbitrary. We will construct a finite signed measure $\sigma \in \MM_{\pi d-2\epsilon}(A,B)$ which will finish the proof. From~\cite{PolGap} we know first that there exists a finite signed measure $\sigma_1$ supported on $\Lambda$ with a spectral gap of the size $2\pi d-\e$. Using the extreme point procedure from above if needed we can assume that this measure $\sigma_1$ oscillates between the consecutive points in its support. More precisely, if we denote the support of $\sigma_1$ by $\L':=\{\l'_n\}\subset \Lambda$, we have $\sigma_1(\l'_n)\sigma_1(\l'_{n+1})<0$ for all $n$. This measure however may not be in $\MM_{\pi d-\epsilon}(A,B)$ since we do not know that $\L'$ alternates between the sets $A$ and $B$.

Consider the intervals $(\l'_n, \l'_{n+1})$ which contain odd number of points from $\Lambda$. These are exactly the bad intervals since in this case $\l'_n$ and $\l'_{n+1}$ are both in $A$ or both in $B$ even though $\sigma_1$ has opposite sign at these points. To fix this form a sequence $\Gamma$ by picking one point from $\L$ in each such interval. Then $D_{BM}^+(\Gamma)\leq D_{BM}^+(\L\setminus \L')\leq \e$. By the Beurling-Malliavin theorem there exists an entire function $F(z)\in \PP\WW_{\e}$ which vanishes at $\Gamma$. We may assume that $F$ is real on the real line (otherwise take $F+F^\#$).

If the real zero set of $F(z)$ is exactly $\Gamma$, then we will be done by taking $d\sigma:=Fd\sigma_1$. However, $F(z)$ may have additional real zeros. Denote by $\Gamma'$ these zeroes. Without loss of generality we can assume that $\Gamma'$ is an infinite sequence with bounded gaps (this condition will be needed to apply the last lemma).
Otherwise, we can add an arithmetic progression of density $\epsilon$ to the zero set of $F(z)$.

Form the sequence $\Delta=\{\delta_n\}$ of the arithmetic means of consecutive pairs in $\Gamma'$, like in the last lemma.
We can now apply the last lemma to replace the unwanted real zeros of $F(z)$ with double zeros and obtain an entire function $G(z)\in\BB_{\e}$ such that $G$ is real and bounded on $\R$,
the zero set of $G$ on $\R$ is exactly  $(\Gamma\setminus\Gamma')\cup \Delta$ and $G$ has double zeros on $\Delta$.
Consider the measure $d\sigma=Gd\sigma_1$. Since $G$ is bounded and real on $\R$ this is a finite signed measure. The fact that $G$ is of exponential type no greater than $\e$ implies that $\sigma$ has a spectral gap of the size $2\pi d-3\e$. In addition, in each of the bad intervals $(\l'_n,\l'_{n+1})$ the function $G$ has exactly one simple zero and the rest of its zeros are double zeros. Therefore, $\sigma$ has opposite sign point masses at $\l'_n$ and $\l'_{n+1}$ and has the same good behavior as $\sigma_1$ on the rest of $\L'$. Thus, $\sigma\in \MM_{\pi d-2\epsilon}(A,B)$.
\end{proof}

As a consequence of the oscillation theorem we derive the following result stated in the introduction.

\DetGap*

\begin{proof}  Let $Det(X)=d$. By definition, for any $\e>0$ there exists a $(d-\e)$-indeterminate measure $\mu$ with a support included in $X$. By Theorem~\ref{indeterminate} we have that $\supp \mu$ contains a $(d-\e)/\pi$-uniform sequence. Therefore, by the gap theorem in~\cite{PolGap} we have that $G(X)\geq 2(d-\e)$. Hence, $G(X)\geq 2Det(X)$.

To prove the other inequality let $G(X)=a$. By the gap theorem, it follows that $X$ contains an $(a-\e)/\pi$-uniform sequence $\G=\{\gamma_n\}$.
 Let $\L=\G\cup \G'$, where $\gamma'_n=(\gamma_n+\gamma_{n+1})/2$. Then $\L$ is a $2(a-\e)/\pi$-uniform sequence and therefore, by theorem~\ref{oscillation},
 $ \MM_{2 a-10\e}(\G,\G')$ contains a non-zero measure. The positive part of this signed measure is $2(a-5\e)/\pi$-determinate and supported on $\G\subset X$. Thus,
  $2Det(X)\geq G(X)$.
\end{proof}

\section{Sign changes of measures with spectral gap}\label{last}

We say that a real measure $\sigma$ has at least one sign change on an interval $(a,b)$ if  there exist Borel sets $P, N\subset [a,b]$ such that $\sigma(P)>0$ and $\sigma(N)<0$. In \cite{EN, EN1}
the number of sign changes of $\sigma$ on an interval $(a,b)$ is  was defined as the minimal degree of a polynomial $p$ for which $pd\sigma$ is a positive measure on $[a,b]$. Clearly one can use either one of these definitions in the statements below.

As a consequence of our oscillation theorem we obtain an improvement of the main result in~\cite{EN, EN1}:

\begin{restatable}{theorem}{Eremenko} \label{Eremenko} If $\sigma$ is a nonzero signed measure with spectral gap $(-a, a)$  then there exists an $a/\pi$-uniform sequence $\{\lambda_n\}$ such that  $\sigma$ has at least one sign change in every $(\lambda_n, \lambda_{n+1})$.
\end{restatable}
\noindent

\begin{proof} Let $A=\supp\mup$, $B=\supp\mum$.  Here, as usual, $\mup$ and $\mum$ denote the positive and negative parts of $\sigma$ in the canonical Jordan decomposition $\sigma=\mup-\mum$. Then $\sigma\in\MM_a(A,B)$. It follows from Lemma~\ref{KrMil} that $\MM_a(A,B)$ contains an extreme point (measure).  By Theorem~\ref{oscillation} this measure is discrete and its support contains a $(a/\pi-\e)$-uniform subsequence for any $\e>0$. Using lemma~\ref{uff}, we can choose an $a/\pi$-uniform subsequence $\L=\{\lambda_n\}$ such that  $\{\lambda_{2n}\}\subset A, \{\lambda_{2n+1}\}\subset B$. Now one can move the points $\lan$, if necessary, so that the sets $A=\supp\mup$ and $B=\supp\mum$ intersect each
interval $[\lambda_n, \lambda_{n+1}]$ and the sequence $\L$ remains $a/\pi$-uniform.
Then $\sigma$ has at least one sign change on each of the intervals $(\lambda_n, \lambda_{n+1})$.
\end{proof}

Next, we  deduce the main result of~\cite{EN, EN1} from corollary~\ref{poissonsummablecor}.

\ErNov*

\begin{proof} Let $\Lambda=\{\lambda_n\}$ be the sequence from corollary \ref{poissonsummablecor}. It is clear that $s(r,\sigma)\geq n_{\Lambda}(r)$, where $n_{\Lambda}(r)$ is the  counting function of $\Lambda$. The fact that $\pi n_\Lambda(x)-\frac a2 x$ is harmonic conjugate of a Poisson-summable function implies that
$$\Pi(\{|\pi n_\Lambda(x)-\frac a2 x|>t\})=o(t)\ \ \textrm{ as }t\to\infty.$$
Since $n_\L$ is a growing function, the last relation implies that $\pi n_\Lambda(x)-\frac a2 x=o(x)$. Therefore $$\lim_{r\to\infty}\frac{n_{\Lambda}(r)}{r}= \frac{a}{2\pi}.$$
\end{proof}
\begin{remark*}
Notice that to prove the last statement we did not use any difficult results of Harmonic Analysis such as the Beurling-Malliavin theory or the
results of \cite{PolGap, PolType}. All in all, our proof seems to be shorter and more elementary than the original proof given in~\cite{EN}.
\end{remark*}

\end{document}


TITLE: How do I show that a distribution is locally p-integrable?
QUESTION [3 upvotes]: Let $\Omega \subseteq \mathbb{R}^n$ (open) and $u \in \mathcal{D}'$
  be a distribution, that has a distributional derivative
  which is in $L^p(\Omega)$ (for some $p \geq 1$).
Show that $u \in L^p_{loc}(\Omega)$

...where $L^p(\Omega)$ is defined as in
https://en.wikipedia.org/wiki/Lp_space#Lp_spaces
and $L^p_{loc}(\Omega)$ is defined as in:
https://en.wikipedia.org/wiki/Locally_integrable_function#Generalization:_locally_p-integrable_functions

My thoughts:
1) For notational simplicity let's call $Du = D_i u$ and $\phi' = \partial_{x_i}\phi$ for some $i \in \lbrace 1,2 \dots n\rbrace$
Then by definition of distributional derivatives, we have that:
$\int_{\Omega} (Du) \phi dx=-\int_{\Omega} u \phi' dx, \forall \phi \in C^{\infty}(\Omega)$ with compact support
2) Let also $K \subset \Omega$ (compact). We need to show that $\int_K \vert u \vert^p dx < \infty$
3) A simple density argument of the test functions should help us go from $K$ to $\Omega$ so this shouldn't be a big deal
4) We can also use something like Holder's inequality (and the fact that $Du \in L^p$) to show something among the lines of:
$\vert \int_{\Omega} (Du) \phi  dx \vert \leq \int_{\Omega} \vert (Du) \phi \vert dx \leq (\int_{\Omega} \vert Du \vert^p)^{\frac{1}{p}} (\int_{\Omega} \vert \phi \vert^q)^{\frac{1}{q}}< \infty$
5) Then from the definition of $Du$ we also have that:
$\vert \int_{\Omega} u \phi'  dx \vert = \vert \int_{\Omega} (Du) \phi  dx \vert$
6) But how do we find an inequality that involves $\int \vert u \vert ^p$ and $\int\vert u \vert$ (in this order) to fill in the chain of inequalities? (Assuming of course, that such a "chain" is the way to go in this problem)
7) Another idea I am toying with (see update below), is to use some sort of a density argument to go from - say - $\int\vert u \vert ^p$ to $\int\vert u \vert \phi_n$ for an appropriate sequence of $\phi_n$. Not sure if this is possible though.
Update:

Question: What density arguments are available for locally integrable
  functions? For instance, since $u \in L^1_{loc}$, is it possible to
  find a sequence $\phi_n$ in $C^{\infty}_0(\Omega)$ or even in
  $C^{\infty}(\Omega)$ such that:
$$\lim\limits_{n \rightarrow \infty}\phi_n = \vert u \vert^{p-1}$$
I think that would complete the chain of inequalities and thus the proof.

REPLY [3 votes]: I will outline the solution for the case $n=1$ and $p=2$, and if you can not generalize it to $p\ge1$ and $n\ge 1$, ask in comments, we will dive into the murky details. Without further ado:
Let $K$ be a non-trivial compact set, $\phi$ be a test function with support in $K$ and $u'\in L^2(K)\cap D'(\Bbb R)$.
We can say that by Cauchy-Bunyakovskiy-Schwarz inequality
$$|\langle u',\phi\rangle| = \left|\int_K u'(x)\phi(x)dx\right|\le \|u',L^2(K)\|\|\phi,L^2(K)\|$$
Then again, by Poincaré's inequality, for a constant $C$ that depends only on $K$, $$\|\phi,L^2(K)\|\le C \|\phi',L^2(K)\|.$$
This gives us
$$|\langle u,\phi'\rangle| \le C\|u',L^2(K)\| \|\phi',L^2(K)\| .$$
Therefore, we can already say that $u$ is a continuous linear operator over the set $L^2(K)\cap \{f: \int_Kf=0\}$. We can extend it to the whole $L^2(K)$ by noticing that if $\psi$ is a test function supported in $K$ with $\int_k\psi=1$, then for any $\phi$ as above we have
$$\langle u,\phi\rangle = \left\langle u,\phi-\psi\int_K\phi\right\rangle + \left(\int_K\phi\right)\langle u,\psi\rangle.$$
The test function $\phi-\psi\int_K\phi$ is a derivative of a test function (easy exercise), hence you can apply the estimations above; and the term $\left(\int_K\phi\right)\langle u,\psi\rangle$ is bounded by $C_1 \|\phi,L^2(K)\|$ (essentially, $C_1$ acts like an arbitrary constant when you find an antiderivative). Thus, $\langle u,\phi\rangle$ is bounded by $$C\|u',L^2(K)\|  \left\| \phi-\psi\int_K\phi,L^2(K)\right\| + C_1 \|\phi,L^2(K)\|.$$
It easy to see that $u$ now can be seen as continuous linear functional over $L^2(K)$. Therefore, $u$ itself belongs to $L^2(K)$.
Notes for the  generalizations: the Poincaré's inequality works for $p\in [1,\infty)$ and $n\ge 1$, so that part is ok. The representation theorem definitely works for $p>1$, some tinkering might be necessary to cover the case $p=1$.


TITLE: What is the analogue of simple prime closed geodesic for prime numbers?
QUESTION [30 upvotes]: The prime geodesic theorem (of Margulis?) states that on a compact surface of (constant?) negative curvature, the number of prime closed geodesics of length at most $L = \log x$ is approximately $e^L/L = x/\log x$ as $x$ grows. This is commonly viewed as an analogue of the prime number theorem.
Reading the report on Mirzakhani's work in relation to her receiving the Fields medal, I learned that she proved that the number of simple (i.e. non-self-intersecting) prime closed geodesics of length at most $L$ is of the order of $L^{6g-6}$, where $g$ is the genus of the surface, as $L$ grows. What would be the analogue of simple prime closed geodesic in the context of prime numbers and what would one expect the analogue of Mirzakhani's result to be?
Edit: My colleague Alan Reid suggested to look at the universal cover to distinguish simple from non-simple closed geodesics. Since I want to keep the analogy with arithmetic, instead it's better to look at finite covers. Is it fair to say that simple closed geodesics are more likely to become non-prime in finite covers? That would suggest that "simple" primes would be those more likely to split in finite extensions.
Edit 2: My history is sloppy. See GH's comment and Igor's answer.

REPLY [15 votes]: First of all, the prime geodesic theorem is due to Sarnak, in Margulis' result the geodesics need not be prime. Secondly, for constant curvature, the result goes back to Huber and Selberg, with precise error terms. Margulis extended this to variable negative curvature (with no error term), and that was sharpened (at least theoretically) by Dolgopyat to show that there is an exponentially decaying error term, but the exponent is not explicit at all.
Thirdly, the order of growth result for simple geodesics is due not to Maryam, but to yours truly:
%0 Journal Article
%A Rivin, Igor
%T Simple curves on surfaces
%J Geom. Dedicata
%V 87
%D 2001
%N 1-3
%P 345--360
%@ 0046-5755
%L MR1866856 (2003c:57018)
%R doi:10.1023/A:1012010721583
%U http://dx.doi.org/10.1023/A:1012010721583
Maryam proved an asymptotic result, which (as is not surprising to number theorists) is weaker in practice (my result has explicit bounds, hers does not), but certainly conceptually better (with modern technology - mixing of Teichmuller geodesic flow - one can presumably show the existence of error term a la Dolgopyat, but nothing explicit).
Lastly, and getting back to the actual question, the connection between prime numbers and closed geodesics goes via the Selberg zeta function, and its analogy to the Riemann zeta function. So, I will not speak of prime numbers, but the lattice point counting. It turns out that the counting of all closed geodesics and simple closed geodesics is conceptually very similar, despite completely different result that one gets: counting closed geodesics corresponds to counting points of an orbit of a Fuchsian group (for example, $PSL(2, Z)$) in a disk in the hyperbolic plane (that was what Selberg counted initially). The insight is that simple closed curves lie in a finite set of orbits of the mapping class group, and so counting simple closed geodesic reduces to lattice point counting for the mapping class group on Teichmuller space. I am not sure what the number-theoretic analogue would be (that is, arithmetic geometers certainly study moduli, but I am not sure what easy-to-understand things this counts). Notice that the way Huber/Selberg count lattice points in symmetric spaces is via harmonic analysis, and so what is needed for a parallel theory for Teichmuller space is developing tools for analysis on Teichmuller/moduli spaces. Mirzakhani made the first step in this, by combining Greg McShane's amazing identity (or an extension thereof) with a simple idea of Siegel. This allowed her to integrate the constant function. A baby step, to be sure (an anecdote: I applied for an NSF grant once to develop analysis on Teichmuller space. It was turned down as "too narrow").


TITLE: Is the domain of a finite set a finite set again without the axiom of choice?
QUESTION [0 upvotes]: I don’t think that the following can be proven without $\sf AC$.

Let $A$ be a finite set (i.e. it has a bijection with the set $\{n\in\mathbb{N} : 1\le n\le N\}$ for some $N\in\mathbb{N}$) of Kuratowski ordered pairs (i.e. $(a,b):=\{\{a\},\{a,b\}\}$ for all $a,b$). Then the set of the first coordinates of ordered pairs in $A$ (i.e. the set $\{a\in\bigcup\bigl(\bigcup A\bigr) : (a,b)\in A \text{ for some b}\}$  is finite.

What do you think?

This is a question that I came up with, so it might lack some precision. Feel free to edit.

REPLY [4 votes]: It is indeed provable without choice. It's useful to rephrase it as follows:

Suppose $X$ is finite and $f:X\rightarrow Y$ is surjective. Then $Y$ is finite.

Fix a bijection $h:X\rightarrow \{1,...,n\}$ for some $n\in\mathbb{N}$. For each $y\in Y$ let $$k(y)=\min\{h(x): f(x)=y\}.$$ Then $k$ is an injection from $Y$ to $\{1,...,n\}$. Now just "collapse" it: consider the map $$\hat{k}:Y\rightarrow\{1,...,n\}: y\mapsto \vert\{z\in Y: k(z)\le k(y)\}\vert.$$ It's not hard to show that $\hat{k}$ is a bijection between $Y$ and $\{1,...,m\}$ for some $m$.


TITLE: Prove that in a triangle $ABC$ $\forall P, P\neq M_{a},M_{b} ,M_{c}:min\{ \frac{|PA|}{|PM_{a}|},\frac{|PB|}{|PM_{b}|},\frac{|PC|}{|PM_{c}} \}\le 2.$
QUESTION [0 upvotes]: Suppose a triangle $ABC$. Define $M_{a},M_{b} ,M_{c}$ as the midpoints of the sides $a,b,c$ respectively. Then prove that for any point $P$ different from $M_{a},M_{b} ,M_{c}$, the smallest ratio of magnitudes of lines from $P$ to $M_{x}$ and from $P$ to $x$ is $\le 2$.
I have had certain success in prooving it for special cases, such as triangle $ABP$ being isosceles, but I cannot see a way to generalize this properly. I would think this is a pretty important (or at least very interesting result) but I have not seen it anywhere before (or its proof) so I'm intrigued to hear your ideas.

REPLY [0 votes]: It seems to me (I may be wrong) that the two geometric solutions given to this problem so far have not been very clear to beginners so I want to give
an algebraic solution.
The triangle $\triangle{ABC}$ can be given by its vertices $A (0,0), B (a, 0), C (b, c)$ with the condition of $c\ne0$. The locus of the points $P(x,y)$ such that $\dfrac{\overline{PA}}{\overline{PM_A}}=2$ is a circle and the equations of the three circles involved, noted $\Gamma_A,\Gamma_B,\Gamma_C$  are
$$\Gamma_A:\hspace 0.5cm 3x^2-4(a+b)x+3y^2-4cy+(a+b)^2+c^2=0..........(1)\\\Gamma_B:\hspace 0.5cm 3x^2-2(2b-a)x+3y^2-4cy+b^2+c^2-a^2=0..........(2)\\\Gamma_C:\hspace 0.5cm3x^2-2(2a-b)x+3y^2+2cy+a^2-b^2-c^2=0..........(3)$$
The set union of intersections $X=(\Gamma_A\cap\Gamma_ B)\cup(\Gamma_A\cap\Gamma_ C)\cup(\Gamma_B\cap\Gamma_ C)$ turns out to be a kind of irregular trifolium and we prove algebraically that the only point belonging to the three foliums is the centroid $G$ of the triangle (which is graphically evident) so that the only solution of the system of the three equations above is given by the known formula $G=\left(\dfrac{a+b}{3}, \dfrac{c}{3}\right)$.
This is straightforward because $(1)-(2)$ gives a single possible value $x=\left(\dfrac{a+b}{3}\right)$ and $(1)-(3)$ and $(2)-(3)$ give a linear system in $x$ and $y$ whose solution is $(x,y)=\left(\dfrac{a+b}{3}, \dfrac{c}{3}\right)$.
Now, noting $\overset{\circ}\Gamma_A$ the interior of the circle $\Gamma_A$,  by construction of $\Gamma_A$ we have $\dfrac{\overline{PA}}{\overline{PM_A}}\gt2\iff P\in\overset{\circ}\Gamma_A$ and this is so for the other two circles. It follows that for two of $\left\{ \dfrac{|PA|}{|PM_{a}|},\dfrac{|PB|}{|PM_{b}|},\dfrac{|PC|}{|PM_{c}} \right\}$ be greater than $2$ it is necessaire that the point $P$ belongs to the interior of the intersection of two circles.
For example, if $P(x,y)\in\overset{\circ}\Gamma_A\cap\overset{\circ}\Gamma_ B$ then $\dfrac{\overline{PA}}{\overline{PM_A}}\gt2$ and $\dfrac{\overline{PB}}{\overline{PM_B}}\gt2$ but in this case $P$ must be outside the interior of $\Gamma_C$ since $\{G\}=\Gamma_A\cap\Gamma_ B\cap\Gamma_ C$ and $P\ne G$ (because $\dfrac{\overline{GA}}{\overline{GM_A}}=2)$. This means that  $\dfrac{\overline{PC}}{\overline{PM_C}}\lt2$.
We are done.


TITLE: Is work done dependent on velocity?
QUESTION [5 upvotes]: If I lift an object off the ground to a particular height at a particular velocity is the work done in this action the same as if I lifted the object to the same height but at a greater/lower velocity? (Ignoring friction but taking into account gravitational pull on the object)
Thanks for any explanations.

REPLY [0 votes]: Let’s say there are 2 boxes, same mass same volume same everything, needing to be brought to the same shelf. Box A will be lifted in 10 seconds while Box B will be lifted in 5 seconds. It may seem as though since the second one needs more force to push the box faster and more, the work done is unrelated.
Force = mass x acceleration
In Box A, let’s say the mass is 10kg, and there is no acceleration, it has a constant speed of 50cm/s maybe but there is no added acceleration, so it remains as 10.
For Box B the mass is the same, but the speed now is 1m/s, however, the acceleration is still the same, it moves at a constant speed so the force in both is still 10kg x 10m/s^2.


TITLE: Algebraic number theory book with answers to selected questions for self-study?
QUESTION [1 upvotes]: Can anyone recommend some easy to follow algebraic number theory books with answers (hints) to selected questions for self-study?
If it has no answers to questions, but if you know that some professors choose the book as a textbook and post answers to selected questions online (as homework solutions), that will be OK too. In this case, can you please give the name of the book and the URL of the course/homework?

REPLY [0 votes]: This is several years too late for the OP, but maybe it will help others with a similar question. Algebraic Number Theory by Frazer Jarvis is a very good introductory text with extensive hints and solutions in the appendix. See the MAA review.


TITLE: Why do lighter isotopes evaporate faster than heavier isotopes?
QUESTION [4 upvotes]: I have just read that we are able to estimate the temperature of the earth thousands of years ago by measuring the ratio of certain isotopes present in ice cores that froze thousands of years ago.
My understanding of this idea is that $\mathrm{H_2O}$ molecules containing lighter isotopes (i.e. with $\mathrm{^{16}O}$) evaporate quicker than those molecules that contain heavier isotopes (i.e. with $\mathrm{^{18}O}$) and that this difference in evaporation rate exhibits a temperature-dependent relationship.
My problem is that the equipartition theorem says that each molecule should contain an energy of $\frac{1}{2}kT$ per degree of freedom regardless of the molecular mass. If we assume that all the liquid $\mathrm{H_2O}$ molecules roughly obey a Maxwell-Boltzmann distribution, then the heavier isotopes will be moving slower than the lighter isotopes (on average at a given temperature) but their greater mass will mean that all isotopes will still possess the same average kinetic energy (in accordance with the equipartition theorem). So if all the water molecules possess the same kinetic energy on average (regardless of the isotopes they contain), why do those molecules with lighter mass preferentially evaporate? I understand that they are moving quicker than their heavier counterparts but the determination of whether a particle escapes a potential well and evaporates is its kinetic energy not its velocity.
In effect, I am asking why the speed of the molecules determines their proclivity to evaporate (to escape the potential well they inhabit due to intermolecular attraction) when it should be their kinetic energy that determines whether they escape their intermolecular potential well or not. If the evaporation rate is determined by kinetic energy (which I think it should be?) as opposed to velocity, then there shouldn't be any difference in the evaporation rates of different water molecules containing different isotopes because they all contain the same kinetic energy on average.

REPLY [6 votes]: You're right, the difference in evaporation rate has little to do with the speed of the molecules.  The correct explanation is that hydrogen bonding is slightly stronger in heavy water than in normal water.  As a result, heavy water molecules are slightly more attracted to one another than molecules of light water, and it is harder for a molecule of heavy water to evaporate.
Why is hydrogen bonding stronger in heavy water?  To get some intuition, let's consider the quantum harmonic oscillator.  If you solve the Schrödinger equation with a potential energy function of $V(x) = \frac12 kx^2$, you get solutions with energy $\hbar \omega (n + \frac12)$ where $n$ is an integer and $\omega = \sqrt{\frac{k}{m_r}}$.  Here $m_r$ is the reduced mass $\frac{m_1m_2}{m_1+m_2}$.  Thus, a quantum harmonic oscillator with higher masses at the endpoints (heavy water molecules) will have lower $\omega$ and a lower ground state energy $\hbar \omega/2$.  This means that more energy is needed to excite the bonding electrons from the ground state to the dissociation energy, so the bond should be slightly stronger.
I should mention that this explanation is rather oversimplified in a lot of important ways.  In fact, surprisingly many important questions about hydrogen bonding in water are still not well understood.  It is known that the hydrogen bond in a heavy water dimer is stronger than the one in a light water dimer (as we'd expect based on the argument above).  There is some theoretical evidence that this conclusion should generalize to similar small systems.  However, it's possible that the relevant physics may be different (and unfortunately, less elegant) for larger amounts of water.  One intriguing experimental study found that the hydrogen bond in light water is actually shorter than the hydrogen bond in heavy water, which would imply a stronger bond.  The study also reported that the number of hydrogen bonds per water molecule drops from $3.76 \pm
 0.1$ in heavy water to $3.62 \pm 0.1$ in light water because the geometric packing of the molecules is slightly different.  Perhaps this effect may be more important than the strength of the individual bonds.


TITLE: Do we have any way of knowing if natural phenomena in the real world follow the "Lipschitz Condition"?
QUESTION [5 upvotes]: Recently, I keep coming across terms containing "Lipschitz" pertaining to statistical models and machine learning. This includes terms such as  "p-lipschitz (rho), lipschitz convexity, lipschitz loss, lipschitz continuity, lipschitz condition, etc.
For example, the (famous) Gradient Descent algorithm that is used heavily in the field of machine learning to optimize the "loss functions" of Neural Networks, allegedly claims that it can provide a solution arbitrarily close to the true solution, provided an infinite number of iterations and that the "loss function" is "Lipschitz Continuous":

Although the mathematical definitions of the Lipschitz Condition are quite detailed and complicated, I think I was able to find a simple enough definition to understand for this question:

I am not sure if this is correct or not, but from an applied perspective : I have heard that the Lipschitz Condition is a desirable property for mathematical functions (and the gradients of these functions) because it suggests that "small changes in the input values of the function can not result in infinite values of the output of the same function".
My intuition is that if a function you are dealing with does not satisfy the Lipschitz Condition and can potentially grow in infinite size provided small changes in the inputs - this would suggest that this function has the ability to display very volatile, erratic and chaotic behaviors, making it fundamentally difficult unpredictable and unstable to deal with:

My Question: When it comes to complicated mathematical functions (e.g. functions that model natural phenomena in the real world), we can use the definitions of "convexity" (e.g. provided here) to find out if the function is convex or non-convex (i.e. if the function does not follow the required convexity conditions, it must be non-convex).
Can the same be said about real world functions obeying the Lipschitz Condition?
For example - if we look at the specific "loss function" being used in a specific neural network (e.g. specified number of layers, neurons, types of activation function, etc.), it might be possible to prove whether this "loss function" obeys the Lipschitz Condition.
But what about the original function that we are trying to approximate with the neural network itself?
At this point, are we just assuming that natural phenomena in the real world (e.g. ocean tides, economic patterns in the market, animal migration trajectories, etc.) are obeying the Lipschitz Condition - either in their su generis form, or for the sake of the machine learning algorithms to work?
Thanks!

REPLY [2 votes]: One condition where Lipschitz continuity naturally arises is if the function is differentiable with a bounded derivative. Let's say the function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and the derivative is bounded: $\left|\frac{df}{dx}\right|\le M$. As per Mean Value Theorem:
$$f(b)-f(a)=\frac{df}{dx}(\xi)(b-a)$$
for some $\xi\in(a,b)$. Now, applying modulus and using the above condition of boundedness, we get:
$$|f(b)-f(a)|=\left|\frac{df}{dx}(\xi)\right||b-a|\le M|b-a|$$
Notice if the condition is satisfied on the whole $[a,b]$, then it is satisfied on any subinterval $[x,y]\subset [a,b]$ for the same $M$, so:
$$|f(y)-f(x)|\le M|y-x|$$
for all $x,y\in[a,b]$, which is the Lipschitz condition.

One special case of this is of the function $f$ that is continuously differentiable on $[a,b]$ because, well, $\frac{df}{dx}$ is then continuous and thus necessarily bounded.
If you can now argue that, in most applications of calculus in physics and natural sciences, all real functions of a real variable are not only continuously differentiable once, but are normally continuously differentiable infinitely many times, in fact they are analytic - then they will all be Lipschitz functions on any closed finite interval due to the previous statement.


TITLE: What is the rate of growth of the sum of reciprocals of square roots of natural numbers?
QUESTION [3 upvotes]: We all know that $$ \sum_{k=1}^n \frac{1}{k} \sim \log(n).$$
What is the analagous result for $\sum_{k=1}^n k^{-1/2}$ ?
Many thanks for your help.

REPLY [0 votes]: $$
\frac \beta{k^\alpha} - \frac \beta{(k+1)^\alpha} \sim \frac {\alpha\beta}{k^{\alpha +  1}} = \frac 1{\sqrt{k}}
$$if $\alpha = -1/2$ and $\beta = -2$
now as $\sum \frac 1{\sqrt{k}} = \infty$:
$$
\sum_{k=1}^N \frac 1{\sqrt{k}}\sim \sum_{k=1}^N \frac \beta{k^\alpha} - \frac \beta{(k+1)^\alpha}
\sim -\frac \beta{N^\alpha} = 2\sqrt{N}
$$