text:
\begin{document}

\bibliographystyle{amsplain}

\author{Tirdad Sharif}
\address{Tirdad Sharif, School of Mathematics, Institute for Studies in
Theoretical Physics and Mathematics, P. O. Box 19395-5746, Tehran,
Iran} \email{sharif@ipm.ir}
\urladdr{http://www.ipm.ac.ir/IPM/people/personalinfo.jsp?PeopleCode=IP0400060}
\thanks{The author was supported by a grant from IPM, (No. 83130311).}

\title[A tight closure approach]{A tight closure approach\\to a result of G. Faltings}

\keywords{Complete Intersection Algebras, Deformation Algebras,
Hecke Algebras, Tight Closure} \subjclass[2000]{13A35, 13H10}

\begin{abstract} Using a result of M. Hochster and C. Huneke on $F$-rational rings
a criterion for complete intersection rings of characteristic $p>0$
is presented. As an application, we give a completely different
proof for an algebraic result of G. Faltings that was used by Taylor
and Wiles in \cite{TW} for a simplification of the proof of the
minimal deformation problem.
\end{abstract}

\maketitle

\section{Introduction}
The theory of tight closure was created by Melvin Hochster and Craig
Huneke. An important notion in the theory of tight closure is the
notion of \emph{$F$-rationality}. There is a close connection
between this notion and \emph{rational singularity}, see \cite{sm}.

The main purpose of this note is to use a result of M. Hochster and
C. Huneke in \cite{HH1} on $F$-rational rings, to give a criterion
for complete intersection rings of characteristic $p>0$, see
Theorem~\ref{TS2} for the precise statement.

In \cite{TW} G. Faltings proved an algebraic result that was used by
Taylor and Wiles to reprove the \emph{minimal deformation problem}
by a simpler method. As an application of Theorem~\ref{TS2} we prove
Proposition~\ref{TS4}, which is an extension of Faltings result. Our
approach to prove this proposition is fundamentally different from
Faltings method where he simplifies Taylor and Wiles argument, see
Remark~\ref{TS3}.

In order to give a view for the reader we recall the minimal
deformation problem very briefly.

The modularity conjecture for semistable elliptic curves depends on
a critical conjecture of Wiles that was proved in \cite{TW,W}, see
Conjecture (2.16) of \cite{W}.

Let $\mathcal{D}$ be a deformation theory and let $R_{\mathcal{D}}$
and $T_{\mathcal{D}}$ be the universal deformation and Hecke
algebras associated to $\mathcal{D}$, respectively. The universal
property of $R_{\mathcal{D}}$ implies that there is a homomorphism
$\varphi_{\mathcal{D}}:R_{\mathcal{D}}\rightarrow T_{\mathcal{D}}$.
Wiles conjecture asserts that $\varphi_{\mathcal{D}}$ is an
isomorphism. In the special case, if $\mathcal{D}$ is a minimal
deformation theory, then the above conjecture is called the minimal
deformation problem. For different types of deformation theories and
their relations see Chapter 1 and 2 of \cite{W}.

Wiles method in \cite{W} to prove the above conjecture, was closely
related to show that certain Hecke algebras are complete
intersection that was shown in \cite{TW}.

There is no need for the reader to know the number theoretic
materials of Wiles proof in detail. We refer the interested readers
to \cite{R} for an elegant exposition of the above matters.

\section{Definitions, Notations and The Main Theorem}\label{TS0}
Throughout this note all rings are commutative and Noetherian of
characteristic $p>0$ and all of modules are finite (that is,
finitely generated).

In this note we use the notations $\nu_R(M)$ and $\ell_R(M)$
respectively for the minimal number of generators of $M$ and the
length of $M$ over $R$.

\begin{Defn}\label{C} Let $(R,\m,k)$ be a local ring and let
$R[[X]]=R[[X_1,...,X_n]]$ be the formal power series of $n$
variables over $R$. Write $A=R[[X]]/J$ for an ideal $J$ of $R[[X]]$.
If the local ring $A\otimes_{R}k$ is complete intersection, then $A$
is called a complete intersection $R$-algebra.
\end{Defn}

\begin{Defn} Let $R$ be a ring and let $J=(a_1,...,a_t)$ be an ideal of $R$.
An element $x\in R$ is said to be in the tight closure of $J$ and
write $x\in J^*$ if there is an element $c\in R^0$ such that for all
large $q=p^n$ we have $cx^q\in J^{[q]}$, wherein $R^0$ is the
complement of the union of all minimal primes of $R$ and
$J^{[q]}=(a_1^q,...,a_t^q)$.

An ideal $J$ is called tightly closed if $J=J^*$. The ring $R$ is
called $F$-rational if every parameter ideal of $R$ is tightly
closed and in particular, if every ideal of $R$ is tightly closed,
then it is called $F$-regular.
\end{Defn}

\begin{fact}\label{F} Let $Q$ be a regular local ring, then it is
$F$-regular, see \cite[(4.4)]{HH}.
\end{fact}

To state the main theorem of this note we will make use of the
following simple lemma the proof of which is omitted.

\begin{lem}\label{TS1} Let I be an ideal of R, then $(I^*)^n\subseteq(I^n)^*$ for
$n\geq1$.
\end{lem}

\begin{thm}\label{TS2} Let $(A,\n,k)$ be a local ring of dimension $t$
and let $\q$ be an ideal of $A$, generated by a system of
parameters. Let $\eta\colon A\to Q$ and $\theta\colon Q\to B$ be
local ring homomorphisms such that $Q$ is regular of dimension $t$,
$\theta$ is surjective, and $\eta$ induces an isomorphism between
the residue fields of $A$ and $Q$. Set $\delta =\theta\eta\colon A
\to B$ and $\ell(B/\q^* B)=d$. Assume for an integer
$q=p^n>d^{t}t^{t-1}$ we have $\ell(B/{\q}^{[q]}B)\geq
\ell(A/{\q}^{[q]})$. If the ideal ${\q}^{[q]}$ is tightly closed,
then the canonical homomorphism $\pi:Q/q^*Q\rightarrow B/\q^* B$ is
an isomorphism.

\noindent In particular, if $A$ is equidimensional homomorphic image
of a Cohen-Macaulay ring, then $\pi$ is an isomorphism between
complete intersection rings.
\end{thm}

\begin{proof}
Write $\m$ for the maximal ideal of $Q$ and $J=\Ker\theta.$ We may
assume that $B=Q/J$. From the assumptions it follows that
$\m^d\subseteq J+{\q}^*Q.$ Therefore we have the next inclusions
$\m^{tqd}\subseteq (J+{\q}^*Q)^{tq}\subseteq (J+(\q^*)^{tq}Q)$.
Lemma ~\ref{TS1} implies that $\m^{tqd}\subseteq J +(\q^{tq})^*Q$.
Since $\q$ is generated by $t$ elements it is easy to see that
${\q}^{tq}\subseteq {\q}^{[q]}$. This yields that $\m^{tqd}\subseteq
J+(\q^{[q]})^*Q$ and from our assumptions we have $\m^{tqd}\subseteq
J+\q^{[q]}Q$.

\noindent We claim that $J\subseteq \m^{d+1}$, otherwise choose $u$
in $J$ not in $\m^{d+1}$ and consider the following exact sequence
$$0\longrightarrow \Ker(\alpha)\longrightarrow Q/\m^{c}\stackrel{\alpha}\longrightarrow
Q/\m^{c}\longrightarrow \coker(\alpha)\longrightarrow
0\hspace{1in}(2.5.1)
$$
wherein the map $\alpha$ is the homothety by element $u$ and
$c=tdq$. Obviously, there is a surjective map as the following
$$\coker(\alpha)=Q/(uQ+\m^c)\longrightarrow
Q/(J+{\q}^{[q]}Q)=B/\q^{[q]}B.
$$
Therefore $\ell_Q(\coker(\alpha))\geq \ell_Q(B/{\q}^{[q]}B)$. It is
clear that $B/{\q}^{[q]}B$ has finite length over $Q$. The ideal
$\q$ is an $\n$-primary ideal of $A$, therefore $\ell_{A}
(B/{\q}^{[q]}B)$ is finite and since $k=A/\n\simeq Q/\m$, clearly
the length of $B/{\q}^{[q]}B$ over both of $Q$ and $A$ are equal. By
assumptions $\ell(B/{\q}^{[q]}B)\geq \ell(A/{\q}^{[q]})$, therefore
we have
$\ell_Q(\coker(\alpha))\geq\ell(B/{\q}^{[q]}B)\geq\ell(A/{\q}^{[q]})$
and from \cite[(14.10)]{M} we find that $\ell(A/\q^{[q]})\geq q^t
e(A)\geq q^t$ wherein the symbol $e(A)$ is the multiplicity of $A$.
Thus $\ell_Q(\coker(\alpha))\geq q^t$.

Let $x_1,x_2,...,x_t$ be a minimal generator for $\m.$ The local
ring $Q$ is regular, so the associated graded ring $gr_{\m}(Q)$ is a
polynomial ring in $t$ variables over $k$, see \cite[Page 76,Theorem
9(d)]{se}. This fact implies that $x_j$ for $1\leq j\leq t$ are
analytically independent in $\m$ and so that
$\nu_Q(\m^{c-i})={c-i+t-1\choose t-1}$ for $i\geq 1$. Since $u$ is
not an element of $\m^{d+1}$ therefore $\ker(\alpha)\subseteq
\m^{c-d}/\m^c$ and hence
$\ell_Q(\ker(\alpha))\leq\ell_Q(\m^{c-d}/\m^c)$.

\noindent The following equality is clear
$$\ell_Q(\m^{c-d}/\m^c)=\displaystyle{\sum_{i=1}^{d}}\nu_Q(\m^{c-i}).\hspace{1in}(2.5.2)$$
Now from (2.5.1) we have $\ell_Q(\ker\alpha)=\ell_Q(\coker\alpha)$
and from (2.5.2) we find that $\ell_Q(\m^{c-d}/\m^c)\leq
d{c+t-2\choose t-1}\leq dc^{t-1}=d(tdq)^{t-1}.$ Therefore $q^t\leq
d^{t}t^{t-1}q^{t-1}$ and this contradicts the choice of $q$. Hence
we must have $J\subseteq \m^{d+1}$ and consequently $\m^d\subseteq
J+{\q}^* Q\subseteq \m^{d+1}+{\q}^*Q.$ Thus
$\m^d+{\q}^*Q=\m^{d+1}+{\q}^*Q$ and by using Nakayama's Lemma
$\m^{d+1}\subseteq {\q}^*Q$. This yields that $J\subseteq {\q}^*Q$
and this shows that the canonical homomorphism $\pi : Q/{\q}^*Q
\rightarrow B/{\q}^*B$ is an isomorphism.

Now let $A$ be an equidimensional ring which is homomorphic image of
a Cohen-Macaulay ring. In this case, since ${\q}^{[q]}$ is generated
by a system of parameters and it is tightly closed it follows from
\cite[(4.2.e)]{H} that $A$ is $F$-rational and hence $\q ={\q}^*$.
Now it is easy to see that $\pi$ is an isomorphism between complete
intersection rings.
\end{proof}

\begin{disc}\label{TS3} Let $(\mathcal{O},\m,k)$ be a complete discrete
valuation ring with finite residue field $k$ and let $\mathcal{D}$
be a minimal deformation theory. Let $R$ and $T$ be the universal
deformation and Hecke $\mathcal{O}$-algebras associated to
$\mathcal{D}$, respectively and let $\varphi=\varphi_{\mathcal{D}}$
be the homomorphism described in Section 1. Taylor and Wiles in the
appendix of \cite{TW} for a simplification of some arguments in
Section 3 of \cite{TW} and Chapter 3 of \cite{W} use a commutative
algebraic result of G. Faltings to reprove that $\varphi$ is an
isomorphism between complete intersection algebras. In
Proposition~\ref{TS4} and Corollary~\ref{TS6} we give an extension
of corresponding results due to Taylor-Wiles and Faltings without
the assumption of finiteness of $k$. Our method to prove the above
results is totally different from their methods.

\noindent In the following we refine the definition of a
\emph{(level) n-structure} due to Wiles and Faltings in \cite{TW}.
\end{disc}

\begin{Defn} Let $(\mathcal{O},\m,k)$ be a local ring
 and let $\mathcal{O}[[Y]]=\mathcal{O}[[Y_1,...,Y_t]]$.
 Set $\q=(Y_1,...,Y_t)$ and assume that for an integer $n\geq 1$ we have a
 commutative diagram of local $\mathcal{O}$-algebras as the
 following in which $\varphi$ is surjective and $T$ is a finite free
$\mathcal{O}$-module.
$$
\begin{array}{llll}
 \mathcal{O}[[Y]]\stackrel{\psi_n}\longrightarrow & R_n&\stackrel{\varphi_n}{\longrightarrow} &T_n\\
&\downarrow &&\downarrow \\
& R&\stackrel{\varphi}{\longrightarrow}&T \\
 \end{array}
 $$
By a level $n$-structure we mean the above diagram with the
following properties:

$(1)$ There is a surjective homomorphism
$\lambda_n:\mathcal{O}[[X_1,X_2,...,X_t]]\longrightarrow R_n$.

 $(2)$ The ring homomorphism
 $R_n\longrightarrow T_n$ is surjective.

 $(3)$ From the vertical homomorphisms we get
 the isomorphisms $R_n/{\q}R_n\rightarrow R$ and
 $T_n/{\q}T_n\rightarrow T$.

 $(4)$ The ring $T_n/{\q^{[p^n]}}T_n$ is
 finite and free as a module over $\mathcal{O}[[Y]]/{\q^{[p^n]}}$.

 \vspace{0.05in}\noindent For convenience we represent a level $n$-structure
 by the notation $L_n(\mathcal{O},\varphi,\varphi_n,\psi_n,\lambda_n)$.

\end{Defn}

\begin{prop}\label{TS4} Let $m\geq 1$ be an integer such that for every $n\geq m$,
the level $n$-structure
$L_n(\mathcal{O},\varphi,\varphi_n,\psi_n,\lambda_n)$ has the
following additional property:

\noindent $(5)$ For each $x\in \m$ we have $\psi_n(x)\in \m R_n$.

Then $\varphi$ is an isomorphism between complete intersection
$\mathcal{O}$-algebras.
\end{prop}

\begin{proof} Set
 $\widehat{\square}=\square\otimes_{\mathcal{O}} k$, $\q_n=\q^{[p^n]}$
 and $\mathcal{O}[[X]]=\mathcal{O}[[X_1,X_2,...,X_t]]$. From (1) there
 is a surjective homomorphism $k[[X]]\rightarrow
\widehat{R_n}$. The fifth property implies that
$\psi_n(\m[[Y]])\subseteq \m R_n$. Thus $\psi_n$ induces a local
homomorphism $\widehat{\psi_n}:k[[Y]]\longrightarrow \widehat{R_n}$.
Since $k[[X]]$ is a complete local ring, using \cite[(7.16)]{E} we
can lift the homomorphism $k[[Y]]\rightarrow \widehat{R_n}$ to a
unique homomorphism $k[[Y]]\rightarrow k[[X]]$. Thus from (1), (2)
and (3) we find that there is a surjective homomorphism as the
following
$$\alpha:k[[X]]/{\q}k[[X]]\longrightarrow \widehat{T_n}/{\q}\widehat{T_n}\simeq \widehat{T}.
$$
From (3) it follows that $\widehat{T_n}/{\q}\widehat{T_n}$ is a
finite vector space over $k$. We may assume that the length of
$\widehat{T_n}/{\q}\widehat{T_n}$ over $k[[X]]$ is equal $d$. Now
choose $n\geq m$ such that $p^n>t^{t-1}d^t$. From the fourth
property it is clear that the length of
$\widehat{T_n}/{\q_n}{\widehat{T_n}}$ over $k[[Y]]$ is equal or
greater than the length of $k[[Y]]/{\q_n}k[[Y]]$.

Now by writing $A=k[[Y]]$, $Q=k[[X]]$ and $B=\widehat{T_n}$ from
Theorem~\ref{TS2} and Fact~\ref{F} we get that $\alpha$ is an
isomorphism between complete intersection rings.

On the other hand, $\varphi_n$ is surjective, from this and the
third property we have the following surjective homomorphisms
$$\beta:k[[X]]/{\q}k[[X]]\longrightarrow \widehat{R_n}/{\q}\widehat{R_n}$$
$$\gamma:\widehat{R_n}/{\q}\widehat{R_n}\longrightarrow
\widehat{T_n}/{\q}\widehat{T_n}
$$
such that $\alpha=\gamma\beta$. It was shown that $\alpha$ is an
isomorphism thus from (3) it is clear that $\widehat{\varphi}$ is an
isomorphism between $\widehat{R}$ and $\widehat{T}$ as complete
intersection rings. Since $T$ is a finite free module and $\varphi$
is surjective, Nakayama's Lemma implies that $\varphi$ is an
isomorphism. It is obvious that the local ring $T$ is a homomorphic
image of $\mathcal{O}[[X]]$. Since $\widehat{T}$ is a complete
intersection ring from Definition~\ref{C} it follows that $T$ and so
$R$ are complete intersection $\mathcal{O}$-algebras.
\end{proof}

\begin{disc} In the proof of the above proposition the fifth property enables us
to reduce the level $n$-structure
$L_n(\mathcal{O},\varphi,\varphi_n,\psi_n,\lambda_n)$ to the level
$n$-structure
$L_n(k,\widehat{\varphi},\widehat{\varphi_n},\widehat{\psi_n},\widehat{\lambda_n})$
and then by lifting property which is based on \cite[(7.16)]{E} we
can prove the proposition as a simple corollary of our main theorem.
If we assume that $\mathcal{O}$ is $\m$-adic complete, then
$\mathcal{O}[[X]]$ is $(\m[[X]]+(X))$-adic complete. From (1) we
have a surjective homomorphism $\mathcal{O}[[X]]\longrightarrow
R_n$. Now again by using \cite[(7.16)]{E} we can lift the local
homomorphism $\psi_n$ to a unique local homomorphism
$\mathcal{O}[[Y]]\longrightarrow \mathcal{O}[[X]]$. Now by a similar
method to our proof we can show that $\varphi$ is an isomorphism
between complete intersection $\mathcal{O}$-algebras. In this case,
we don't need to use the fifth property directly. However, in the
following corollary as a simple application of the level
$n$-structures, we make clear the relation between the completeness
of $\mathcal{O}$ and the fifth property. In a sense, we can say that
the assumption of completeness of our base ring implies the fifth
property. We use the next simple lemma the proof of which is
omitted.
\end{disc}

\begin{lem}\label{TS5} Let $L_n(\mathcal{O},\varphi,\varphi_n,\psi_n,\lambda_n)$
be a level $n$-structure, then there is a level $n$-structure
$L'_n(\mathcal{O},\varphi, \varphi'_n, \psi'_n,\lambda'_n)$ such
that $\lambda'_n$ is the canonical epimorphism.
\end{lem}

\begin{cor}\label{TS6} Let $(\mathcal{O},\m,k)$ be an $\m$-adic complete local ring, then
$\varphi$ is an isomorphism between complete intersection
$\mathcal{O}$-algebras.
\end{cor}
\begin{proof} From Lemma~\ref{TS5} it follows that there is a level
$n$-structure such that the homomorphism
$\lambda_n:\mathcal{O}[[X]]\longrightarrow \mathcal{O}[[X]]/J_n=R_n$
is a canonical epimorphism. Therefore for each $g\in
\mathcal{O}[[X]]$ we have $\lambda_n(g)=g+J_n$, where
$J_n=\Ker\lambda_n$. Since $\mathcal{O}$ is $\m$-adic complete hence
$\mathcal{O}[[X]]$ is $(\m[[X]]+(X))$-adic complete. Thus from
\cite[(7.16)]{E} we can lift $\psi_n$ to a unique homomorphism
$\xi:\mathcal{O}[[Y]]\longrightarrow \mathcal{O}[[X]]$ such that
$\xi(x)=x$ for each $x\in \m$. On the other hand, we have
$\lambda_n\xi=\psi_n$. Hence $\psi_n(x)=x+J_n$ and this shows that
$\psi_n(x)\in \m R_n$. Now from Proposition~\ref{TS4} assertion
holds.
\end{proof}

\section*{Acknowledgments}
The author is grateful to Sean Sather-Wagstaff and Irena Swanson for
their useful comments on this note.

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TITLE: Covering of Intervals
QUESTION [9 upvotes]: For a real interval $I=[x-r,x+r]$ and a number $\alpha>0$ we write $\alpha I:=[x-\alpha r,x+\alpha r]$, i.e. $\alpha I$ is the interval $I$ blown up by a factor $\alpha$ around its center.
For example, the Vitali Covering Lemma says that for any finite collection of intervals $([a_j,b_j])_{1\leq j\leq n}$ there is an index set $J\subseteq\{1,\ldots,n\}$ such that $([a_j,b_j])_{j\in J}$ is a subcollection of pairwise disjoint intervals with
$$
\bigcup_{1\leq j\leq n}[a_j,b_j]\subseteq\bigcup_{j\in J}3[a_j,b_j]
$$
Setting: I have two finite collections of intervals $([a_j,b_j])_{1\leq j\leq n}$ and $([c_j,d_j])_{1\leq j\leq n}$ with the following three properties.

$|a_j-c_j|\leq\delta$ and $|b_j-d_j|\leq \delta$ for $1\leq j\leq n$,
$\sum_{j=1}^n|(a_j-c_j)-(b_j-d_j)|\leq\delta$,
$\left|\bigcup_{j=1}^n[a_j,b_j]\right|\leq\varepsilon$, where $|A|$ means the Lebesgue measure of a set $A$.

Here, $\delta$ and $\varepsilon$ are two positive "small" numbers.
Problem: I would like to show (or disprove) that the Lebesgue measure of $\bigcup_{j=1}^n[c_j,d_j]$ is, up to a constant that does not depend on the intervals or $n$, like that of $\bigcup_{j=1}^n[a_j,b_j]$ also small.
My attempt: I could prove that this is true if the $[a_j,b_j]$ are mutually disjoint: Indeed, by Vitali's Covering Lemma there is an index set $J\subseteq\{1,\ldots,n\}$ such that $([c_j,d_j])_{j\in J}$ is a subcollection of pairwise disjoint intervals with
$$
\bigcup_{1\leq j\leq n}[c_j,d_j]\subseteq\bigcup_{j\in J}3[c_j,d_j].
$$
This implies with the help of 2. and 3. that
\begin{align*}
\left|\bigcup_{j=1}^n[c_j,d_j]\right|
&\leq\left|\bigcup_{j\in J}3[c_j,d_j]\right|
 \leq\sum_{j\in J}\Big|3[c_j,d_j]\Big|
  =3\sum_{j\in J}|c_j-d_j| \\
&\leq 3\sum_{j\in J}|(a_j-b_j)-(c_j-d_j)|+3\sum_{j\in J}|a_j-b_j| \\
&\leq 3\delta+3\left|\bigcup_{j=1}^n[a_j,b_j]\right| \\
&\leq 3(\delta+\varepsilon).
\end{align*}
However, I have not used 1. In case that the intervals $[a_j,b_j]$ are not pairwise disjoint, I don't know how to proceed. Any ideas are highly appreciated. Thanks in advance!

EDIT: (Thanks to mathworker21): Vitali's Covering Lemma is not necessary here, we can use the estimate
\begin{align*}
\left|\bigcup_{j=1}^n[c_j,d_j]\right|
&\leq\sum_{j=1}^n|c_j-d_j|
\end{align*}
instead.

REPLY [1 votes]: (Disclaimer: this is more of a pedantic comment than an answer!)
Does it change the meaning of the question essentially if one drops
the stipulation that $\delta$ and $\epsilon$ are "small" - I
confess to being unable to understand what that statement means in
the present context - and asks instead:

Do there exist a number $h > 0$, and functions $f, g$ (of one and
  two range-limited real variables, respectively), such that if
  $0 < \epsilon < h$, and condition 3 holds, then for all $\delta$
  such that $0 < \delta < f(\epsilon)$, and all $n$ and
  $a_j, b_j, c_j, d_j$ ($1 \leqslant j \leqslant n$) such that
  conditions 1 and 2 hold, we have
  $\left\lvert\bigcup_{j=1}^n[c_j,d_j]\right\rvert \leqslant
g(\epsilon, \delta)$?

A more demanding (but I think less plausible) reformulation of the
question is as follows:

Do there exist numbers $h, h' > 0$, and a function $g$ (of two
  range-limited real variables), such that if $0 < \epsilon < h$ and
  $0 < \delta < h'$, then for all $n$ and $a_j, b_j, c_j, d_j$
  ($1 \leqslant j \leqslant n$) such that conditions 1 to 3 hold, we
  have $\left|\bigcup_{j=1}^n[c_j,d_j]\right| \leqslant
g(\epsilon, \delta)$?

If the second reformulation accurately represents the meaning of the
question, then a special case of mathworker21's argument still gives
the answer "no".
On the other hand, if the first reformulation is satisfactory, then
mathworker21's construction cannot be applied, and the question
remains open.
[Not so! See the addendum below.]
Disposing of the second reformulation first:
Given $h, h' > 0$, take any $\epsilon$ such
that  $0 < \epsilon < \min\{h, h', 1\}$. Let $\delta = \epsilon$, let
$k$ be a positive integer, and, in mathworker21's construction, let
$L = R = k$, $n = k^2$. Then
$\left\lvert\bigcup_{j=1}^n[c_j,d_j]\right\rvert = k\epsilon$.
Because this measure is not bounded above by any function of
$\epsilon$ and $\delta$, the required function $g$ cannot exist.
$\square$
On the first reformulation (which I think is a more reasonable
interpretation of the question), we may still choose any value of
$\epsilon$ that is "small enough", i.e. $\epsilon < h$, and try
to derive a contradiction; but now we must allow $\delta$ to take
on any "sufficiently small" value, i.e. any value less than some
unspecified $f(\epsilon)$. In these circumstances, I don't see how
mathworker21's construction can be carried out.
(I don't know what the answer is in this case, but it seems best to
check my understanding of the question before possibly barking up
the wrong tree!)

Even if the first reformulation is valid, the same general
construction does still apply. This "answer" is therefore
probably best viewed as a long-winded comment in confirmation of
mathworker21's answer.
Given $h > 0$, and a function $f$, take any $\epsilon$ such that
$0 < \epsilon < \min\{h, 1\}$, and any positive integer $k$ such
that $\frac{\epsilon}{k} < f(\epsilon)$.
Define $\delta = \frac{\epsilon}{k}$, In mathworker21's
construction, let $R = k$, $L = k^2$, $n = k^3$. Then
$\left\lvert\bigcup_{j=1}^n[c_j,d_j]\right\rvert = k\epsilon$, much
as before. Again, the required function $g$ cannot exist.
$\square$


text:
\begin{document}
\begin{abstract}
For every $\tau\in\R$ and every integer $N$, let $\m_N(\tau)$ be the minimum of the distance of $\tau$
from the sums $\sum_{n=1}^N s_n / n$, where $s_1, \ldots, s_n \in \{-1, +1\}$. We prove that $\m_N(\tau)
< \exp\!\big(-C(\log N)^2\big)$, for all sufficiently large positive integers $N$ (depending on $C$ and
$\tau$), where $C$ is any positive constant less than $1/\log 4$.
\end{abstract}

\maketitle

\begin{center}
To appear in C. R. Math. Acad. Sci. Paris 2018.\\

\end{center}

\section{Introduction}
For each positive integer $n$, let
\begin{equation*}
H_n := 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}
\end{equation*}
be the $n$th \emph{harmonic number}. Harmonic numbers have long been an active area of research. For
instance, Wolstenholme~\cite{Wol62} proved that for any prime number $p \geq 5$ the numerator of $H_{p -
1}$ is divisible by $p^2$; while Taeisinger~\cite[p.~3115]{Wei02} showed that $H_n$ is never an integer
for $n > 1$. This latter result has been generalized by Erd\H{o}s~\cite{Erd32} to sums of inverses of
numbers in arithmetic progression. Also, the $p$-adic valuation of $H_n$ has been studied by
Boyd~\cite{MR1341721}, Eswarathasan and Levine~\cite{MR1129989}, Wu and Chen~\cite{MR3608179}, and
Sanna~\cite{MR3486261}. Moreover, harmonic numbers are special cases of \emph{Egyptian fractions}
(rational numbers which are sums of distinct unit fractions), themselves an active area of
research~\cite[\S{D11}]{MR2076335}.

It is well known that $H_n \to +\infty$ as $n \to +\infty$. More precisely,
\begin{equation}\label{equ:harmonic}
H_n = \log n + \gamma + O\!\left(1/{n}\right)
\end{equation}
for all positive integers $n$, where $\gamma$ is the \emph{Euler--Mascheroni constant}.
On the other hand, the alternating signs harmonic number
\begin{equation*}
H_n^\prime = 1 - \frac{1}{2} + \frac{1}{3} - \cdots + \frac{(-1)^{n + 1}}{n}
\end{equation*}
converges to $\log 2$ as $n \to +\infty$. Building on earlier work by Morrison~\cite{Morrison2,Mor},
Schmuland~\cite{MR2040884} proved that the random harmonic series
\begin{equation*}
X:=\sum_{n \,=\, 1}^\infty \frac{s_n}{n} ,
\end{equation*}
where $s_1, s_2, \ldots$ are independent uniformly distributed random variables in $\{-1,+1\}$, converges
almost surely to a random variable with smooth density function $g$ supported on the whole real line.
Interestingly, $g(0)$ and $g(2)$ are extremely close to, but slightly smaller than, $\frac{1}{4}$ and
$\frac{1}{8}$ respectively (the error being of the order of $10^{-6}$ and $10^{-43}$ respectively).
We refer to~\cite[p.~101]{MR2051473} and~\cite{MR2040884} for some more information on these constants and
to~\cite{Crandall,Offner} for more information on the random variable $X$.

\begin{figure}[h]
\centering
\includegraphics[width=0.50\textwidth]{BMS_Figure1.pdf}
\caption{The graph of the distribution function $g(x)$ of $X$.}
\label{Fig:1}
\end{figure}

In this paper we are interested in the set
\begin{equation*}
\SS_N := \left\{\sum_{n \,=\, 1}^N \frac{s_n}{n} : s_1, \ldots, s_N \in \{-1,+1\}\right\} .
\end{equation*}
Clearly, $\SS_N$ is symmetric respect to the origin and
\begin{equation*}
\max \SS_N = H_N \sim \log N
\end{equation*}
as $N \to +\infty$, by~\eqref{equ:harmonic}.
On the other hand, the quantity
\begin{equation*}
\m_N := \min\left\{|\s| : \s \in \SS_N\right\}
\end{equation*}
is much more mysterious. It is not difficult to prove (see Proposition~\ref{pro:nonzero} below) that
$\m_N\neq 0$ for all $N\in\N$. In particular, estimating the least common multiple of the denominators
using the Prime Number Theorem, one easily obtains the following lower bound for $\m_N$,
\begin{equation}\label{equ:simpleintro}
\m_N > \exp\!\left(-N + o(N) \right),
\end{equation}
as $N \to +\infty$.

More generally, we shall study the function
\begin{equation*}
\m_N(\tau):= \min\left\{|\s-\tau|\colon \s \in \SS_N\right\}, \qquad\tau\in\R.
\end{equation*}
Using an easy argument, in Proposition~\ref{pro:lower} below we show that for almost every $\tau$,
\begin{equation}\label{equ:A3}
\m_N(\tau) > \exp\!\left(-0.665\,N + o(N) \right) ,
\end{equation}
as $N \to +\infty$ (notice that $0.665<\log 2= 0.693\ldots$). This bound holds for almost every $\tau$,
but not for all of them: in fact, $\m_N(\tau)$ can be arbitrary small infinitely often. Precisely, given
any $f \colon \N \to \R_{>0}$ we can construct $\tau_f \in \R$ such that $\m_N(\tau_f) < f(N)$ for
infinitely many $N$ (see~\cite[Proposition~5.9]{BettinMolteniSanna2}). The bound in~\eqref{equ:A3} is not
optimal, and some minor variations of our proof are already able to produce some small improvement.

In this paper we are mainly interested in the opposite direction where the upper bound for $\m_N(\tau)$
is sought. Our main result is the following.

\begin{thm}\label{thm:upper}
For every $\tau\in \R$ and for any positive constant $C$ less than $1/\log 4$, we have
\begin{equation}\label{equ:upbound}
\m_N(\tau) < \exp\!\left(-C (\log N)^2\right) ,
\end{equation}
for all sufficiently large $N$, depending on $C$ and $\tau$.
\end{thm}

Notice that a sequence of signs $s_{1},\dots,s_{N}$ realizing the minimum in the definition of
$\m_{N}(\tau)$ does not come from a ``universal'' infinite sequence $(s_n)_{n\geq1}$ such that, setting
$\sigma_N :=\sum_{n\,=\,1}^N s_n / n$, we have $\m_N(\tau) = |\sigma_N - \tau|$ for all $N$. Indeed,
$|\sigma_N-\sigma_{N-1}|=1/N$ and so $\sigma_N$ and $\sigma_{N-1}$ cannot both be less then $1/(2N)$ away
from $\tau$.

The upper and lower bounds given in the inequalities~\eqref{equ:simpleintro} and~\eqref{equ:upbound} are
quite distant and thus they do not indicate clearly what is the real size of $\m_N$. A heuristic argument
suggests that the inequality $\m_N > \exp\!\left(-\frac{1}{2}N + o(N)\right)$ is satisfied for infinitely
many $N$, and numerical computations (cf.~Figure~\ref{Fig:2}) might suggest that actually $\m_N =
\exp\!\left(-\frac{1}{2}N + o(N)\right)$. However, because of the exponential nature of the problem, we
were able to compute only the first $\np$ values of $\m_N$, which are clearly not enough to draw a solid
conclusion. We shall give these values of $\m_N$ in the appendix. Despite the limited amount on data at
disposal, some interesting observations can be drawn from them. For example, $\m_N$ is not a decreasing
function of $N$ and there are several repeated values. One can then perhaps expect that there are
infinitely many values of $N$ such that $\m_N=\m_{N+1}$ or even such that $\m_N=\cdots=\m_{N+k}$ for any
fixed $k\in\N$.

\begin{figure}[h]
\centering
\includegraphics[width=0.60\textwidth]{BMS_Figure2.pdf}
\caption{The first $\np$ values of $\frac{\log \m_N}{\log L_N}$, where $L_N= \lcm\{1, \ldots,
N\}=e^{N+o(N)}$, plotted against the constant line $-\frac{1}{2}$.} \label{Fig:2}
\end{figure}

We prove Theorem~\ref{thm:upper} using a probabilistic argument. More precisely, in
Theorem~\ref{thm:loclimthm} below we shall prove a small scale distribution result for
$X_N:=\sum_{n\,=\,1}^N \frac{s_n}{n}$,  where $s_1,\dots,s_N$ are independently uniformly distributed
random variables in $\{-1, +1\}$. Theorem~\ref{thm:upper} will follow immediately from this result
(cf.~Corollary~\ref{cor:maincorollary}). Interestingly, this distribution problem for $X_N$ will lead us
to another classical number theoretic problem: that of bounding a short average of the number of divisors
in a prescribed small interval. We will attack this problem in two different ways, first using Rankin's
trick together with a bound for the divisor function $\sigma_s(n)$ proved in Ramanujan's lost
notebook~\cite{MR1606180}, and then using a more complicated arithmetic construction. Surprisingly, the
two methods both lead to the same bound~\eqref{equ:upbound}, albeit with different constants.

While the probabilistic approach has the advantage of showing the existence of several $N$-tuples of
signs $s_1, \ldots, s_N$ giving small values for $|\sigma_N - \tau|$, this approach does
not produce any explicit instance of these $N$-tuples. If one is interested in exhibiting explicit
sequences, then one can construct some special signed harmonic series converging to $\tau$ and estimate
the absolute value of their partial sums. A natural candidate is the ``greedy'' sequence obtained by
setting $s_{N+1} := +1$ if $\sigma_N \leq \tau$, and $s_{N+1} := -1$ otherwise. It is clear that $\sigma_N$
converges to $\tau$, since at each step one chooses the sign which makes $\sigma_N$ closer to $\tau$ and more
precisely one has $|\sigma_N-\tau| \leq 1/N$ for all $N$ large enough (depending on $\tau$). On the other
hand, as observed above, $\sigma_N$ cannot be always very close to $\tau$ and in fact the inequality
$|\sigma_N - \tau| \geq 1/(N+1)$ is satisfied infinitely often. However, it is still possible to prove that for any $A
> 0$ one has $|\sigma_N - \tau|\ll_A N^{-A}$ for infinitely many positive integers $N$.
In fact we can show that for almost all $\tau$ one has
\begin{equation*}
\liminf_{n \to +\infty} \frac{\log|\sigma_N - \tau|}{(\log N)^2} = -\frac1{\log 4} .
\end{equation*}
It is quite remarkable that this ``greedy'' algorithm and the probabilistic method developed in this
paper both give a decay rate of $\exp\!\big(-\big(\tfrac1{\log 4} + o(1)\big) (\log N)^2\big)$. The study
of this ``greedy'' sequences needs completely different tools from those employed here, thus we leave its
study to another paper~\cite{BettinMolteniSanna2}.

\begin{acknowledgements}
S.~Bettin is member of the INdAM group GNAMPA.
G.~Molteni and C.~Sanna are members of the INdAM group GNSAGA.
The work of the first and second author is partially supported by PRIN 2015 ``Number Theory and Arithmetic Geometry''.
The authors would also like to thank D.~Koukoulopoulos and M.~Radziwi\l\l{} for several useful discussions.
The authors would also like to thank the anonymous referee for carefully reading the paper.
\end{acknowledgements}

\subsection*{Notation}
We employ the Landau--Bachmann ``Big Oh'' and ``little oh'' notations $O$ and $o$, as well as the
associated Vinogradov symbols $\ll$ and $\gg$, with their usual meanings. Any dependence of the implied
constants is explicitly stated or indicated with subscripts. As usual, we write $\mathbb{E}[X]$ for the
expected valued of a random variable $X$, and $\mathbb{P}[E]$ for the probability of an event $E$. Also,
we indicate with $\mathcal{C}_c(\R)$ the space of continuous functions with compact support on $\R$ and
with $\mathcal{C}^\infty_c(\R)$ the subspace of  $\mathcal{C}_c(\R)$ consisting of smooth functions.
Finally, for each $\Phi \in \mathcal{C}_c(\R)$ we let $\widehat{\Phi}$ denote its Fourier transform, here
defined by
\begin{equation*}
\widehat{\Phi}(x) := \int_{\R} \Phi(y) e^{-2\pi ixy} \,\mathrm{d} y
\end{equation*}
for all $x \in \R$.

\section{The small scale distribution of $X_N$}
We start with stating our result on the small scale distribution of $X_N$. We remind that $X_N$ is the
random variable defined by $X_N:=\sum_{n\,=\,1}^N s_n / n$, where $s_n$ are taken uniformly and
independently at random in $\{-1,+1\}$.
\begin{thm}\label{thm:loclimthm}
Let $C$ be any positive constant less than $1/\log 4$.
Then, for all intervals $I\subseteq\R$ of length
$|I|>\exp\!\left(-C (\log N)^2\right)$ one has
\begin{equation*}
\mathbb{P}[X_N\in I] = \int_I g(x) \, \mathrm{d} x+o(|I|),
\end{equation*}
as $N\to\infty$, where
\begin{equation*}
g(x):=2\int_{0}^{+\infty}\cos(2\pi u x) \prod_{n=1}^{+\infty}\cos(2\pi u/n)\,\mathrm{d} u.
\end{equation*}
\end{thm}
\begin{remark}\label{rem:distribfcn}
As shown by Schmuland~\cite{MR2040884}, $g(x)$ is a smooth strictly positive function which is
$O_A\!\left(x^{-A}\right)$ as $x \to \pm\infty$, for any $A > 0$.
\end{remark}
\begin{corol}\label{cor:maincorollary}
Let $C$ be any positive constant less than $1/\log 4$.
Then, for all $\tau\in\R$ one has
\begin{equation*}
\#\left\{(s_1, \ldots, s_N )\in \{-1,+1\}^N : \left|\tau - \sum_{n \,=\, 1}^N \frac{s_n}{n} \right| < \delta \right\}
\sim  2^{N+1} g(\tau) \delta(1 +o_{C,\tau}(1))
\end{equation*}
as $N\to\infty$ and $\delta\to 0$, uniformly in $\delta \geq \exp\!\left(-C (\log N)^2\right)$.
In particular, for all large enough $N$ one has $\m_N (\tau)<\exp\!\left(-C (\log N)^2\right)$.
\end{corol}
\begin{proof}
The result follows immediately from Theorem~\ref{thm:loclimthm} and Remark~\ref{rem:distribfcn}.
\end{proof}

We now proceed to proving Theorem~\ref{thm:loclimthm}.
For each $N\in\N\cup\{\infty\}$ and for any real number $x$, define the product
\begin{equation*}
\varrho_N(x) := \prod_{n\,=\,1}^N \cos\!\Big(\frac{\pi x}{n}\Big)
\end{equation*}
and let $\varrho(x):=\varrho_\infty(x)$.
\begin{lem}\label{lem:expected}
We have
\begin{equation*}
\mathbb{E}[\Phi(X_N)] = \int_{\R} \widehat{\Phi}(x) \varrho_N(2x) \, \mathrm{d} x
\end{equation*}
for all $\Phi \in \mathcal{C}_c^1(\R)$.
\end{lem}
\begin{proof}
By the definition of expected value and by using inverse Fourier transform, we get
\begin{align*}
\mathbb{E}[\Phi(X_N)]
&= \frac{1}{2^N} \sum_{s_1,\ldots,s_N\,\in\,\{-1,+1\}} \Phi\!\left(\sum_{n\,=\,1}^N \frac{s_n}{n}\right) \\
&= \frac{1}{2^N} \sum_{s_1, \ldots, s_N \in \{-1, +1\}}
   \int_{\R} \widehat{\Phi}(x) \exp\!\left(2\pi ix \sum_{n\,=\,1}^N \frac{s_n}{n}\right) \mathrm{d} x \\
&= \frac{1}{2^N}
   \int_{\R} \widehat{\Phi}(x) \sum_{s_1, \ldots, s_N \,\in\, \{-1, +1\}} \exp\!\left(2\pi ix \sum_{n\,=\,1}^N \frac{s_n}{n}\right) \mathrm{d} x \\
&= \int_{\R} \widehat{\Phi}(x) \varrho_N(2x) \,\mathrm{d} x,
\end{align*}
as desired.
\end{proof}
In the following lemma, whose proof we postpone to Section~\ref{sec:rho}, we collect some results on $\varrho_N$.
\begin{lem}\label{lem:rhoN}
For all $N\in\N$ and $x\in\big[0,\sqrt{N}\big]$ we have
\begin{equation}\label{equ:rho2}
\varrho_N(x) =\varrho(x) (1+O(x^2/N)).
\end{equation}
Moreover, there exist absolute constants $B,C,E> 0$ such that
\begin{equation}\label{equ:rho3}
|\varrho_N(x)| < \exp\!\left(-B \exp\!\left(E\sqrt{\log x}\right)\right)
\end{equation}
for all sufficiently large positive integers $N$ and for all $x \in \left[1, \exp\!\left(C(\log N)^2\right)\right]$.
In particular, $C$ can be taken as any positive real number less than $1/\log 4$.
\end{lem}

\begin{figure}[h]
\centering
\includegraphics[width=0.60\textwidth]{BMS_Figure3.pdf}
\caption{Plot of $\varrho_{N}(x)/x$ for $N=40$ and $x\in[0,N^3]$.}
\label{Fig:3}
\end{figure}

We are now in a position to prove Theorem~\ref{thm:loclimthm}. Let $C$ be any positive constant less than
$1/\log 4$. Pick any small $\eps>0$, and set $\xi_{N,\pm \eps} := \exp\!\left(-(1\pm\eps)C(\log
N)^2\right)$ and $\xi_N:=\xi_{N,0}$. For an interval $I=[a,b]$ with $b-a>2\xi_N$, let $\Phi_{N,\eps}^+$
(respectively $\Phi_{N,\eps}^-$) be a smooth function $\R\to[0,1]$ with support contained in $[a -
\xi_{N,-\eps}, b + \xi_{N,-\eps}]$ (resp. $I$) and with $\Phi_{N,\eps}^+(x)=1$ if $x\in I$, (resp. $x \in
[a + \xi_{N,-\eps}, b - \xi_{N,-\eps}]$). Also, suppose $\Phi_{N,\eps}^{\pm(j)}(x)\ll_j
(\xi_{N,-\eps})^{-j}$ for all $j\geq 0$. It is not difficult to construct such functions. It follows that
the Fourier transform of $\Phi_{N,\eps}^{\pm}$ satisfies
\begin{equation}\label{equ:phibound}
\widehat{\Phi}_{N,\eps}^{\pm}\ll_A (1+|x|\xi_{N,-\eps})^{-A}
\end{equation}
for all $A>0$ and all $x\in\R$. Since
\begin{equation*}
\mathbb{E}[\Phi_{N,\eps}^{-}(X_N)] \leq \mathbb{P}[X_N\in I] \leq \mathbb{E}[\Phi_{N,\eps}^{+}(X_N)] ,
\end{equation*}
it suffices to show
\begin{equation*}
\mathbb{E}[\Phi_{N,\eps}^{\pm}(X_N)] = \int_\R \Phi_{N,\eps}^{\pm}(x) g(x) \, \mathrm{d} x+o_\eps(|I|) ,
\end{equation*}
because this quantity is evidently equal to $\int_I g(x) \, \mathrm{d} x + o(|I|)$.
From now on, let
$\Phi_{N,\eps}$ be one of $\Phi_{N,\eps}^{+}$, $\Phi_{N,\eps}^{-}$.
By Lemma~\ref{lem:expected} we have
\begin{equation*}
\mathbb{E}[\Phi_{N,\eps}(X_N)]
= \frac{1}{2}\int_{\R} \widehat{\Phi_{N,\eps}}(x/2) \varrho_N(x) \, \mathrm{d} x
= I_1+I_2+I_3,
\end{equation*}
where $I_1$, $I_2$, $I_3$ are the integral supported in $|x| < N^\eps$, $|x|\in[N^\eps, M_N^{1+\eps}]$,
and $|x|> M_N^{1+\eps}$, respectively, where $M_N := \xi_{N,-\epsilon}^{-1}$. Note that $M_N^{1+\eps}>
N^\eps$ for $N$ large enough, that $M_N^{1+\eps}=\xi_N^{-(1-\eps^2)}< \xi_N^{-1}$, and that
$M_N^{1+\eps}\xi_{N,-\eps} = \xi_{N,-\eps}^{-\eps} =\xi_{N}^{-\eps(1-\eps)}$ goes to infinity when $N$
goes to infinity.

Offner~\cite{Offner} showed that $\varrho(x)$ decays double exponentially. In particular, using
also~\eqref{equ:rho2}, we have
\begin{align*}
I_1
&=\frac{1}{2}\int_{-N^\eps}^{N^\eps} \widehat{\Phi_{N,\eps}}(x / 2) \varrho_N(x) \, \mathrm{d} x
 =\frac{1}{2}\int_{-N^\eps}^{N^\eps} \widehat{\Phi_{N,\eps}}(x / 2) \varrho(x)\, \mathrm{d} x
  + O\!\left(\|\widehat{\Phi_{N,\eps}}\|_\infty N^{-1+3\eps}\right) \\
&=\int_\R \widehat{\Phi_{N,\eps}}(x) \varrho(2x)\, \mathrm{d} x+O_\eps\!\left(\|{\Phi_{N,\eps}}\|_1 N^{-1+3\eps}\right).
\end{align*}
By~\eqref{equ:rho3} if $N$ is sufficiently large we have
\begin{align*}
|I_2|
&\leq \|\widehat{\Phi_{N,\eps}}\|_\infty\int_{N^\eps}^{M_N^{1+\eps}} |\varrho_N(x)| \,\mathrm{d} x \\
&\leq \|\Phi_{N,\eps}\|_1 \int_{N^\eps}^{+\infty} \!\!\exp\!\left(-B\exp\!\left(E\sqrt{\log x}\right)\right)\,\mathrm{d} x
\ll_\eps \|\Phi_{N,\eps}\|_1 N^{-1}.
\end{align*}
Now, by~\eqref{equ:phibound} we easily have
\begin{align*}
|I_3|
&\leq \int_{|x| > M_N^{1+\eps}} |\widehat{\Phi_{N,\eps}}(x)| \,\mathrm{d} x
 \ll_A \int_{M_N^{1+\eps}}^{+\infty} (1+ x\xi_{N,-\eps})^{-A} \,\mathrm{d} x\\
&\ll_A (1+ M_N^{1+\eps}\xi_{N,-\eps})^{1-A}
\ll_A \xi_{N}^{\eps(1-\eps)(A-1)}
= o_\eps(\xi_N)
= o_\eps(|I|) ,
\end{align*}
where in the last steps we have chosen $A=1+2/\eps$.
Thus, collecting the above results
\begin{align*}
\mathbb{E}[\Phi_{N,\eps}(X_N)]
&= \int_{\R} \widehat{\Phi_{N,\eps}}(x ) \varrho(2x) \, \mathrm{d} x+O_\eps(\|{\Phi_{N,\eps}}\|_1 N^{-1+3\eps})+ o_\eps(|I|)\\
&= \int_{\R} \Phi_{N,\eps}(x) g(x) \, \mathrm{d} x+O_\eps(\|{\Phi_{N,\eps}}\|_1 N^{-1+3\eps}) + o_\eps(|I|),
\end{align*}
by Parseval's theorem and the proof of Theorem~\ref{thm:loclimthm} is completed, because
$\|{\Phi_{N,\eps}}\|_1 = O_\eps(|I|)$.
\smallskip

We conclude the section with the following propositions which prove the bounds~\eqref{equ:simpleintro}
and~\eqref{equ:A3}.
\begin{pro}\label{pro:nonzero}
We have $\m_N \neq 0$ for each positive integer $N$. Moreover, as $N\to\infty$,
\begin{equation}\label{equ:simple}
\m_N > \exp\!\left(-N + o(N)\right).
\end{equation}
\end{pro}
\begin{proof}
For each positive integer $N$, define $L_N := \lcm\{1, \ldots, N\}$. Let $k$ be the unique nonnegative
integer such that $2^k \leq N < 2^{k+1}$. Then, for all $n \in \{1, \ldots, N\}$, we have that $L_N / n$
is an integer which is odd if and only if $n = 2^k$. As a consequence, for all $s_1, \ldots, s_N \in
\{-1, +1\}$, we have that
\begin{equation*}
\sum_{n \,=\, 1}^N \frac{L_N}{n} s_n
\end{equation*}
is an odd integer and, in particular, the sum $\sigma_N := \sum_{n \,=\, 1}^N s_n / n$ is nonzero, so that
$\m_N > 0$.
Furthermore, $|\sigma_N| \geq 1 / L_N$.
Thanks to the Prime Number Theorem, we have
\begin{equation*}
L_N = \exp(\psi(N)) = \exp(N + o(N))
\end{equation*}
as $N \to +\infty$, where $\psi$ is Chebyshev's function, and~\eqref{equ:simple} follows.
\end{proof}

\begin{pro}\label{pro:lower}
For almost all $\tau\in\R$, as $N \to +\infty$ we have
\begin{equation*}
\m_N(\tau) > \exp\!\left(-0.665N\right) .
\end{equation*}
\end{pro}
\begin{proof}
The claim follows by the Borel--Cantelli lemma: suppose we have an upper bound $\#\SS_N \leq e^{\alpha
N}$ for some $\alpha>0$, for all large enough $N$. Then for any fixed $\eps>0$
\begin{align*}
\mathcal{E} &:= \{\tau \in \R : \m_N(\tau) \leq e^{-(\alpha+\eps) N} \text{ for infinitely many } N\} \\
&= \bigcap_{M = 1}^\infty \bigcup_{N \geq M} \{\tau \in \R : \m_N(\tau) \leq e^{-(\alpha+\eps) N}\} .
\end{align*}
The Lebesgue measure of $\mathcal{E}$ is bounded by
\begin{align*}
|\mathcal{E}|
\leq \inf_{M} \sum_{N \geq M} 2e^{-(\alpha+\eps) N} \#\SS_N
\leq \inf_{M} \sum_{N \geq M} 2e^{-\eps N}
= \inf_{M} \frac{2e^{-\eps M}}{1 - e^{-\eps}}
= 0.
\end{align*}
This implies that for almost every $\tau$, the lower bound $\m_N(\tau) > e^{-(\alpha+\eps) N}$ holds for
all $N$ large enough. The upper bound for $\#\SS_N$ with $\alpha = \log 2$ is trivial, since $\#\SS_N
\leq 2^N$. The claim will follow from a slightly better estimation for this quantity. In fact, the sum
\[
\frac{s_1}{1} + \frac{s_2}{2} + \frac{s_3}{3} + \frac{s_4}{4} + \frac{s_6}{6} + \frac{s_{12}}{12}
\]
takes only $29$ different values when $s_j\in\{\pm 1\}$.

Thus, let
\[
F := \{\,\{k,2k,3k,4k,6k,12k\}\colon k\in D\}
\]
with
\[
D := \{2^{3a}3^{2b}m\colon m\geq 1,\ a,b\geq 0,\ 2,3\nmid m\}.
\]
With this choice for $D$ any natural number $n$ can be contained in at most one $6$-tuple. Indeed, the
numbers in $F$ associated with a given $k=2^{3a}3^{2b}m$ are
\[
\{2^{3a}3^{2b}m,    \
  2^{3a+1}3^{2b}m,  \
  2^{3a}3^{2b+1}m,  \
  2^{3a+2}3^{2b}m,  \
  2^{3a+1}3^{2b+1}m,\
  2^{3a+2}3^{2b+1}m\}
\qquad
2,3\nmid m
\]
and comparing the evaluations in $2$ and $3$ we see that no number of this family can be produced twice.
The cardinality of the union of all $5$-tuples in $F$ containing numbers $\leq N$ is $6$ times the number
of $k\in D$ which are $\leq N/12$. The number of such $k$ can be easily seen to be
\[
\frac{1+o(1)}{(1-2^{-3})(1-3^{-2})}\frac{\varphi(6)}{6}\frac{N}{12} = \Big(\frac{1}{28}+o(1)\Big)N.
\]

As said, any $6$-tuple gives rise to only $17$ different values, not $64$, thus the inequality  $\#\SS_N
\leq e^{\alpha N}$ holds for any
\[
\alpha > \Big(1-\frac{5}{28}\Big)\log 2 + \frac{1}{28}\log 29
= 0.6648\ldots,
\]
and the result follows.
\end{proof}

\section{The bounds for $\varrho$ and $\varrho_N$}\label{sec:rho}
In this section we prove Lemma~\ref{lem:rhoN}. We observe that for $0\leq x\leq \sqrt{N}$ we have
\begin{equation*}
\prod_{n=N+1}^{\infty}\cos(\pi x/n)
=\prod_{n=N+1}^{\infty}\big(1+O(( x/n)^2)\big)
=\exp(O(x^2/N))
=1+O(x^2/N),
\end{equation*}
which proves~\eqref{equ:rho2}.

We now move to the proof of~\eqref{equ:rho3}. We remark that it is sufficient to prove such inequality
for $x \in \left[N, \exp\!\left(C(\log N)^2\right)\right]$; indeed, one can reduce to this case also when
for $x<N$ since $|\varrho_{N}(x)|\leq |\varrho_{\lfloor x\rfloor}(x)|$.

For positive integers $k, N$ and for real $\delta, x \geq 0$, define
\begin{equation}\label{equ:definSk}
\mathcal{S}_k(N, \delta, x)
:= \big\{n \in \{1, \ldots, N\} : \|x / n^k\| \geq \delta \big\},
\end{equation}
where $\|y\|$ denotes the distance of $y \in \R$ from its nearest integer. By the following lemma, the
set $\mathcal{S}_1(N, \delta, x)$ plays a crucial role in the proof of~\eqref{equ:rho3}.
\begin{lem}\label{lem:varrhoNexp}
We have
\begin{equation*}
|\varrho_N(x)| \leq \exp\!\left(-\frac{\pi^2 \delta^2}{2} \cdot \#\mathcal{S}_1(N, \delta, x)\right)
\end{equation*}
for each positive integers $N$ and for all $x, \delta \geq 0$.
\end{lem}
\begin{proof}
The claim follows easily from the inequality
\begin{equation*}
|\cos(\pi x)| \leq \exp\!\left(-\frac{\pi^2\|x\|^2}{2}\right),
\end{equation*}
holding for all $x \in \R$, and from the definitions of $\varrho_N(x)$ and $\mathcal{S}_1(N,\delta,x)$.
\end{proof}
In the next two subsections we will prove a bound for $\varrho_N$ by giving two lower bounds for
$\mathcal{S}_1(N, \delta, x)$ for some suitable values of $\delta$. More precisely, in
Section~\ref{sec:arithmetic} we will complete the proof of Lemma~\ref{lem:rhoN}, showing
that~\eqref{equ:rho3} holds for all $x \in \left[N, \exp\!\left(C(\log N)^2\right)\right]$.
However, before doing this, in the next subsection we give a simpler argument proving that in the range
$x \in \left[N, \exp\!\left(C'(\log N)^2\right)\right]$ one has $|\varrho_N(x)|<1/x^2$. We remark that
this weaker inequality would still be sufficient for our application for Theorems~\ref{thm:upper}
and~\ref{thm:loclimthm}. If optimized, this argument would lead to the constant $C'=(4e)^{-2}+o(1)$.

\subsection{A short average of the number of divisors in a prescribed interval}
\label{sec:shortaverage}
In this subsection we prove the following proposition.
\begin{pro}\label{pro:secbound}
There exists $C'> 0$ such that $|\varrho_N(x)|<1/x^2$ for all sufficiently large positive integers $N$
and for all $x \in \left[N, \exp\!\left(C'(\log N)^2\right)\right]$.
\end{pro}
We start with the following lemma, which shows that the size of $\mathcal{S}_1(N, \delta, x)$ is strictly
related to the size of a certain divisor sum.
\begin{lem}\label{lem:divisoraverage}
For any $0<\delta<\frac{1}{2}$, $x\in\R$ and $N\in\N$ we have
\begin{equation*}
\frac{N}{2} - \sum_{x -\delta N \,<\, m \,<\, x + \delta N} \sum_{\substack{n \,\mid\, m \\ N / 2 \,\leq\, n \,\leq\, N}} 1
< \#\mathcal{S}_1(N, \delta, x)
< N- \sum_{x -\frac{\delta}{2} N \,<\, m \,<\, x + \frac{\delta}{2} N} \sum_{\substack{n \,\mid\, m \\ N / 2 \,\leq\, n \,\leq\, N}} 1.
\end{equation*}
\end{lem}
\begin{proof}
First we observe that
\begin{align*}
\tfrac{N}{2} - \#\big\{n \in \Z \cap [\tfrac{N}{2}, N] : \|\tfrac{x}{n}\| < \delta\big\}
<\#\mathcal{S}_1(N, \delta, x)
< N - \#\big\{n \in \Z \cap [\tfrac{N}{2}, N] : \|\tfrac{x}{n}\| < \delta\big\}.
\end{align*}
Now,
\begin{align*}
\#\big\{n \in \Z \cap [\tfrac{N}{2}, N] : \|\tfrac{x}{n}\| < \delta\big\}
&= \#\big\{n \in \Z \cap [\tfrac{N}{2}, N] : \exists \ell \in \Z \;\; \ell - \delta < x / n < \ell + \delta\big\} \nonumber \\
&= \#\big\{n \in \Z \cap [\tfrac{N}{2}, N] : \exists \ell \in \Z \;\; x - \delta n < \ell n < x + \delta n\big\} \nonumber \\
&< \#\big\{n \in \Z \cap [\tfrac{N}{2}, N] : \exists \ell \in \Z \;\; x - \delta N < \ell n < x + \delta N \big\} \nonumber \\
&= \sum_{x - \delta N \,<\, m \,<\, x + \delta N} \sum_{\substack{n \,\mid\, m \\ N / 2 \,\leq\, n \,\leq\, N}} 1
\end{align*}
and the lower bound for $\#\mathcal{S}_1(N, \delta, x) $ follows.
Similarly one obtains the upper bound.
\end{proof}
We take $\delta=\frac{4\sqrt{\log x}}{\pi} N^{-\frac{1}2}$ and assume $x\in[N,e^{N/8}]$ so that
$0<\delta<\frac{1}{2}$ and $\delta N<2x$. In particular, by Lemmas~\ref{lem:varrhoNexp}
and~\ref{lem:divisoraverage} we obtain $|\varrho_N(x)|<1/x^2$ whenever the inequality
\begin{equation}\label{equ:Dineq}
D(x,N) < N/4
\end{equation}
is satisfied, where
\begin{equation*}
D(x,N):=\sum_{x -\frac{4}{\pi} \sqrt{N \log x} \,<\, m \,<\, x + \frac{4}{\pi} \sqrt{N \log x}}
        \sum_{\substack{n \,\mid\, m \\ N / 2 \,\leq\, n \,\leq\, N}} 1.
\end{equation*}
Now, we take $w \in(0,\frac{1}{2})$ and use Rankin's trick to bound the inner sum:
\begin{align}
D(x,N)
&<   \frac{9}{\pi} \sqrt{N \log x} \cdot \max_{m\leq 2x}\sum_{\substack{n \,\mid\, m \\ N / 2 \,\leq\, n \,\leq\, N}} 1
\leq \frac{9}{\pi} \sqrt{N \log x} \cdot \max_{m\leq 2x}\sum_{\substack{n \,\mid\, m \\ N / 2 \,\leq\, n \,\leq\, N}} \left(\frac{N}{n}\right)^w\notag\\
&<   \frac{9}{\pi}  N^{\frac{1}{2}+w} \sqrt{\log x}\cdot \max_{m\leq 2x}\sigma_{-w}(m),                                             \label{equ:boundforD}
\end{align}
where, for any $s\in\R$, $\sigma_s(m)$ is defined as the sum of the $s$-th powers of the divisors of $m$.
In his lost notebook~\cite{MR1606180}, Ramanujan studied the large values of $\sigma_{-s}(n)$ for any
$s\in[0,1]$. We state his result in a slightly weaker form in the following Lemma.
\begin{lem}\label{lem:highlycomposite}
For each fixed $\varepsilon > 0$ there exists $C_1 > 0$ such that
\begin{equation*}
\sigma_{-s}(m) < \exp\!\left(C_1\frac{(\log m)^{1-s}}{\log \log m}\right) ,
\end{equation*}
for all integers $m \geq 3$ and for all $s \in [\varepsilon, 1 - \varepsilon]$.
\end{lem}
\begin{proof}
This is a consequence of \cite[(380)--(382)]{MR1606180} (see the remark before (383) on how to make the
inequalities unconditional).
See also \cite[Ch.~3, \S3, 1b]{MR2186914}.
\end{proof}
Applying the bound given in this lemma in~\eqref{equ:boundforD}, we obtain
\begin{align*}
D(x,N)
&< \frac{9}{\pi} N^{\frac{1}{2}+w} \sqrt{\log x}\cdot \exp\!\left(C_1\frac{(\log 2x)^{1-w}}{\log \log 2x}\right)
\end{align*}
for some $C_1>0$ and any $\frac{1}{4}<w<\frac{1}{2}$, $N\in\N$ and $x\in[N,e^{N/8}]$. Picking $w =
\frac{1}{2} - 1/\log\log 2x$, so that $\frac{1}{4}<w<\frac{1}{2}$ for sufficiently large $N$, this
inequality becomes
\begin{align*}
D(x,N)< \frac{9}{\pi} N\sqrt{\log x}\cdot \exp\!\left(- \frac{\log N-C_1 e\,(\log 2x)^{\frac{1}{2}}}{\log \log 2x}\right).
\end{align*}
If $x<\exp\left(C'(\log N)^2\right)$, with $C':=(2 C_1 e)^{-2}$, then this is $o(N)$ and
so~\eqref{equ:Dineq} holds for $N$ large enough. In particular, we obtain $|\varrho_N(x)|<1/x^2$ for
$x\in[N,\exp(C'(\log N)^2)]$, and the proof of Proposition~\ref{pro:secbound} is completed.

\subsection{An arithmetic construction}\label{sec:arithmetic}
Here we complete the proof of Lemma~\ref{lem:rhoN}. More specifically, we show the following proposition.
\begin{pro}\label{pro:mainbound}
For every positive $C<1/\log 4$ there exists a positive constant $E$ depending on $C$, such that
\begin{equation*}
|\varrho_N(x)| \leq \exp\!\left(- \frac{\pi^2}{400^2} \exp\!\left(E\sqrt{\log x}\right)\right)
\end{equation*}
for all $x \in [N, \exp(C(\log N)^2)]$, for all sufficiently large $N$.
\end{pro}
We start by giving a lower bound for $\#\mathcal{S}_k(N, \delta, x)$. We remind that $\mathcal{S}_k$ was
defined in~\eqref{equ:definSk}.
\begin{lem}\label{lem:SkNdxbound}
For all $a>0$, $\delta \in (0,1/2)$ and $x \in [e^{a k^2}, N^k]$, we have
\begin{equation*}
\#\mathcal{S}_k(N, \delta, x) \geq \big((1/2-\delta)(2^{-1/k}-(3/2)e^{-a}) -(2/3)^k\big)x^{1/k}
\end{equation*}
when $k$ is large enough (depending on $a$), and $N> e^{ak}$.
\end{lem}
\begin{proof}
Let $b>1$ be a parameter that will be chosen later. If $\ell$ and $n$ are integers such that
\begin{equation*}
1 \leq \ell \leq b^k-\tfrac{1}{2}
\quad\text{ and }\quad
\Big(\frac{x}{\ell + 1/2}\Big)^{1/k} < n \leq \Big(\frac{x}{\ell + \delta}\Big)^{1/k},
\end{equation*}
then it follows easily that $n \in \mathcal{S}_k(N, \delta, x)$. As a consequence,
\begin{equation}\label{eq:A1}
\#\mathcal{S}_k(N,\delta, x)
\geq \sum_{1\leq\ell\leq b^k-\tfrac{1}{2}} \Big(\frac{x}{\ell + \delta}\Big)^{1/k} - b^k.
\end{equation}
For $0 \leq s < t \leq 1$, we have the lower bounds
\begin{equation*}
\frac{1}{(1 + s)^{1/k}} - \frac{1}{(1 + t)^{1/k}}
= \frac{1}{k}\int_s^t \frac{\,\mathrm{d} y}{(1+y)^{1+1/k}}
\geq \frac{t - s}{k(1+t)^{1+1/k}}.
\end{equation*}
Applying these inequalities in~\eqref{eq:A1} with $s = \delta / \ell$ and $t = 1/(2\ell)$, we get
\begin{equation*}
\#\mathcal{S}_k(N, \delta, x)
\geq \Big(\frac{1}{2}-\delta\Big)\frac{1}{k}
      \sum_{1\leq\ell\leq b^k-\tfrac{1}{2}} \frac{x^{1/k}}{(\ell+1/2)^{1+1/k}}
    - b^k.
\end{equation*}
Since
\begin{equation*}
\sum_{1\leq\ell\leq b^k-\tfrac{1}{2}} \frac{1}{(\ell+\frac{1}{2})^{1 + 1/k}}
\geq \int_1^{b^k-\frac{1}{2}}\frac{\,\mathrm{d} y}{(y+1/2)^{1+1/k}}
= k((3/2)^{-1/k}-b^{-1})
\geq k(2^{-1/k}-b^{-1}),
\end{equation*}
this bound show that
\begin{align*}
\#\mathcal{S}_k(N, \delta, x)
\geq (1/2-\delta)\big(2^{-1/k} - b^{-1}\big)x^{1/k} - b^k.
\end{align*}
From the assumption $x \geq e^{a k^2}$ we get the claim setting $b:=2e^a/3$.
\end{proof}
Now we state a well-known identity~(see, e.g., \cite[Ch.~1, Problem~5]{MR554488}).
\begin{lem}\label{lem:identity}
For all integers $m \geq 0$, the identity
\begin{equation*}
\sum_{j \,=\, 0}^m (-1)^j \binom{m}{j} \frac{1}{x + j}
= \frac{m!}{x(x+1) \cdots (x + m)}
\end{equation*}
holds in $\mathbb{Q}(x)$.
\end{lem}
\begin{proof}
By induction on $m$.
\end{proof}
The next lemma is a simple inequality which will be useful later.
\begin{lem}\label{lem:ntokbound}
We have
\begin{equation*}
0 \leq \frac{1}{n^k} - \frac{1}{n(n + 1)\cdots(n + k - 1)}
< \frac{k^2}{2n^{k+1}},
\end{equation*}
for all positive integers $n$ and $k$.
\end{lem}
\begin{proof}
Since $1 + x \leq e^x$ for all real number $x$, we have
\begin{equation*}
0 \leq 1 - \prod_{j\,=\,0}^{k - 1} \left(1 + \frac{j}{n}\right)^{-1}
\leq 1 - \exp\!\left(-\sum_{j \,=\, 0}^{k - 1} \frac{j}{n}\right)
< 1 - e^{-k^2/(2n)}
< \frac{k^2}{2n},
\end{equation*}
and dividing everything by $n^k$ we get the desired claim.
\end{proof}
Next, using Lemmas~\ref{lem:identity} and~\ref{lem:ntokbound}, we deduce a bound for $\#\mathcal{S}_1$
from the bound for $\#\mathcal{S}_k$ given by Lemma~\ref{lem:SkNdxbound}.

Next, we use the previous lemmas to deduce a bound for $\mathcal{S}_1$ from Lemma~\ref{lem:SkNdxbound}.
\begin{lem}\label{lem:S1Ndxbound}
For all $\delta \in (0, 2^{-k}/20]$ and $x \in [4^{k^2}, N^k/(k-1)!]$, we have
\begin{equation*}
\#\mathcal{S}_1(N, \delta, x) \geq \frac{x^{1/k}}{200}
\end{equation*}
when $k$ is large enough and $N \geq k4^{k}$.
\end{lem}
The assumption $N \geq k4^{k}$ is an easy way to ensure that $4^{k^2}<N^k/(k-1)!$.
\begin{proof}
We set $\delta=d\cdot 2^{-(k+1)}$ for some $d$ that we fix later.
First, we have
\begin{align}\label{eq:A2}
\#\mathcal{S}_1(N, \delta, x)
&\geq \frac{1}{k} \cdot \#\big\{n \in \{1, \ldots, N\} \colon \exists j \in \{0, \ldots, k - 1\}\text{ with }\|x/(n + j)\| \geq \delta \big\} \\
&= \frac{1}{k} \cdot \big(N - \#\mathcal{T}_k(N, \delta, x)\big), \nonumber
\end{align}
where
\begin{equation*}
\mathcal{T}_k(N, \delta, x) := \big\{n \in \{1, \ldots, N\} \colon \|x/(n + j)\| < \delta\text{ for all } j \in \{0, \ldots, k - 1\}\big\}.
\end{equation*}
If $n \in \mathcal{T}_k(N, \delta, x)$, then for all $j \in \{0, \ldots, k - 1\}$ there exists an integer
$\ell_j$ such that
\begin{equation*}
\Big|\frac{x}{n + j} - \ell_j \Big| < \delta.
\end{equation*}
Therefore, setting
\begin{equation*}
\ell := \sum_{j=0}^{k-1} (-1)^j \binom{k-1}{j} \ell_j
\end{equation*}
and using Lemma~\ref{lem:identity}, we obtain
\begin{equation*}
\Big|\frac{x (k-1)!}{n(n+1)\cdots(n+k-1)} - \ell\Big|
\leq \sum_{j=0}^{k-1} \binom{k-1}{j} \Big|\frac{x}{n + j} - \ell_j\Big|
< 2^{k-1} \delta
= d/4.
\end{equation*}
Furthermore, assuming $n \geq \eta k (x/d)^{1/(k+1)}$ for some $\eta>0$, thanks to
Lemma~\ref{lem:SkNdxbound} we have that
\begin{align*}
\Big|\frac{x (k-1)!}{n^k} - \ell\Big|
&\leq \frac{d}{4} + \Big|\frac{x (k-1)!}{n^k} - \frac{x (k-1)!}{n(n+1)\cdots(n+k-1)}\Big|
\leq  \frac{d}{4} + \frac{x k^2(k-1)!}{2n^{k+1}}\\
&\leq \frac{d}{4} + \frac{x d k^2k^k}{2e^k\eta^{k+1}k^{k+1}x}
= \frac{d}{4}\Big(1 + \frac{2k}{\eta (\eta e)^k}\Big).
\end{align*}
Choosing $\eta> e^{-1}$ this quantity becomes $\leq 3d/10$ if $k$ is large enough (depending on the
choice of $\eta$). Choosing $d<5/3$ we ensure that this quantity is strictly smaller than $1/2$.
Therefore, under these hypotheses
\begin{equation*}
\Big\|\frac{x (k-1)!}{n^k}\Big\| \leq \frac{3d}{10}.
\end{equation*}
Summarizing, we have proved that for all $n \in \mathcal{T}_k(N, \delta, x)$, but at most $\eta
k(x/d)^{1/(k+1)}$ exceptions, it holds $n \notin \mathcal{S}_k(N, 3d/10, x(k-1)!)$. As a consequence,
\begin{equation*}
\#\mathcal{T}_k(N, \delta, x) - \eta k(x/d)^{1/(k+1)} \leq N - \#\mathcal{S}_k(N, 3d/10, x(k-1)!).
\end{equation*}
Hence, recalling~\eqref{eq:A2} and thanks to Lemma~\ref{lem:SkNdxbound}, since by hypothesis $x \geq
e^{ak^2}$ for some $a \geq 1$, we obtain
\begin{align*}
\#\mathcal{S}_1(N, \delta, x)
&\geq \frac{1}{k} \#\mathcal{S}_k(N, 3d/10, x(k-1)!) - \eta(x/d)^{1/(k+1)}\\
&\geq \Big(\Big(\frac{1}{2}-\frac{3d}{10}\Big)(2^{-1/k}-\frac{3}{2}e^{-a}) -(2/3)^k\Big)\frac{(k-1)!^{1/k}}{k} \,x^{1/k}
    - \eta(x/d)^{1/(k+1)}.
\end{align*}
Collecting $x^{1/k}$ and using the inequality $k!\geq (k/e)^k$ we get
\begin{align*}
\#\mathcal{S}_1(N, \delta, x)
&\geq x^{1/k} \Big(\Big(\Big(\frac{1}{2}-\frac{3d}{10}\Big)(2^{-1/k}-\frac{3}{2}e^{-a}) -(2/3)^k\Big)\frac{(1/k)^{1/k}}{e}
            - \frac{\eta}{d^{1/(k+1)}x^{1/(k^2 + k)}}\Big),
\end{align*}
and recalling the assumption $x\geq e^{ak^2}$, we obtain
\begin{align*}
\#\mathcal{S}_1(N, \delta, x)
&\geq x^{1/k} \Big(\Big(\Big(\frac{1}{2}-\frac{3d}{10}\Big)(2^{-1/k}-\frac{3}{2}e^{-a}) -(2/3)^k\Big)\frac{(1/k)^{1/k}}{e}
            - \frac{\eta}{d^{1/(k+1)}e^{ak/(k + 1)}}\Big).
\end{align*}
For $k$ large enough, this quantity is positive as soon as
\begin{equation*}
\Big(\frac{1}{2}-\frac{3d}{10}\Big)(1-\frac{3}{2}e^{-a}) > \frac{\eta e}{e^a}.
\end{equation*}
If $\eta$ is very close to $e^{-1}$ and $d$ is very small, this inequality is satisfied by any $a$ with
$(1-(3/2)e^{-a})>2e^{-a}$, i.e. $a > \log (7/2)$. We set $a=\log 4$, allowing the choice $\eta=0.4$
and $d=0.1$, when $k$ is large. An explicit computation shows that with these values for the parameters
the lower bound is larger than $x^{1/k}/200$ as soon as $k$ is larger than $400$.

\end{proof}
We are now ready to prove Proposition~\ref{pro:mainbound}. Let $C$ be any positive constant, $C< 1/\log
4$, and pick any $C'$ with $C<C'<1/\log 4$. We take $\delta := 2^{-k}/20$, and $k :=
\intpart{\sqrt{C'\log x}}$ for every $x$ in the given range. Then $x$ is in the interval $[4^{k^2},
N^k/(k-1)!]$. In fact, the inequality $4^{k^2} \leq x$ is evident, and
\begin{gather*}
x \leq \frac{N^k}{(k-1)!}
\Leftarrow
x \leq \Big(\frac{eN}{k}\Big)^k
\iff
\log x \leq k\log\Big(\frac{eN}{k}\Big).
\end{gather*}
Since $\sqrt{C'\log x} - 1 \leq k = \intpart{\sqrt{C'\log x}} \leq \sqrt{C'\log x}$, the last inequality
is implied by
\begin{equation*}
\frac{\log x}{\sqrt{C'\log x}-1} + \log(\sqrt{\log x}) \leq \log(eN/\sqrt{C'}).
\end{equation*}
As a function of $x$ this can be written as
\begin{equation*}
\sqrt{\frac{\log x}{C'}} + \log(\sqrt{\log x}) \leq \log N  + O_{C'}(1).
\end{equation*}
We are assuming that $\log N \leq \log x \leq C(\log N)^2$, hence this is implied by
\begin{equation*}
\sqrt{C/C'}\log N + \log\log N \leq \log N  + O_{C,C'}(1)
\end{equation*}
which is true as soon as $N$ is large enough.
This proves that we can apply Lemma~\ref{lem:S1Ndxbound}, getting
\begin{align*}
\delta^2 \cdot \#\mathcal{S}_1(N, \delta, x)
&\geq \frac{1/200}{400} \cdot 4^{-k} x^{1/k}
 =    \frac{2}{400^2} \exp\Big(\frac{\log x}{k} - k\log 4\Big) \\
&\geq \frac{2}{400^2} \exp(E\sqrt{\log x}),
\end{align*}
where $E:=\tfrac{1}{\sqrt{C'}} - \sqrt{C'}\log 4$.
Hence, applying Lemma~\ref{lem:varrhoNexp}, we get
\begin{align*}
|\varrho_N(x)|
&\leq \exp\!\Big(-\frac{\pi^2 \delta^2}{2} \cdot \#\mathcal{S}_1(N, \delta, x)\Big)
 \leq \exp\!\Big(-\frac{\pi^2}{400^2}\exp(E\sqrt{\log x})\Big),
\end{align*}
which is the claim.

\section*{Appendix}
The time needed for the computation of $\m_N$ with a direct exhaustive computation grows exponentially
with $N$ and becomes unpractical already for $N\approx 30$. Thus, for computing $\m_N$ for larger $N$ we
used the following idea. Let
\begin{equation*}
A:=\left\{\sum_{n \,=\, 1}^R \frac{s_n}{n} \colon s_1,\dots, s_R\in\{-1,+1\}\right\}
\quad
B:=\left\{\sum_{n \,=\, R+1}^N \frac{s_n}{n} \colon s_{R+1},\dots, s_N\in\{-1,+1\}\right\},
\end{equation*}
for any intermediate parameter $R\in[1, N]$. Then $\m_N = \min\{|a-b|\colon a\in A,\ b\in B\}$. The
algorithm producing this minimal distance is very fast if one preorders the lists $A$ and $B$. In this
way we were able to compute all $\m_N$ with $N\leq \np$; see the table below. The need of a large
quantity of RAM for storing the lists prevents us to compute significantly larger values of $N$. For the
computations we have used PARI/GP~\cite{PARI2}.

\begin{center}
\begin{tabular}{ rr| rr| rr| rr}
$N$&    $\m_N L_N$ & $N$&    $\m_N L_N$ & $N$&    $\m_N L_N$ & $N$&    $\m_N L_N$  \\
  \hline
 1 &             1 & 17 &            97 & 33 &        902339 & 49 &    421936433719\\
 2 &             1 & 18 &            97 & 34 &       7850449 & 50 &    175378178867\\
 3 &             1 & 19 &          3767 & 35 &       7850449 & 51 &      8643193037\\
 4 &             1 & 20 &          3767 & 36 &       7850449 & 52 &      8643193037\\
 5 &             7 & 21 &          3767 & 37 &      10683197 & 53 &    461784703049\\
 6 &             3 & 22 &          2285 & 38 &      68185267 & 54 &    461784703049\\
 7 &            11 & 23 &         24319 & 39 &      37728713 & 55 &    461784703049\\
 8 &            13 & 24 &         24319 & 40 &      37728713 & 56 &    461784703049\\
 9 &            11 & 25 &         71559 & 41 &     740674333 & 57 &    514553001783\\
10 &            11 & 26 &          4261 & 42 &     740674333 & 58 &    116096731427\\
11 &            23 & 27 &         13703 & 43 &    1774907231 & 59 &   2810673355099\\
12 &            23 & 28 &         13703 & 44 &    1774907231 & 60 &   2810673355099\\
13 &           607 & 29 &        872843 & 45 &    1774907231 & 61 &   4723651835663\\
14 &           251 & 30 &        872843 & 46 &    1699239271 & 62 & 136420009515743\\
15 &           251 & 31 &      17424097 & 47 &    3103390393 & 63 & 136420009515743\\
16 &           125 & 32 &      13828799 & 48 &    3103390393 & 64 &  23093515509397\\
\end{tabular}
\end{center}

\bibliographystyle{amsplain}

config: arxiv


TITLE: Variant of Picard-Lindelöf theorem
QUESTION [4 upvotes]: Question
Let $I=[0,a]$ and define the norm $||f||_{\lambda}=\sup_I |e^{-\lambda x}f(x)|$ for $f\in C(I)$. Let $\phi:\;\mathbb{R}^2\to\mathbb{R}$ satify $|\phi(x,u)-\phi(y,v)|\leq\rho |u-v|$ for all $x,y,u,v\in\mathbb{R}$ and some $\rho >0$. Define $\tau:\;f\mapsto \int_0^x \phi(t,f(t))\;dt$
I need to find a $\lambda$ such that $\tau$ is a contraction under the norm $||\cdot||_{\lambda}$
Thoughts
I am not too sure how to do this; my first line of thought was:
$$\begin{aligned}||\tau (f)-\tau (g)||_{\lambda} &=\sup_I\Big| e^{-\lambda x} \int_0^x \phi(t,f(t))-\phi(t,g(t))\;dt\Big| \\ &\leq \sup_I e^{-\lambda x} \int_0^x |\phi(t,f(t))-\phi(t,g(t))|\;dt\\ &\leq\sup_I \rho e^{-\lambda x} \int_0^x |f(t)-g(t)|\;dt \end{aligned}$$
But I can't see how to get $\cdots \leq \alpha\sup_I |e^{-\lambda x}(f(x)-g(x))|$ for some $\alpha<1 $ and some $\lambda$ from this. Any help would be appreciated.

REPLY [0 votes]: In general you can prove that if $w : I := [a,b] \longmapsto \mathbb{R}$ is strictly positive and continuos, it defines a norm equivalent to $\|f\|_{\infty}$ on $C^{0}(I)$, defined as $\|f\|_{w} := \sup\limits_{t \in I} \left\lbrace w(t) |f(t)|\right\rbrace$
Defining $w(t) := e^{-2L |t-t_{0}|}$ we define the norm $\|f\|_{w} := \sup\limits_{t \in I} \left\lbrace  e^{-2L |t-t_{0}|} |f(t)|\right\rbrace$
Let's prove that with this choice, $T$ is a contraction on $I = [t_{0}-\delta,t_{0}+\delta]$ :
$$e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t}[f(s,v(s))-f(s,u(s))]ds| \leq e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t}|f(s,v(s))-f(s,u(s))|ds| \leq$$
$$e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t} L |v(s)-u(s)|ds| \leq Le^{-2L|t-t_{0}|} |\int_{t_{0}}^{t} \|v-u\|_{w}e^{2L|s-t_{0}|} ds|$$
$$ = \frac{\|v-u\|_{w}}{2}(1-e^{-2L|t-t_{0}|}) \leq \frac{1}{2}\|v-u\|_{w}$$
So, taking the supremum for $t \in I$ we get exactly
$$\|T(v)(t)-T(u)(t)\|_{w} \leq \frac{1}{2}\|v-u\|_{w}$$
In other words $T$ is a contraction $\hspace{0.2cm} \Box.$
In you case $I$ becomes $[0,a]$, $T$ becomes $\tau$, $L$ becomes $\rho$ and $\lambda$ is the constant chosen in the definition of the weighted norm.


TITLE: Prove: For every non-empty finite set $S$ we have that: $|P_{0}(S)|=|P_{1}(M)|$
QUESTION [1 upvotes]: Task from old exam:

$|S|$ is the amount of elements in a finite set $S$ and $$P_{0}(S)=
\left\{N \subseteq S: |N| \text{ is even}\right\}$$ $$P_{1}(S)=
\left\{N \subseteq S: |N| \text{ is uneven}\right\}$$
Prove by induction: For every non-empty set S we have that:
  $|P_{0}(S)|=|P_{1}(S)|$

I wouldn't be sure how to prove this but I would go for induction. First show that the satement works for $S$ and then if it worked for $S$ show that it will work for every other, so for
$$S'= S\cup \left\{x\right\}, x \notin S$$ but then there is a problem, on which set we need to look at to prove this via induction? Or maybe this way is just bad/impossible here..?
Maybe on these sets would work? $$\left\{N \in P_{j}(S'): x \in N\right\}$$ and then also for $$\left\{N \in P_{j}(S'): x \notin N\right\}$$ where $j \in \left\{0,1\right\}$
This would be the idea how I would do it but there is a better, easier way maybe?

REPLY [4 votes]: Sketch: Induction is the right way to go.
Base case: If $|S|=1$, then $S=\{x\}$ and there are two subsets of $S$, $\emptyset$ and $\{x\}$.  We see that $P_0(S)=\{\emptyset\}$ and $P_1(S)=\{\{x\}\}$ since the empty set has zero elements (an even number).
Inductive case: Suppose that the claim is true for sets of size $k\geq 1$ and let $|S|=k+1>1$.  Fix $x\in S$ and let $S'=S\setminus\{x\}$.  By the inductive hypothesis, $|P_0(S')|=|P_1(S')|$.
Now, what is $P_0(S)$ and $P_1(S)$?  Let's focus on $P_0(S)$.  Suppose that $T\in P_0(S)$; then $|T|$ is even.  If $x\not\in T$, then $T\subseteq S'$, so $T\in P_0(S')$.  On the other hand, if $x\in T$, then let $T'=T\setminus \{x\}$.  We know that $T'\subseteq S'$ and $|T'|=|T|-1$, so $|T'|$ is odd, so $T'\in P_1(S')$.  Therefore, $|P_0(S)|=|P_0(S')|+|P_1(S')|$.
By a similar argument, $|P_1(S)|=|P_0(S')|+|P_1(S')|$.


TITLE: Shortest curve that divides circle into two regions of equal area
QUESTION [4 upvotes]: Of all the curves that divide the circle into two regions with the same area, is the diameter the shortest one?

REPLY [3 votes]: It can be shown, using a variational argument, that the curve should intersect the circle perpendicularly and has constant curvature. It's not hard to show that only the diameter is the only curve of this type dividing the disk in two equal halfs.


TITLE: Evaluate $\int_0^{\pi/4}{(4\cot x\ln\sec x-x)\ln^2\tan xdx}$
QUESTION [29 upvotes]: Question: How can we prove $$I=\int_0^{\pi/4}{(4\cot x\ln\sec x-x)\ln^2\tan xdx}=\frac5{2304}\pi^4?$$

I confirmed, numerically, that it holds for 1000 decimal places. This integral came up when I was calculating $\sum_{n=1}^\infty\frac{H_{4n}}{n^3}$.
Attempt
Note that $\tan x=i\frac{1-e^{2ix}}{1+e^{2ix}}$, substituing $t=e^{ix}$ gives
$$I=\int_1^{e^{i\pi/4}}\frac it\left(\frac{1+t^2}{1-t^2}\ln\frac{2t}{1+t^2}-\frac1i\ln t\right)\left(\frac{\pi i}2+\ln\frac{1-t^2}{1+t^2}\right)^2dt,$$
then I tried $t^2\mapsto t$,
$$I=\frac12\int_1^{i}\frac it\left(\frac{1+t}{1-t}(\ln2+(\ln t)/2-\ln(1+t))-\frac1{2i}\ln t\right)\left(\frac{\pi i}2+\ln\frac{1-t}{1+t}\right)^2dt,$$
For the next step I substituted $u=\frac{1-t}{1+t}$, but it makes the integral more complex. Moreover, the substitution seem to turn the integral into the original form. I got stuck here.

REPLY [13 votes]: $\DeclareMathOperator{\arctanh}{arctanh}$
We may write
\begin{align*}
I =&\int_0^{\frac \pi 4} \Big(4\cot x\log (\sec x) -x \Big)\log^2 (\tan x) \ dx\\
 =&  4\underbrace{\int_0^{\frac \pi 4} \cot x\log (\sec x) \log^2 (\tan x) \ dx}_{\tan x \mapsto x} -\underbrace{\int_0^{\frac \pi 4} x \log^2 (\tan x) \ dx}_{=:I_2}\\
=& 2\underbrace{\int_0^1 \frac{\log^2 x\log(1+x^2)}{x(1+x^2)} dx}_{x^2 \to x} - I_2\\
=&\underbrace{\frac 1 4 \int_0^1 \frac{\log^2 x \log(1+x)}{x(1+x)}dx}_{=:I_1} - I_2\\
=&I_1 - I_2.
\end{align*} Using the Maclaurin series of $\frac{\log(1+x)}{1+x} = -\sum_{k=1}^\infty (-1)^{k}H_k x^k,$ we get
\begin{align*}I_1 =&- \frac 1 4 \sum_{k=1}^\infty (-1)^{k}H_k \int_0^1 x^{k-1} \log^2 x\ dx\\
=&\boxed{-\frac 1 2 \sum_{k=1}^\infty \frac{(-1)^{k}H_k}{k^3}}
\end{align*} as can be found in @omegadot's answer. (In fact, it holds that
$$
 \sum_{k=1}^\infty  \frac{(-1)^{k}H_k}{k^3}=2\text{Li}_4(1/2) + \frac {7\ln 2 \zeta(3)} 4  - \frac {\pi^2 \ln^2 2 }{12}  -\frac {11\pi^4}{360} + \frac {\ln^4 2}{12},
$$ though we don't really need this fact for our purpose now.)

Next, we derive that:
\begin{align*}
\boxed{I_2 = \frac 1 8 \int_0^1 \frac{\log x \log^2\left(\frac {1+x}{1-x}\right)}{x}dx +\frac{\pi^4}{256}.}
\end{align*}
Then, using the Maclaurin series of $$\displaystyle\frac 1 4
\log^2\left(\frac {1+x}{1-x}\right)=\arctanh^2 x =\sum_{k= 1}^\infty \frac {H_{2k}-\frac 1 2 H_k}{k}x^{2k} $$ we get:
\begin{align*}
\frac 1 8 \int_0^1 \frac{\log x \log^2\left(\frac {1+x}{1-x}\right)}{x}dx =&\sum_{k=1}^\infty \frac{H_{2k}-\frac 1 2 H_k}{2k}\int_0^1 x^{2k-1} \log x\ dx\\
=- &\sum_{k=1}^\infty \frac{H_{2k}-\frac 12  H_k}{8k^3}  \\
=&-\frac 1 2\sum_{k=1}^\infty\frac{(1+(-1)^k)H_k}{k^3} +\frac 1 {16} \sum_{k=1}^\infty\frac{H_k}{k^3}\\
=&-\frac 1 2\sum_{k=1}^\infty\frac{(-1)^kH_k}{k^3} - \frac 7 {16} \sum_{k=1}^\infty\frac{H_k}{k^3}\\
=& I_1 - \frac 7 {16} \sum_{k=1}^\infty\frac{H_k}{k^3}.
\end{align*}
This implies that
\begin{align*}
I_2 =& I_1 - \frac 7 {16} \sum_{k=1}^\infty\frac{H_k}{k^3} +\frac{\pi^4}{256}\\
\end{align*} hence
$$
\boxed{I = I_1 - I_2 = \frac 7{16}\sum_{k=1}^\infty\frac{H_k}{k^3}-\frac {\pi^4}{256} = \frac 7 {16}\frac {\pi^4} {72} - \frac {\pi^4}{256}= \frac{5\pi^4}{2304}.}
$$ Here, Euler's formula $$ 2\sum_{k=1}^\infty\frac{H_k}{k^n} = (n+2)\zeta(n+1) - \sum_{k=1}^{n-2}\zeta(k+1)\zeta(n-k),\qquad n\ge 2$$ can be used to evaluate $\sum_{k=1}^\infty\frac{H_k}{k^3} = \frac{\pi^4}{72}$.

Evaluation of $I_2$: We note by the symmetry that
\begin{align*}
\int_0^{\frac \pi 4} \log^2(\tan(x)) dx= \frac 1 2\int_0^{\frac \pi 2} \log^2\left(\tan\left(\frac x 2\right)\right) dx= \frac 1 4 \int_{0}^{\pi} \log^2\left(\tan\left(\frac x 2\right)\right) dx = \frac{\pi^3}{16}.
\end{align*} This follows from $\displaystyle \log\left(\tan\left(\frac x 2\right)\right) = \sum_{k=1}^\infty \frac{(-1)^k -1}{k} \cos(kx)$ and Parseval's identity. Thus it follows
\begin{align*}
I_2 =& \underbrace{\int_0^{\frac\pi 4} \left(x-\frac \pi 4\right) \log^2(\tan(x)) dx }_{\frac \pi 4 -x\mapsto x}+ \frac{\pi^4}{64}\\
=& -\underbrace{\int_0^{\frac \pi 4} x \log^2\left(\frac {1+\tan x}{1-\tan x}\right) dx}_{\tan x = y} + \frac{\pi^4}{64} \\
=& -\underbrace{4\int_0^1 \frac{\arctan y}{1+y^2}\arctanh^2 y\ dy}_{=:J}+ \frac{\pi^4}{64}\\=&-J +\frac{\pi^4}{64}.
\end{align*}

Now let us define
$$
f(z) = \frac{\arctan z}{1+z^2}\left( \arctanh^2 z +\frac {\pi^2} {16}\right)
$$ so that $f$ is analytic on $|z|<1$ and continuous on $|z|\le 1$. We will use the symmetric nature of $f$ to evaluate the $J$.
We first notice that
$$
\int_{0}^1 f(z)dz =\frac 1 4 J +\frac{\pi^2}{16}\int_0^1\frac{\arctan z}{1+z^2}dz =\frac 1 4 J + \frac{\pi^4}{512} .
$$
On the other hand, we have that
\begin{align*}
\int_{0}^i f(z)dz=&i\int_0^1 f(ix)dx \\
=&i\int_0^1 \frac{\arctan(ix)}{1-x^2}\left(\arctanh^2(ix) + \frac{\pi^2}{16}\right) dx\\
=&  \int_0^1 \frac{\arctanh x}{1-x^2} \left(\arctan^2 x - \frac{\pi^2}{16}\right) dx\\
\underset{\text{IBP}}{=}&-\int_0^1 \arctanh^2 x\ \frac{\arctan x}{1+x^2} dx\\
=&-\frac 1 4 J
\end{align*} where we have used the relations $\arctan(ix) = i\arctanh x$ and $\arctanh(ix) = i \arctan x$.
Combining these, we know that
\begin{align*}
J =& \frac 12 J +\frac 1 2 J\\
=&2\left(\int_{0}^1f(z)dz -\frac{\pi^4}{512}\right) -2 \int_{0}^i f(z)dz\\=&-\underbrace{2 \int_1^i f(z)dz}_{=:K} -\frac{\pi^4}{256}\\
=&-K-\frac{\pi^4}{256}.
\end{align*}

For $K$, we parametrize $[1,i]$ by $z=e^{i\theta}, 0\le \theta \le \frac \pi 2$. Then exploiting the fact that  $$\arctan(e^{i\theta}) = -\frac i 2 \Big[\log\big(\cot\big(\scriptsize{\frac \theta 2+\frac\pi 4} \normalsize\big)\big)+ \frac {\pi i} 2\Big]$$
$$
\arctanh(e^{i\theta}) = \frac 1 2 \Big[\log\big(\cot\big(\scriptsize \frac \theta 2 \normalsize\big)\big) +\frac{\pi i}2\Big]
$$ we can see that (by Cauchy's integral theorem)
\begin{align*}
K=&2 \int_0^{\frac \pi 2} f(e^{i\theta}) ie^{i\theta} d\theta \\
=&2i\int_0^{\frac \pi 2} \frac{\arctan(e^{i\theta})}{1+e^{i2\theta}} \left(\arctanh^2(e^{i\theta}) + \frac{\pi^2}{16}\right)e^{i\theta}d\theta \\
=&\frac 1 {8} \int_0^{\frac\pi 2} \frac{\log\big(\cot\big(\frac \theta 2 +\frac \pi 4\big)\big)+\frac{\pi i}2}{\cos \theta}\Big[\log^2\big(\cot\big(\scriptsize \frac \theta 2 \normalsize\big)\big)+\pi i \log\big(\cot\big(\scriptsize \frac \theta 2 \normalsize\big)\big)\Big]d\theta.
\end{align*} Since $K$ is real, we can take the real part of the integral and perform the change of variable $\tan(\frac \theta 2) = y$ to get
\begin{align*}
K =&\frac 1 {8}\int_0^{\frac\pi 2} \frac{\log^2\big(\cot\big(\frac \theta 2\big)\big)\log\big(\cot\big(\frac \theta 2 +\frac \pi 4\big)\big) -\frac{\pi^2}2 \log\big(\cot\big(\frac\theta 2\big)\big)}{\cos \theta}d\theta\\
=&\frac 1 {4} \int_0^1 \frac{1}{1-y^2}\left(\log^2 y\log\left(\frac {1-y}{1+y}\right) +\frac{\pi^2}2 \log y\right)dy \\
=&\frac 1 {4} \underbrace{\int_0^1 \frac{\log^2 y\log\left(\frac {1-y}{1+y}\right)}{1-y^2}dy}_{\frac{1-y}{1+y} = x} +\frac{\pi^2}{8} \int_0^1 \frac{\log y}{1-y^2}dy\\
=&\frac 1 {8}\int_0^1 \frac{\log x\log^2\left(\frac {1+x}{1-x}\right)}{x}dx +\frac{\pi^2}{8} \sum_{n=0}^\infty \int_0^1 y^{2n}\log y\ dy \\
=&\frac 1 {8}\int_0^1 \frac{\log x\log^2\left(\frac {1+x}{1-x}\right)}{x}dx -\frac{\pi^2}{8} \underbrace{\sum_{n=0}^\infty\frac 1{(2n+1)^2}}_{=\frac 3 4 \zeta(2) = \frac{\pi^2}{8}} \\
=&\frac 1 {8}\int_0^1 \frac{\log x\log^2\left(\frac {1+x}{1-x}\right)}{x}dx -\frac{\pi^4}{64}.
\end{align*}

This gives
\begin{align*}
J =& -K -\frac{\pi^4}{256}\\
=&-\frac 1 {8}\int_0^1 \frac{\log x\log^2\left(\frac {1+x}{1-x}\right)}{x}dx +\frac{3\pi^4}{256}
\end{align*} and
\begin{align*}
I_2 =& -J +\frac{\pi^4}{64}\\
=&\boxed{\frac 1 8 \int_0^1 \frac{\log x \log^2\left(\frac {1+x}{1-x}\right)}{x}dx +\frac{\pi^4}{256}}
\end{align*} as wanted.
Thus it holds
$$
I = \frac {5\pi^4}{2304}
$$ by the previous argument.