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\begin{document}
\title{On the Growth of Averaged Weyl Sums for Rigid Rotations}
\author{S. De Bi\`evre\thanks{E-mail: debievre@mathp7.jussieu.fr }\and
G. Forni\thanks{E-mail: forni@dm.unibo.it and forni@dpmms.cam.ac.uk}}
\maketitle
%{\centerline \today}
\centerline{\it $^{\ast}$ UFR
de Math\'ematiques et Laboratoire de Physique Th\'eorique et Math\'ematique}
\centerline{\it Universit\'e Paris VII, 2 place Jussieu F-75251 Paris Cedex 05,
FRANCE}
\centerline{\it $^{\dag}$ Dipartimento di Matematica, Universit\`a di Bologna,
Bologna I-40127, ITALY}
{{\it\& Newton Institute, Cambridge, UK}\hfil}
\begin{abstract}
{\em Let $\omega\in\R\setminus\Q$ and $f\in L^2(\To)$. We study the asymptotic
behaviour of the {\it Weyl sums } $S(m,\omega)f(x) =\sum^{m-1}_{k = 0} f (x+k
\omega)$
and their averages ${\hat S}(m,\omega)f(x) ={1\over m}\sum^{m}_{j = 1}
S(j,\omega)f(x)$, in the $L^2$-norm. In particular, for a suitable
class of Liouville rotation numbers $\omega\in \R\setminus\Q$,
we are able to construct examples of functions $f\in H^s(\To)$,
$s>0$, such that, for all $\epsilon>0$, $||{\hat S}(m,\omega)f||_2
\geq C_{\epsilon} m^{{1\over {1+s}}-\epsilon}$ as $m\to \infty$.
In addition, for all $f\in H^s(\To)$, $\liminf m^{-{1\over{1+s}}}
(\log m)^{-1/2} \parallel {\hat S}(m,\omega)f \parallel_2<\infty$,
for all $\omega\in\R\setminus\Q$. }
\end{abstract}
\sect{Introduction and statement of the results}
We study the asymptotic behaviour of the following skew products on the
cylinder:
\begin{equation}
(x_0, v_0)\in \To \times \R\equiv T^*\To\to (x_1=x_0+\omega, v_1=v_0+f(x_0)).
\end{equation}
Here $\omega\in\R\setminus\Q$ and $f:\To\to\R,\quad \To = \R/\Z$.
After $m$ iterations
of the map we have
$$
x_m = x_0 + m\omega,\quad v_m = v_0 + S(m,\omega)f(x_0),
$$
where
the {\it Weyl sums } $S(m,\omega)f$ of the
function $f$ for the rigid rotation by $\omega$ are defined to be
\begin{equation}
S(m,\omega)f(x) =\sum^{m-1}_{k = 0} f (x+k\omega)\,\,.
\end{equation}
The asymptotic behaviour of the $v$-variable is therefore completely
determined by the asymptotic behaviour of the Weyl sums.
It is well known [KN] that, if $\{p_k/q_k\}_{k\geq 1}$ denotes the sequence
of approximants of the continued fraction expansion of the irrational
number $\omega$, then, for any $f\in BV(\To)$ of zero mean,
\begin{equation}
\parallel
S(q_k,\omega)f\parallel_\infty < {\em Var} (f) \quad {\rm for\ all}\
k\in \N\,.
\end{equation}
The upper bound $(1.3)$, known as the Denjoy-Koksma inequality, rules out
the possibility of having growth to infinity of Weyl sums for smooth
functions of zero mean. However, the set of ``times" where the bound $(1.3)$
holds is rather small since the $q_k$'s grow at least exponentially. Perhaps
then
$S(m,\omega)f$ still gets to be sufficiently large at intermediate times
resulting in growth of the averaged Weyl sums
\begin{equation}
{\hat S}(m,\omega)f(x) ={1\over m}\sum^{m}_{j = 1}
S(j,\omega)f(x).
\end{equation}
We will show this picture is essentially correct and obtain bounds on the growth
of
the $\hat S(m, \omega) f$. More precisely, we are interested in the $L^2$
growth, as $m\to +\infty$, of the averaged Weyl sums $(1.4)$
for $L^2$-functions $f$.
We write $f_n$ for the Fourier coefficients of the function $f\in L^2(\To)$,
i.e.
$$
f(x)=\sum_{n\in \Z} f_n \exp (2\pi inx)\, ,
$$
and recall that one can express the smoothness properties of $f$
in terms of its Fourier coefficients using the habitual Sobolev spaces
$H^s(\To)$:
\begin{equation}
H^s(\To)=\{f\in L^2(\To)\mid \parallel f\parallel_s^2:=
\sum_{n\in \Z} n^{2s}\,|f_n|^2\,<\,+\infty\}.
\end{equation}
To state and interpret our main results, we need some notation. We start by
recalling
the basic properties of the {\it continued fraction expansion }of irrational
numbers,
explained in detail in the book [K]. Let
$$\omega=[a_1,...,a_k,...]$$
be the continued fraction expansion of $\omega\in [0, 1[\setminus\Q$.
Then the sequence of approximants $\{p_k/q_k\}_{k\in {\N}}$ (where
$p_k/q_k=[a_1,...a_k]$) is given by the following recursive
identities [K,Th.1]: $p_0=0, q_0=1$, $p_1=1, q_1=a_1$ and, for $k\geq1$,
\begin{eqnarray}
p_{k+1}=a_kp_k+p_{k-1} \nonumber\\
q_{k+1}=a_kq_k+q_{k-1}\,.
\end{eqnarray}
The sequence of denominators $\{q_k\}$ has the following equivalent
definition [K, Ths.16 and 17]:
\begin{equation}
q_{k+1}:=\min\{j>q_k\,\mid\,\parallel j\omega\parallel<
\parallel q_k\omega\parallel\}\,\,,
\end{equation}
where $\parallel\cdot\parallel$ denotes the distance from the
nearest integer. As a consequence we have that
\begin{equation}
\parallel j\omega\parallel\geq \parallel q_k\omega\parallel
\quad {\rm for\ all}\ j0\,$ such that
\begin{equation}
\parallel q_k\omega\parallel\leq
{R \over {q_k ^{1+\gamma}}}\qquad\forall k\in \N.
\end{equation}
By the definition (1.10), ${\cal R}_{\gamma'}\subset {\cal R}_{\gamma}$
if $\gamma'\geq \gamma$ and by (1.9) ${\cal R}_0=\R\setminus\Q$.
It is a straigthforward consequence of $(1.6)$ that
${\cal R}_{\gamma}\not=\emptyset$ for any $\gamma>0$. In fact, a number
$\omega\in {\cal R}_{\gamma}$ is easily produced by constructing
recursively its continued fraction expansion according to
$(1.6)$ and by choosing at each step:
$
a_k\geq [q_k^{\gamma}]\,\,,
$
where $[\cdot]$ denotes the {\it integer part}. A recursive
estimate then shows that $\omega=[a_0,...,a_k,...]\in {\cal R}_{\gamma}$.
This construction produces Diophantine irrational numbers
$\omega\in {\cal R}_{\gamma}$
as well as Liouville ones. Actually, it is easy to see that
all ${\cal R}_\gamma$ are $G_\delta$ sets.
The intersection ${\cal R}_{\infty}$ of all sets
${\cal R}_{\gamma}$, for $\gamma\geq 0$, is a subset
of the
set of Liouville irrationals, and is of course still a $G_\delta$ dense set.
Note that it follows from (1.9)-(1.10) that
if $\omega\in {\cal R}_\gamma$, then
\begin{equation}
q_{k+1}\geq \frac{1}{2R}q_k^{1+\gamma}\,.
\end{equation}
This clearly shows the sequence $q_k$ is lacunary when $\gamma >0$
and this all the more so as $\gamma $ is larger. It will finally be useful
to consider, for $\omega \in {\cal R}_\gamma$ ($\gamma\geq 0$) and for $r\geq1$
the set
\begin{equation}
A_r(\omega)=
\{m\in\N\mid\exists \ell\in\N {\ \rm so \ that\ }
2q_{\ell +1}\leq m \leq 2r q_{\ell +1}\}.
\end{equation}
The points in $A_r(\omega)$ form a divergent subsequence of $\N$.
For $\gamma >0$, they
should be thought of as lying ``just above" one
of the $q_{\ell +1}$ and ``far below" the next one.
We can now state our main results. The first one gives lower bounds
on the averaged Weyl sums. To simplify the formulations, we will always
assume that
\begin{equation}
f_{0}=\int_{\To} f dx = 0.
\end{equation}
\begin{th} (i) Suppose $|f_n| \geq c |n|^{-\nu}$ for some $\nu>1/2$.
Let $\omega \in {\cal R}_\gamma$ for some $\gamma\geq 0$. Then for all $r\geq 1$
there exists a $C_r>0$ so that for all $m\in A_r(\omega)$
\begin{equation} \parallel {\hat S}(m,\omega)f\parallel_2\geq C_r
m^{(1-\frac{\nu}{1+\gamma})}. \end{equation}
Moreover, there exists a constant $C_{\gamma}>0$ so that $\forall
m\in\N$ \begin{equation}
\parallel {\hat S}(m,\omega)f\parallel_2\geq C_{\gamma}
m^{(1-\frac{\nu}{1+\gamma})\frac{1+\gamma}{1+\gamma+\gamma\nu}}.
\end{equation}
(ii) Let $\omega \in {\cal R}_\gamma$ and $s\geq 0$. Then there exists
a function $f\in H^s(\To)$ with the following properties. First,
for all $r\geq1$ and for all $\epsilon>0$, there is a constant
$C_{r,\epsilon}>0$ so that for all $m\in A_r(\omega)$
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel_2 \geq
C_{r,\epsilon}\, m^{1-{\,\,\,\,s\over{1+\gamma}}-\epsilon}\,\,.
\end{equation}
Moreover, there exists a $C_{\gamma,\epsilon}>0$ so that
{\em for all} $m\in \N$
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel_2\geq C_{\gamma,\epsilon}\,
m^{(1-{\,\,\,s\over{1+\gamma}}){{1+\gamma}\over{1+\gamma+\gamma s}}-\epsilon}.
\end{equation}
\end{th}
Note that (1.15) is optimised, for $\nu>1/2$ fixed, as $\gamma\to +\infty$,
so that for any Liouville irrational numbers $\omega \in
{\cal R}_{\infty}$ you get, $\forall \epsilon>0$:
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel_2
\geq C_\epsilon m^{{1\over {1+\nu}}-\epsilon},
\quad \forall m\in\N.
\end{equation}
Similarly, for $s>0$ fixed, (1.17) is optimised as $\gamma \to +\infty$.
Hence we get the following statement. For all $s\geq 0$ and for any Liouville
rotation number $\omega\in
{\cal R}_{\infty}$ there exists an
$f\in H^s(\To)$ such that, for all $\epsilon>0$,
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel_2 \geq C_\epsilon m^{{1\over
{1+s}}-\epsilon},\quad\forall m\in\N.
\end{equation}
Note that the lower bounds get smaller as $f$ is taken smoother. To see
that this is not due to a bad choice of rotation number, we prove the
following upper bound, which shows that the above lower bounds
are close to optimal.
\begin{th}
Let $\omega\in\R\setminus \Q$.
\begin{itemize}
\item[(i)] Let $s\geq 0$. Then there exists
a diverging
sequence of natural numbers $\{m_k\}_{k\in \N}$ such that, for
any $f\in H^s (\To)$
\begin{equation}
\parallel {\hat S}(m_k,\omega)f\parallel_2\leq\, C\parallel f
\parallel_s\, m_k^{1\over {1+s}}\sqrt{\log m_k}\,\,,
\end{equation}
where $\parallel \cdot\parallel_s$ denotes the $H^s$ norm.
In addition, if $\omega\in \R\setminus\Q$ satisfies $\parallel q_k\omega
\parallel \geq r/q_k^{1+\gamma},\quad \forall k\in\N$ and for some
$r>0$, then for all $m\in\N$
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel _2\,
\leq\, C\parallel f\parallel_s\, m^{1-{s\over {1+\gamma}}}
\sqrt{\log m}
\end{equation}
\item[(ii)] Let $\nu>1/2$ and $\epsilon>0$. Then there exists
a diverging sequence of natural numbers $\{m_k\}_{k\in \N}$ such that,
if $\mid f_n\mid\leq C \mid n\mid^{-\nu}$, then
\begin{equation}
\parallel{\hat S}(m_k,\omega)f\parallel _2\,
\leq\, C_{\epsilon}\, m_k^{{1\over {1/2+\nu}}+\epsilon}\,.
\end{equation}
In addition, if $\omega\in \R\setminus\Q$ satisfies $\parallel q_k\omega
\parallel \geq r/q_k^{1+\gamma},\quad \forall k\in\N$ and for some
$r>0$, then $\forall \epsilon, \ \exists C_\epsilon$, so that for all
$m\in\N$
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel _2\,
\leq\, C_\epsilon\, m^{1-{\nu-1/2\over 1+\gamma}+\epsilon}\,.
\end{equation}
\end{itemize}
\end{th}
Note that (1.20) and (1.22) should be compared to (1.19) and (1.18)
respectively, while (1.21) and (1.23) to (1.16) respectively (1.14).
We can now explain the general picture emerging from these
results. To fix ideas, let $|f_n| = |n|^{-\nu} (\nu > 1/2)$.
Then $f\in H^s(\To)$ for any $s<\nu-1/2$. We wish to find irrational
$\omega$ so that $\parallel {\hat S}(m,\omega)f\parallel_2$ goes to
infinity as fast as possible. At first, (1.14) suggests that by taking
$\gamma$ large we could achieve a growth exponent arbitrarily close
to $1$. However (1.22) shows this is illusory since it gives an
$\omega$-independent exponent ${1\over {\nu+1/2}}+\epsilon$, decreasing
with $\nu$, in the upper bound. The best lower bound we were able to get
is (1.15), which is optimized as $\gamma \to +\infty$, yielding (1.18). Note
that there also the exponent tends to zero as $\nu$ tends to infinity.
Similarly, (1.20) shows that the growth of $\parallel {\hat
S}(m,\omega)f\parallel_2 $ is increasingly slow as $f$ is taken
smoother and smoother, which shows (1.19) is close to optimal.
In fact, as the proofs will show, the behaviour
of $\parallel {\hat S}(m,\omega)f\parallel_2 $ is controlled by two
competing effects. Note first that, if $\omega\in\Q$, then both
$\parallel S(m,\omega)f\parallel_2 $ and $\parallel{\hat S}(m,\omega)
f\parallel_2 $ will typically grow like $m$. On the other hand, if
$\omega\in\R\setminus\Q$, the ergodic theorem tells us that
$\parallel{ S}(m,\omega)f\parallel_2 ={\it o}(m)$, hence $\parallel{
\hat S}(m,\omega)f\parallel_2 ={\it o}(m)$. This suggests that
irrationals well approximated by rationals ($\gamma$ large) are the
best candidates for producing growth in $\parallel{\hat S}
(m,\omega)f\parallel_2 $. This is further corroborated by (1.14)
and (1.23), which show there is a competing effect between the smoothness
of $f$ and the Diophantine properties of the rotation number $\omega$ and
that for $\gamma$ sufficiently large the growth exponent approaches $1$.
However, as $\gamma$ becomes larger, so do the gaps between successive
$q_k$, since $q_{k+1}$ is at least of the order of $q_k^{1+\gamma}$. This
is at the origin of the exponent of (1.15) which gives a weaker growth than
expected.
For $\omega \in {\cal R}_\infty$, one sees, comparing (1.14) and (1.22), that
the averaged Weyl sums continue to fluctuate between
$m^{{1\over 1/2+\nu}+\epsilon}$ and $m^{1-\epsilon}$, pushing up against
the upper bound imposed by the ergodic theorem. So, although averaging did
allow us to beat Denjoy-Koksma, the averaged Weyl sums still fluctuate and
do not have an asymptotic power law behaviour, for example.
In fact, our lower bounds predict superdiffusive behaviour, i.e.
$$ \parallel{\hat S}(m,\omega)f\parallel_2 \geq C m^\tau $$
for some $\tau > 1/2$, for $1/2<\nu<1$ and for suitable rotation numbers
$\omega\in {\cal R}_{\gamma}$, for $\gamma$ sufficiently
large. On the other hand, the upper bounds in Theorem 1.2, in particular
$(1.20)$, rule out diffusive or superdiffusive behaviour for all $f\in H^s(S^1)$
with $s>1$ and for any choice of an irrational rotation number.
There are several reasons for our interest in the model (1.1).
Note that it is a symplectic transformation of $T^*\To$ with symplectic
form $dx\wedge dv$, that can be seen as a perturbation of
$$
(x, v)\in T^* \To \to(x+\alpha, v)\in T^*\To
$$
which is completely integrable with invariant tori $v = c$
[G][Be][B]. If $f$ is in $C^k(\To)$ and $\omega$ sufficiently poorly
approximated by the rationals (i.e. Diophantine of sufficiently high order),
it is easy to see that those tori
are preserved. It suffices to solve
\begin{equation}
f(x) = g(x+\omega)
- g(x) \end{equation} for a smooth $g$.
Our results above deal with the opposite extreme, when $\omega$ is
sufficiently well approximated by the rationals, i.e. weakly Diophantine
or Liouville. The tori are then broken and the motion is unbounded. Actually,
unboundedness of the Weyl sums $\parallel{ S}(m,\omega)f\parallel_2 $
(i.e. $\sup_m \parallel{ S}(m,\omega)f\parallel_2 =\infty$) is known to be
equivalent to the non-solvability of (1.24) for $g\in L^2(\R, dx)$ [L].
The only previous work that we are aware of discussing the asymptotic
behaviour of $v_m$ is [B]. There an argument is proposed claiming to
show that the Weyl sums $\parallel S(m,\omega)f\parallel_2 $ themselves,
before averaging, with $f_n \cong
|n|^{-\nu}$ satisfy the lower bound (1.14) {\em for all values of} $m$
and $\nu>1/2$. As we already pointed out, this can not be true for $\nu
>1$ because of Denjoy-Koksma. As it turns out, Theorem 1.2 shows that the
result is not true for the averaged Weyl sums either, as one might have
hoped. Only when $\nu<1$, when Denjoy-Koksma is no longer an
obstruction, does the following result indeed give a non-trivial lower bound:
\begin{th}
Suppose $|f_n|\geq c|n|^{-\nu}$ for some $\nu>1/2$. Then
there exists a $C>0$ so that for all $\omega\in\R\setminus\Q$ and
for all $m\in\N$
\begin{equation}
\parallel{ S}(m,\omega)f\parallel_2 \geq C
m^{1-\nu}.
\end{equation}
\end{th}
Once the invariant tori are broken, it is legitimate to ask whether the
map (1.1) is ergodic on the cylinder $T^* \To$ with respect to Lebesgue measure.
Note that (1.1) can be viewed as a map on $\To\times\To$ as well. The Anzai
theorem
[P1] shows that it is ergodic on $\To\times\To$ as soon as (1.24) is not
soluble.
Ergodicity on $T^*\To$ is more delicate. In [HL], the following result is shown.
If $f\in C^1([0,1]),\ f(0)\not= f(1)$ and if $\omega$ is irrational, then the
map
(1.1) is ergodic on the cylinder
$\To\times\R$, so that
$S(m,\omega)f(x)$ is unbounded for almost all $x$.
In this case, the Fourier coefficients of $f$ satisfy
$|f_n|>c |n|^{- 1}$ so that Theorem 1.1(i) as well as (1.18) apply with $\nu=1$.
A particular example that has been much studied is
$$
f_\beta (x) = \chi_{[0,\beta]}(x) -\beta,
$$
for some $0<\beta <1$. It is a classical result of Kesten [Ke][P2] that
$S(m,\omega)f$ is bounded iff $\beta\in\Z\omega$ (mod $1$). The asymptotic
behaviour of $\hat S(m,\omega)f$ is important in the study
of incommensurate one-dimensional structures, where the following model was
proposed [AG]. With $u_m$ denoting the $m$-th
atomic position, define the sequence $u_m$ recursively by $(0<\xi\betac>0$ such that the
following inequalities hold:
\begin{equation}
c/\sin^2\pi x\leq G_m(x)\leq C/\sin^2\pi x\qquad \forall\
1/m\leq x \leq 1/2.
\end{equation}
and
\begin{equation}
c m^2\leq G_m(x)\leq C m^2\qquad\forall \ 0\leq x\leq 1/m
\end{equation}
In addition, the upper bounds in (2.3) and (2.4) hold for all $x\in\R$.
\end{lem}
\noindent {\bf Proof. } We know from basic calculus that
\begin{equation}
{2\over \pi} \leq {{\sin \pi x}\over {\pi x}}\leq 1
\quad {\rm for\ all}\,\, 0\leq x\leq 1/2\,\,,
\end{equation}
where the upper bound holds for all $x\in\R$. It follows from
(2.5) that
\begin{equation}
{2\over \pi}\leq |{{\sin \pi mx}\over{m\sin \pi x}}|=|{{\sin \pi mx}
\over {mx}}{x\over{\sin \pi x}}|\leq {\pi\over 2}\,\,,
\end{equation}
where the lower inequality holds for $0\leq x\leq 1/2m$ and the
upper inequality for all $x\in \R$, by evenness and periodicity.
Thus the inequality
$$
G_m(x)\leq C_1/\sin^2\pi x \quad \forall
\, x\in\R
$$
holds by choosing $C_1$ large enough:
$$
{1\over 4}\Bigl(1\,+\,\max_{x\in\To}|{{\sin \pi mx} \over{m\sin \pi x}}|
\Bigr)^2\leq{1\over 4}\Bigl(1\,+\,{\pi\over 2}\Bigr)^2\leq C_1.
$$
Secondly, the inequality
\begin{equation}
G_m(x)\geq c_1/\sin^2\pi x \quad \forall
1/2\geq x\geq 1/m
\end{equation}
holds since by $(2.5)$,
$$
|{{\sin \pi mx}\over {m\sin \pi x}}|\leq {1\over {m\sin\pi x}}
\leq {1\over 2} \quad \forall\, {1\over m}\leq x
\leq 1/2\,\,,
$$
which implies
$$
|1\,-\,e^{\pi i(m+1)x}\,
{{\sin \pi mx}\over {m\sin \pi x}}|^2\geq \Bigl(1-
|{{\sin \pi mx}\over {m\sin \pi x}}|\Bigr)^2\geq 1/4\,\, ,
$$
so that we can take $c_1 \leq 1/16$ in (2.7). This proves (2.3).
In order to prove (2.4), write
$$ u_m(x):=1\,-\,{{\sin \pi m x}\over {m\sin \pi x}}\,\,
e^{\pi i(m+1)x}\,. $$
There exist $C>c>0$ such that,
$$ c mx\leq |u_m(x)|\leq C mx \qquad \forall 0\leq x\leq 1/m. $$
In fact, the estimate from above can be obtained as follows. Since
$$ u_m(x)= 1-{{\sin \pi m x}\over {\pi mx}}+{{\sin \pi m x}\over {\pi mx}}\,
(1-{\pi x\over \sin \pi x})+{{\sin \pi m x}\over {m\sin \pi x}}
\,(1-e^{\pi i(m+1)x})\,,$$
the conclusion follows by basic calculus, (2.5) and (2.6).
As to the estimate from below, we will consider separately
the two intervals $0\leq x\leq 1/2m$ and $1/2m\leq x\leq 1/m$.
In the first case, since $\,u_m(0)=0\,$ and, by a straigthforward
computation,
$$
u_m'(x)=\Bigl(\pi {{\cos\pi x}\over {\sin \pi x}}-
\pi m {{\cos \pi mx}\over {\sin \pi mx}}-i\pi(m+1)\Bigr)
{{\sin \pi m x}\over {m\sin \pi x}}\,e^{\pi i(m+1)x}\,\,,
$$
the conclusion follows by the mean value theorem and (2.6).
In the second case, it can be noticed that the above argument
proving (2.7) holds true for $x\geq 1/2m$.
Since $\,G_m(x)=|u_m(x)|^2/4\sin^2 \pi x\,$, (2.4) follows. \endproof
\vskip10pt
The functions $\frac{\sin^2\pi mx}{\sin^2\pi x}$
and $G_m(x)$ are traced in Fig. 2.1. Note that the behaviour of
both functions is similar in the region $0\leq x\leq {1\over 4m}$,
where they are both of order $m^2$. For bigger values of $x$ the
difference manifests itself in that
unlike
$\frac{\sin^2\pi mn\omega}{\sin^2\pi n\omega}$, $G_m(x)$ has no zeros
(see (2.3)).
This eventually explains the lower bounds of Theorem 1.1
which we can now start proving.
\vglue0.5cm
\noindent {\bf Proof of Theorem 1.1.}
(i) Let $\omega\in {\cal R}_\gamma$ and denote as before by $\{p_k/q_k\}
_{k\in \N}$ the sequence of truncations of its continued fraction
expansion. We can assume that $1+\gamma-\nu>0$, since otherwise the bounds
in (1.14)-(1.15) are trivially satisfied. By Lemma 2.1 a lower bound
for the averaged Weyl
sum ${\hat S}(m,\omega)f$ can be obtained as follows:
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel^2_2\geq
c \Bigl(\sum_{\parallel n\omega\parallel \geq 1/m}
|f_n|^2 \parallel n\omega\parallel ^{-2}
\,+\, m^2\sum_{\parallel n\omega\parallel < 1/m}
|f_n|^2\Bigr)\,.
\end{equation}
Suppose now that $m\in A_r(\omega)$, so that, for some $\ell\in\N$
$$
\frac{1}{2rq_{\ell+1}} \leq \frac{1}{m} \leq \frac{1}{2 q_{\ell +1}}.
$$
Hence (1.9) implies
$$
\frac{1}{m}\leq\parallel q_\ell\omega\parallel\leq \frac {2r}{m}.
$$
Using (2.8) we therefore get
$$
\parallel{\hat S}(m,\omega)f\parallel^2_2\geq
c
|f_{q_\ell}|^2 \parallel q_\ell\omega\parallel ^{-2},
$$
from which (1.14) follows upon using (1.11).
To prove (1.15), we proceed as follows.
Let $(k_m)_{m\in\N}$ be the sequence of natural numbers defined in the
following way:
\begin{equation}
k_m=\max\{k\in {\N}\,|\, \parallel q_k\omega\parallel\geq 1/m\}\,.
\end{equation}
Then
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel^2_2\geq
c\Bigl(|f_{q_{k_m}}|^2 q_{k_m+1}^2\,+\, m^2|f_{q_{k_m}+1}|^2\Bigr)
\end{equation}
and, by the hypothesis on the Fourier coefficients of $f$,
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel_2\geq
c\Bigl(q_{k_m}^{-\nu} q_{k_m+1}\,+\, m q_{k_m+1}^{-\nu}\Bigr)\,.
\end{equation}
Consider now for all
$k\in\N$ the functions
\begin{equation}
S_k(x)=q_k^{-\nu} q_{k+1}+x q_{k+1}^{-\nu}\,\,\,\,\hbox{ and }\,\,
\,\, S_{k,\alpha}(x)=x^{-\alpha} S_k(x)\,,\,\,\,\,x\in \R^+\,.
\end{equation}
The idea is to find the exponent $\alpha\in (0,1)$ so that all the functions
$S_{k,\alpha}$ have a common strictly positive lower bound on $\R^+$. In fact, basic
calculus shows that $S_{k,\alpha}$ has a unique minimum on $\R^+$ and that
\begin{equation}
\min S_{k,\alpha}=c_{\alpha} q_k^{-\nu(1-\alpha)}
q_{k+1}^{1-(1+\nu)\alpha}\,\,.
\end{equation}
where $c_{\alpha}>0$ and depends only on $\alpha\in (0,1)$. Since
$\omega\in {\cal R}_{\gamma}$, by (1.11),
\begin{equation}
\min S_{k,\alpha}\geq c_{\alpha}
q_{k+1}^{(1-\alpha)(1-{\nu\over {1+\gamma}})-\nu\alpha}\,\,.
\end{equation}
Thus, we may choose $\alpha\in (0,1)$ such that
\begin{equation}
(1-\alpha)(1-{\nu\over {1+\gamma}})-\nu\alpha=0\,
\end{equation}
thereby obtaining
\begin{equation}
S_{k,\alpha}(x)\geq c_{\alpha}>0\,\,\,\,\forall x\in \R^+\,.
\end{equation}
A short computation of the exponent given by (2.15) yields (1.15) since,
by (2.11) and (2.16),
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel_2\geq
c S_{k_m}(m)\geq c m^{\alpha}S_{k_m,\alpha}(m)\geq c'_{\alpha} m^{\alpha}\,.
\end{equation}
(ii) Let $\omega\in {\cal R}_\gamma$. Consider
\begin{equation}
f(x) = \sum_{k\in\N} {1\over k} q_k^{-s} \exp 2\pi i q_k x.
\end{equation}
Clearly $f\in H^s(\To)$. Also, for all $s'>s,\, \exists c_{s'}>0$ so that
$|f_{q_k}| \geq c_{s'}\,q_k^{-s'}$
for all $k$. The reasoning is now identical to the one in (i),
with $s'$ replacing $\nu$.
\endproof
\vglue 0.5cm
We now turn to the proof of Theorem 1.2 which deals with upper bounds
along subsequences for the averaged Weyl sums. We start with a lemma.
\begin{lem}
Suppose that $\omega\in \R\setminus\Q$
satisfies the following property for some $\gamma>0$: there exist a constant
$R>0$ and an infinite set ${\cal K}\subset\N$ such that,
for any $k\in {\cal K}$,
\begin{equation}
\parallel q_k\omega\parallel <
{R\over {q_k^{1+\gamma}}}\,\,.
\end{equation}
Define, for any $0<\gamma^-\leq\gamma,
\ \,m_k=[q_k^{1+\gamma^-}/R]\,$ for $k\in {\cal K}$. Then:
\begin{itemize}
\item[(i)] If $f\in H^s(\To)$ for some $s\geq 0$, then
\begin{equation}
\parallel {\hat S}(m_k,\omega)f\parallel_2 \leq C
\parallel f\parallel_s\,\Bigl(m_k^{1-{s\over
{1+\gamma^-}}}\,+\,m_k^{1\over {1+\gamma^-}}\Bigr)\,\,;
\end{equation}
\item[(ii)] If, in particular,
$\mid f_n\mid \leq C \mid n\mid^{-\nu}$, then
\begin{equation}
\parallel {\hat S}(m_k,\omega)f\parallel_2 \leq C
\,\Bigl(m_k^{1-{{\nu - 1/2}\over
{1+\gamma^-}}}\,+\,m_k^{1\over {1+\gamma^-}}\Bigr)\,\,.
\end{equation}
\end{itemize}
\end{lem}
\noindent {\bf Proof.} We split the series $(2.2)$ as follows:
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel_2^2=\sum_{n1/2$, then $f\in H^{\nu^-}(S^1)$, for all
$\nu^-<\nu-1/2$.
\endproof
\vglue 0.5cm
\noindent {\bf Proof of Theorem 1.2.} Lemma 2.2 already contains
the main part of the proof of Theorem 1.2. In fact, suppose $f\in H^s(\To)$.
Then, Lemma 2.1 and (1.8) imply
\begin{equation}
\parallel{\hat S}(m,\omega)f\parallel_2^2\leq C
\Bigl(\sum_{jFrom (2.34) and (2.36) equation (1.20) now follows.
The proof of (1.22) follows from (1.20)
by noticing that if the Fourier coefficients of $f$ satisfy the estimates
$|f_n|\leq C n^{-\nu}$, $\nu>1/2$, then $f\in H^{\nu^-}(S^1)$, for all
$\nu^-<\nu-1/2$.
Finally, to prove (1.21) and (1.23), it is enough to use the hypothesis
on $\omega$ in (2.31).
\endproof
\vglue 0.5cm
\noindent {\bf Proof of Theorem 1.3.} We have
\begin{eqnarray*}
\parallel S(m,\omega)f\parallel_2^2&\geq&c\sum_{n\in \Z}
{1\over |n|^{2\nu}} \frac{\sin^2\pi mn\omega}{\sin^2\pi n\omega}\\
& \geq &c\sum_{k}{1\over {q_k}^{2\nu}}
\frac{\sin^2\pi m\parallel q_k\omega\parallel}{\sin^2\pi
\parallel q_k\omega\parallel}\,\,. \end{eqnarray*}
Let $k'_m=\min\{k\,\mid\, \parallel q_k\omega\parallel<{1\over 2m}\}$,
then
$$
\parallel S(m,\omega)f\parallel_2\geq C\ m\ q_{k'_m}^{-\nu}.
$$
But, since $\parallel q_{k'_m-1}\omega \parallel\geq {1\over 2m}$,
(1.9) implies $2m\geq q_{k'_m}$, so that
$$
\parallel S(m,\omega)f\parallel_2\geq C_\nu\ m^{1-\nu},
$$
proving the result.
\endproof
\noindent{\bf Acknowledgements:}
Stephan De Bi\`evre would like to thank the
University and the Research Centre of Crete, where part of
this work was performed, for their hospitality. Giovanni Forni
is grateful to the UFR de Math\'ematiques et LPTM, Universit\'e
de Paris VII, for their very kind hospitality during Spring 1995
while most of this work was performed.
\vskip25pt
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\end{document}