# Spectral clumping for functions and distributions decreasing rapidly on a
half-line

Bartosz Malman  Division of Mathematics and Physics, Mälardalen University,
Västerås, Sweden<EMAIL_ADDRESS>
###### Abstract.

We demonstrate a phenomenon of condensation of the Fourier transform
$\widehat{f}$ of a function $f$ defined on the real line $\mathbb{R}$ which
decreases rapidly on one half of the line. For instance, we prove that if $f$
is square-integrable on $\mathbb{R}$, then a one-sided estimate of the form

$\rho_{f}(x):=\int_{x}^{\infty}|f(t)|\,dt=\mathcal{O}\big{(}e^{-c\sqrt{x}}\big{)},\quad
x>0$

for some $c>0$, forces the non-zero frequencies
$\sigma(f):=\\{\zeta\in\mathbb{R}:|\widehat{f}(\zeta)|>0\\}$ to clump: this
set differs from an open set $U$ only by a set of Lebesgue measure zero, and
$\log|\widehat{f}|$ is locally integrable on $U$. In particular, if $f$ is
non-zero, then there exists an interval on which $\log|\widehat{f}|$ is
integrable. The roles of $f$ and $\widehat{f}$ above may be interchanged, and
the result extends also to a large class of tempered distributions. We show
that the above decay condition is close to optimal, in the following sense: a
non-zero entire function $f$ exists which is square-integrable on
$\mathbb{R}$, for which $\sigma(f)$ is a subset of a compact set $E$
containing no intervals, and for which the estimate
$\rho_{f}(x)=\mathcal{O}\big{(}e^{-x^{a}}\big{)}$, $x>0$, holds for every
$a\in(0,1/2)$.

##  1\. Introduction

###  1.1. Fourier transform, its support and size

This note studies a certain manifestation of the uncertainty principle in
Fourier analysis, where a smallness condition on a function $f$ forces its
Fourier transform $\widehat{f}$ to be, in some sense, large. Vice versa,
smallness of $\widehat{f}$ forces $f$ to be large. In our context, the
smallness is defined in terms of a one-sided decay condition, and the
largeness in terms of the existence of a clump. This will be our moniker for
an interval on which the function has an integrable logarithm. We emphasize
that our results concern functions with a spectrum which might vanish on an
interval (commonly referred to as functions with a spectral gap), but for
which the spectrum should be large on some other interval.

We will use the following definition of the transform:

(1.1)
$\widehat{f}(\zeta):=\int_{\mathbb{R}}f(x)e^{-ix\zeta}\,d\lambda(x),\quad\zeta\in\mathbb{R}.$

Here $d\lambda(x)=dx/\sqrt{2\pi}$ is a normalization of the Lebesgue measure
$dx$ on $\mathbb{R}$. Then, the inversion formula is given by

(1.2) $f(x):=\int_{\mathbb{R}}\widehat{f}(\zeta)e^{i\zeta
x}\,d\lambda(\zeta),\quad x\in\mathbb{R}.$

For $p>0$, let $\mathcal{L}^{p}(\mathbb{R},dx)$ be the usual Lebesgue space of
functions $f$ for which $|f|^{p}$ is integrable with respect to $dx$. The
formula (1.1) can be interpreted literally only for
$f\in\mathcal{L}^{1}(\mathbb{R},dx)$. It is interpreted in terms of
Plancherel’s theorem in the case $f\in\mathcal{L}^{2}(\mathbb{R},dx)$, and in
order to state our most general results we will later need to interpret the
transform in the sense of distribution theory. The spectrum $\sigma(f)$ of a
function $f$ is the subset of $\mathbb{R}$ on which $\widehat{f}$ lives. Since
$\widehat{f}\in\mathcal{L}^{2}(\mathbb{R},dx)$ is defined only up to a set of
Lebesgue measure zero, so is the spectrum $\sigma(f)$ in this case. If we
accept making errors of measure zero (which we will), we may define the
spectrum as

$\sigma(f):=\\{\zeta\in\mathbb{R}:|\widehat{f}(\zeta)|>0\\},\quad
f\in\mathcal{L}^{1}(\mathbb{R},dx)\cup\mathcal{L}^{2}(\mathbb{R},dx).$

Note specifically that our definition of $\sigma(f)$ might not coincide with
the usual notion of closed support of the distribution $\widehat{f}$.

The uncertainty principle in Fourier analysis presents itself in plenty of
ways, and the excellent monograph [6] of Havin and Jöricke describes many of
its most interesting interpretations. One of them is the following statement,
well-known to function theorists. If $f\in\mathcal{L}^{2}(\mathbb{R},dx)$ is
non-zero and $\mathbb{R}_{-}$ is the negative half-axis, then we have the
implication

(1.3) $f(x)\equiv 0\text{ on
}\mathbb{R}_{-}\quad\Rightarrow\quad\int_{\mathbb{R}}\frac{\log|\widehat{f}(\zeta)|}{1+\zeta^{2}}\,d\zeta>-\infty.$

Here the extreme decay (indeed, vanishing) of $f$ on a half-axis implies
global integrability of $\log|\widehat{f}|$ against the Poisson measure
$d\zeta/(1+\zeta^{2})$. A fortiori, $\log|\widehat{f}|$ is integrable on every
interval $I$ of $\mathbb{R}$. Naturally, this is not typical. By Plancherel’s
theorem, every function in $\mathcal{L}^{2}(\mathbb{R},dx)$ is the Fourier
transform of some other function in the same space. So on the other extreme,
plenty of functions $f\in\mathcal{L}^{2}(\mathbb{R},dx)$ have a Fourier
transform which lives on sparse sets containing no intervals. This forces the
divergence of the logarithmic integral of $\widehat{f}$ over any interval. In
other words, plenty of functions admit no spectral clumps. The results of this
note give conditions under which such clumps form.

###  1.2. Condensation and sparseness of spectra and supports

We shall introduce our results at first in the context of the Hilbert space
$\mathcal{L}^{2}(\mathbb{R},dx)$. Here we can prove a claim which symmetric in
$f$ and $\widehat{f}$, and also we can argue for near-optimality of the
result. This is the content of CondensationTheorem and SparsenessTheorem. The
more general distributional clumping result is presented in
DistributionalClumpingTheorem.

######  Theorem A.

CondensationTheorem If $f\in\mathcal{L}^{2}(\mathbb{R},dx)$ satisfies the
estimate

(1.4)
$\rho_{f}(x):=\int_{x}^{\infty}|f(t)|\,dt=\mathcal{O}\big{(}e^{-c\sqrt{x}}\big{)},\quad
x>0$

for some constant $c>0$, then there exists an open set $U$ which coincides
with $\sigma(f)$ up to a set of Lebesgue measure zero, and for every $x\in U$
there exists an interval $I$ containing $x$ such that

$\int_{I}\log|\widehat{f}(t)|\,dt>-\infty.$

In other words, the one-sided decay condition (1.4) implies that $\widehat{f}$
lives on the union of the spectral clumps of $f$. Since the Fourier transform
is a unitary operation on $\mathcal{L}^{2}(\mathbb{R},dx)$, the roles of $f$
and $\widehat{f}$ may obviously be interchanged in the statement of
CondensationTheorem. Thus a one-sided spectral decay condition of $f$ implies
local integrability properties of $\log|f|$ on the set where $f$ lives. In
this form, the result encourages us to extend it to tempered distributions. We
shall do so in a moment.

The integrand in (1.4) may seem a bit unnatural in the context of square-
integrable functions $f$. It is more natural in the context of functions of
tempered growth appearing in DistributionalClumpingTheorem. Anyhow, we note
that one can prove that an estimate of the form
$\int_{x}^{\infty}|f(t)|^{2}\,dt=\mathcal{O}\big{(}e^{-c\sqrt{x}}\big{)}$ in
fact implies (1.4) for some slightly smaller $c$.

We can prove also that the condition (1.4) on the decay of $\rho_{f}$
appearing in CondensationTheorem is close to optimal. We do so by exhibiting a
non-zero function with rapid one-sided decay but sparse spectrum.

######  Theorem B.

SparsenessTheorem For every $b>0$, there exists a compact set
$E\subset\mathbb{R}$ contained in $[0,b]$ which contains no intervals, and a
non-zero entire function $f\in\mathcal{L}^{2}(\mathbb{R},dx)$ which satisfies

$\rho_{f}(x)=\mathcal{O}\big{(}e^{-x^{a}}\big{)},\quad x>0$

for every $a\in(0,1/2)$, and such that $\sigma(f)$ is contained within $E$.

After an initial reduction, the proof of this result follows ideas of
Khrushchev from [8]. Note that the function $f$ appearing in SparsenessTheorem
is entire by the virtue of having a spectrum $\sigma(f)$ of compact support.
More importantly, the condition on $E$ implies that $I\setminus E$ has
positive Lebesgue measure for every interval $I$, so we obtain

$\int_{I}\log|\widehat{f}(t)|\,dt=-\infty$

for every interval $I\subset\mathbb{R}$. This is in contrast to the conclusion
of CondensationTheorem. It follows that the exponent $a=1/2$ in estimates of
the form $\rho_{f}(x)=\mathcal{O}\big{(}e^{-x^{a}}\big{)}$ is critical for the
spectral clumping phenomenon.

As mentioned above, clumping statements makes sense for objects in a class
much wider than $\mathcal{L}^{2}(\mathbb{R},dx)$. Here is our distrbutional
result.

######  Theorem C.

DistributionalClumpingTheorem Let $f$ be a tempered distribution on
$\mathbb{R}$ which is a measurable function satisfying

$\int_{\mathbb{R}}\frac{|f(x)|}{(1+|x|)^{n}}\,dx<\infty$

for some $n>0$. If the distributional Fourier transform $\widehat{f}$ is an
integrable function on some interval $[A,\infty)$, and the estimate
$\rho_{\widehat{f}}(\zeta)=\mathcal{O}\big{(}e^{-c\sqrt{\zeta}}\big{)}$ holds
for all sufficiently large positive $\zeta$, then there exists an open set $U$
such that $f$ vanishes almost everywhere outside of $U$, and for each $x\in U$
there exists an interval $I$ containing $x$ satisfying

(1.5) $\int_{I}\log|f(t)|\,dt>-\infty.$

For instance, the result shows that a function
$f\in\mathcal{L}^{1}(\mathbb{R},dx)$ which lives on a sparse set containing no
intervals cannot satisfy even a one-sided spectral decay condition of the form
$\rho_{\widehat{f}}(\zeta)\lesssim e^{-c\sqrt{\zeta}}$. Note also that in this
extended form, our result includes the trivial but important examples such as
$f=1$ and $\widehat{f}=\delta_{0}$ (Dirac delta), the trigonometric functions
and the polynomials.

###  1.3. A converse result

The Beurling-Malliavin theory implies a partial converse result. If $f$ is a
locally integrable function on $\mathbb{R}$ which has a clump $I$ as in (1.5),
and a constant $c>0$ is given, then a bounded multiplier $m$ exists for which
$mf$ has a Fourier transform satisfying
$\rho_{\widehat{mf}}(\zeta)=\mathcal{O}\big{(}e^{-c\sqrt{\zeta}}\big{)}$ for
$\zeta>0$. To see this, recall that a smooth function $g$ supported in $I$
exists which satisfies the bilateral spectral decay $|\widehat{g}(\zeta)|\leq
e^{-c\sqrt{|\zeta|}}$, $\zeta\in\mathbb{R}$ (this simpler version of the
famous Beurling-Malliavin theorem is proved in [6, p. 276-277], and in fact we
may ensure an even faster bilateral spectral decay of $g$). There exists also
a bounded function $h\in\mathcal{L}^{1}(\mathbb{R},dx)$ which satisfies
$\sigma(h)\subset(0,\infty)$ and $|h(x)|=\min(|f(x)|,1)$ on $I$ (we use the
assumption that $I$ is a clump for $f$ and construct $h$ as in (2.7) below).
Then an argument similar to the one used in the proof of
ConvolutionFourierDecayLemma below shows that the function $\overline{h}g$
will satisfy the desired one-sided spectral decay, and clearly
$\overline{h}g=mf$ for some bounded function $m$ supported in $I$.

###  1.4. Clumping in other parts of analysis

The motivation for the research presented in this note was a desire to produce
a self-contained exposition of the clumping phenomenon which was observed in
two other contexts, both somewhat more esoteric than Fourier analysis on the
real line.

The first of these is a polynomial approximation problem in the unit disk
$\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$. Here we are presented with a measure

$d\mu=G(1-|z|)dA(z)+w(z)d\textit{m}(z),$

where $dA$ and $d\textit{m}$ are the area and arc-length measures on
$\mathbb{D}$ and $\mathbb{T}:=\partial\mathbb{D}=\\{z\in\mathbb{C}:|z|=1\\}$.
The functions $G$ and $w$ are non-negative weights, and one would like to
understand under which conditions splitting occurs. Namely, when is the
weighted space $\mathcal{L}^{2}(\mathbb{T},w\,d\textit{m})$ contained in the
closure of analytic polynomials in the $\mathcal{L}^{2}$-norm induced by the
measure $\mu$? In the case that $G(1-|z|)$ decays exponentially as $|z|\to
1^{-}$, the necessary and sufficient condition is that $w$ has no clumps, or
in other words that the integral of $\log w$ diverges over any arc on
$\mathbb{T}$. The lack-of-clumping condition was conjectured by Kriete and
MacCluer in [9] and confirmed in [11]. Some of the techniques used in the
proofs of the results in the present note are adaptations of the ideas from
[11].

The other context is a circle of ideas surrounding the Aleksandrov-Clark
measures appearing in spectral theory, and spaces $\mathcal{H}(b)$ defined by
de Branges and Rovnyak, well-known to operator theorists. To any positive
finite Borel measure $\nu$ on $\mathbb{T}$ we may associate a so-called Clark
operator $\mathcal{C}_{\nu}$ which takes a function
$g\in\mathcal{L}^{2}(\mathbb{T},d\nu)$ to the analytic function in
$\mathbb{D}$ given by the formula

$\mathcal{C}_{\nu}g(z):=\frac{\int_{\mathbb{T}}\frac{g(x)}{1-\overline{x}z}d\nu(x)}{\int_{\mathbb{T}}\frac{1}{1-\overline{x}z}d\nu(x)},\quad
z\in\mathbb{D}.$

The operator $\mathcal{C}_{\nu}$ maps $\mathcal{L}^{2}(\mathbb{T},d\nu)$ onto
a space of analytic functions denoted by $\mathcal{H}(b)$, the symbol function
$b:\mathbb{D}\to\mathbb{D}$ itself being related to $\nu$ by the formula

$\frac{1}{1-b(z)}=\int_{\mathbb{T}}\frac{1}{1-\overline{x}z}d\nu(x),\quad
z\in\mathbb{D}$

in the case that $\nu$ is a probability measure, with a similar formula in the
general case. For many choices of $\nu$ (or equivalently, choices of $b$), the
space $\mathcal{H}(b)$ is somewhat mysterious, with the distinctive feature of
containing very few functions extending analytically to a disk larger than
$\mathbb{D}$. This extension property is characterized by the exponential
decay of the Taylor series of the function, and the clumping of the absolutely
continuous part of $\nu$ is decisive for existence and density of functions in
$\mathcal{H}(b)$ which have a Taylor series decaying just a bit slower than
exponentially. Results of this nature are contained in [12]. In fact, a
Fourier series version of CondensationTheorem is a consequence of the results
in [12].

###  1.5. Other forms of the uncertainty principle

The implication (1.3) has a well-known Fourier series version. If a function
$f$ defined on the circle $\mathbb{T}:=\\{z\in\mathbb{C}:|z|=1\\}$ is
integrable with respect to arc-length $ds$ on $\mathbb{T}$, and the negative
portion of the Fourier series of $f$ vanishes, then
$\int_{\mathbb{T}}\log|f|ds>-\infty$, unless $f$ is the zero function..
Volberg derived the same conclusion from the weaker hypothesis of nearly-
exponentially decaying negative portion of the Fourier series (see [15] and
the exposition in [16]). Work of Borichev and Volberg [3] contains related
results.

The decay condition (1.4) on $f\in\mathcal{L}^{2}(\mathbb{R},dx)$ prohibits
$\widehat{f}$ from living on a set $S$ containing no intervals. Somewhat
related are uniqueness statements in which one seeks to give examples of pairs
of sets $(E,S)$ for which the following implication is valid: if $f$ in a
certain class lives on $E$ and $\widehat{f}$ lives on $S$, then $f\equiv 0$.
One says that $(E,S)$ is then a uniqueness pair for the corresponding class. A
famous result of Benedicks presented in [2] (see also [1]) says that $(E,S)$
is a uniqueness pair for integrable $f$ if both sets have finite Lebesgue
measure, and the result holds not only for the real line $\mathbb{R}$ but also
for the $d$-dimensional Euclidean space $\mathbb{R}^{d}$. Hedenmalm and
Montes-Rodríguez worked with the hyperbola
$H=\\{(x,y)\in\mathbb{R}^{2}:xy=1\\}$ and th class of finite Borel measures
$\mu$ supported on $H$ which are absolutely continuous with respect to
arclength on $H$. They proved in [7] that if $\widehat{\mu}$ vanishes on
certain types of discrete sets $\Lambda\subset\mathbb{R}^{2}$, then $\mu\equiv
0$, thus exhibiting interesting uniqueness pairs of the form
$(H,\mathbb{\mathbb{R}}^{2}\setminus\Lambda)$. Recent work of Radchenko and
Viazovska on interpolation formulas for Schwartz functions in [13] gives
examples of pairs of discrete subsets $E$ and $S$ of $\mathbb{R}$ for which
$(\mathbb{R}\setminus E,\mathbb{R}\setminus S)$ is a uniqueness pair for
functions in the Schwartz class. Kulikov, Nazarov and Sodin exhibit similar
interpolation formulas, and consequently new uniqueness pairs, in their recent
work in [10].

###  1.6. Notation

For a set $E\subset\mathbb{R}$ and a measure $\mu$ defined on $\mathbb{R}$,
the space $\mathcal{L}^{p}(E,d\mu)$ denotes the usual Lebesgue space
consisting of equivalence classes of functions living only on $E$ and
satisfying the integrability condition $\int_{E}|f(x)|^{p}d\mu(x)<\infty$. The
containment $\mathcal{L}^{p}(E,d\mu)\subset\mathcal{L}^{p}(\mathbb{R},d\mu)$
is interpreted in the natural way. The symbols such as $dx$, $dt$ and $d\zeta$
denote the usual Lebesgue measure of the real line, while
$d\lambda=dx/\sqrt{2\pi}$ will be the normalized version used in formulas
involving Fourier transforms. If $E$ is a subset of $\mathbb{R}$, then $|E|$
denotes its usual Lebesgue measure. The positive half-axis of $\mathbb{R}$ is
denoted by $\mathbb{R}_{+}:=\\{x\in\mathbb{R}:x\geq 0\\}$, and we set also
$\mathbb{R}_{-}:=\mathbb{R}\setminus\mathbb{R}_{+}$. The notions of almost
everywhere and of measure zero are always to be interpreted in the sense of
Lebesgue measure on $\mathbb{R}$. The indicator function of a measurable set
$E$ is denoted by $\mathbbm{1}_{E}$. Finally, we put
$\log^{+}(x):=\max(\log(x),0)$.

##  2\. Preliminaries

Our proofs will use Hilbert space techniques and the complex method. In
particular, we will use the complex interpretation of the Hardy classes of
functions on the line with positive spectrum. In this section, we recall those
basic facts of the theory of the Hardy classes
$\mathcal{H}^{1}(\mathbb{R}),\mathcal{H}^{2}(\mathbb{R})$ and
$\mathcal{H}^{\infty}(\mathbb{R})$ which will be important in the coming
sections. We discuss also properties of the shift operators $f(t)\mapsto
e^{its}f(t)$ on weighted spaces on the real line, and their invariant
subspaces.

###  2.1. Hardy classes

For $p$ equal to $1$ or $2$, we denote by $\mathcal{H}^{p}(\mathbb{R})$ the
subspace of $\mathcal{L}^{p}(\mathbb{R},dx)$ consisting of those functions $f$
for which the Fourier transform $\widehat{f}$ vanishes on the negative part of
the real axis:

$\mathcal{H}^{p}(\mathbb{R}):=\\{f\in\mathcal{L}^{p}(\mathbb{R},dx):\widehat{f}|\mathbb{R}_{-}\equiv
0\\}.$

It is a well-known fact that functions in the Hardy classes
$\mathcal{H}^{1}(\mathbb{R})$ and $\mathcal{H}^{2}(\mathbb{R})$ admit a type
of analytic extension to the upper half-plane

$\mathbb{H}:=\\{x+iy\in\mathbb{C}:y>0\\}.$

We recall what exactly is meant by this extension and how it can be
constructed. The Poisson kernel of the upper half-plane

$\mathcal{P}(t,x+iy):=\frac{1}{\pi}\frac{y}{(x-t)^{2}+y^{2}},\quad y>0,$

admits a decomposition

(2.1) $\mathcal{P}(t,z)=\operatorname{Re}\Bigg{(}\frac{1}{\pi
i(t-z)}\Bigg{)}=\frac{1}{2\pi
i}\Bigg{(}\frac{1}{t-z}-\frac{1}{t-\overline{z}}\Bigg{)},\quad
z=x+iy\in\mathbb{H}.$

Since

(2.2)
$\frac{1}{t-\overline{z}}=-i\int_{0}^{\infty}e^{-i\overline{z}s}e^{its}\,ds$

we may use Fubini’s theorem to compute, in the case
$f\in\mathcal{H}^{1}(\mathbb{R})$, that

(2.3)
$\int_{\mathbb{R}}\frac{f(t)}{t-\overline{z}}\,dt=-i\int_{0}^{\infty}\Bigg{(}\int_{\mathbb{R}}f(t)e^{its}\,dt\Bigg{)}e^{-i\overline{z}s}ds=0,$

where the vanishing of the integral follows from

$\int_{\mathbb{R}}f(t)e^{its}\,ds=\widehat{f}(-s)=0,\quad s>0,$

which holds for any $f\in\mathcal{H}^{1}(\mathbb{R})$ by the definition of the
class. In the case $f\in\mathcal{H}^{2}(\mathbb{R})$ this argument does not
work, but what instead works is an application of Plancherel’s theorem and
FourierTransformCauchyKernel below to the first integral in (2.3), which again
shows that this integral vanishes. Consequently, whenever
$f\in\mathcal{H}^{p}(\mathbb{R})$ for $p=1,2$, the formula

$f(z):=\int_{\mathbb{R}}f(t)\mathcal{P}(t,z)\,dt=\int_{\mathbb{R}}\frac{f(t)}{t-z}\frac{dt}{2\pi
i},\quad z\in\mathbb{H}$

defines, by the second integral expression above, an analytic extension of $f$
to $\mathbb{H}$. By the first expression, and classical properties of the
Poisson kernel (see [4, Chapter I]), this extension satisfies

(2.4) $\lim_{y\to 0^{+}}f(x+iy)=f(x),\text{ for almost every }x\in\mathbb{R}.$

Moreover, we have

(2.5) $\sup_{y>0}\int_{\mathbb{R}}|f(x+iy)|^{p}\,dx<\infty$

and

(2.6) $\lim_{y\to 0^{+}}\int_{\mathbb{R}}|f(x+iy)-f(x)|^{p}\,dx=0.$

if $f\in\mathcal{H}^{p}(\mathbb{R})$. The property (2.5) follows readily from
the Poisson integral formula for the extension of $f$ and Fubini’s theorem.
The property (2.6) is a bit tricky to establish, and is proved in [4, Chapter
I, Theorem 3.1]. In fact, the above listed properties characterize the
functions in the Hardy classes.

######  Proposition 2.1.

H1CharacterizationProp For $p=1$ and $p=2$, a function
$f\in\mathcal{L}^{p}(\mathbb{R},dx)$ is a member of
$\mathcal{H}^{p}(\mathbb{R})$ if and only if there exists an analytic
extension of $f$ to $\mathbb{H}$ which satisfies the three properties in
(2.4), (2.5) and (2.6).

The proposition is not hard to derive from H1FourierTransformFormula below.
Anyway, a careful proof can be found in [6, p. 172].

The following restriction on smallness of the modulus $|f|$ of a function
$f\in\mathcal{H}^{1}(\mathbb{R})$ will be of crucial importance to us.

######  Proposition 2.2.

HadyClassLogIntProp If $f\in\mathcal{H}^{1}(\mathbb{R})$, then

$\int_{\mathbb{R}}\frac{\log|f(x)|}{1+x^{2}}dx>-\infty$

unless $f$ is the zero function.

A proof of the proposition can be found in [6, p. 35].

We shall also need to use the corresponding Hardy class of functions which are
merely bounded on $\mathbb{R}$, and not necessarily integrable or square-
integrable on $\mathbb{R}$. We use directly the complex interpretation of the
class. Namely, we define $\mathcal{H}^{\infty}(\mathbb{R})$ to consist of
those functions $f\in\mathcal{L}^{\infty}(\mathbb{R},dx)$ which can be
realized as limits

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

for almost every $x\in\mathbb{R}$, where $f$ is bounded and analytic in
$\mathbb{H}$. It can be checked that such $f$ has a distributional spectrum
which vanishes on $\mathbb{R}_{-}$. Another important point is that if
$f\in\mathcal{H}^{\infty}(\mathbb{R})$, then

$\frac{f(x)}{(i+x)^{2}}\in\mathcal{H}^{1}(\mathbb{R}),$

since we may apply H1CharacterizationProp to the analytic function

$z\mapsto\frac{f(z)}{(i+z)^{2}},\quad z\in\mathbb{H}.$

A function $h\in\mathcal{H}^{\infty}(\mathbb{R})$ of a given (bounded,
measurable) modulus $|h|=W$ on $\mathbb{R}$ may be constructed by setting

(2.7) $\log h(z):=\frac{1}{\pi
i}\int_{\mathbb{R}}\Big{(}\frac{1}{t-z}-\frac{t}{1+t^{2}}\Big{)}\log
W(t)\,dt,\quad z\in\mathbb{H},$

and $h(z):=e^{\log h(z)}$. The integral above converges if

$\int_{\mathbb{R}}\frac{\log W(t)}{1+t^{2}}\,dt>-\infty$

which is a necessary condition for the construction to be possible. Then

$\log|h(z)|=\int\mathcal{P}(t,z)\log W(t)\,dt,$

so that the equality $\lim_{y\to 0^{+}}|h(x+iy)|=|h(x)|=W(x)$ for almost every
$x\in\mathbb{R}$ is a consequence of the well-known properties of the Poisson
kernel.

###  2.2. A formula and an estimate for the Fourier transform of a Hardy class
function

If $f\in\mathcal{H}^{1}(\mathbb{R})$, then the values of $\widehat{f}(\zeta)$
may be computed using a formula different from (1.1). To wit, denote by $f(z)$
the extension of $f$ to $\mathbb{H}$ which was discussed in Section 2.1. The
function

$G_{\zeta}(z):=f(z)e^{-iz\zeta}=f(z)e^{-ix\zeta+y\zeta},\quad
z=x+iy\in\mathbb{H}$

is analytic in $\mathbb{H}$, and for this reason Cauchy’s integral theorem
implies

(2.8) $\int_{R(\epsilon,y,a)}G_{\zeta}(z)dz=0,$

where $dz$ denotes the complex line integral, $\epsilon,y,a$ are all positive
numbers, $\epsilon<y$, and $R(\epsilon,y,a)$ denotes the rectangular contour
having as corners the four points with coordinates $(-a,\epsilon)$,
$(a,\epsilon)$, $(a,y)$, $(-a,y)$, oriented counter-clockwise.

Fix $y>0$ and let $S_{y}$ denote the horizontal strip in $\mathbb{H}$
consisting of all complex numbers with imaginary part between $0$ and $y$.
Then it follows from Fubini’s theorem and (2.5) that

$\displaystyle\int_{S}|G_{\zeta}(z)|dA(z)\leq
e^{y|\zeta|}\int_{\mathbb{R}}\int_{0}^{y}|f(x+is)|dsdx<\infty,$

where $dA(z)$ denotes the area measure on the complex plane. This expresses
the integrability on $\mathbb{R}$ of the continuous function

$x\mapsto\int_{0}^{y}|G_{\zeta}(x+is)|ds.$

Hence there exists a positive sequence $\\{a_{n}\\}_{n}$ which satisfies

$\lim_{n\to\infty}a_{n}=+\infty$

and for which

$\lim_{n\to\infty}\int_{0}^{y}|G_{\zeta}(a_{n}+is)|ds+\int_{0}^{y}|G_{\zeta}(-a_{n}+is)|ds=0.$

This means that

$\displaystyle 0$
$\displaystyle=\lim_{n\to\infty}\int_{R(\epsilon,y,a_{n})}G_{\zeta}(z)\,dz$
$\displaystyle=-\int_{\mathbb{R}}G_{\zeta}(x+iy)dx+\int_{\mathbb{R}}G_{\zeta}(x+i\epsilon)dx$

Moreover, equation (2.6) quite easily implies

$\lim_{\epsilon\to
0^{+}}\int_{\mathbb{R}}G_{\zeta}(x+i\epsilon)dx=\int_{\mathbb{R}}f(x)e^{-ix\zeta}dx=\sqrt{2\pi}\widehat{f}(\zeta).$

We have proven the following formula by combining the above two expressions.

######  Proposition 2.3.

H1FourierTransformFormula For $f\in\mathcal{H}^{1}(\mathbb{R})$ we may compute
the Fourier transform $\widehat{f}(\zeta)$ using the formula

$\widehat{f}(\zeta)=e^{y\zeta}\int_{\mathbb{R}}f(x+iy)e^{-ix\zeta}\,d\lambda(x)$

for any choice of $y>0$, where $f(x+iy)$ denotes the values of the analytic
extension of $f$ to $\mathbb{H}$.

This formula has the following simple corollary which will be of critical
importance below.

######  Corollary 2.4.

FourierDecayFromExtensionGrowthCorollary If
$h\in\mathcal{H}^{\infty}(\mathbb{R})$ has an analytic extension to
$\mathbb{H}$ which satisfies, for some constant $c>0$, an estimate of the form

$\sup_{x\in\mathbb{R}}\,|h(x+iy)|\leq e^{c/y},\quad\text{ for all }y>0,$

then the Fourier transform $\widehat{h_{*}}$ of the function

$h_{*}(x):=\frac{h(x)}{(i+x)^{2}}\in\mathcal{H}^{1}(\mathbb{R})$

satisfies

$|\widehat{h_{*}}(\zeta)|\leq\sqrt{\frac{\pi}{2}}e^{2\sqrt{c}\sqrt{\zeta}},\quad\zeta>0.$

###### Proof.

It was mentioned in Section 2.1 that $h_{*}\in\mathcal{H}^{1}(\mathbb{R})$.
Therefore, we may use H1FourierTransformFormula to estimate

$\displaystyle|\widehat{h_{*}}(\zeta)|$ $\displaystyle\leq
e^{y\zeta}\int_{\mathbb{R}}\frac{|h(x+iy)|}{|i+x+iy|^{2}}\,d\lambda(x)$
$\displaystyle\leq
e^{y\zeta}\int_{\mathbb{R}}\frac{e^{c/y}}{1+x^{2}}\,d\lambda(x)$
$\displaystyle=\sqrt{\frac{\pi}{2}}e^{y\zeta+c/y}.$

Since $y>0$ can be freely chosen, we may now set it to $y=\sqrt{c/\zeta}$ to
obtain the desired estimate. ∎

###  2.3. A semigroup of operators and its invariant subspaces

If $w\in\mathcal{L}^{1}(\mathbb{R},dx)$ and $s\in\mathbb{R}$, the operator
$U^{s}:\mathcal{L}^{2}(\mathbb{R},w\,dx)\to\mathcal{L}^{2}(\mathbb{R},w\,dx)$
given by

$U^{s}f(x):=e^{isx}f(x)$

is unitary on $\mathcal{L}^{2}(\mathbb{R},w\,dx)$. We shall be interested in
subspaces of $\mathcal{L}^{2}(\mathbb{R},w\,dx)$ which are invariant for the
operators in the semigroup $\\{U^{s}\\}_{s>0}$. Given any element
$f\in\mathcal{L}^{2}(\mathbb{R},w\,dx)$, we denote by $[f]_{w}$ the smallest
closed linear subspace of $\mathcal{L}^{2}(\mathbb{R},w\,dx)$ which contains
$f$ and also all the functions $U^{s}f$, $s>0$.

######  Proposition 2.5.

UsInvSubspacesProp Let $f\in\mathcal{L}^{2}(\mathbb{R},w\,dx)$ be a non-zero
element which satisfies

$\int_{\mathbb{R}}\frac{\log\big{(}|f(x)|^{2}w(x)\big{)}}{1+x^{2}}dx=-\infty.$

Then the subspace $[f]_{w}$ coincides with $L^{2}(E,w\,dx)$, where
$E=\\{x\in\mathbb{R}:|f(x)|>0\\}$.

######  Remark 2.6.

As usual, the set $E$ above is defined in a bit imprecise way. Since $f$ is,
strictly speaking, merely a representative of an equivalence class of
measurable functions in $\mathcal{L}^{2}(\mathbb{R},w\,dx)$, the set $E$ is
not well-defined pointwise. However, it is well-defined up to a set of
Lebesgue measure zero, and so the initial choice of the representative is
unimportant.

###### Proof of UsInvSubspacesProp.

Since the function $f$ vanishes almost everywhere outside of the set $E$, then
so does $U^{s}f$ for any $s>0$. Consequently,
$[f]_{w}\subset\mathcal{L}^{2}(E,w\,dx)$. Conversely, let us consider an
element $g\in\mathcal{L}^{2}(E,w\,dx)$ with the property that

$\int_{\mathbb{R}}U_{s}f(x)\overline{g(x)}w(x)dx=\int_{\mathbb{R}}e^{isx}f(x)\overline{g(x)}w(x)dx=0,\quad
s>0.$

Setting $h:=f\overline{g}w\in\mathcal{L}^{1}(\mathbb{R},dx)$, we note that the
vanishing of the integrals above is equivalent to $h$ being a member of the
Hardy class $\mathcal{H}^{1}(\mathbb{R})$. We note also that

$\int_{\mathbb{R}}\frac{\log|h(x)|}{1+x^{2}}dx=\frac{1}{2}\int_{\mathbb{R}}\frac{\log\big{(}|f(x)|^{2}w(x)\big{)}}{1+x^{2}}dx+\frac{1}{2}\int_{\mathbb{R}}\frac{\log\big{(}|g(x)|^{2}w(x)\big{)}}{1+x^{2}}dx.$

The above equality is to be interpreted in a generalized sense: the first
integral on the right-hand side is divergent by our assumption, and so may the
second, but their positive parts are certainly finite by the assumption that
$f,g\in\mathcal{L}^{2}(\mathbb{R},w\,dx)$. This implies that

$\int_{\mathbb{R}}\frac{\log|h(x)|}{1+x^{2}}=-\infty.$

HadyClassLogIntProp now shows that $h=f\overline{g}w$ must be the zero
function. Since $|f(x)|w(x)>0$ on $E$ and $g$ vanishes outside of $E$, this
means that $g\equiv 0$. So $[f]_{w}$ is a closed and dense subspace of
$\mathcal{L}^{2}(E,w\,dx)$, which means that the two spaces are equal. ∎

######  Corollary 2.7.

invSubspaceCorollary If $f\in\mathcal{L}^{2}(\mathbb{R},w\,dx)$ is also a
member of $\mathcal{L}^{p}(\mathbb{R},w\,dx)$ for some $p>2$, and if

$\int_{\mathbb{R}}\frac{\log w(x)}{1+x^{2}}\,dx=-\infty,$

then $[f]_{w}$ coincides with $\mathcal{L}^{2}(E,w\,dx)$, where
$E=\\{x\in\mathbb{R}:|f(x)|>0\\}$.

###### Proof.

To prove the corollary we need to verify the condition in UsInvSubspacesProp.
Note that, pointwise, we have

$\log\big{(}|f|^{2}w\big{)}=(2/p)\log\big{(}|f|^{p}w\big{)}+(1-2/p)\log w.$

The coefficients $2/p$ and $1-2/p$ are positive. The inequality $\log(x)\leq
x$ for $x>0$ shows that

$\int_{\mathbb{R}}\frac{\log\big{(}|f(x)|^{p}w(x)\big{)}}{1+x^{2}}\,dx\leq\int_{\mathbb{R}}|f(x)|^{p}w(x)dx<+\infty.$

Note that the integral on the left might very well be equal to $-\infty$, but
that is of no concern to us: we conclude from the assumption, and the
pointwise inequality above, that

$\int_{\mathbb{R}}\frac{\log\big{(}|f(x)|^{2}w(x)\big{)}}{1+x^{2}}\,dx=-\infty$

and apply UsInvSubspacesProp. ∎

##  3\. A product space and its Hardy subspace

Let $\rho:\mathbb{R}_{+}\to\mathbb{R}_{+}$ be a bounded, continuous, non-
negative and decreasing function, and
$w\in\mathcal{L}^{1}(\mathbb{R},dx)\cap\mathcal{L}^{\infty}(\mathbb{R},dx)$ be
a non-negative function. We consider the product space
$\mathcal{L}^{2}(\mathbb{R},w\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$.
Inside of this space we embed the linear manifold
$\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$ in the following
way:

$Jf:=(f,\widehat{f})\in\mathcal{L}^{2}(\mathbb{R},w\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx),\quad
f\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R}).$

The tuple $Jf$ is well-defined as an element of the product space, since both
$f$ and $\widehat{f}$ are members of $\mathcal{L}^{2}(\mathbb{R},dx)$ and both
$\rho$ and $w$ are bounded. We define the Hardy subspace $\mathcal{H}(w,\rho)$
as the norm-closure of the linear manifold

$\\{Jf:f\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})\\}$

inside of the product space
$\mathcal{L}^{2}(\mathbb{R},w\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$.
Thus each tuple $(h,k)\in\mathcal{H}(w,\rho)$ has the property that there
exists some sequence $\\{f_{n}\\}_{n}$ of functions in
$\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$ such that

$h=\lim_{n\to\infty}f_{n}$

in the space $\mathcal{L}^{2}(\mathbb{R},w\,dx)$, and simultaneously

$k=\lim_{n\to\infty}\widehat{f_{n}}$

in the space $\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$.

We could have used a set of tuples $Jf$ with $f\in\mathcal{H}^{2}(\mathbb{R})$
in the definition of the Hardy subspace, and arrived at the same space.
Indeed, we have the following proposition.

######  Proposition 3.1.

KernelContainmentHardySubspace With $w$ and $\rho$ as above, the Hardy
subspace $\mathcal{H}(w,\rho)$ contains all tuples of the form
$(f,\widehat{f})$, $f\in\mathcal{H}^{2}(\mathbb{R})$. Moreover, tuples $Jf$
where $f\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$ and $f$
extends analytically to a half-space $\\{z=x+iy\in\mathbb{C}:y>-\delta\\}$,
$\delta=\delta(f)>0$, are norm-dense in $\mathcal{H}(w,\rho)$.

###### Proof.

Fix $f\in H^{2}(\mathbb{R})$, and consider the functions $f_{\epsilon}(x)$
defined by the formula

$f_{\epsilon}(x):=\frac{if(x+i\epsilon)}{\epsilon x+i},\quad
x\in\mathbb{R},\epsilon>0.$

These functions are contained in
$\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$ for each
$\epsilon>0$, and they are analytic in a half-space larger than $\mathbb{H}$.
Note that

$\Bigg{|}\frac{i}{\epsilon x+i}\Bigg{|}\leq 1,\quad x\in\mathbb{R}$

and that

$\lim_{\epsilon\to 0^{+}}\frac{i}{\epsilon x+i}=1.$

We readily see from H1CharacterizationProp and the dominated convergence
theorem that we have

$\lim_{\epsilon\to 0^{+}}\int_{\mathbb{R}}|f_{\epsilon}-f|^{2}dx=0.$

By Plancherel’s theorem, we therefore also have

$\lim_{\epsilon\to
0^{+}}\int_{\mathbb{R}_{+}}|\widehat{f_{\epsilon}}-\widehat{f}|^{2}d\zeta=0.$

A fortiori, we have

$\lim_{\epsilon\to 0^{+}}\int_{\mathbb{R}}|f_{\epsilon}-f|^{2}w\,dx=0$

and

$\lim_{\epsilon\to
0^{+}}\int_{\mathbb{R}_{+}}|\widehat{f_{\epsilon}}-\widehat{f}|^{2}\rho\,d\zeta=0.$

Thus, as $\epsilon\to 0$, the tuples $Jf_{\epsilon}\in\mathcal{H}(w,\rho)$
converge in the norm of the space to the tuple $(f,\widehat{f})$, which is
therefore contained in $\mathcal{H}(w,\rho)$. This proves the first statement
of the proposition. The second has the same proof, we merely start with
$f\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$ and run the
same argument. ∎

The shift operators

$U^{s}f(x)=e^{ixs}f(x),\quad f\in\mathcal{L}^{2}(\mathbb{R},w\,dx)$

are unitary. Using the convention that $g(x)\equiv 0$ for $x<0$ and
$g\in\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$, the translation operators

$\widehat{U^{s}}g(x):=g(x-s),\quad
g\in\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$

are contractions on $\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$, whenever
$s>0$. This fact is a consequence of the assumption that $\rho$ is decreasing:

$\displaystyle\int_{\mathbb{R}_{+}}|g(x-s)|^{2}\rho(x)dx$
$\displaystyle=\int_{s}^{\infty}|g(x-s)|^{2}\rho(x)dx$
$\displaystyle\leq\int_{s}^{\infty}|g(x-s)|^{2}\rho(x-s)dx$
$\displaystyle=\int_{\mathbb{R}_{+}}|g(x)|^{2}\rho(x)dx.$

We used that $g(x-s)$ vanishes for $x\in(0,s)$. Consequently, the operators

$U^{s}_{*}:=U^{s}\oplus\widehat{U^{s}},\quad s>0$

are bounded on the space
$\mathcal{L}^{2}(\mathbb{R},w\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$.
Moreover, the Hardy subspace $\mathcal{H}(w,\rho)$ is invariant for these
operators. Indeed, if
$f\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$, then by the
well-known property of the Fourier transform
$\widehat{U^{s}}\widehat{f}=\widehat{U^{s}f}$, we obtain

$U_{*}^{s}Jf=(U^{s}f,\widehat{U^{s}}\widehat{f})=(U^{s}f,\widehat{U^{s}f})=JU^{s}f.$

The function $U^{s}f$ is contained in
$\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$, and so the above
relation shows that a dense subset of $\mathcal{H}(w,\rho)$ is mapped into
$\mathcal{H}(w,\rho)$ under each of the bounded operators $U^{s}_{*}$. The
mentioned invariance follows.

##  4\. Strategy of the proofs

This section outlines the strategy of the proofs of CondensationTheorem and
SparsenessTheorem.

###  4.1. Two easy computations

In our strategy, we will need to use the results of the following two
computations.

######  Lemma 4.1.

FourierTransformCauchyKernel Let

$\psi_{z}(x):=\frac{i}{\sqrt{2\pi}(x-\overline{z})},\quad z\in\mathbb{H}.$

The Fourier transform $\widehat{\psi_{z}}$ equals

$\widehat{\psi_{z}}(\zeta)=e^{-i\overline{z}\zeta}\mathbbm{1}_{\mathbb{R}_{+}}(\zeta),$

and the Fourier transform $\widehat{\overline{\psi_{z}}}$ of the conjugate of
$\psi_{z}$ equals

$\widehat{\overline{\psi_{z}}}(\zeta)=e^{-iz\zeta}\mathbbm{1}_{\mathbb{R}_{-}}(\zeta).$

###### Proof.

It is perhaps easiest to apply the Fourier inversion formula to the asserted
formula for $\widehat{\psi_{z}}$. We readily compute

$\displaystyle\int_{\mathbb{R}}\widehat{\psi_{z}}(\zeta)e^{i\zeta
x}\,d\lambda(\zeta)$
$\displaystyle=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{(-i\overline{z}+ix)\zeta}\,d\zeta$
$\displaystyle=\psi_{z}(x).$

The other formula follows from
$\widehat{\overline{\psi_{z}}}(\zeta)=\overline{\widehat{\psi_{z}}(-\zeta)}$,
which is an easily established property of the Fourier transform. ∎

######  Proposition 4.2.

ConvolutionFourierDecayLemma Assume that $f\in\mathcal{L}^{2}(\mathbb{R},dx)$
satisfies

$\rho_{\widehat{f}}(x)=\int_{x}^{\infty}|\widehat{f}(\zeta)|d\zeta=\mathcal{O}\big{(}e^{-c\sqrt{x}}\big{)},\quad
x>0$

for some $c>0$, and let

$s(x):=\overline{\psi_{i}(x)}=\frac{-i}{\sqrt{2\pi}(x-i)}.$

Then

$\big{|}\widehat{fs}(\zeta)\big{|}=\mathcal{O}\big{(}e^{-c\sqrt{\zeta}}\big{)},\quad\zeta>0.$

###### Proof.

Note that $fs\in\mathcal{L}^{1}(\mathbb{R},dx)$, and recall that the Fourier
transform $\widehat{fs}$ is thus a continuous function given by the
convolution of the Fourier transforms of $f$ and $s$. By
FourierTransformCauchyKernel, we obtain

$\widehat{s}(\zeta)=e^{\zeta}\mathbbm{1}_{\mathbb{R}_{-}}(\zeta),$

and so

$\widehat{fs}(\zeta)=\int_{\mathbb{R}}\widehat{f}(x)e^{\zeta-x}\mathbbm{1}_{\mathbb{R}_{-}}(\zeta-x)\,d\lambda(x)=\int_{\zeta}^{\infty}\widehat{f}(x)e^{\zeta-x}\,d\lambda(x).$

The exponential term in the last integral is bounded by $1$. Therefore
$\big{|}\widehat{fs}(\zeta)\big{|}\leq\rho_{\widehat{f}}(\zeta)$, and the
desired estimate follows from the decay assumption on $\rho_{\widehat{f}}$. ∎

###  4.2. Strategy of the proof of CondensationTheorem

Given a function $f\in\mathcal{L}^{2}(\mathbb{R},dx)$ we consider the set

$E:=\\{x\in\mathbb{R}:|f(x)|>0\\},$

which is well-defined up to a set of Lebesgue measure zero. Let $\mathcal{F}$
denote the family of all finite open intervals $I$ which satisfy

$\int_{I}\log|f(x)|\,dx>-\infty,$

and set

$U:=\cup_{I\in\mathcal{F}}I.$

Since $\log|f|\equiv-\infty$ on $I\setminus E$ for every interval $I$, it
follows that if $\log|f|$ is integrable on $I$, then the set difference
$I\setminus E$ must have measure zero. Consequently, since one can easily
argue that we can express $U$ as a countable union of intervals $I$ on which
$\log|f|$ is integrable, the Lebesgue measure of the set difference
$U\setminus E$ must be zero. However, the set difference $E\setminus U$ might
have positive measure. We set

$\text{res}(f):=E\setminus U$

and call this set the residual of $f$. The residual is well-defined up to a
set of Lebesgue measure zero.

######  Claim 1.

claim1 Under the assumption that
$\rho_{\widehat{f}}(\zeta)=\mathcal{O}\big{(}e^{-c\sqrt{\zeta}}\big{)}$ for
some $c>0$, the set $\text{res}(f)$ has Lebesgue measure zero.

CondensationTheorem follows immediately from the above claim. Indeed, the
roles of $f$ and $\widehat{f}$ may obviously be interchanged in the statement
of CondensationTheorem, and the above claim implies that the open set $U$
equals $E$ up to an error of measure zero. Local integrability of $\log|f|$ on
the set $U$ follows from its construction.

We set

$w(x):=\min(|f(x)|^{2},1).$

Note that $\text{res}(w)=\text{res}(f)$ and that
$w\in\mathcal{L}^{1}(\mathbb{R},dx)$. Our claim1 will follow from the next
assertion.

######  Claim 2.

claim2 Let $\rho:\mathbb{R}_{+}\to\mathbb{R}_{+}$ be a bounded, continuous,
non-negative and decreasing function which satisfies
$\rho(x)=\mathcal{O}\big{(}e^{-d\sqrt{x}}\big{)}$ for some $d>0$ and $x>0$.
Then, every tuple of the form

$(h,0)\in\mathcal{L}^{2}(\mathbb{R},w\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx),$

where $h$ is any function in $\mathcal{L}^{2}(\mathbb{R},dx)$ which lives only
on the set $\text{res}(w)$, is contained in the Hardy subspace
$\mathcal{H}(w,\rho)$.

To prove claim1 from claim2 we will use a trick involving Plancherel’s
theorem. We set $\rho(x)=e^{-c\sqrt{x}}$, where $c>0$ is the constant
appearing in claim1. Let $h$ be as in claim2, and

$s(x):=\overline{\psi_{i}}(x)=\frac{-i}{\sqrt{2\pi}(x-i)}$

be as in ConvolutionFourierDecayLemma. We will show that

$\int_{\mathbb{R}}h\overline{fs}\,dx=0.$

This implies, by the generality of $h$, that $fs$ is zero on the set
$\text{res}(w)=\text{res}(f)$. Since $s$ is non-zero everywhere on
$\mathbb{R}$, in fact $f$ is zero on $\text{res}(f)$. Since
$\text{res}(f)\subset E=\\{x\in\mathbb{R}:|f(x)|>0\\}$, it follows that the
residual has Lebesgue measure zero. Thus establishing the vanishing of the
above integral is sufficient to prove claim1 from claim2. We do so next.

Because $(h,0)\in\mathcal{H}(w,\rho)$, there exists a sequence $\\{g_{n}\\}$
of functions
$g_{n}\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$ such that
$g_{n}\to h$ in the norm of $\mathcal{L}^{2}(\mathbb{R},w\,dx)$ and
$\widehat{g_{n}}\to 0$ in the norm of
$\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$. Consider the quantities

$\int_{\mathbb{R}}(h-g_{n})\overline{fs}\,d\lambda=\int_{E}(h-g_{n})\sqrt{w}\frac{\overline{fs}}{\sqrt{w}}d\lambda.$

We passed from domain of integration $\mathbb{R}$ into $E$, since $f$ vanishes
outside of $E$ anyway (note also that $w>0$ almost everywhere on $E$). By the
Cauchy-Schwarz inequality, we obtain

$\displaystyle\Bigg{|}\int_{\mathbb{R}}(h-g_{n})\overline{fs}\,d\lambda\Bigg{|}\leq\sqrt{\int_{\mathbb{R}}|h-g_{n}|^{2}w\,d\lambda}\sqrt{\int_{E}\frac{|f|^{2}|}{w}|s|^{2}d\lambda}.$

Note that the first of the factors on the right-hand side of the inequality
above converges to $0$. The other factor is finite. Indeed, since
$|f|^{2}/w\equiv 1$ on the set where $|f|<1$, and $|f|^{2}/w=|f|^{2}$ on the
set where $|f|\geq 1$, we obtain $\frac{|f|^{2}}{w}\leq|f|^{2}+1$, and
consequently

$\frac{|f|^{2}}{w}|s|^{2}\leq|f|^{2}|s|^{2}+|s|^{2}\in\mathcal{L}^{1}(\mathbb{R},dx).$

This computation implies the formula

$\int_{\mathbb{R}}h\overline{fs}\,d\lambda=\lim_{n\to\infty}\int_{\mathbb{R}}g_{n}\overline{fs}\,d\lambda=\lim_{n\to\infty}\int_{\mathbb{R}_{+}}\widehat{g_{n}}\overline{\widehat{fs}}\,d\lambda,$

where we used Plancherel’s theorem in the last step. Now, recall that by
ConvolutionFourierDecayLemma we may estimate

$\Bigg{|}\int_{\mathbb{R}_{+}}\widehat{g_{n}}\overline{\widehat{fs}}\,d\lambda\Bigg{|}\leq
A\int_{\mathbb{R}_{+}}|\widehat{g_{n}}(\zeta)|e^{-c\sqrt{\zeta}}\,d\lambda(\zeta)$

for some positive constant $A$. By again using Cauchy-Schwarz inequality, we
obtain

$\displaystyle\Bigg{|}\int_{\mathbb{R}}\widehat{g_{n}}\overline{\widehat{fs}}\,d\lambda\Bigg{|}\leq
A\sqrt{\int_{\mathbb{R}_{+}}|\widehat{g_{n}}(\zeta)|^{2}e^{-c\sqrt{\zeta}}\,d\lambda(\zeta)}\sqrt{\int_{\mathbb{R}_{+}}e^{-c\sqrt{\zeta}}\,d\lambda(\zeta)}.$

The second factor on the right-hand side is certainly finite. Since
$\rho(x)=e^{-c\sqrt{x}}$ and $g_{n}\to 0$ in
$\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$, the first factor above converges
to $0$, as $n\to\infty$. All in all, we have obtained that

$\int_{\mathbb{R}}h\overline{fs}\,d\lambda=\lim_{n\to\infty}\int_{\mathbb{R}_{+}}\widehat{g_{n}}\overline{\widehat{fs}}\,d\lambda=0.$

By the earlier discussion, this is sufficient to establish claim1 from claim2.
We need to prove claim2 in order to prove CondensationTheorem. We will do so
in the coming sections.

###  4.3. Strategy of the proof of SparsenessTheorem

We will derive SparsenessTheorem from the following claim.

######  Claim 3.

claim3 There exists a compact set $E\subset\mathbb{R}$ of positive Lebesgue
measure, and an increasing function $M:\mathbb{R}_{+}\to\mathbb{R}_{+}$ which
satisfies

(4.1) $\lim_{x\to\infty}\frac{M(x)}{x^{a}}=\infty$

for every $a\in(0,1/2)$, such that if

$\rho(x):=e^{-M(x)},\quad x>0,$

then the Hardy subspace $\mathcal{H}(\mathbbm{1}_{E},\rho)$ is properly
contained in
$\mathcal{L}^{2}(E,\mathbbm{1}_{E}\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$.

We proceed to show how one proves SparsenessTheorem from this claim. Let $E$
and $\rho=e^{-M}$ be as in claim3, and assume that the non-zero tuple
$(h,k)\in\mathcal{L}^{2}(E,\mathbbm{1}_{E}\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$
is orthogonal to $\mathcal{H}(\mathbbm{1}_{E},\rho)$. We shall soon see that
$h$ in fact is non-zero. The function $h$ lives only on the set $E$, and we
will show that it has the required spectral decay. Recall
FourierTransformCauchyKernel, let $\psi_{z}\in\mathcal{H}^{2}(\mathbb{R})$ be
as in that lemma, and let $g$ be the inverse Fourier transform of $k\rho$, so
that $\widehat{g}=k\rho\in\mathcal{L}^{2}(\mathbb{R}_{+},dx)$. In fact
$g\in\mathcal{H}^{2}(\mathbb{R})$, since its spectrum is positive. The
orthogonality means that

$\displaystyle 0$
$\displaystyle=\int_{E}h\overline{\psi_{z}}\,d\lambda+\int_{\mathbb{R}_{+}}k\overline{\widehat{\psi_{z}}}\rho\,d\lambda$
$\displaystyle=\int_{E}h\overline{\psi_{z}}\,d\lambda+\int_{\mathbb{R}}g\overline{\psi_{z}}\,d\lambda.$

We used Plancherel’s theorem. The above relation shows that the function
$G:=h+g\in\mathcal{L}^{2}(\mathbb{R},dx)$ is orthogonal in
$\mathcal{L}^{2}(\mathbb{R},dx)$ to each of the functions $\psi_{z}$. Let
$P:\mathcal{L}^{2}(\mathbb{R},dx)\to\mathcal{H}^{2}(\mathbb{R})$ be the
orthogonal projection. In terms of Fourier transforms, we have

$\widehat{Ph}(\zeta)=\widehat{h}(\zeta)\mathbbm{1}_{\mathbb{R}_{+}}(\zeta).$

Then

$G_{0}:=PG=Ph+g$

is orthogonal not only to $\psi_{z}$, but also to $\overline{\psi_{z}}$, since
by FourierTransformCauchyKernel the functions $\overline{\psi_{z}}$ have
spectrum contained in $\mathbb{R}_{-}$. But then the decomposition formula for
the Poisson kernel in (2.1) shows that

$\int_{\mathbb{R}}G_{0}(t)\mathcal{P}(t,z)\,dt=0$

for each $z\in\mathbb{H}$, and it is an elementary fact about the Poisson
kernel that we must, in this case, have $G_{0}\equiv 0$. So $Ph=-g$. We can
now argue that $h\neq 0$. Indeed, if $h=0$, then $g=0$. Since
$\widehat{g}=k\rho$, that would mean $k=0$, contradicting that the tuple
$(h,k)$ is non-zero. Having established that $h\neq 0$, we proceed by taking
Fourier transforms to obtain

$\widehat{h}\mathbbm{1}_{\mathbb{R}_{+}}=\widehat{Ph}=-\widehat{g}=-k\rho.$

Using Cauchy-Schwarz inequality, we may now estimate

$\displaystyle\rho_{\widehat{h}}(x)$
$\displaystyle=\int_{x}^{\infty}|k|\rho\,d\zeta$
$\displaystyle\leq\sqrt{\int_{x}^{\infty}\rho\,d\zeta}\sqrt{\int_{x}^{\infty}|k|^{2}\rho\,d\zeta},\quad
x>0.$

The second factor is finite, and the growth of $M$ asserted in claim3 implies
that for every fixed $a\in(0,1/2)$ there exists a constant $C(a)>0$ for which
we have

$\rho(\zeta)=e^{-M(\zeta)}\leq e^{-C(a)\zeta^{a}},\quad\zeta>0.$

It follows that the integral inside the square root of the first factor above
satisfies

$\int_{x}^{\infty}\rho\,d\zeta\leq\int_{x}^{\infty}e^{-C(a)\zeta^{a}}\,d\zeta=\mathcal{O}\big{(}e^{-C(a)x^{a}}\big{)},\quad
x>0.$

Since $a\in(0,1/2)$ was arbitrary, we conclude that
$\rho_{\widehat{h}}(x)=\mathcal{O}\big{(}e^{-x^{a}})$ for every $a\in(0,1/2)$
and $x>0$. This easily implies SparsenessTheorem.

It follows that SparsenessTheorem is implied by claim3.

##  5\. Proof of CondensationTheorem

Our proof is an adaptation to the half-plane setting of the authors’ technique
from [11], and in fact the two proofs are very similar. The problem studied in
[11] is different, but in both the present work and in the reference, the main
trick consists of constructing a highly oscillating sequence of functions
which simultaneously obey appropriate spectral bounds.

###  5.1. A sufficient construction

We start by reducing our task to a construction of a certain sequence of
bounded functions. Recall that
$w(x)=\min(|f(x)|^{2},1)\in\mathcal{L}^{1}(\mathbb{R},dx)$ and that $\rho$ has
the decay $\rho(\zeta)=\mathcal{O}\big{(}e^{-d\sqrt{\zeta}}\big{)}$ for
$\zeta>0$ and some $d>0$. Note that we may assume throughout that

$\int_{\mathbb{R}}\frac{\log w(x)}{1+x^{2}}\,dx=-\infty.$

Indeed, if on the contrary this integral converges, then
$\text{res}(w)=\text{res}(f)$ is void, and both claim2 of Section 4 and
CondensationTheorem (with $f$ and $\widehat{f}$ playing opposite roles) hold
trivially.

We may decompose $\text{res}(w)$ as

$\text{res}(w)=\cup_{m\geq 1}\,F_{m}$

where

(5.1) $F_{m}:=[-m,m]\cap\\{x\in\mathbb{R}:w(x)>1/m\\}\cap\text{res}(w).$

The set equality above holds up to an error of measure zero. The sets $F_{m}$
are bounded, and on each of them $w$ is bounded from below.

######  Proposition 5.1.

hnSplittingSequence In order to establish claim2, it suffices to construct,
for any fixed $m\in\mathbb{N}$ and $c>0$, a sequence of functions
$\\{h_{n}\\}_{n}$ in $\mathcal{H}^{\infty}(\mathbb{R})$ which has the
following properties.

  1. (i)

The analytic extensions of the functions $h_{n}$ to $\mathbb{H}$ satisfy the
bound $|h_{n}(x+iy)|\leq e^{\frac{c}{y}}$ for $y>0$,

  2. (ii)

$\lim_{n\to\infty}h_{n}(x)=0$ for almost every $x\in F_{m}$,

  3. (iii)

$\lim_{n\to\infty}h_{n}(z)=1$ for every $z\in\mathbb{H}$,

  4. (iv)

there exists an $A>0$ and $p>2$ such that $|h(x)|^{p}w(x)<A$ for almost every
$x\in\mathbb{R}$.

###### Proof.

Property $(i)$, together with FourierDecayFromExtensionGrowthCorollary,
implies that the functions
$g_{n}(x)=\frac{h_{n}(x)}{(i+x)^{2}}\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$
obey the spectral bound
$|\widehat{g_{n}}(\zeta)|\leq\sqrt{\frac{\pi}{2}}e^{2\sqrt{c}\sqrt{\zeta}},\zeta>0$.
Since $\rho(\zeta)=\mathcal{O}\big{(}e^{-d\sqrt{\zeta}}\big{)}$ for some
$d>0$, the spectral bound implies that

$\sup_{n}\int_{\mathbb{R}_{+}}|\widehat{g_{n}}|^{2}\rho\,d\zeta<\infty$

if $c$ is small enough. Together with $(iv)$, we see that
$Jg_{n}=(g_{n},\widehat{g_{n}})$ forms a bounded subset of the Hardy subspace
of product space
$\mathcal{L}^{2}(\mathbb{R},w\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$.
We may thus assume, by passing to a subsequence, that $\\{Jg\\}_{n}$ tends
weakly in $\mathcal{H}(w,\rho)$ to some tuple $(h,k)$. In fact, $(iv)$ implies
that $\\{g_{n}\\}_{n}$ is a sequence bounded in
$\mathcal{L}^{p}(\mathbb{R},w\,dx)$, so we have that
$h\in\mathcal{L}^{p}(\mathbb{R},w\,dx)$ for some $p>2$. The fact that $h\equiv
0$ on $F_{m}$ is a consequence of the weak convergence of $g_{n}$ to $h$ and
the condition $(ii)$, which implies $\lim_{n\to\infty}g_{n}(x)=0$ for almost
every $x\in F_{m}$. Moreover, by the formula in H1FourierTransformFormula, we
have

$\widehat{g_{n}}(\zeta)=e^{y\zeta}\int_{\mathbb{R}}\frac{h_{n}(x+iy)}{(i+x+iy)^{2}}e^{-ix\zeta}d\lambda(x)$

for any $y>0$. The integrand converges pointwise to
$\frac{e^{-ix\zeta}}{(i+x+iy)^{2}}$ by $(iii)$, and it is dominated pointwise
by the integrable function

$x\mapsto\frac{e^{\frac{c}{y}}}{1+x^{2}},\quad x\in\mathbb{R}.$

The dominated convergence theorem implies that

$\lim_{n\to\infty}\widehat{g_{n}}(\zeta)=e^{y\zeta}\int_{\mathbb{R}}\frac{1}{i+x+iy}e^{-ix\zeta}\,d\lambda(x)=\int_{\mathbb{R}}\frac{1}{(i+x)^{2}}e^{-ix\zeta}\,d\lambda(x)=\widehat{\Psi}(\zeta),$

where
$\Psi(x):=\frac{1}{(i+x)^{2}}\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$.
In the next-to-last equality we used H1FourierTransformFormula backwards. Weak
and pointwise convergence implies, as previously, that $k=\widehat{\Psi}$.
Since $J\Psi\in\mathcal{H}(w,\rho)$, we have that
$(h,k)-J\Psi=(h-\Psi,0)\in\mathcal{H}(w,\rho)$. The function $h-\Psi$ is non-
zero almost everywhere on $F_{m}$. Indeed, $h$ vanishes on $F_{m}$, and
$\Psi(x)$ is non-zero everywhere on $F_{m}$. Also, since
$h\in\mathcal{L}^{p}(\mathbb{R},w\,dx)$, we have
$h-\Psi\in\mathcal{L}^{p}(\mathbb{R},w\,dx)$. The conditions to apply
invSubspaceCorollary are thus satisfied, and by the invariance of
$\mathcal{H}(w,\rho)$ under the operators $U^{*}_{s}$ defined in Section 3, we
conclude that

$\mathcal{L}^{2}(E,w\,dx)\oplus\\{0\\}\subset\mathcal{H}(w,\rho),$

where $F_{m}\subset E:=\\{x\in\mathbb{R}:|h(x)-\Psi(x)|>0\\}$. Since $m$ is
arbitrary, we conclude that

$\mathcal{L}^{2}(\cup_{m}F_{m},w\,dx)\oplus\\{0\\}=\mathcal{L}^{2}(\text{res}(w),w\,dx)\oplus\\{0\\}\subset\mathcal{H}(w,\rho).$

This is sufficient to conclude the validity of claim2. ∎

###  5.2. An estimate for Poisson integrals

Let $\mu$ be a finite real-valued measure on $\mathbb{R}$. The Poisson
integral of $\mu$ is the harmonic function
$\mathcal{P}_{\mu}:\mathbb{H}\to\mathbb{R}$ which is given by the formula

$\mathcal{P}_{\mu}(z):=\int_{\mathbb{R}}\mathcal{P}(t,z)\,d\mu(t)=\frac{1}{\pi}\int_{\mathbb{R}}\frac{y}{(x-t)^{2}+y^{2}}\,d\mu(t),\quad
z=x+iy\in\mathbb{H}.$

By the triangle inequality, and an estimation of $\mathcal{P}(t,z)$ by its
supremum $\frac{1}{\pi y}$ over $\mathbb{R}$, we easily obtain the inequality

$\big{|}\mathcal{P}_{\mu}(z)\big{|}\leq\frac{|\mu|(\mathbb{R})}{\pi y},\quad
z=x+iy\in\mathbb{H},$

and where $|\mu|$ denotes the usual variation of the measure $\mu$. We obtain
a much better inequality for measures $\mu$ which are oscillating rapidly. The
following lemma is the half-plane version of an estimate in [11, Lemma 3.2].

######  Lemma 5.2.

PsnIntegralEstimate Let $\mu$ be a finite real-valued measure on $\mathbb{R}$
which has the following structure: there exists a finite sequence of disjoint
intervals $\\{I_{j}\\}_{j}$ of $\mathbb{R}$, and a decomposition
$\mu=\sum_{j}\mu_{j}$, where $\mu_{j}$ is a real-valued measure supported
inside $I_{j}$, $\mu_{j}(I_{j})=0$, and $|\mu_{j}|(I_{j})\leq C$ for some
$C>0$ which is independent of $j$. Then

$\big{|}\mathcal{P}_{\mu}(x+iy)\big{|}\leq\frac{C}{\pi y},\quad
x+iy\in\mathbb{H}.$

###### Proof.

Since $-\mu$ satisfies the same conditions as $\mu$, and so does any
translation of $\mu$, it suffices to prove that
$\mathcal{P}_{\mu}(y)\leq\frac{C}{\pi y}$ for any $y>0$. We have

$\mathcal{P}_{\mu}(y)=\sum_{j}\frac{1}{\pi}\int_{\mathbb{R}}\frac{y}{t^{2}+y^{2}}d\mu_{j}(t).$

If $\mu_{j}=\mu_{j}^{+}-\mu_{j}^{-}$ is the decomposition of $\mu_{j}$ into
its positive and negative parts, then we have the estimate

$\displaystyle\mathcal{P}_{\mu}(y)$
$\displaystyle\leq\frac{1}{\pi}\sum_{j}\sup_{t\in
I_{j}}\frac{y}{t^{2}+y^{2}}\cdot\mu^{+}_{j}(I_{j})-\inf_{t\in
I_{j}}\frac{y}{t^{2}+y^{2}}\cdot\mu^{-}_{j}(I_{j})$
$\displaystyle\leq\frac{C}{2\pi}\sum_{j}\sup_{t\in
I_{j}}\frac{y}{t^{2}+y^{2}}-\inf_{t\in I_{j}}\frac{y}{t^{2}+y^{2}}$
$\displaystyle:=\frac{C}{2\pi}S$

In the second step we used that
$\mu_{j}(I_{j})=\mu_{j}^{+}(I_{j})-\mu_{j}^{-}(I_{j})=0$ and that
$|\mu_{j}|(I_{j})=\mu_{j}^{+}(I_{j})+\mu_{j}^{-}(I_{j})\leq C$. Since the
intervals $\\{I_{j}\\}_{j}$ are disjoint, and the function
$t\mapsto\frac{y}{t^{2}+y^{2}}$ is increasing for $t<0$, decreasing for $t>0$,
and attains a maximum value of $1/y$ at $t=0$, the sum $S$ in the above
estimate cannot be larger than $\frac{2}{y}$ (it is easily seen to be bounded
by twice the height of the graph of $t\mapsto\frac{y}{t^{2}+y^{2}}$). The
estimate follows. ∎

###  5.3. The construction

In accordance with the earlier discussion in Section 5.1, we recall the
decomposition (5.1) and assume below that $F:=F_{m}$ is a bounded subset of
$\text{res}(w)\cap\\{x\in\mathbb{R}:w(x)>\delta\\}$ for some $\delta>0$. The
set $F$ inherits the following property from $\text{res}(w)$: if $I$ is an
interval, and $|I\cap F|>0$, then $\int_{I}\log w\,dx=-\infty.$

######  Lemma 5.3.

pieceLemma If $|I\cap F|>0$ for some finite interval $I$, then given any $c>0$
and any $p>0$, there exists $D>0$ and a measurable subset $E_{I}\subset I$
disjoint from $F$ for which we have

(5.2) $\int_{E_{I}}\min\big{(}p^{-1}\log^{+}(1/w),D\big{)}\,dx=c.$

###### Proof.

On the set $F$, $\log(w)$ is bounded from below by $\log\delta$. Hence
$\int_{I\setminus F}\log w\,dx=-\infty$. Consequently,

$\lim_{D\to+\infty}\int_{I\setminus
F}\min\big{(}p^{-1}\log^{+}(1/w),D\big{)}\,dx=\int_{I\setminus
F}p^{-1}\log^{+}(1/w)\,dx=+\infty.$

So for $D$ sufficiently large we will have $\int_{I\setminus
F}\min\big{(}p^{-1}\log^{+}(1/w),D\big{)}\,dx>c$, and then by the absolute
continuity of the finite positive measure
$\min\big{(}p^{-1}\log^{+}(1/w),D\big{)}\,dx$ we may choose a set
$E_{I}\subset I\setminus F$ for which (5.2) holds. ∎

We will now construct the sequence in hnSplittingSequence. Let $p>2$,
$\\{c_{n}\\}_{n\geq 1}$ be some sequence of positive numbers which tends to
$0$ slowly enough so that $c_{n}2^{n}$ tends to $+\infty$, and let $F$ be as
above. Fix some integer $n>0$, cover $\mathbb{R}$ by a sequence of (say, half-
open) disjoint intervals of length $2^{-n}$ and let $\\{\ell_{k}\\}_{k}$ be
those intervals for which $|\ell_{k}\cap F|>0$. Apply pieceLemma with
$c=c_{n}$ to each of the intervals $\ell_{k}$ to obtain a corresponding
constant $D_{k}>0$ and a set $E_{k}:=E_{\ell_{k}}\subset\ell_{k}$ for which
(5.2) holds. We set $d\mu(t)=\log W(t)\,dt$, where

$\log
W(t)=\sum_{k}\min\big{(}p^{-1}\log^{+}(1/w(t)),D_{k}\big{)}\mathbbm{1}_{E_{k}}(t)-\frac{c_{n}}{|\ell_{k}\cap
F|}\mathbbm{1}_{\ell_{k}\cap F}(t).$

Then $\mu$ is an absolutely continuous real-valued measure with bounded
density $\log W$, $\mu(\ell_{k})=0$ and $|\mu|(\ell_{k})=2c_{n}$. We construct
$h_{n}\in\mathcal{H}^{\infty}(\mathbb{R})$ by letting the logarithm $\log
h_{n}(z)$ of its extension to $\mathbb{H}$ be given by the right-hand side of
the formula (2.7). Then, by PsnIntegralEstimate, we have the inequalities

$|\mathcal{P}_{\mu}(x+iy)|\leq\frac{2c_{n}}{\pi y},\quad z=x+iy\in\mathbb{H}$

and consequently, since $|h(z)|=e^{\mathcal{P}_{\mu}(z)}$, we have the bounds

(5.3) $e^{-2c_{n}/\pi y}\leq|h_{n}(z)|\leq e^{2c_{n}/\pi y},\quad
z=x+iy\in\mathbb{H}.$

Since $c_{n}\to 0$, by multiplying $h_{n}$ by an appropriate unimodular
constant, we may assume by (5.3) that

(5.4) $\lim_{n\to\infty}h_{n}(z)=1$

for every $z\in\mathbb{H}$ (possibly after passing to a subsequence). Also,
for almost every $x\in F$, we have

(5.5) $|h_{n}(x)|=e^{-c_{n}/|\ell_{k}\cap F|}\leq e^{-c_{n}2^{n}}.$

Our assumption on $c_{n}$ then implies that $\lim_{n\to\infty}h_{n}(x)=0$ for
almost every $x\in F$. For almost every $x\in\mathbb{R}\setminus F$, we have
instead

(5.6) $|h_{n}(x)|^{p}w(x)=e^{p\log W(x)}w(x)\leq e^{\log^{+}(1/w(x))}w(x)\leq
1.$

The equations (5.3), (5.4), (5.5) and (5.6) show that the sequence
$\\{h_{n}\\}_{n}$ satisfies the conditions stated in hnSplittingSequence. Thus
claim2 holds, and consequently the proof of CondensationTheorem is complete.

##  6\. Proof of SparsenessTheorem

###  6.1. A bit of concave analysis

Let $M:\mathbb{R}_{+}\to\mathbb{R}_{+}$ be an increasing and concave function
which is differentiable for $x>0$. We assume that $M(0)=0$, and that

(6.1) $M(x)\leq\sqrt{x},\quad x>0.$

For a function $M$ with the above properties, the integrals

(6.2) $I_{M}(y):=\int_{0}^{\infty}e^{M(x)-2yx}\,dx$

converge for every $y>0$, and estimation of the growth of $I_{M}(y)$ as $y\to
0^{+}$ will be of importance in the proof of SparsenessTheorem. To estimate
$I_{M}$, we define a function $M_{*}$ in the following way. Since $M$ is
increasing, concave and differentiable, the derivative $M^{\prime}(x)$ is
defined for $x>0$, and it is a positive and decreasing function. The concavity
of $M$ implies that

(6.3) $M^{\prime}(x)\leq M(x)/x$

from which it follows by (6.1) that

$\lim_{x\to\infty}M^{\prime}(x)=0.$

It is only the asymptotic behaviour of $M$ as $x\to\infty$ that concerns us,
so we will also assume for convenience that $\lim_{x\to
0^{+}}M^{\prime}(x)=+\infty$. In this case, the inverse function

(6.4) $K(y):=(M^{\prime})^{-1}(y),\quad y\in(0,\infty)$

is well-defined and positive. It is decreasing, and satisfies $\lim_{y\to
0^{+}}K(y)=+\infty$. We set

(6.5) $M_{*}(y):=M(K(y)),\quad y\in(0,\infty)$

The function $M_{*}$ is decreasing, and satisfies $\lim_{y\to
0^{+}}M_{*}(y)=+\infty$. The integrals $I_{M}(y)$ can be estimated in terms of
$M_{*}$.

######  Proposition 6.1.

IMestimate For $M$ as above, we have

$I_{M}(y)\leq\frac{2e^{M_{*}(y)}}{y^{2}}$

for all sufficiently small $y>0$.

###### Proof.

We will need the following observation. The inequality (6.3) together with
(6.1) implies that $M^{\prime}(x)\leq\frac{1}{\sqrt{x}}$ if $x>0$ is
sufficiently large. We want to show the inequality

$K(y)\leq\frac{1}{y^{2}},$

for sufficiently small $y>0$. Set $K(y)=x$ and $\frac{1}{y^{2}}=x_{*}$. Then

$M^{\prime}(x)=y=\frac{1}{\sqrt{x_{*}}}\geq M^{\prime}(x_{*})$

if $y>0$ is sufficiently small (and consequently $x_{*}$ is sufficiently
large). Since $M^{\prime}$ is a decreasing function, the above inequality
shows that $x_{*}\geq x$, which is the same as the desired inequality.

We split the integral (6.2) at $x=K(y)$. Both pieces can be estimated very
crudely. For the first piece, we have

$\displaystyle\int_{0}^{K(y)}e^{M(x)-2yx}\,dx$
$\displaystyle\leq\int_{0}^{K(y)}e^{M(K(y))}\,dx$
$\displaystyle=K(y)e^{M_{*}(y)}$
$\displaystyle\leq\frac{e^{M_{*}(y)}}{y^{2}}.$

We used our initial observation in the last step. For the second piece, we
note that

$\sup_{x>0}\,M(x)-yx=M(K(y))-yK(y)\leq M_{*}(y).$

Indeed, the supremum is attained at the point $x$ where $M^{\prime}(x)=y$,
which by definition is $x=K(y)$. Thus

$\displaystyle\int_{K(y)}^{\infty}e^{M(x)-2yx}\,dx\leq$
$\displaystyle\int_{K(y)}^{\infty}e^{M_{*}(y)-xy}\,dx$ $\displaystyle\leq$
$\displaystyle e^{M_{*}(y)}\int_{0}^{\infty}e^{-yx}\,dx$ $\displaystyle=$
$\displaystyle\frac{e^{M_{*}(y)}}{y}$ $\displaystyle\leq$
$\displaystyle\frac{e^{M_{*}(y)}}{y^{2}}.$

In the last inequality we require that $y\in(0,1)$. We obtain the desired
estimate by combining the estimates for the two pieces of the integral. ∎

In the proof of SparsenessTheorem, a point will come up where we will need
integrability of $M_{*}$ near the origin. The next proposition describes the
functions $M$ corresponding to $M_{*}$ which are integrable in this way.

######  Proposition 6.2.

MstarIntegrabilityProp For $M$ as above, the following two statements are
equivalent.

  1. (i)

$\int_{0}^{\delta}M_{*}(y)\,dy<\infty$ for some $\delta>0$.

  2. (ii)

$\int_{1}^{\infty}(M^{\prime}(x))^{2}\,dx<\infty$.

###### Proof.

We start with the integral in $(i)$ above and implement the change of variable
$y=M^{\prime}(x)$. Then $dy=M^{\prime\prime}(x)\,dx$ and $M_{*}(y)=M(x)$, and
by next using integration by parts, we obtain

$\displaystyle\int_{0}^{\delta}M_{*}(y)\,dy$
$\displaystyle=-\int_{K(\delta)}^{\infty}M(x)M^{\prime\prime}(x)\,dx$
$\displaystyle=\lim_{R\to+\infty}-M(R)M^{\prime}(R)+M(K(\delta))M^{\prime}(K(\delta))$
$\displaystyle+\lim_{R\to+\infty}\int_{K(\delta)}^{R}(M^{\prime}(x))^{2}\,dx.$

The assumptions (6.1) and (6.3) together imply that the quantity
$M^{\prime}(R)M(R)$ stays bounded, as $R\to+\infty$. Thus the desired
equivalence follows from the above computation. ∎

######  Remark 6.3.

ConcFuncRemark We should point out that a concave function indeed exists which
satisfies all of our conditions. For instance, note that any concave function
$M$ which for large positive $x$ coincides with

$M(x):=\frac{\sqrt{x}}{\log x}$

satisfies the equivalent conditions of the above proposition. Indeed, one
verifies by differentiation that the right-hand side is concave on some
interval $[A,\infty)$ and that the integral in $(ii)$ of
MstarIntegrabilityProp converges. Such a function $M$ can be easily chosen to
satisfy all assumptions made in this section. Moreover, clearly it satisfies
(4.1).

###  6.2. A growth estimate for Hardy class functions

######  Proposition 6.4.

growthestimateUpperHalfplane Let
$f\in\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$ have a
Fourier transform $\widehat{f}$ satisfying

$\int_{\mathbb{R}_{+}}|\widehat{f}|^{2}\rho\,d\zeta\leq C$

for some constant $C>0$ and $\rho(\zeta)=e^{-M(\zeta)}$. Then there exist a
positive constant $\delta$ such that the analytic extension of $f$ to
$\mathbb{H}$ satisfies the estimate

$|f(x+iy)|\leq\sqrt{2C}\frac{e^{M_{*}(y)}}{y},\quad y\in(0,\delta).$

###### Proof.

By developments of Section 2.1, we have

$f(z)=\int_{\mathbb{R}}\frac{f(t)}{t-z}\frac{dt}{2\pi
i}=\int_{\mathbb{R}}f(t)\overline{\psi_{z}(t)}d\lambda(t),\quad
z=x+iy\in\mathbb{H},$

and where $\psi_{z}\in\mathcal{H}^{2}(\mathbb{R})$ is as in
FourierTransformCauchyKernel. By Plancherel’s theorem and the lemma, we obtain

$\displaystyle f(z)$
$\displaystyle=\int_{\mathbb{R}_{+}}\widehat{f}(\zeta)e^{iz\zeta}\,d\lambda(\zeta)$
$\displaystyle=\int_{\mathbb{R}_{+}}\widehat{f}(\zeta)\sqrt{\rho(\zeta)}\frac{e^{-y\zeta+ix\zeta}}{\sqrt{\rho(\zeta)}}\,d\lambda(\zeta)$

An application of Cauchy-Schwarz inequality leads to

(6.6)
$|f(z)|\leq\sqrt{C}\sqrt{\int_{0}^{\infty}e^{M(\zeta)-2y\zeta}\,d\zeta}.$

Now IMestimate applies to obtain the desired estimate. ∎

###  6.3. Construction of the compact set

If $M$ satisfies the equivalent conditions of MstarIntegrabilityProp, then the
logarithm of the right-hand side in the inequality of
growthestimateUpperHalfplane, namely

(6.7) $H(y):=\frac{\log(2C)}{2}+M_{*}(y)-\log y,\quad y\in(0,\delta),$

is, for small enough $\delta>0$, positive and integrable over the interval
$y\in(0,\delta)$. To $H$ and any $A>0$ we will associate a Cantor-type compact
set $E$ contained in $[0,A]$ which contains no intervals and for which the
integral

(6.8) $\int_{[0,A]\setminus E}H(\text{dist}(x,E))\,dx$

converges. Here $\text{dist}(x,E)$ denotes the distance from the point
$x\in[0,A]$ to the closed set $E$. Let $\mathcal{U}=\\{\ell\\}$ be the system
of maximal disjoint open intervals, union of which constitutes the complement
of $E$ within $(0,A)$. The convergence of the integral above is easily seen to
be equivalent to the convergence of the sum

(6.9) $\sum_{\ell\in\mathcal{U}}\int_{0}^{|\ell|}H(x)\,dx$

where $|\ell|$ denotes the length of the interval $\ell$. To construct $E$, we
choose a sequence of numbers $\\{L_{n}\\}_{n\geq 1}$ which is so quickly
decreasing that

(6.10) $\sum_{n=1}^{\infty}2^{n}\int_{0}^{L_{n}}H(x)\,dx<\infty$

and

(6.11) $\sum_{n=1}^{\infty}2^{n}L_{n}\leq A/2.$

From such a sequence, we construct $E$ as in the classical Cantor set
construction. We set $E_{0}=[0,A]$, and recursively define a compact set
$E_{n+1}$ contained in $E_{n}$. The set $E_{n+1}$ consists of $2^{n+1}$ closed
intervals in $[0,A]$ which we obtain by removing from the $2^{n}$ closed
intervals $\\{E_{n,i}\\}_{i=1}^{2^{n}}$ constituting $E_{n}$ an open interval
of length $L_{n}$ lying in the middle of $E_{n,i}$. Thus splitting each
$E_{n,i}$ into two new closed intervals. The above summation condition (6.11)
ensures that $|E_{n}|>A/2$, and so $E:=\cap_{n=0}^{\infty}E_{n}$ has positive
Lebesgue measure which is not less than $A/2$. The integral condition (6.8)
holds by its equivalence to (6.9) and by (6.10). Clearly $E$ contains no
intervals.

###  6.4. Collapse of the Fourier transforms

We are now ready to prove claim3. We will do so by showing that
$\mathcal{H}(\mathbbm{1}_{E},\rho)$ does not contain any non-zero tuple of the
form $(0,k)$, $k\in\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$, where $M$ is as
in ConcFuncRemark, for instance, and where $E$ is as in Section 6.9. We set
$\rho=e^{-M}$.

######  Lemma 6.5.

FourierCollapseLemma Let $E$, $M$and $\rho$ be chosen as above. Assume that
$\\{f_{n}\\}_{n}$ is a sequence of functions in
$\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$, each of which
has an analytic extension to a half-space larger than $\mathbb{H}$. If
$\lim_{n\to\infty}f_{n}=0$ in the norm of
$\mathcal{L}^{2}(E,\mathbbm{1}_{E}\,dx)$ and the sequence of Fourier
transforms $\\{\widehat{f_{n}}\\}_{n}$ satisfies

$\sup_{n}\int_{\mathbb{R}_{+}}|\widehat{f_{n}}|^{2}\rho\,d\zeta<\infty,$

then we have

$\lim_{n\to\infty}f_{n}(z)=0,\quad z\in\mathbb{H}.$

The convergence is uniform on compact subsets of $\mathbb{H}$.

In the proof of FourierCollapseLemma given below we will use a technique of
Khrushchev from [8] for estimating harmonic measures on certain domains. For
general background on the theory of harmonic measures, see [5] or [14].

Let $\mathcal{U}=\\{\ell\\}$ be the collection of finite open intervals
complementary to $E$, and let

$T_{\ell}:=\\{x+iy\in\mathbb{H}:x\in\ell,y\leq\text{dist}(x,E)\\}$

be a triangle with base at $\ell$. We define
$\Omega=R\setminus\big{(}\cup_{\ell\in\mathcal{U}}T_{\ell}\big{)}$ to be the
bounded domain in $\mathbb{H}$ which consists of a rectangle $R$, with a base
being the shortest closed interval containing the set $E$, with the triangles
$T_{\ell}$ removed from $R$. See Figure 1. An observation that Khrushchev made
regarding this type of domains is the following property of their harmonic
measure.

######  Lemma 6.6.

KhruschevsEstimateLemma Let $E$, $M$ and $\rho$ be chosen as above, and let
$H$ be given by (6.7). Let $\Omega$ be the domain described above. If
$\omega_{z}$ is the harmonic measure of the domain $\Omega$ at any point
$z\in\Omega$, then

$\int_{\partial\Omega\cap\mathbb{H}}H(\operatorname{Im}t)\,d\omega_{z}(t)<\infty.$

We emphasize that $\partial\Omega\cap\mathbb{H}$ equals
$\partial\Omega\setminus\mathbb{R}$.

###### Proof.

The proof is very similar to the one given by Khrushchev in [8], only minor
details differ. If $\ell=(a,b)$ is one of the finite intervals complementary
to $E$, and $T_{\ell}$ is the triangle standing on top of it, then we denote
by $A(s)$ the part of the boundary of $T_{\ell}$ which lies above the interval
$(a,a+s)\subset\mathbb{R}$, $0<s<|\ell|/2$. If $u$ is the harmonic measure in
$\mathbb{H}$ of the interval $(a,a+s)$, then it is easy to see from the
explicit formula

$u(x+iy)=\frac{1}{\pi}\int_{a}^{a+s}\frac{y}{(x-t)^{2}+y^{2}}\,dt,\quad
x+iy\in\mathbb{H}$

that $u(x+iy)\geq\frac{1}{2\pi}$ for $x+iy\in A(s)$. Since $u$ is harmonic and
continuous in the closure of $\Omega$ except possibly at the two points $a$
and $a+s$, the reproducing formula $\int_{\partial\Omega}u\,d\omega_{z}=u(z)$
holds, and so

$\omega_{z}(A(s))=\int_{A(s)}d\omega_{z}\leq\int_{\partial\Omega}2\pi
u\,d\omega_{z}=2\pi u(z)\leq\frac{2s}{y},\quad z=x+iy\in\mathbb{H}.$

We have used the positivity of $u$ and $\omega_{z}$ in the first inequality,
and the second one is an easy consequence of the explicit formula for $u$
above. Set $A_{1}:=A(|\ell|/2)$, which is the left side of the boundary of
$T_{\ell}$, and further set $A_{n}:=A(|\ell|/2^{n})$, $n\geq 1$. Then, since
$H$ is decreasing,

$\displaystyle\int_{\partial
T_{\ell}\cap\mathbb{H}}H(\operatorname{Im}t)d\omega_{z}(t)$
$\displaystyle=2\int_{A_{1}}H(\operatorname{Im}t)\,d\omega_{z}(t)$
$\displaystyle\leq
2\sum_{n=1}^{\infty}\int_{A_{n}-A_{n+1}}H(|\ell|/2^{n+1})\,d\omega_{z}(t)$
$\displaystyle\leq
2\sum_{n=1}^{\infty}H(|\ell|/2^{n+1})\frac{2|\ell|}{2^{n}y}$
$\displaystyle\leq\frac{16}{y}\int_{0}^{|\ell|}H(t)\,dt.$

Now the desired claim follows from (6.9). ∎

[scale=1]OmegaDomain

Figure 1. The domain $\Omega$ in the proof of FourierCollapseLemma. There is a
triangular tent between $\Omega$ and each complementary interval of $E$, and
$E$ lives on $\mathbb{R}$ inbetween the tents.

###### Proof of FourierCollapseLemma .

Note that it is sufficient to establish the claim that the sequence
$\\{f_{n}\\}_{n}$ contains a subsequence which converges pointwise in $\Omega$
to $0$. Indeed, the proof of growthestimateUpperHalfplane shows that our
assumption on the Fourier transforms $\widehat{f}$ implies pointwise
boundedness of the sequence $\\{f_{n}\\}_{n}$ on each half-plane
$\\{x+iy\in\mathbb{H}:y>\delta\\}$, $\delta>0$. Hence the sequence
$\\{f_{n}\\}_{n}$ forms a normal family on $\mathbb{H}$. If we establish the
above claim, then every subsequences of $\\{f_{n}\\}_{n}$ contains a further
subsequence convergent to $0$ in $\mathbb{H}$. This is equivalent to
convergence of the entire initial sequence $\\{f_{n}\\}_{n}$ to $0$.

Fix $z\in\Omega$. Since $\log|f_{n}|$ is a subharmonic function and
$\max(-N,\log|f_{n}|)$ is a bounded continuous function on $\partial\Omega$,
we obtain by the maximum principle for subharmonic functions that

$\log|f_{n}(z)|\leq\int_{\partial\Omega}\max(-N,\log|f_{n}(t)|)\,d\omega_{z}(t).$

We let $N\to+\infty$ and, by the monotone convergence theorem, obtain

$\displaystyle\log|f_{n}(z)|$
$\displaystyle\leq\int_{\partial\Omega}\log|f_{n}(t)|\,d\omega_{z}(t)$
$\displaystyle=\int_{\partial\Omega\cap\mathbb{H}}\log|f_{n}(t)|\,d\omega_{z}(t)+\int_{E}\log|f_{n}(t)|\,d\omega_{z}(t)$

The assumption in the lemma, growthestimateUpperHalfplane, the definition of
$H$ in (6.7) and KhruschevsEstimateLemma show that

$\int_{\partial\Omega\cap\mathbb{H}}\log|f_{n}(t)|\,d\omega_{z}(t)\leq\int_{\partial\Omega\cap\mathbb{H}}H(\operatorname{Im}t)d\omega_{z}(t)<A$

where $A$ is some positive constant which is independent of $n$. By Egorov’s
theorem, we may pass to a subsequence (the same subsequence for each
$z\in\Omega)$ and assume that $f_{n}$ converge uniformly to $0$ on some subset
$E^{\prime}$ of $E$ which is of positive Lebesgue measure. On $E\setminus
E^{\prime}$ we have the estimate

$\int_{E\setminus
E^{\prime}}\log|f_{n}(t)|\,d\omega_{z}(t)\leq\int_{E\setminus
E^{\prime}}|f_{n}(t)|^{2}\,d\omega_{z}(t)\leq\frac{1}{\pi\operatorname{Im}z}\int_{E\setminus
E^{\prime}}|f_{n}(t)|^{2}\,dx.$

The last inequality follows from monotonicity of the harmonic measure with
respect to domains (see [14, Corollary 4.3.9]), which applied to
$\Omega\subset\mathbb{H}$ leads to the inequality

$\omega_{z}(B)\leq\int_{B}\mathcal{P}(t,z)\,dt\leq\frac{|B|}{\pi\operatorname{Im}z}$

for any Borel subset $B$ of $E$. Thus
$d\omega_{z}\leq\frac{dx}{\pi\operatorname{Im}z}$, attesting the integral
inequality above. By the convergence of $f_{n}$ to $0$ in the norm of
$\mathcal{L}^{2}(E,\mathbbm{1}_{E}\,dx)$, the integrals $\int_{E\setminus
E^{\prime}}|f_{n}|^{2}dx$ are uniformly bounded by some constant $C>0$, and so
the above inequalities give

$\log|f_{n}(z)|\leq A+C+\int_{E^{\prime}}\log|f_{n}(t)|d\omega_{z}(t).$

But $\omega_{z}(E^{\prime})>0$, since the harmonic measure and the arc-length
measure on the rectifiable curve $\partial\Omega$ are mutualy absolutely
continuous (see [5, Theorem 1.2 of Chapter VI]), so since $|f_{n}|$ converge
uniformly to $0$ on $E^{\prime}$, the integral on the right-hand side above
converges to $-\infty$ as $n\to\infty$. Thus $|f_{n}(z)|\to 0$, and since
$z\in\Omega$ was arbitrary, the desired claim follows. ∎

The following proposition implies claim3 of Section 4, and so it also implies
SparsenessTheorem.

######  Proposition 6.7.

FourierCollapseProposition Let $E$, $M$and $\rho$ be chosen as above. If a
tuple of the form
$(0,k)\in\mathcal{L}^{2}(E,\mathbbm{1}_{E}\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho)$
is contained in the Hardy subspace $\mathcal{H}(\mathbbm{1}_{E},\rho)$, then
$k\equiv 0$.

###### Proof.

By the containment $(0,k)\in\mathcal{H}(\mathbbm{1}_{E},\rho)$ and
KernelContainmentHardySubspace, there exists a sequence $\\{f_{n}\\}_{n}$ of
functions in $\mathcal{H}^{1}(\mathbb{R})\cap\mathcal{H}^{2}(\mathbb{R})$
extending analytically across $\mathbb{R}$ and for which the tuples
$Jf_{n}=(f_{n},\widehat{f}_{n})$ converge in the norm of the product space
$\mathcal{L}^{2}(\mathbb{R},\mathbbm{1}_{E}\,dx)\oplus\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$
to $(0,k)$. By passing to a subsequence, we may assume that the Fourier
transforms $\widehat{f_{n}}$ converge pointwise almost everywhere on
$\mathbb{R}_{+}$ to $k$. One might attempt to prove the proposition by using
the formula in H1FourierTransformFormula, and observing that

$k(\zeta)=\lim_{n\to\infty}\widehat{f_{n}}(\zeta)=\lim_{n\to\infty}e^{y\zeta}\int_{\mathbb{R}}f_{n}(x+iy)e^{-ix\zeta}\,d\lambda(x)$

holds for almost every $\zeta\in\mathbb{R}_{+}$ and for any $y>0$. By
FourierCollapseLemma the integrand converges pointwise to $0$. However, an
appeal to the usual convergence theorems for integrals is not justified, and
we have to proceed more carefully. Note that since
$k\in\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$ and $\rho$ is bounded from
below on compact subsets of $\mathbb{R}_{+}$, in fact $k$ is locally
integrable on $\mathbb{R}$. It follows that we can interpret $k$ as a
distribution on $\mathbb{R}_{+}$. Thus to show that $k\equiv 0$, it suffices
to establish that $\int_{\mathbb{R}_{+}}k\phi\,d\lambda=0$ for every smooth
function $\phi$ which is compactly supported in $\mathbb{R}_{+}$.

Let $\phi$ be as above. Since we have that $\widehat{f_{n}}\to k$ in
$\mathcal{L}^{2}(\mathbb{R}_{+},\rho\,dx)$ and $\rho$ is bounded from below on
compact subsets of $\mathbb{R}_{+}$, we obtain

(6.12)
$\int_{\mathbb{R}_{+}}k\phi\,d\lambda=\lim_{n\to\infty}\int_{\mathbb{R}_{+}}\widehat{f_{n}}\phi\,d\lambda.$

Fix some small $y>0$. By H1FourierTransformFormula, we get that

$\widehat{f_{n}}(\zeta)=e^{y\zeta}\int_{\mathbb{R}}f_{n}(x+iy)e^{-ix\zeta}\,d\lambda(x).$

Plugging this formula into (6.12) and noting that the use of Fubini’s theorem
is permitted, we obtain

$\displaystyle\int_{\mathbb{R}_{+}}k\phi\,d\lambda$
$\displaystyle=\lim_{n\to\infty}\int_{\mathbb{R}}\Big{(}\int_{\mathbb{R}_{+}}\phi(\zeta)e^{y\zeta}e^{-ix\zeta}d\lambda(\zeta)\Big{)}f_{n}(x+iy)\,d\lambda(x)$
(6.13)
$\displaystyle=\lim_{n\to\infty}\int_{\mathbb{R}}D(x)f_{n}(x+iy)\,d\lambda(x)$

where

$D(x):=\int_{\mathbb{R}_{+}}\phi(\zeta)e^{y\zeta}e^{-ix\zeta}d\lambda(\zeta)$

is the Fourier transform of the compactly supported smooth function
$\zeta\mapsto\phi(\zeta)e^{y\zeta}$. As such, $D$ is certainly integrable on
$\mathbb{R}$. By FourierCollapseLemma, we have
$\lim_{n\to\infty}f_{n}(x+iy)=0$, and

$\sup_{n}\,\sup_{x\in\mathbb{R}}|f_{n}(x+iy)|<\infty$

holds by growthestimateUpperHalfplane. Therefore, this time, the dominated
convergence theorem applies to (6.13), and we conclude that

$\int_{\mathbb{R}_{+}}k\phi\,d\lambda=0.$

Thus $k$ is the zero distribution on $\mathbb{R}_{+}$, and therefore $k\equiv
0$. ∎

##  7\. Clumping for tempered distributions

In this last section, we indicate how one can derive
DistributionalClumpingTheorem from CondensationTheorem. We will skip most of
the details of the necessary computations, which are in any case standard.

Let $f$ be a function which satisfies

(7.1) $\int_{\mathbb{R}}\frac{|f(x)|}{(1+|x|)^{n}}\,dx<\infty$

for some positive integer $n$. Then $f$ can be interpreted as a tempered
distribution on $\mathbb{R}$ in the usual way, and so $f$ has a distributional
Fourier transform $\widehat{f}$. Our hypothesis is that $\widehat{f}$ is an
integrable function on some half-axis $[\zeta_{0},\infty)$ and that

(7.2)
$\rho_{\widehat{f}}(\zeta)=\mathcal{O}\big{(}e^{-c\sqrt{\zeta}}\big{)},\quad\zeta>\zeta_{0}.$

We may assume that $\zeta_{0}=0$. In order to prove
DistributionalClumpingTheorem, we will construct an appropriate multiplier
$m:\mathbb{R}\to\mathbb{C}$ with the property that $mf$ is a function to which
CondensationTheorem applies. In particular, the following properties will be
satisfied by $m$:

  1. (i)

$m(x)$ is a bounded function of $x\in\mathbb{R}$ which is non-zero for almost
every $x\in\mathbb{R}$.

  2. (ii)

$mf\in\mathcal{L}^{2}(\mathbb{R},dx)$,

  3. (iii)

$\rho_{\widehat{mf}}(\zeta)=\mathcal{O}\big{(}e^{-c\sqrt{\zeta}}\big{)}$ for
some $c>0$ and $\zeta>0$,

  4. (iv)

$\int_{I}\log|m|\,dx>-\infty$ for every interval $I\subset\mathbb{R}$.

If we construct such a multiplier $m$, then $(ii)$, $(iii)$ and
CondensationTheorem imply that $\log|mf|$ is locally integrable on an open set
$U$ which coincides, up to a set of measure zero, with
$\\{x\in\mathbb{R}:|f(x)m(x)|>0\\}$. By $(i)$, $U$ differs from
$\\{x\in\mathbb{R}:|f(x)|>0\\}$ at most by a set of measure zero. Moreover,
the formula $\log|f|=\log|fm|-\log|m|$ and $(iv)$ show that $\log|f|$ is
locally integrable on $U$. This proves DistributionalClumpingTheorem, as a
consequence of existence of a multiplier satisfying the above conditions. We
now show how to construct such a multiplier.

We set

$\Phi(x):=\frac{(-i)^{n}n!}{\sqrt{2\pi}(x-i)^{n}}$

and let $h$ be defined by the equation (2.7), with

$\log|h(x)|=\log\min(1,|f(x)|^{-1}),\quad x\in\mathbb{R}.$

The condition (7.1) ensures that $h$ is well-defined, and it is a member of
$\mathcal{H}^{\infty}(\mathbb{R})$. We put

$h_{*}(x):=\frac{h(x)}{(x+i)^{2}},\quad x\in\mathbb{R}.$

and finally

$m(x):=\Phi(x)\overline{h_{*}(x)},\quad x\in\mathbb{R}.$

Clearly, $m$ is bounded. Since $|f|$ is locally integrable on $\mathbb{R}$,
the set $\\{x\in\mathbb{R}:|f|=\infty\\}$ has measure zero. Consequently,
$|h_{*}|>0$ almost everywhere in $\mathbb{R}$, and so the desired property
$(i)$ of $m$ holds. The choice of $h_{*}$ and $\Phi$ ensures that $mf$ is both
bounded and integrable on $\mathbb{R}$, implying
$mf\in\mathcal{L}^{2}(\mathbb{R},dx)$, so that $(ii)$ above holds. Property
$(iv)$ holds by HadyClassLogIntProp, since
$h_{*}\in\mathbb{H}^{1}(\mathbb{R})$. So the critical property left to be
verified is the spectral estimate of $mf$ in $(iii)$ above.

######  Lemma 7.1.

With notation and definitions as above, the Fourier transform $\widehat{mf}$
satisfies

$\rho_{\widehat{mf}}(\zeta)=\mathcal{O}\big{(}e^{-c\sqrt{\zeta}}\big{)},\quad\zeta>0$

for some $c>0$.

###### Proof.

A standard argument shows that $\widehat{f\Phi}$ must coincide on
$\mathbb{R}_{+}$ with the convolution $\widehat{f}\ast\widehat{\Phi}$ (which,
note, is a function on $\mathbb{R}_{+}$). Indeed, let $s$ be a Schwartz
function which has a Fourier transform $\widehat{s}$ supported on some compact
interval $[a,b]$, $0<a<b$. Note that the function $\Phi s$ is also of Schwartz
class. It follows immediately from the integral definition of the Fourier
transform (1.1) that
$\widehat{\overline{\Phi}s}=\widehat{\overline{\Phi}}\ast\widehat{s}$, and
that $\widehat{\overline{\Phi}s}$ is supported on the interval $[a,\infty)$.
Hence, by the definition of the distributional Fourier transform, we obtain

$\int_{\mathbb{R}_{+}}\widehat{f\Phi}\,\overline{\widehat{s}}\,d\lambda=\int_{\mathbb{R}}f\Phi\overline{s}\,d\lambda=\int_{\mathbb{R}}f\overline{\overline{\Phi}s}\,d\lambda=\int_{\mathbb{R}_{+}}\widehat{f}\overline{(\widehat{\overline{\Phi}}\ast\widehat{s})}\,d\lambda.$

Fubini’s theorem and the computational rule
$\widehat{\overline{\Phi}}(x)=\overline{\widehat{\Phi}}(-x)$ shows that the
last integral above equals

$\int_{\mathbb{R}_{+}}(\widehat{f}\ast\widehat{\Phi})\,\overline{\widehat{s}}\,d\lambda,$

proving our claim about the structure of $\widehat{f\Phi}$ on
$\mathbb{R}_{+}$.

Hence $\widehat{f\Phi}$ is a bounded continuous function which coincides with

$\widehat{f\Phi}(\zeta)=\int_{\mathbb{R}}\widehat{f}(x)\widehat{\Phi}(\zeta-x)\,d\lambda(x)$

for $\zeta>0$. By a computation similar to the one in the proof of
FourierTransformCauchyKernel one sees that $\Phi$ has the Fourier transform

$\widehat{\Phi}(\zeta)=|\zeta|^{n}e^{\zeta}\mathbbm{1}_{\mathbb{R}_{-}}(\zeta).$

For such $\zeta$, we estimate

$\displaystyle\big{|}\widehat{f\Phi}(\zeta)\big{|}$
$\displaystyle\leq\int_{\mathbb{R}}|\widehat{f}(x)||\zeta-x|^{n}e^{\zeta-x}\mathbbm{1}_{\mathbb{R}_{-}}(\zeta-x)\,d\lambda(x)$
$\displaystyle=\int_{\zeta}^{\infty}|\widehat{f}(x)||\zeta-x|^{n}e^{\zeta-x}\,d\lambda(x)$
$\displaystyle=\sum_{k=0}^{\infty}\int_{\zeta 2^{k}}^{\zeta
2^{k+1}}|\widehat{f}(x)||\zeta-x|^{n}e^{\zeta-x}\,d\lambda(x).$

We now make the rather rough estimate

$|\zeta-x|^{n}e^{\zeta-x}\leq\zeta^{2(k+1)n},\quad x\in[\zeta 2^{k},\zeta
2^{k+1}],$

which gives

$\big{|}\widehat{f\Phi}(\zeta)\big{|}\leq\sum_{k=0}^{\infty}\rho_{\widehat{f}}(\zeta
2^{k})\zeta^{2(k+1)n}\leq
A\sum_{k=0}^{\infty}e^{-c\sqrt{\zeta}\sqrt{2}^{k}}\zeta^{2(k+1)n}$

for some $A>0$. The above sum can be readily estimated to be of order
$\mathcal{O}\big{(}e^{-d\sqrt{\zeta}})$ for some $d>0$ slightly smaller than
$c$. Since $fm$ is the product of two integrable functions $f\Phi$ and
$\overline{h_{*}}$, we have

$\widehat{fm}=\widehat{f\Phi}\ast\widehat{\overline{h_{*}}},$

where $\widehat{\overline{h_{*}}}(\zeta)=\overline{\widehat{h_{*}}(-\zeta)}$
is non-zero only for $\zeta<0$. Note that $h_{*}$ is integrable on
$\mathbb{R}$, and so $\widehat{h_{*}}$ is bounded. We obtain

$\displaystyle|\widehat{fm}(\zeta)|$
$\displaystyle\leq\int_{\mathbb{R}}\Big{|}\widehat{f\Phi}(x)\overline{\widehat{h_{*}}(x-\zeta)}\Big{|}\,d\lambda(x)$
$\displaystyle\leq
B\int_{\zeta}^{\infty}e^{-d\sqrt{x}}\,d\lambda(x)=\mathcal{O}\big{(}e^{-d\sqrt{\zeta}}\big{)},$

where $B$ is some positive constant. The desired estimate on
$\rho_{\widehat{fm}}$ follows readily from this estimate. ∎

By the above discussion, we have proved DistributionalClumpingTheorem.

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