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%Mi tema: 
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\title{Bishop operators and Diophantine approximation}
\institute{Rutgers Number Theory Seminar}
\date{December 3, 2019}
\author[Fernando Chamizo]{Fernando Chamizo {\footnotesize (UAM-ICMAT)}}
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\begin{document}


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\begin{frame}[noframenumbering]
\titlepage
\end{frame}
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\begin{frame}
\tableofcontents
\end{frame}
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\begin{frame}

\vskip15pt

\begin{miblock}
{\footnotesize\color{gray}\textbf{Short abstract.} One of the oldest unsolved problems in functional analysis is the invariant subspace problem.  E. Bishop proposed a family of potential counterexamples depending on an irrational parameter and 20 years later A. M. Davie strongly contradicted Bishop's intuition proving that it only could be a counterexample for Liouville numbers. 
The purpose of this talk is to illustrate this noticeable interplay between number theory and functional analysis and to present a recent joint work with E.~Gallardo, M.~Monsalve and A.~Ubis.

}
\end{miblock}

\vskip10pt

\begin{miblock}[colframe=ct0,boxrule=.5pt,no shadow,standard jigsaw,opacityback=.5]
{\textbf{\color{ct00}Warning.} The emphasis here is in the \emph{application} of number theory not in the strength of the number theoretical results proven with this purpose.

}
\end{miblock}

\end{frame}
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\section{Invariant subspaces}
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% SLIDE  3 %
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\begin{frame}
\frametitle{The invariant subspace problem}

\begin{miblock}
\textbf{\color{ct0}Does every bounded linear operator on a (complex, $\infty$-dim., separable)
Hilbert space have a nontrivial closed invariant subspace?}
\end{miblock}

\medskip

These are natural hypotheses

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\node[anchor=west,rectangle,rounded corners,draw=ct2,fill=white,line width=.5pt,text width=1.5cm,align=center] (a3) at (-0.3,-3) {\strut separable};
\only<2->{%
\node[anchor=west] (b1) at (2,-1) {Rotations on $\R^2$};
\draw[-latex,line width=.8pt,ct2] (a1)--(b1);}
\only<3->{%
\node[anchor=west] (b2) at (2,-2) {otherwise too easy {\small(linear algebra)}};
\draw[-latex,line width=.8pt,ct2] (a2)--(b2);}
\only<4>{%
\node[anchor=west] (b3) at (2,-3) {otherwise too easy {\small(take the closure of $\text{span}\{T^n x_0\}$)}};
\draw[-latex,line width=.8pt,ct2] (a3)--(b3);}
\end{tikzpicture}

\medskip

\footnotesize{Recall: invariant means $T(M)\subset M$. We focus on closed subspaces $M$ to avoid silly examples ``erasing'' the boundary of the full Hilbert space.}


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\begin{frame}

Every known Hilbert space satisfying the hypotheses has an invariant subspace. Should we bet that invariant subspaces always exist?

\medskip

\begin{miblock}
\punto\ \textbf{\color{ct0}Yes?} Lomonosov (1973). True for any operator commuting with a compact operator.

\smallskip

\punto\ \textbf{\color{ct0}No?} Enflo (1976-1987), Read (1984). There are counterexamples in Banach spaces. In fact there exists an operator on $\ell^1$ without invariant subspaces (Read, 1985). 
\end{miblock}

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% SLIDE  5 %
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\begin{frame}

Gossiping about the invariant subspace problem:

{\setbeamertemplate{itemize item}{\color{ct11}$\blacktriangleright$}
\begin{itemize}
\item Lomonosov's proof of his theorem is astonishingly simple and elegant. His paper is only two pages long.
\item It made obsolete the previous partial results on the problem. Even during some years it was unclear if Lomonosov's proof had fully solved it because any known operator satisfied the hypotheses.
\item Enflo's paper is 101 pages long, took 7 years between  submission and publication and 10 years since the initial announcement.
\item Read's first paper is 64 pages long and it was published while Enflo's paper was in the refereeing process. 
\end{itemize}}

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% SLIDE  6 %
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\begin{frame}

\vskip10pt

\begin{miblock}
\color{gray}

\punto[gray] \textbf{Yes?} Lomonosov (1973). True for any operator commuting with a compact operator.

\smallskip

\punto[gray] \textbf{No?} Enflo (1976-1987), Read (1984). There are counterexamples in Banach spaces. In fact there exists an operator on $\ell^1$ without invariant subspaces (Read, 1985).
\end{miblock}

The counterexamples are too complicated to give any insight.

Let's ask for a third opinion

\begin{miblock}
\punto\ \textbf{\color{ct0}Maybe (not)?} Bishop (ca. 1950) suggested that the operator
\[
f(t)\mapsto t\,f\big(\{t+\alpha\}\big)
\]
on $L^2[0,1)$ with $\alpha\in\R-\Q$ and $\{\cdot\}=$ fractional part, could be a counterexample. 
\end{miblock}

\end{frame}
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% SLIDE  7 %
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\begin{frame}
\frametitle{Bishop operators}

Let us define the \textbf{Bishop operators} with a certain scaling
\begin{miblock}[halign=flush center,math,colframe=ct2,boxrule=1pt]
Tf(t) =  et\, f\big(\{t+\alpha\}\big)
\qquad\text{\footnotesize $e=2.718182\dots$}
\end{miblock}

{\footnotesize 
It is the composition of two well-known operators: a shift operator (with a continuous spectrum) and a multiplication operator (not invertible and no eigenfunctions). 
If $\alpha\in\Q$ the subspace of functions with small support around $k\alpha$ is invariant. }

\

Possible rationale for it:

\begin{miblock}
The equidistribution of $\{k\alpha\}$ for 
$\alpha\not\in\Q$ mixes the support and the multiplication avoid eigenfunctions. If this implies that $\{T^kf\}$ gives everything we have solved the problem in the negative.
\end{miblock}

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\begin{frame}

\vskip10pt

\begin{miblock}[halign=flush center,math,colframe=ct2,boxrule=1pt]
Tf(t) =  et\, f\big(\{t+\alpha\}\big)
\qquad\text{\footnotesize $e=2.718182\dots$}
\end{miblock}

\begin{tcolorbox}[empty,halign=flush center,size=minimal]
$f=\sin^2(\pi x)$,\qquad $\alpha=\sqrt{2}$

\vskip5pt

\begin{tikzpicture}
\node[bloque,fill=white] (a) {\includegraphics[height=65.9485pt, keepaspectratio=true]{./images/T10.eps}};
\node[bloque,fill=white,above=-5pt] at (a.north) {$T^{10} f$};
\end{tikzpicture}\quad
\begin{tikzpicture}
\node[bloque,fill=white] (a) {\includegraphics[height=65.9485pt, keepaspectratio=true]{./images/T20.eps}};
\node[bloque,fill=white,above=-5pt] at (a.north) {$T^{20} f$};
\end{tikzpicture}

\vskip5pt

\begin{tikzpicture}
\node[bloque,fill=white] (a) {\includegraphics[height=65.9485pt, keepaspectratio=true]{./images/T29.eps}};
\node[bloque,fill=white,above=-5pt] at (a.north) {$T^{29} f$};
\end{tikzpicture}\quad
\begin{tikzpicture}
\node[bloque,fill=white] (a) {\includegraphics[height=65.9485pt, keepaspectratio=true]{./images/T30.eps}};
\node[bloque,fill=white,above=-5pt] at (a.north) {$T^{30} f$};
\end{tikzpicture}
\end{tcolorbox}

\end{frame}
%%%%%%%%%%%%
\section{Davie theorem}
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% SLIDE  9 %
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\begin{frame}
\frametitle{Davie theorem}

More than 20 years later, A.M. Davie dealt Bishop's intuition a heavy blow. 

\medskip

\begin{miblock}[colframe=ct0,boxrule=1pt]
\textbf{{\color{ct00}Theorem} (Davie, 1974)}.
If $\alpha$ is not a Liouville number then $T$ has nontrivial invariant subspaces.
\end{miblock}

\medskip

Nevertheless many authors have continued the research on $T$ and its variants. It seems that it is a natural operator to consider.

\begin{miblock}
The folklore conjecture is the opposite of the original one: 
$T$ has nontrivial invariant subspaces for every $\alpha$.
\end{miblock}

\end{frame}
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% SLIDE 10 %
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\begin{frame}

\vskip10pt

Gossiping about Davie's paper:

{\setbeamertemplate{itemize item}{\color{ct11}$\blacktriangleright$}
\begin{itemize}
\item Halmos (1974) 
was rather tough in Mathematical reviews. ``Several canons of both the pedantic grammarian and the mathematical expositor are violated.  There is no hint to what motivates the proof [\dots] the techniques make use of nontrivial facts about subjects as diverse as diophantine approximations, Banach algebras, and quasi-analytic classes. The proofs appear, however, to be correct [\dots]'' 
\item Halmos (1987) praises Davie and the result in his photo book. 
\item Flattot (2008) says that the proof is ``very elliptic''.
\end{itemize}}

\begin{miblock}
The proof is based on a method introduced by Wermer and extended by Atzmon. When it is applied, Davie had to confront some basic questions about Diophantine approximation. 
\end{miblock}

\end{frame}
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\section{Atzmon theorem}
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% SLIDE 11 %
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\begin{frame}
\frametitle{Atzmon theorem {\footnotesize (weak version)}}

\begin{miblock}[colframe=ct0,boxrule=1pt]
\textbf{{\color{ct00}Theorem} (Atzmon, 1984)}.
$L:H\longrightarrow H$, $f,g\in H-\{0\}$ with 
\vspace{-5pt}
\[
\|L^{\pm n} f\|,
\|(L^*)^{\pm n} g\|
\le
W(n),
\qquad
W(n)=
\exp\Big(
\dfrac{n}{(\log n)^{1+\epsilon}}
\Big)
\]

\vskip-5pt
\hskip1.5cm {\small $+$ technical condition (easy for Bishop's operators)}

\vskip5pt
\hskip2cm \structure{$\Rightarrow$} $L$
has a nontrivial invariant subspace.
\end{miblock}

{\small 
\textbf{Sketch}

\vskip1mm 

{\color{ct0}\hrule}

\vskip2mm

Take
$\phi(z)=\sum a_kz^k$, $\psi(z)=\sum b_kz^k$ with $\psi\cdot\phi=0$  on $S^1$ and $a_kW(|k|),b_kW(|k|)$ small {\footnotesize (possible by Denjoy-Carleman Theorem)}.

\medskip

Define $u=\phi(L)f$, $v=\psi(L^*)g$ \quad{\footnotesize ($\ne 0$ by the technical condition)}.

$M=\text{span}\{L^n u\}$ is invariant and it is nontrivial because $v\in M^\perp$:
\[
\langle L^n u,v\rangle
=
\langle \phi(L)L^nf,\psi(L^*)g\rangle
=
\langle (\psi\phi)(L)L^nf,g\rangle
=0.\qed
\]

}

\end{frame}
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\section{Results}
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% SLIDE 12 %
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\begin{frame}

\vskip15pt

\begin{miblock}[colframe=ct0,boxrule=.5pt,no shadow]
\textbf{\color{ct00}Folklore conjecture}. The Bishop operators 
$Tf(t) = et\, f\big(\{t+\alpha\}\big)$
have nontrivial invariant subspaces for every $\alpha\not\in\Q$.
\end{miblock}

\textbf{Davie (1974)}. 
True for 
$\big|\alpha-\frac{a}{q}\big|>q^{-N}$, \quad {\small ($N$ arbitrary)}.

{\footnotesize\begin{center}
{\color{ct00}\dotfill(MacDonald, other authors)\dotfill}
\end{center}}

\textbf{Flattot (2008)}. 
True for 
$\big|\alpha-\frac{a}{q}\big|>\exp\big(-q^{1/3}\big)$, \quad {\small ($\frac 13\leftrightarrow\frac 12-\epsilon$)}.

\medskip

\begin{miblock}
\textbf{Our contribution} (joint with Gallardo, Monsalve, Ubis 2018).
\smallskip

$\bullet$ Short proof of Flattot result.
\smallskip

$\bullet$ True for 
$\big|\alpha-\frac{a}{q}\big|>\exp\big(-q^{1-\epsilon}\big)$.
\smallskip

$\bullet$ For 
$\big|\alpha-\frac{a}{q}\big|\not >\exp\big(-\frac{Cq}{\log q}\big)$
$\nexists f,g$ to which Atzmon theorem can be applied {\footnotesize (no possibility of improvement with the known methods)}.
\end{miblock}

\end{frame}
%%%%%%%%%%%%
\section{Diophantine approx.}
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% SLIDE 13 %
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\begin{frame}
\frametitle{Diophantine approximation}

\begin{miblock}[halign=flush center,math,colframe=ct0,boxrule=1pt]
Tf(t) =  et\, f\big(\{t+\alpha\}\big)
\qquad\text{\footnotesize $e=2.718182\dots$}
\end{miblock}

A direct substitution proves
\[
T^nf(t)
=e^{L_n(t)}f\big(\{t+n\alpha\}\big),
\quad
T^{-n}f\big(\{t+n\alpha\}\big)
=e^{-L_n(t)}f(t)
\]
and similar formulas for $T^*$ with
\[
L_n(t)=\sum_{k=0}^{n-1}\big(1+\log\{t+k\alpha\}\big)
\qquad \text{\footnotesize note $\int_0^1\log=-1$}.
\]

\begin{miblock}[colframe=ct0,boxrule=.5pt,no shadow,standard jigsaw,opacityback=.5]
\textbf{\color{ct00}Idea} (Davie). Choose the support of $f$ in such a way that $t$ is never close ($\text{mod. }1$) to $m\alpha$, $m\in\Z$.
\end{miblock}

\begin{miblock}[colframe=ct0,boxrule=.5pt,no shadow,standard jigsaw,opacityback=.5]
\textbf{\color{ct00}Bad guys}.  If $\alpha$ is very close to a rational, $L_n$ is amplified.
\end{miblock}

\end{frame}
%%%%%%%%%%%%
\section{Proofs}
%%%%%%%%%%%%
% SLIDE 14 %
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\begin{frame}
\frametitle{Matching Flattot's result}

Take $f=\textbf{1}_{\mathcal{D}}$ with 
$
\mathcal{D}
=
\big\{t\,:\,
\langle t\pm n\alpha\rangle >\frac{1}{2019n^2},\ n\in\Z^+
\big\}
$ (Davie)
\smallskip

$a/q =$\, convergent of $\alpha$
\vspace{-10pt}
\[
L_n(t)
=
\sum_{k=0}^{n-1}\big(1+\log\{t+k\alpha\}\big)
\ll
q+\frac{n+q}{q}\log(n+q).
\]

\vskip-10pt

\begin{miblock}[colframe=ct0,boxrule=.5pt,no shadow,standard jigsaw,opacityback=.5]
\textbf{\color{ct00}Idea}. Each $q$-block contributes $O(\log n+\log q)$.
\end{miblock}

\vskip10pt

{\small 
\textbf{Short proof of the result by Flattot}

{\color{ct0}\hrule}

\vskip2mm
 
Atzmon theorem requires 
$L_n(t)\ll \frac{n}{(\log n)^{1+\epsilon}}$. It is achieved if
$\frac{\log n}{q} \ll \frac{1}{(\log n)^{1+\epsilon}}$
for $n>q^{3/2}$. 

$A/Q=$\, next convergent, $q^{3/2}<n\le Q^{3/2}$ $\rightsquigarrow$
$\log Q\ll q^{1/2-\epsilon}$ and it is fulfilled for $\big|\alpha-\frac{a}{q}\big|>\exp(-q^{1/2-\epsilon})$. 
\qed
}

\end{frame}
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% SLIDE 15 %
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\begin{frame}
\frametitle{Going beyond}

$t\in\mathcal{D}$ $\Rightarrow$
$\langle t\pm n\alpha\rangle >\epsilon_n\asymp n^{-2}$
and
essentially 
\[
L_n(t)\ll
q+nq^{-1}\log\big(\epsilon_n^{-1}+q\big)
\quad\text{is best possible}.
\]

\begin{miblock}[colframe=ct0,boxrule=.5pt,no shadow,math,standard jigsaw,opacityback=.5]
\begin{aligned}
\text{\textbf{\color{ct00}Pessimistic view}. \ Increase $\epsilon_n$}
& \ \structure{\Rightarrow}\  |\mathcal{D}|=0 \ \text{we ran out of points}\\
& \ \structure{\overset{(?)}{\Rightarrow}}\  \text{No chance of improvement.}
\end{aligned}
\end{miblock}

\medskip

\begin{miblock}[colframe=ct0,boxrule=.5pt,no shadow,standard jigsaw,opacityback=.5]
\textbf{\color{ct00}Dreaming}. 
With the nonsensical choice $\epsilon_n=1$, Atzmon theorem imposes
$\frac{n}{q}\log q\ll \frac{n}{(\log n)^{1+\epsilon}}$
and for $q^{3/2}<n\le Q^{3/2}$ gives $\log Q\ll q^{1-\epsilon}$ (Our result).
\end{miblock}

{\footnotesize\textbf{Hope}. The dream only requires $\log(n+q)\to\log q$ to come true.} 

\end{frame}
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% SLIDE 16 %
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\begin{frame}

\[
L_n(t)=\sum_{k=0}^{n-1}\big(1+\log\{t+k\alpha\}\big)
\qquad
\frac{a}{q},\ \frac{A}{Q}
\text{ conv. }
\alpha\not\in\Q
\]
\[
\mathcal{D}
=
\big\{t\,:\,
\langle t\pm n\alpha\rangle \gg\frac{1}{n^2},\ \forall n\in\Z^+
\big\}
\]

\begin{miblock}[colframe=ct0,boxrule=.5pt,no shadow,standard jigsaw,opacityback=.5]
\strut \textbf{\color{ct00}Dreamed bound}.
$L_n(t)\ll q+\frac nq\log q$ for $t\in\mathcal{D}$
\end{miblock}

\begin{miblock}[colframe=ct0,boxrule=1pt]
\textbf{\color{ct00}Proposition}.
For $Q\gg q^4$, $n\le Q^{3/2}$, $t\in\mathcal{D}$
\[
L_n(t)\ll
q+\frac{n'}{q}\log q+\frac{n+Q}{Q}\log n
\]
$n' =\,$remainder when $n$ is divided by $Q$.

\end{miblock}

\end{frame}
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% SLIDE 17 %
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\begin{frame}
\frametitle{Idea of the proof}
\vspace{-10pt}

\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-0.35,3) grid (10.35,-3.5);}
\node at (5,3) {$\displaystyle{L_n(t)=\sum_{k=0}^{n-1}\big(1+\log\{t+k\alpha\}\big)}$};
\draw[bloque,fill=white] (-.3,2.2) rectangle (4.9,-3);
\node[inner sep=0pt] at (2.3,0)
{\includegraphics[height=120.1567pt, keepaspectratio=true]{./images/circ_q.eps}};
\node[inner sep=0pt] at (2.3,-2.44)
{$t\in\mathcal{D} \ \structure{\Rightarrow}\ \mu\gg q^{-2} \ \structure{\Rightarrow}\ \log q$};
\draw[bloque,fill=white] (5.1,2.2) rectangle (10.3,-3);
\node[inner sep=0pt] at (7.7,0)
{\includegraphics[height=120.1567pt, keepaspectratio=true]{./images/circ_n.eps}};
\node[inner sep=0pt] at (7.7,-2.5)
{$t\in\mathcal{D} \ \structure{\overset{q\mid n}{\Rightarrow}}\ \mu\gg n^{-2} \structure{\Rightarrow} \dfrac{n}{q}\log n$};
\node at (5,-3.5) {\textbf{Target}. Replace $\log n$ by $\log q$.};
\end{tikzpicture}

\end{frame}
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% SLIDE 18 %
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\begin{frame}

\vskip10pt

\tcbox[on line,size=small,colframe=ct3,colback=white,boxrule=.5pt,enhanced,drop fuzzy shadow]{Range to keep in mind $\log Q\ll q^{1-\epsilon}$}
\hfill{\footnotesize\color{ct3} ($q^{3/2}<n<Q^{3/2}$)}

\vskip10pt

\textbf{Case $n\ge Q$}
\quad {\color{ct3}\bf\footnotesize [Use next convergent]}


\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-0.35,2.85) grid (10.35,-3);}
\draw[bloque,fill=white] (-.3,2.5) rectangle (4.9,-3.3);
\node[inner sep=0pt] at (2.3,0)
{\includegraphics[height=135.1567pt, keepaspectratio=true]{./images/circ_n_ne.eps}};
\node[inner sep=0pt] at (2.3,-2.8)
{$\dfrac{n}{q}\log n$};
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\node[inner sep=0pt] at (7.7,0)
{\includegraphics[height=135.1567pt, keepaspectratio=true]{./images/circ_n_neQ.eps}};
\node[inner sep=0pt] at (7.9,-2.8)
{$\dfrac{n}{Q}\log Q \ll \dfrac{n}{q}\log q \quad\text{{\color{ct3}\checkmark}}$};
\end{tikzpicture}

\end{frame}
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% SLIDE 19 %
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\begin{frame}

\vskip12pt

\tcbox[on line,size=small,colframe=ct3,colback=white,boxrule=.5pt,enhanced,drop fuzzy shadow]{Range to keep in mind $\log Q\ll q^{1-\epsilon}$}
\hfill{\footnotesize\color{ct3} ($q^{3/2}<n<Q^{3/2}$)}

\vskip7.5pt

\textbf{Case $n\le Q$}
\quad {\color{ct3}\bf\footnotesize [Perturbation]}
\qquad 
$\alpha=\frac{a}{q}+\frac{\delta}{qQ}$,
\quad $n=qk+j$
\[
\{t+n\alpha\}
=
\big\{t+j\alpha+\frac{k\delta}{Q}\big\},
\qquad
t\in\mathcal{D}\Rightarrow\{t+j\alpha\}\gg \frac{1}{q^2}
\]
If $n/qQ\ll q^{-2}$ then $k/Q\ll q^{-2}$ $\to$ perturbed version of $L_q(t)$
\[
L_n(t) \to
\frac{n}{q}\widetilde{L}_q(t).
\]
Then essentially
\[
n\ll \frac{Q}{q}
\quad\Rightarrow\quad
L_n(t)
\ll
\frac{n}{q}\log q\quad\text{{\color{ct3}\checkmark}}
\]

It remains to consider $n\gg Q/q$

\end{frame}
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% SLIDE 20 %
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\begin{frame}

\vskip-20pt

\tcbox[on line,size=small,colframe=ct3,colback=white,boxrule=.5pt,enhanced,drop fuzzy shadow]{Range to keep in mind $\log Q\ll q^{1-\epsilon}$}
\hfill{\footnotesize\color{ct3} ($q^{3/2}<n<Q^{3/2}$)}

\vskip7.5pt

\textbf{Case $Q/q\ll n\le Q$}
\ {\color{ct3}\bf\footnotesize [$j\ne 0\to$ small scale]}
\quad 
$\alpha=\frac{a}{q}+\frac{\delta}{qQ}$, $n=qk+j$
\[
L_n(t)\to
\sum_{k=0}^{n/q-1}
\sum_{j=0}^{q-1}
\Big(
1+
\log\big(
\mu+\frac{j}{q}+\frac{k\delta}{Q}
\big)
\Big)
,\qquad\log\mu\ll \log n.
\]
If $j\ne 0$, $k/Q$ is in a smaller scale than $j/q$ $\Rightarrow$ no interference.
Essentially
\[
L_n(t)-\text{(contrib. $j=0$)}\ll
\log n+\frac{n}{q}\log q
\ll
\frac{n}{q}\log q
\quad\text{{\color{ct3}\checkmark}}
\]

\end{frame}
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% SLIDE 21 %
%%%%%%%%%%%%
\begin{frame}

\vskip9.5pt

\tcbox[on line,size=small,colframe=ct3,colback=white,boxrule=.5pt,enhanced,drop fuzzy shadow]{Range to keep in mind $\log Q\ll q^{1-\epsilon}$}
\hfill{\footnotesize\color{ct3} ($q^{3/2}<n<Q^{3/2}$)}

\vskip7.5pt

\textbf{Case $Q/q\ll n\le Q$}
\ {\color{ct3}\bf\footnotesize [$j=0\to$ direct computation]}
\quad 
$\alpha=\frac{a}{q}+\frac{\delta}{qQ}$

The contribution of $j=0$ is 
\[
\sum_{k=0}^{n/q-1}
\Big(
1+
\log\big(
\mu+\frac{k\delta}{Q}
\big)
\Big)
\]
with $\log\mu\ll \log n$. Then
\vspace{-15pt}
\begin{eqnarray*}
\text{(contrib. $j=0$)}
&\ll&
\frac{n}{q}+
\log n
+
\sum_{k=1}^{n/q-1}
\log\frac{qQ}{n}
+
\sum_{k=1}^{n/q-1}
\log\frac{n}{qk}
\\
&\ll&
\frac{n}{q}+
\log n
+
\quad\text{(trivial)}\hskip.45cm
+
\quad
\text{(Stirling)}
\\
&\ll&
\frac{n}{q}+
\log n
+
\frac{n}{q}\log q
+
\frac{n}{q}
\ll
\frac{n}{q}\log q.
\quad\text{{\color{ct3}\checkmark}}
\end{eqnarray*}

\end{frame}
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% SLIDE 22 %
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\begin{frame}
\frametitle{Sharpness}

\begin{miblock}[halign=flush center,math,colframe=ct0,boxrule=1pt]
\Big|
\alpha-\frac{a}{q}
\Big|
\not>
\exp\Big(
-C\frac{q}{\log q} 
\Big)
\ \Rightarrow\ 
\nexists f
\text{ under Atzmon theorem}.
\end{miblock}

The proof employs the result:
\[
\frac{q^2\log q}{\epsilon^2}
<n<
\epsilon^2\frac{Q}{q}
\ \Rightarrow\ 
L_n(t-n\alpha), -L_n(t)>\epsilon \frac{n}{q}\log q
\]
for $t\in [0,1)$ except for a set $S$ of measure $O(\epsilon)$. 

\begin{miblock}[halign=flush center,math,colframe=ct0,boxrule=1pt]
\|T^nf\|
+
\|T^{-n}f\|
\gg
\inf_{t\not\in S}
\Big(
e^{L_n(t-n\alpha)}
+
e^{-L_n(t)}
\Big)\|f\|
\ \to\ \text{too large}.
\end{miblock}

\end{frame}
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% SLIDE 23 %
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\vskip-17pt}
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\begin{frame}
\vskip12.5pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,-8);}
%\node[anchor=west,text width=10cm,align=left] at (0,-2.5) {Colgaré esta presentación en:};
\node at (5,-2.8) {A copy of these slides is available in};
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\node[ct00,font=\large] at (5,-4) {\url{https://www.uam.es/fernando.chamizo}};
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\node at (5,-7) {\bf Thank you for your attention!};
\end{tikzpicture}
\end{frame}
%%%%%%%%%%%%

\end{document}
