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\title[Partition identities and Groebner basis]{Partition identities and application to infinite dimensional Groebner basis and viceversa}

\author{Pooneh Afsharijoo, Hussein Mourtada}

\begin {document}

\maketitle
\begin{abstract} In the first part of this article, we consider a  Groebner basis of the differential ideal $[x_1^2]$ with respect to "the" weighted
lexicographical monomial order and show that its computation is related with an identity involving the partitions that
appear in the first Rogers-Ramanujan identity. We then prove that a Grobener basis of this ideal is not differentially finite in contrary
with the case of "the" weighted reverse lexicographical order. In the second part, we give a simple and direct proof of a theorem of Nguyen Duc Tam about the Groebner basis of the differential ideal $[x_1y_1];$ we then obtain identities involving partitions with $2$ colors.

\end{abstract}

\footnote{{\textbf{2010 Mathematics Subject Classification.}
05A17,12H05,13D40,13P10.\\
\textbf{Keywords} Rogers-Ramanujan identities, Arc spaces, Hilbert series.}}

\section*{Intoduction}

An integer partition of a positive integer number $n$ is a decreasing sequence of positive integers $$\lambda=(\lambda_1\geq\lambda_2\geq \ldots\geq\lambda_l),$$
such that $\lambda_1+\lambda_2+ \cdots+\lambda_l=n.$ The $\lambda_i$'s are called the parts of $\lambda$ and $l$ is its size. The book \cite{A} is a classic in the theory
of integer partitions. A famous identity related to partitions and which plays an important role in  this paper is the  \textbf{First Rogers-Ramanujan Identity:}

\begin{center}

\textit{ The number of partitions of $n$ with no consecutive parts, neither equal parts is equal to the number of of partitions of $n$ whose parts are congruent to $1$ or $4$ modulo $5.$}
 \end{center}
In \cite{BMS} (see also \cite{BMS1}) we got to this identity by considering the space of arcs  of the double point $X=\mbox{Spec} (\k[x]/x^2)$  centred at the origin; denote it
by $X^0_\infty.$ The coordinate ring $A$ of $X^0_\infty$ is naturally graded and we  associate with it its Hilbert-Poincar\'e series
that we call the Arc Hilbert-Poincar\'e series; we denote it by $AHP_{X,0}(t).$ Note that this is an invariant of singularities of algebraic varieties \cite{M,BMS}. We prove in \cite{BMS} that
$AHP_{X,0}(t)$ is equal to the generating sequence of the number
of partitions appearing in the Rogers-Ramanujan identities. The proof uses a Groebner basis computation associated with a monomial ordering (reverse lexicographical). Note
that we have
$$A=\frac{\k[x_i,i \in \textbf{Z}_{>0}]}{[x_1^2]},$$
where $[x_1^2]$ is the differential ideal generated by $x_1^2$ and its iterated derivative by the derivation
$D$ which is determined by $D(x_i)=x_{i+1}.$ So
$$[x_1^2]= (x_1^2,2x_1x_2,2x_1x_3+2x_2^2,\ldots).$$

The grading of $A$ is determined by giving to $x_i$ the weight $i.$ \\

In the frist part of this article, we consider a different monomial ordering (lexicographical) and we find that the Groebner basis computation with respect to this monomial
ordering is related with  an other identity involving the partitions that appear in the first Rogers-Ramanujan identity. We then use this other member in the Rogers-Ramanujan
identitiy to carry the computations of the Groebner-basis in small degrees (usual degree) but in all weights. This leads us to prove, in contrast with the case of the reverse
lexicographical ordering, that the Groebner basis of our ideal with respect to the lexicographical ordering is not differentially finite.\\

In the other part of this article, we will give a direct and simpler proof of a theorem by Nguyen Duc Tam \cite{N} where he computes a Groebner basis of the arc space of
$X=\mbox{Spec} (\k[x,y]/(xy))$ or simply of the differential ideal generated by $xy.$  We then use this theorem to obtain identities of partitions with two colors.

\section{Hilbert series and integer partitions}

In this section, we will begin by considering the Hilbert-Poincar\'e  series of some graded algebras that are inspired by Groebner basis computations; we then interpret these
series as generating sequences of partitions having special properties. At the end of the section, we will give more explanations on how we have found these graded
algebras.\\

We will denote by $\k$ an algebraically closed field of characteristic $0.$ Recall that for a graded $\k-$algebras, $A=\oplus_{i \in \n}A_i$ (we assume that
$\mbox{dim}_\k (A_i) <\infty$) the Hilbert-Poincar\'e series of $A,$ that
we denote by $HP(A),$ is by definition $$HP(A)=\sum_{i\in \n}\mbox{dim}_\k (A_i)q^i,$$
where $q$ is a variable.
For more about Hilbert-Poincar\'e series, see the appendix in \cite{BMS} and the
references there.

Let $n,l\geq 1$ be integer numbers. We consider the graded algebra $\k[x_l,x_{l+1},\ldots]$ where we give $x_i$ the weight $i,$ for
$i\geq l.$  This grading induces a grading on the $\k$-algebra $\frac{\k[x_l,x_{l+1},\ldots]}{(x_{i_1}\cdots x_{i_n},~~i_j\geq l)}.$ \\

We consider the Hilbert-Poincar\'e series $$H_n^l=HP\left(\frac{\k[x_l,x_{l+1},\ldots]}{(x_{i_1}\cdots x_{i_n},~~i_j\geq l)}\right).$$

\begin{lem}\label{rec} We have
$$H_n^l=q^lH_{n-1}^l+H_n^{l+1}.$$

\end{lem}

\begin{proof} Using corollary 6.2 in \cite{BMS}, we have
$$H_n^l=HP\left(\frac{\k[x_l,x_{l+1},\ldots]}{(x_{i_1}\cdots x_{i_n},~~i_j\geq l)}\right)=$$
$$HP\left(\frac{\k[x_l,x_{l+1},\ldots]}{(x_l,x_{i_1}\cdots x_{i_n},~~i_j\geq l)}\right)+q^lHP\left(\frac{\k[x_l,x_{l+1},\ldots]}{\left((x_{i_1}\cdots x_{i_n},~~i_j\geq l): x_l\right)}\right)$$
The last term is exactly $H_n^{l+1}+q^lH_{n-1}^l.$

\end{proof}

\begin{prop}\label{hnl}
$$H_n^l=1+\frac{q^l}{1-q}+\frac{q^{2l}}{(1-q)(1-q^2)}+\cdots+ \frac{q^{(n-1)l}}{(1-q)(1-q^2)\cdots (1-q^{n-1})}.$$

\end{prop}
\begin{proof} The proof is by induction on the integer $n.$ Notice that for $n=1,$
$H_1^l=HP(\k)=1.$ For $n=2,$ the weighted-homogeneous components of
$$\frac{\k[x_l,x_{l+1},\ldots]}{(x_{i_1}x_{i_2},~~i_j\geq l)}$$
are generated by $1$ in degree $0$ and $x_i$ in degree $i$ for $i\geq l;$
for $i=1,\ldots l-1,$ the weighted-homogeneous components of degree  $i$ is the null vector space. Let us assume that the formula is true for $H_{j}^l,j\leq n-1$ and
prove it for $H_n^l.$ Using lemma \ref{rec} repetitively, we obtain
$$H_n^l=q^lH_{n-1}^l+H_n^{l+1}=q^lH_{n-1}^l+q^{l+1}H_{n-1}^{l+1}+H_n^{l+2}=\cdots=$$
$$q^lH_{n-1}^l+q^{l+1}H_{n-1}^{l+1}+\cdots+q^mH_{n-1}^{m}+H_n^{m+1}.$$
But we have $$\lim_{m \to \infty} H_n^m = 1,$$
where the limit is considered for the $q$-adic topology in $\c[[q]];$ hence we can write
$$ H_n^l=1+ q^lH_{n-1}^l+q^{l+1}H_{n-1}^{l+1}+\cdots+q^mH_{n-1}^{m}+q^{m+1}H_{n-1}^{m+1}+\cdots=$$
$$1+\sum_{m\geq l}q^mH_{n-1}^{m}. $$
By the induction hypothesis we obtain
$$ H_n^l=1+$$
$$q^l+\frac{q^{2l}}{1-q}+\frac{q^{3l}}{(1-q)(1-q^2)}+\cdots+ \frac{q^{(n-1)l}}{(1-q)(1-q^2)\cdots (1-q^{n-2})}+$$
$$q^{l+1}+\frac{q^{2(l+1)}}{1-q}+\frac{q^{3(l+1)}}{(1-q)(1-q^2)}+\cdots+ \frac{q^{(n-1)(l+1)}}{(1-q)(1-q^2)\cdots (1-q^{n-2})}+
\cdots=$$
(summing by columns)
$$1+(q^l+q^{l+1}+\cdots) + (\frac{q^{2l}}{1-q}+\frac{q^{2(l+1)}}{1-q}+\cdots)+\cdots +$$
$$(\frac{q^{(n-1)l}}{(1-q)(1-q^2)\cdots (1-q^{n-2})}+
\frac{q^{(n-1)(l+1)}}{(1-q)(1-q^2)\cdots (1-q^{n-2})}+\cdots).$$
The formula for $H_n^l$ follows from the formula
$$q^{jl}+q^{j(l+1)}+q^{j(l+2)}+\cdots=\frac{q^{jl}}{1-q^{j}},~j=1,\ldots,n-1.$$

\end{proof}

\begin{cor}Let $l,n\geq 1$ be integers. The generating series of the integer partitions with parts
greater or equal to $l$ and size (number of parts) less or equal to $n-1$ is
$$1+\frac{q^l}{1-q}+\frac{q^{2l}}{(1-q)(1-q^2)}+\cdots+ \frac{q^{(n-1)l}}{(1-q)(1-q^2)\cdots (1-q^{n-1})}.$$

\end{cor}
\begin{proof}
 This is just the combinatorial interpretation of proposition \ref{hnl}: A basis of the $i-$th weighted-homogeneuous component of
 $\frac{\k[x_l,x_{l+1},\ldots]}{(x_{i_1}\cdots x_{i_n},~~i_j\geq l)}$ is given by the monomials $x_{h_1}\cdots x_{h_r}$ such that
 $h_1+\cdots+h_r=i,h_j\geq l$ and  $x_{h_1}\cdots x_{h_r} \not\in (x_{i_1}\cdots x_{i_n},~~i_j\geq l);$ this corresponds
 to the partitions of $i$  with parts
greater or equal to $l$ and size (number of parts) less or equal to $n-1.$
\end{proof}

We now look at the Hilbert series that makes the link to the first Rogers-Ramanujan identity. For $j\geq 1,$ let
\begin{equation}\label{H1}
 HP\left(\frac{\k[x_j,x_{j+1},\ldots]}{(x_{i_1}\cdots x_{i_k}x_k,~~i_1 \geq i_2\geq \ldots \geq i_k\geq k\geq j)}\right).
\end{equation}

\begin{lem}\label{recp} We have
$$H_j=q^jH^j_{j}+H_{j+1}.$$

\end{lem}

\begin{proof} Using corollary 6.2 in \cite{BMS}, we have
$$H_j=HP\left(\frac{\k[x_j,x_{j+1},\ldots]}{(x_{i_1}\cdots x_{i_k}x_k,~~i_1\geq i_2\geq\ldots\geq i_k\geq k\geq j)}\right)=$$
$$HP\left(\frac{\k[x_j,x_{j+1},\ldots]}{(x_j,x_{i_1}\cdots x_{i_k}x_k,~~i_1\geq i_2\geq\ldots\geq i_k\geq k\geq j)}\right)+$$
$$q^jHP\left(\frac{\k[x_j,x_{j+1},\ldots]}{((x_{i_1}\cdots x_{i_k}x_k,~~i_1\geq i_2\geq\ldots\geq i_k\geq k\geq j):x_j)}\right).$$
The last term is exactly $H_{j+1}+q^jH^j_{j}.$

\end{proof}

\begin{theo}\label{HP}We have
$$H_1=1+\frac{q}{1-q}+ \frac{q^4}{(1-q)(1-q^2)}+ \frac{q^9}{(1-q)(1-q^2)(1-q^3)}+\cdots=$$
$$1+\sum_{n\geq1}\frac{q^{n^2}}{(1-q)(1-q^2)\cdots(1-q^n)}.$$
\end{theo}
\begin{proof}Applying lemma \ref{recp} for $j=1$, we obtain that
$H_1=qH_1^1+H_2;$ applying the same lemma for $j=2$ we have that $H_1=qH_1^1+q^2H^2_2
+H_3;$ Applying repetitively and in the same way  lemma \ref{recp}, we obtain that
for $m\geq 2,$
$$H_1= qH^1_{1}+q^2H^2_{2}+\cdots+q^mH^m_{m}+H_{m+1}.$$
Noticing that $$\lim_{m \to \infty} H_m = 1$$ (where the limit is considered for the $q$-adic topology in $\c[[q]], \c$ being the field of complex numbers) we can write
 $$H_1=1+qH^1_{1}+q^2H^2_{2}+\cdots+q^mH^m_{m}+\cdots.$$ Using proposition \ref{hnl}, we obtain that $H_1$ is equal to \\

 \xymatrix{
      1 & +q &  \\
       & +q^2  & +\frac{q^4}{1-q} & \\
      &  +q^3 & +  \frac{q^6}{1-q} & +\frac{q^9}{(1-q)(1-q^2)}\\
      &  +q^4 & +  \frac{q^8}{1-q} & +\frac{q^{12}}{(1-q)(1-q^2)} &+~~~\cdots\\ }

 Summing by columns we obtain the result.

\end{proof}

An equivalent statement of the theorem the following.

\begin{theo}\label{RRNM1} Let $n\geq 1$ be a positive integer.
The number of partitions of $n$ with size less than or equal to the smallest part is equal to the number  of partitions of $n$ without consecutive nor equal parts.
\end{theo}
\begin{proof}
 This follows from the known fact (\cite{A,BMS}) that the series obtained in theorem \ref{HP} is also the generating sequence of the  partitions
 without consecutive nor equal parts.
\end{proof}

\begin{theo}\label{RRNM2} Let $n\geq k$ be a positive integer.
The number of partitions of $n$ with parts larger or equal to $k$ and size less than or equal to (the smallest part minus $k-1$) is equal to the number of  partitions of $n$ with parts larger or equal to $k$ and without consecutive nor equal parts.
\end{theo}
\begin{proof}Let denote by $H_k$ the Hilbert series of
$$\frac{\k[x_k,x_{k+1},\ldots]}{(x_{i_1}\cdots x_{i_{j-k+1}}x_j,~~i_1\geq i_2\geq\ldots\geq i_{j-k+1}\geq j\geq k)}.$$
By using the same method as used in the proof of the theorem \ref {HP} we have
$$H_k=1+q^k H^k_{1}+q^{k+1}H^{k+1}_{2}+\cdots+q^{k+m}H^{k+m}_{m+1}+\cdots.$$

By proposition \ref {hnl} we have that

 \xymatrix{
 H_k=     1 & +q^k &  \\
       & +q^{k+1}  & +\frac{q^{2k+2}}{1-q} & \\
      &  +q^{k+2} & +  \frac{q^{2k+4}}{1-q} & +\frac{q^{3k+6}}{(1-q)(1-q^2)}\\
      &  +q^{k+3} & +  \frac{q^{2k+6}}{1-q} & +\frac{q^{3k+9}}{(1-q)(1-q^2)} &+~~~\cdots\\ }

summing by columns we obtain that

$$H_k=1+\sum_{n\geq 1} \frac{q^{n(n+k-1)}}{(1-q)(1-q^2)\cdots(1-q^n)}.$$
So $H_{k+1}+q^k H_{k+2}$ equals to

 \xymatrix{
    1 & +\frac{q^{k+1}}{1-q} & +\frac{q^{2k+4}}{(1-q)(1-q^2)}& +\frac{q^{3k+9}}{(1-q)(1-q^2)(1-q^3)} & +~~~\cdots\\
      &  +q^k & +  \frac{q^{2k+2}}{1-q} & +\frac{q^{3k+6}}{(1-q)(1-q^2)} &+~~~\cdots\\ }

Again summing by columns we obtain that

$$H_{k+1}+q^k H_{k+2}=1+\sum_{n\geq 1} \frac{q^{n(n+k-1)}}{(1-q)(1-q^2)\cdots(1-q^n)}=H_{k}.$$

Using this equation repetitively proves that for each $k\geq 1$  we can write following recursion series

$$H_k=A_{k+i}H_{k+i}+B_{k+i+1}H_{k+i+1}.$$

With initial conditions $A_{k+1}=1$ and $B_{k+2}=q^k$, and for all $i\geq 2$ we have $A_{k+i}=A_{k+i-1}+B_{k+i}$ and $B_{k+i+1}=q^{k+i-1}A_{k+i-1}.$

But if we denote by $H'_k$ the Hilbert series of $\frac{\k[x_k,x_{k+1},\ldots]}{(x_i^2,x_{i+1}^2,i\geq k)},$ then by \cite{BMS}, $H'_k$ satisfies in the following recursion formula

$$H'_k=A_{k+i}H'_{k+i}+B_{k+i+1}H'_{k+i+1}.$$
Where $A_j,B_j\in\k[[q]]$ are the same as before. So in the $(q)$-adic topology we have

$$H_k=\lim\big( A_{k+i}H_{k+i}+B_{k+i+1}H_{k+i+1}\big)=lim A_{k+i}=\lim\big( A_{k+i}H'_{k+i}+B_{k+i+1}H'_{k+i+1}\big)=H'_k.$$
This equation gives us the result.
\end{proof}
Theorem \ref{RRNM1} is inspired from a Groebner basis computation of the differential ideal $[x_1^2]:$ By  \cite{BMS}, the initial ideal of $[x_1^2]$
with respect to the reverse lexicographical ordering is $(x_i^2,x_ix_{i+1},i\geq 1),$ while we can guess that its initial ideal
with respect to the lexicographical ordering is $((x_{i_1}\cdots x_{i_k}x_k,~~i_1 \geq i_2\geq \ldots \geq i_k\geq k\geq 1)$ (we make use of this
guess in the next section). Hence the Hilbert series of the quotient rings of  $\k[x_1,x_2,\ldots]$ by these ideals are equal. The Hilbert series
of the quotient by $(x_i^2,x_ix_{i+1},i\geq 1),$  is the generating series of the partitions  without consecutive nor equal parts;
The Hilbert series
of the quotient by $((x_{i_1}\cdots x_{i_k}x_k,~~i_1 \geq i_2\geq \ldots \geq i_k\geq k\geq 1)$  is the generating series of the partitions
 with size less or equal to the smallest part. Theorem \ref{RRNM2} can be guessed in the same way by considering the ideal
 $[x_k^2]$ in $\k[x_k,x_{k+1},\ldots].$

\section{On the lex Groebner basis of $[x_1^2]$ }

Again, let $\k$ be a field of characteristic $0,$ and consider the
graded ring $\k[x_1,x_2,\dots],$ where the weight of $x_i$ is  $i$. So the weight of the monomial $x^{\alpha}:= x_{i_1}^{\alpha_{1}} x_{i_2}^{\alpha_{2}} \dots x_{i_n}^{\alpha_{n}} $, is equal to $\sum_{j=1}^{n} i_j {\alpha_{j}} $, and its (usual) degree is equal to $n$.\\
Let $f_2=x_1^2$ and for $i\geq 3,f_i=D^{i-2}(f_2):=D(f_{i-1}),$ where $D$ is the derivation determined by $D (x_i)=x_{i+1};$  then $I=[f_2]:=(f_2,f_3,\dots)\subset  \k[x_1,x_2,\dots],$ is the defining ideal  (up to isomorphism) of the space of arcs centred at the origin of  $X=\mbox{Spec}(\k[x]/(x^2)).$ The ideal $I$ is a differential ideal, i.e. we have $D(I)\subset I.$ We are interested in the possibility that $I$ have a differentially finite Groebner basis with respect to a monomial ordering ; see the following definition. \\

\begin{defi} Let $J\subset \k[x_1,x_2,\dots]$ be a differential ideal with respect to $D$ (i.e. $D(J)\subset J).$ Let $"<"$ be a total monomial order defined on $\k[x_1,x_2,\dots].$
We say that $J$ has a differentially finite Groebner basis with respect to $"<",$ if  there exist a finite number of polynomials $h_1,\ldots,h_r \in \k[x_1,x_2,\dots]$ such that $J=[h_1,\ldots,h_r]$ and the initial ideal $In_<(J)$ of $J$ with respect to $"<"$ satisfies
$$ In_<(J)=(In_<(D^i(h_j)),j=1,\ldots,r;i\geq 0),$$
where $D^i$ denotes the $i-$th iterated derivative and $D^0$ is the identity.

\end{defi}

Note that the notation $[h_1,\ldots,h_r]$ in the definition denotes the differential ideal generated by the $h_i, i=1,\ldots,r$ and by all their iterated derivatives. . Note that there might be different notions of differential Groebner basis, see \cite{CF}, \cite{O} and their bibliography.\\

In this section, we prove that no Groebner basis of $I$ with respect to
the weighted lexicographical order is  differentially finite. Note that, in contrary
with this case, it follows for \cite{BMS} that in the case of the weighted reverse lexicographical order $I$ has a differentially finite Groebner basis. \\

We denote the $n$-th derivative of a polynomial $f_i$ by $f_i^{(n)}$, so we have

$f_n=f_2^{(n-2)}=(x_1^2)^{(n-2)}=\sum\limits_{i=0}^{n-2} \binom {n-2}{i} x_1^{(i)}x_1^{(n-2-i)}=\sum\limits_{i=0}^{n-2} \binom {n-2}{i} x_{1+i}x_{n-i-1}.$

 Denote the leading term of $f_n$ with respect to the weighted lexicographical order by $LT(f_n).$ So $LT(f_n)=2 x_1 x_{n-1}$ for all $n\geq 2.$ \\

Recall that the S-polynomial of $f,g \in  \k[x_1,x_2,\dots]$ is by definition
$$S(f,g):=\frac{x^\gamma}{LT(f)}f-\frac{x^\gamma}{LT(g)}g,$$
where $x^\gamma$ is the least common multiple of the leading monomials of $f$ and $g.$
A possible reference about S-polynomials and Groebner basis is \cite{GP}.\\

A direct computation of  the S-polynomial of $f_3$ and $f_4$ gives
$$S(f_3,f_4)=x_2^3.$$ We set  $F_{x_2^3}:=S(f_3,f_4).$ For $k>2,$
we recursively define
$$F_{x_2x_k^2}:=S(F_{x_2 x_{k-1} ^2}^{(2)} , S(f_{k+1},f_{k+2})).$$
We then have the following lemma.

\begin{lem}\label{lead}
With respect to the weighted lexicographic order, for $k>2,$
the leading monomial of $F_{x_2x_k^2}$ is $x_2x_k^2.$
\end{lem}

\begin{proof}The proof is by induction on the integer $k$. Notice that for $k=3$ we have $F_{x_2^3}^{(1)}=3x_2^2  x_3$, $F_{x_2^3}^{(2)}=3x_2^2  x_4 + 6x_2  x_3^2$ and $S(f_4,f_5)=x_2^2x_4 - 3x_2x_3^2$. So we have
 $$ S(F_{x_2^3}^{(2)},S(f_4,f_5))=5x_2 x_3^2  :=F_{x_2 x_3^2}.$$
 For  $k=4$ we have $F_{x_2 x_3^2}^{(1)}=10 x_2 x_3 x_4+5 x_3^3$, $F_{x_2 x_3^2}^{(2)}=10 x_2 x_3 x_5+10 x_2 x_4^2 +25 x_3^2 x_4$ and $S(f_5,f_6)=3 x_2 x_3 x_5- 4x_2 x_4^2 -3 x_3^2 x_4$. So
 $$S(F_{x_2 x_3^2}^{(2)},S(f_5,f_6))=\frac{7}{3} x_2 x_4^2 +\frac{7}{2} x_3^2 x_4:=F_{x_2 x_4^2}.$$
Now assume that claim holds for $k-1\geq 4$. This means that for $k-1\geq 4$ we have

$$F_{x_2 x_{k-1} ^2}:=S(F_{x_2 x_{k-2} ^2}^{(2)} , S(f_{k},f_{k+1}) ).$$
Since the leading monomial of $F_{x_2 x_{k-1} ^2}$ is $x_2 x_{k-1} ^2$, we can assume that  $F_{x_2 x_{k-1} ^2}=a x_2 x_{k-1}^2 + g_3$, for some rational number $a$ and some  polynomial $g_3$ with the monomials of the form $ x_{i_1} x_{i_2} x_{i_3}$ such that $3\leq i_1 \leq i_2 \leq i_3$. So, on one hand, the second derivative of $F_{x_2 x_{k-1} ^2}$ will be as follow
$$F_{x_2 x_{k-1} ^2}^{(2)}=2a x_2x_{k-1}x_{k+1}+2a x_2 x_k^2 +h_3.$$
Where $h_3=4a x_3x_{k-1}x_k +a x_4x_{k-1}^2 +g_3^{(2)}$. On the other hand, we have

$S(f_{k+1},f_{k+2})=S(\sum\limits_{i=0}^{k-1} \binom {k-1}{i} x_{1+i}x_{k-i},\sum\limits_{i=0}^{k} \binom {k}{i} x_{1+i}x_{k+1-i})$

$=\frac{1}{2}\sum\limits_{i=0}^{k-1} \binom {k-1}{i} x_{1+i}x_{k-i} x_{k+1}-\frac{1}{2}\sum\limits_{i=0}^{k} \binom {k}{i} x_{1+i}x_{k+1-i} x_k$

$=\frac{1}{2}\sum\limits_{i=1}^{k-2} \binom {k-1}{i} x_{1+i}x_{k-i} x_{k+1}-\frac{1}{2}\sum\limits_{i=1}^{k-1} \binom {k}{i} x_{1+i}x_{k+1-i} x_k$.

So by the above equation we obtain that $LT(S(f_{k+1},f_{k+2}))=(k-1)x_2 x_{k-1} x_{k+1}$.
\\Now we can compute $S(F_{x_2 x_{k-1} ^2}^{(2)} , S(f_{k+1},f_{k+2}))$.

$S(F_{x_2 x_{k-1} ^2}^{(2)} , S(f_{k+1},f_{k+2}) )$
\\$=\frac{1}{2a}(2a x_2 x_{k-1} x_{k+1}+ 2a x_2 x_k^2 + h_3)- \frac{1}{(k-1)}(\frac{1}{2}\sum\limits_{i=1}^{k-2} \binom {k-1}{i} x_{1+i}x_{k-i} x_{k+1}-\frac{1}{2}\sum\limits_{i=1}^{k-1} \binom {k}{i} x_{1+i}x_{k+1-i} x_k)$
\\$=x_2 x_k^2 +\frac{1}{2a} h_3- \frac{1}{(k-1)}(\frac{1}{2}\sum\limits_{i=2}^{k-3} \binom {k-1}{i} x_{1+i}x_{k-i} x_{k+1}-\frac{1}{2}\sum\limits_{i=1}^{k-1} \binom {k}{i} x_{1+i}x_{k+1-i} x_k)$
\\$=x_2 x_k^2 +\frac{1}{2a} h_3+\frac{k}{k-1} x_2 x_k^2- \frac{1}{(k-1)}(\frac{1}{2}\sum\limits_{i=2}^{k-3} \binom {k-1}{i} x_{1+i}x_{k-i} x_{k+1}-\frac{1}{2}\sum\limits_{i=2}^{k-2} \binom {k}{i} x_{1+i}x_{k+1-i} x_k)$
$=\frac{2k-1}{k-1}x_2 x_k^2+\frac{1}{2a} h_3- \frac{1}{(k-1)}(\frac{1}{2}\sum\limits_{i=2}^{k-3} \binom {k-1}{i} x_{1+i}x_{k-i} x_{k+1}-\frac{1}{2}\sum\limits_{i=2}^{k-2} \binom {k}{i} x_{1+i}x_{k+1-i} x_k).$

In the first sum, $2\leq i \leq k-3$ and in the second one $2\leq i\leq k-2$. So each monomial that appears in $S(F_{x_2 x_{k-1} ^2}^{(2)} , S(f_{k+1},f_{k+2}))$ is of the form $x_3 x_{i_1}  x_{i_2}  x_{i_3}$ such that $3\leq i_1 \leq i_2 \leq i_3$, except $x_2 x_k^2$ and hence
$$LT(S(F_{x_2 x_{k-1} ^2}^{(2)}, S(f_{k+1},f_{k+2})))=\frac{2k-1}{k-1}x_2x_k^2.$$
\end{proof}

\begin{theo}\label{NDF}  A Groebner  basis of the ideal $I,$ with respect to the weighted lexicographic order, is not  differentially finite.
\end{theo}
\begin{proof}

 For proving this fact, we will use the idea of Buchberger's algorithm (mainly that any cancellation of initial monomials comes from an $S-$ polynomial \cite{CLO}) to construct a part of a Groebner  basis of the ideal $I$ with respect to the weighted lexicographic order, which is differentially infinite.\\

By lemma \ref{lead}, we have that for every integer $n\geq 3,$ the initial monomial of the polynomial $F_{x_2x_n^2}$  is included in the initial ideal of $I.$\\

Let $G= \{f_i,F_{x_2x_n^2}, F_{x_2x_n^2}^{(m)} | i \geq 2, m\geq 1, n\geq 3 \} $. By Buchberger's algorithm $G$ may be a part of a Groebner  basis of the ideal $I$ but it is not a Groebner  basis of $I$  because

$$S(F_{x_2x_3^2},F_{x_2x_3^2}^{(1)})=S(F_{x_2x_3^2},F_{x_2x_3x_4})
=5x_3^4.$$

But the monomial $x_3^4$ which is a member of the ideal $I$ is not divisible by the leading terms of any element of $G.$ \\

Note that the (usual) degree of the $S-$polynomial of two polynomials is at least equal to the maximum of degrees of these two polynomials. On the other hand, the derivative of a polynomial has the same degree as itself. Hence
the  degree of the $f_i$'s is equal to two, and other elements of $G$ have degree strictly bigger than $2.$
\\This means that the monomials of degree two that appear as the leading terms of elements of Groebner basis, are of the form $x_1x_i$ for $i\geq1$.
\\So we do not have any polynomial in a Groebner basis whose leading monomial is $x_2 x_n $ for some $n\geq 2$, and so  a polynomial having the same initial monomial of $F_{x_2x_n^2}$ should be included in this Groebner
basis for each integer $n\geq 3.$ Since the initial monomial of the polynomial $F_{x_2x_n^2}$ is not the initial of the derivative of any other element of $G,$ the Groebner  basis of the ideal $I$ with respect to the weighted lexicographic order will not be
differentially finite: it should contain  polynomials whose initial monomials are the initials of $F_{x_2x_n^2}, n \geq 3$ and no one of these initial monomials is the derivative of an other initial monomial of an element in $G.$

\end{proof}

\section{Two colors partitions and the node}

Let $S:=\k[x_1,x_2,\ldots,y_1,y_2,\dots]$ be the graded polynomial ring where
$x_i,y_i$ have the weight $i$ for every $i\geq1;$ the order of appearance of the variables
is important since we will use below a reverse lexicographical ordering. We consider the derivation
on $S$ defined by $D(x_i)=x_{i+1}$ and $D(y_i)=y_{i+1}.$ Let $f_2=x_1y_1,$ and let
$$I=[f_2]=(x_1y_1,x_2y_1+x_1y_2,\ldots)$$
be the ideal generated by $xy$ and its iterated derivatives $f_i,i\geq 3$ by $D:$
for $i\geq 3, f_i=D(f_{i-1}).$  Note that the scheme defined by $I$ is the space of arcs centred at the origin of the node $X=\{xy=0\} \subset \textbf{A}^2.$\\
In this section, we are interested in determining a Groebner basis of $I$ with
respect to the weighted reverse lexicographical order and then to apply this result
to integer partitions. Note that the Groebner basis below was found by Nguyen Duc Tam \cite{N}; he has a beautiful but very long and difficult proof that this is  actually a Groebner basis. Below we give a simpler
and very short proof.
\\We will begin by defining elements of $I,$ and we will show later that these elements give the Groebner basis cited above.

\begin{defi}(\cite{N})
For $1 \leq i_1 \leq i_2 \leq \dots \leq i_k $ and for $k\geq 2,$ we set
$$G_{i_1 ,i_{2}+1,i_{3}+2,\dots ,i_{k}+k-1}:=
\det{\begin{bmatrix}
 x_{i_1 -k+2}&x_{i_1 -k+3}&\dots &x_{i_{1}}& f_{i_1 +1}\\
x_{i_2 -k+3}&x_{i_2 -k+4}&\dots &x_{i_{2}+1}& f_{i_2 +2}\\
x_{i_3 -k+4}&x_{i_3 -k+5}&\dots &x_{i_{3}+2}& f_{i_3 +3}\\
\vdots & \vdots & \ddots & \vdots &\vdots\\
x_{i_k +1}&x_{i_k +2}&\dots &x_{i_{k}+k-1}& f_{i_k +k}
\end{bmatrix}}$$
where det stands for determinant.
\end{defi}
Expanding the determinant with respect to the last column, we see that these are elements of $I.$ A direct computation using the definition of the $f_i$ gives the
following:
\begin{lem}\label{IN}\cite{N}
The leading term of  $G_{i_1 ,i_{2}+1,i_{3}+2,\dots ,i_{k}+k-1}$ with respect to weighted reverse lexicographic order, is $x_{i_1} x_{i_2} x_{i_3}\dots x_{i_k}y_k$.
\end{lem}
 We denote by $\mathbb{G}$ the set whose elements are the $ G_{i_1 ,i_{2}+1,i_{3}+2,\dots ,i_{k}+k-1}$ and the $f_i.$ It follows from lemma \ref{IN} that the ideal generated
 by the initials of the elements of $\mathbb{G}$ is
$$J:=(x_{i_1} x_{i_2}\dots x_{i_k} y_k |~~ i_j ,k \geq 1). $$

First, we are interested in computing the Hilbert-Poincar\'e series of $S/J.$ For that we introduce for $n\geq 1$  the Hilbert-Poincar\'e series
$$HP_n=HP\left(\frac{\k [x_i,y_j | i\geq 1 , j\geq n]}{(x_{i_1} x_{i_2}\dots x_{i_k} y_k |~~ i_j\geq 1 ,k \geq n)}\right).$$ So $HP(S/J)=HP_1.$ We will use the following form of $H_n^1$ from
section 1:

\begin{lem}\label{HN1} For any  $n\geq 2$ we have
$$H^1_n = \frac{1}{(1-q)(1-q^2)\dots (1-q^{n-1})}.$$
\end{lem}

\begin{proof}
By proposition $\ref{hnl}$ we have
$$H^1_n = 1+ \frac{q}{1-q} +\frac{q^2}{(1-q)(1-q^2)} +\dots + \frac{q^{n-1}}{(1-q)\dots(1-q^{n-1})}$$
We prove the expression in the lemma by induction on the integer $n$. For $n=2$
$$H^1_2 = 1+\frac{q}{1-q} =\frac{1}{1-q}$$
Assume that $H^1_n= \frac{1}{(1-q)\dots (1-q^{n-1})}$.
Now  we have
$$H^1_{n+1} = 1 +\frac{q}{1-q} + \dots +\frac{q^{n-1}}{(1-q)\dots (1-q^{n-1})}+ \frac{q^n}{(1-q)\dots (1-q^{n})}$$
$$=H^1_n +\frac{q^n}{(1-q)\dots (1-q^n)}$$
By induction hypothesis we obtain
$$H^1_{n+1} = \frac{1}{(1-q)\dots (1-q^{n-1})} +\frac{q^n}{(1-q)\dots (1-q^n)}=\frac{1}{(1-q)\dots (1-q^n)}$$
\end{proof}

\begin{lem}\label{rec1}
$$HP_n =HP_{n+1} + q^n \prod_{i\geq1} \frac{1}{1-q^i}.$$
\end{lem}
\begin{proof}Using corollary 6.2 in \cite{BMS}, we have
$$HP_n=HP\left (\frac{\k [x_i,y_j | i\geq 1 , j\geq n]}{(x_{i_1} x_{i_2}\dots x_{i_k} y_k |~~ i_j \geq 1,k \geq n)}\right)=$$
$$HP\left(\frac{\k [x_i,y_j | i\geq 1 , j\geq n+1]}{(x_{i_1} x_{i_2}\dots x_{i_k} y_k |~~ i_j\geq 1 ,k \geq n)}\right)+q^n HP\left(\frac{\k [x_i,y_j | i\geq 1 , j\geq n]}{(x_{i_1} x_{i_2}\dots x_{i_n}  | ~~i_j \geq 1)}\right)=$$
$$HP_{n+1} + q^n HP\left(\frac{\k [x_1,x_2,\cdots]}{(x_{i_1} x_{i_2}\dots x_{i_n}  | ~~i_j \geq 1)}\right) ~~HP\left(\k[y_{n},y_{n+1},\cdots]\right)= $$
$$HP_{n+1} +q^n H_n^1 \prod_{i\geq n} \frac{1}{1-q^i} ;$$
by lemma \ref{HN1} this is equal to
$$HP_{n+1} +\frac{q^n}{(1-q)\cdots (1-q^{n-1})} \prod_{i\geq n} \frac{1}{1-q^i}= $$
$$HP_{n+1} +q^n \prod_{i\geq 1} \frac{1}{1-q^i}.$$
\end{proof}
\begin{prop}\label{HPn} We have
$$HP(S/J)=HP_{1}=\frac{1}{1-q}~~ \prod_{i\geq 1} \frac{1}{1-q^i}.$$

\end{prop}

\begin{proof}

Using lemma \ref{rec1} repetitively we obtain that, for $m\geq 2$
$$HP_1= q  \prod_{i\geq 1} \frac{1}{1-q^i} + q^2  \prod_{i\geq 1} \frac{1}{1-q^i} +\cdots q^m  \prod_{i\geq 1} \frac{1}{1-q^i}+ HP_{m+1}. $$
On the other hand
$$\lim_{m \to \infty} HP_m =  \prod_{i\geq 1} \frac{1}{1-q^i},$$
where the limit is considered for the $q$-adic topology in $\c[[q]];$ so we have,
$$HP_1 =  \prod_{i\geq 1} \frac{1}{1-q^i} + q  \prod_{i\geq 1} \frac{1}{1-q^i} + q^2  \prod_{i\geq 1} \frac{1}{1-q^i} + \cdots$$
$$=(1+q+q^2+\cdots )\prod_{i\geq 1} \frac{1}{1-q^i}$$
$$=\frac{1}{1-q}~~ \prod_{i\geq 1} \frac{1}{1-q^i}.$$

\end{proof}
We now are ready to prove:

\begin{theo}(\cite{N}) We have that $\mathbb{G}$ is a Groebner basis of $I.$

\end{theo}

    \begin{proof}
    Let $In(I)$ be the initial ideal of $I$ with respect to the weighted reverse
lexicographical order. Since all the elements of $\mathbb{G}$ are also in $I,$
we have that $J\subset In(I);$ to prove that $\mathbb{G}$ is a Groebner basis of $I,$ we need to prove that $J=In(I).$\\
Noticing that $(f_2,f_3,\ldots)$ is a regular sequence \cite{GS} (Note that this is rarely the case \cite{M1}) and that $f_i$ is of weight $i,$ we deduce that
$$HP(S/I)=\frac{1}{1-q}~~ \prod_{i\geq 1} \frac{1}{1-q^i},$$
which is equal by proposition  \ref{HPn} to $HP(S/J).$ But since we have a flat
deformation with generic fibre $S/I$ and special fibre $S/In(I),$ we have
that $HP(S/I)=HP(S/In(I)),$ hence $HP(S/In(I))=HP(S/J).$ We deduce
that the homogeneous components of the same weight of $S/(In(I)$ and $S/J$ have the same (finite) dimension, and since we have an inclusion in one sense because $J\subset In(I),$ they
are equal. Hence $J=In(I).$

  \end{proof}

We will interpret the above results in terms of two color partions: consider that we have two copies of each positive integer
number $m,$ one is blue and the other is red; we denote these copies by $m_b$ and $m_r.$   We define an order between the colored integers by $m_b>m_r$ (so that we do
not count in a partition  $m_b+m_r$ and $m_r+m_b$ as different); if $m>k,$ we say $m_c>k_{c'}$ for $c,c'\in\{b,r\}.$\\

An integer partition of a positive integer number $n$ is a decreasing sequence (with respect to the order that we have just defined) of positive integers of one color or an other
$$\lambda=(\lambda_{1,c_1}\geq\lambda_{2,c_2}\geq \ldots\geq\lambda_{l,c_l}),$$
where $c_i \in\{b,r\}$ and such that $\lambda_{1,c_1}+\lambda_{2,c_2}+ \cdots+\lambda_{l,c_l}=n.$
For example, the two colors integer partitions of $2$ are:
\begin{center}
 $2_b$\\
 $2_r$\\
 $1_b+1_b$\\
 $1_r+1_r$\\
 $1_b+1_r.$

\end{center}

Colored partitions has already appeared in the work of Andrews and Agarwal \cite{APS}.\\

On one hand, we can interpret the series $$\frac{1}{1-q}~~ \prod_{i\geq 1} \frac{1}{1-q^i}$$
as the generating series of the partitions with $2$ colors of $1$ and only the red color of any other positive integer. So the patitions of $2$ of this type
are all the partitions appearing in the above example except the first one.\\

On the other hand, the monomials in $S/J$ of weight $n$ are in bijection with the partitions with $2$ colors of $n$ whose number of blue parts is strictly less
than its smallest red part (if this latter exists). In the above example of partions of $2,$ all the partitions but the last one are of this type.
The Hilbert-Poincar\'e series $HP(S/J)$ is then the generating sequence of this type of partitions. Hence proposition \ref{HPn} gives:

\begin{theo}\label{2c}
The number of partitions of $n$  with $2$ colors of $1$ and only the red color of any other positive integer is equal to the number of
partitions with $2$ colors of $n$ whose number of blue parts is strictly less
than its smallest red part (if this latter exists).
\end{theo}

Playing the same game with the ideal $[x_jy_j]$ instead of $[x_1y_1]$ we can prove the following generalization of theorem \ref{2c}:

\begin{theo}
Let $j$ be a positive integer number. The number of partitions of $n$  with $2$ colors of $1,\ldots,2j-1$ and only the red color of any other positive integer is equal to the number of
partitions with $2$ colors of $n$ whose number of blue parts is strictly less
than its smallest red part (if this latter exists) minus $(j-1).$

\end{theo}
We recover theorem \ref{2c} by putting $j=1.$


text:
\section{Proof of Proposition~\ref{prop:traceQ_bound_0}}
\label{sec:proof_estimation_trQb2}
In order to estimate $\tr\Qb_M^2(\theta)$, we first estimate the norm of the Toeplitz matrix $\Rb_M$. The following lemma is a direct corollary of Theorem~2.3 in \cite{tian2019joint}, so the proof is omitted.

\begin{lemma} \label{lem:norm_toeplitz} If $(\Rb_M)$ is a sequence of Toeplitz matrices satisfying \ref{ass:spectral_density} and \ref{ass:spec_den_behav_0}, then
$$\|\Rb_M\|\asymp M^aL(M).$$
\end{lemma}

We also need the following properties of functions regularly varying at $0$.
\begin{lemma} \label{lem:density_property} If $f$ satisfies \ref{ass:spec_den_behav_0}, then
\begin{enumerate}[label=(\alph*)]
    \item \label{item:density_sup} $\sup \{f(t) \tq x\le t\le \pi\}\sim f(x)$ as $x\to 0^+$.
    \item \label{item:density_inf} $\inf \{f(t) \tq 0< t\le x\}\sim f(x)$ as $x\to 0^+$.
    \item \label{item:density_int_0} $\int_0^x f(t)\dd t \sim \frac{x^{1-a}L(x^{-1})}{1-a}$ as $x\to 0^+$.
    \item \label{item:density_overt_int_xpi} $\int_x^\pi \frac{f(t)}{t}\dd t \sim \frac{L(x^{-1})}{ax^a}$ as $x\to 0^+$.
\end{enumerate}
\end{lemma}
\begin{proof}
By changing the variable $u=x^{-1}$, \ref{item:density_sup} and \ref{item:density_inf} follow from Theorem~1.5.3 of \cite{bingham1989regular}, \ref{item:density_int_0} from Proposition~1.5.10 of \cite{bingham1989regular}, and \ref{item:density_overt_int_xpi} from Proposition~1.5.8 of \cite{bingham1989regular}.
\end{proof}

The bound of $\tr\Qb_M^2(\theta)$ will be estimated in different ways according to the location of $\theta$.  From Lemma~\ref{lem:norm_toeplitz} above, and (11) in \cite{vinogradova2015estimation}, we get a global estimation
\begin{equation}
    \tr\Qb_M^2(\theta)=O(M^{2a}L^2(M)\log M) \label{eq:trQ_bound_trival}
\end{equation}
for any $\theta\in [0,2\pi]$. We note that this bound is much larger than $\log^2 M$, and even much larger than $M$ when $a>1/2$. Thus if $f(\theta)$ is finite for some $\theta$, this bound does not imply the uniform boundedness of \eqref{eq:trQ2_log2}. However this bound is sharp only for $\theta$ very close to $0$. For $\theta\in[\delta, 2\pi-\delta]$ with $\delta>0$ an arbitrary small number, we can find a sharper bound. Let $f$ be the spectral density of $\Rb_M$.

For $x$ in the definition domain of $f$ and $\delta>0$, we define a local $\infty$-norm $\|\cdot\|_{(x,\delta)}$ as
\begin{equation}
    \|f\|_{(x,\delta)}:=\esssup_{t\in (x-\delta,x+\delta)}\{|f(t)|\}. \label{eq:def_local_norm}
\end{equation}

\begin{prop} \label{prop:bound_trQ2} Let $\Qb_M(\theta)$ be defined in \eqref{eq:def_QT} with $\Rb_M$ having positive spectral density $f\in L^1(-\pi,\pi)$. Then there exists an absolute constant $K>0$ such that for any $\theta\in\R$ and $\delta\in (0,\pi)$,
\begin{equation}
    \frac{\tr \Qb_M^2(\theta)}{2\log M}\le \|f\|_{(\theta,\delta)}^2 + \frac{K\|f\|_1(\|f\|_1+\|f\|_{(\theta,\delta)})}{\delta^4\log M} \label{eq:bound_trQ2}
\end{equation}
Moreover if $f$ is continuous at $\theta$, then
\begin{equation}
    \lim_{M\to\infty}\frac{\tr\Qb_M^2(\theta)}{2\log M}=f^2(\theta). \label{eq:lim_trQ2}
\end{equation}
\end{prop}
\begin{proof}
Let $\Rb_M=(r_{i-j})$. Using the integral expression of $r_{i-j}$, we write
\begin{equation}\begin{aligned}\tr \Qb_M^2(\theta) &= \sum_{i,j,k,l}r_{i-j}\frac{e^{\im (j-k)\theta}}{M-|j-k|}r_{k-l}\frac{e^{\im(l-i)\theta}}{M-|l-i|} \\
    &=\sum_{i,j,k,l}\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-\im (i-j)x}\dd x\frac{e^{\im (j-k)\theta}}{M-|j-k|} \frac{1}{2\pi}\int_{-\pi}^\pi f(y)e^{-\im (k-l)y}\dd y\frac{e^{\im(l-i)\theta}}{M-|l-i|} \\
    &=\frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(x+\theta)f(y+\theta)\sum_{i,j,k,l}\frac{e^{\im(j-k)x+\im(l-i)y}}{(M-|i-j|)(M-|k-l|)}\dd x\dd y \\
    &=\frac{1}{4\pi^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(x+\theta)f(y+\theta)\left|\sum_{i,j}\frac{e^{\im(jx-iy)}}{M-|i-j|}\right|^2\dd x\dd y\,.\end{aligned} \label{eq:trQ2_int}
\end{equation}
Denote
$$g(x,y):=\sum_{1\le i,j\le M}\frac{e^{\im(jx-iy)}}{M-|i-j|}.$$
Note that the equality \eqref{eq:trQ2_int} also holds for $\Rb_M=\Ib$ and correspondingly $f\equiv 1$, then we get
\begin{equation}
    \frac{1}{4\pi^2}\int_{-\pi}^\pi\int_{-\pi}^\pi |g(x,y)|^2\dd x\dd y=\tr\Bb_M^2=1+2\sum_{k=1}^{M-1}\frac{1}{k}\sim 2\log M. \label{eq:g2_density_log}
\end{equation}
If we consider $|g(x,y)|^2$ as a measure density on $[-\pi,\pi]^2$, the total mass of this measure is asymptotically $8\pi^2\log M$.

For $\delta\in (0,\pi)$, let $E_\delta:=[-\pi,\pi]\backslash (-\delta,\delta)$. Then we have
\begin{equation}
    \begin{aligned} & 4\pi^2\tr\Qb_M^2(\theta) = \int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(x+\theta)f(y+\theta)|g(x,y)|^2\dd x\dd y \\
    = & \left(\int_{E_\delta}\int_{E_\delta}+\int_{-\delta}^{\delta}\int_{E_\delta}+\int_{E_\delta}\int_{-\delta}^{\delta}+\int_{-\delta}^{\delta}\int_{-\delta}^{\delta}\right)f(x+\theta)f(y+\theta)|g(x,y)|^2\dd x\dd y \\
    \le & \left(\int_{E_{\delta/2}}\int_{E_{\delta/2}}+\int_{-\delta/2}^{\delta/2}\int_{E_{\delta}}+\int_{E_{\delta}}\int_{-\delta/2}^{\delta/2}+\int_{-\delta}^{\delta}\int_{-\delta}^{\delta}\right)f(x+\theta)f(y+\theta)|g(x,y)|^2\dd x\dd y \\
    =: & P_1+P_2+P_3+P_4, \end{aligned} \label{eq:trQ2_integral}
\end{equation}
where we have the inequality because on the RHS the integral on the region $\{(x,y)\in [-\pi,\pi]\tq \delta/2\le |x|, |y|\le \delta\}$ is repeated.  We will show that $P_1+P_2+P_3$ is bounded, and the main contribution to the integral is $P_4$. Also note that the integrated functions are $2\pi$-periodic on $x$ and $y$, so $E_\delta$ and $E_{\delta/2}$ can be replaced by $[\delta,2\pi-\delta]$ and $[\delta/2,2\pi-\delta/2]$ in the integrals.

We can rewrite $g(x,y)$ as
\begin{equation}
    \begin{aligned} g(x,y) =& \frac{1}{M}\sum_{j=0}^{M-1}e^{\im j(x-y)}+ \sum_{m=1}^{M-1}\frac{\sum_{i=0}^{M-m-1}e^{\im (ix+mx-iy)} + \sum_{j=0}^{M-m-1}e^{\im (jx-my-jy)}}{M-m}\\
    =& \frac{1}{M}\frac{1-e^{\im M(x-y)}}{1-e^{\im (x-y)}}  + \frac{e^{\im Mx}}{1-e^{\im (x-y)}}\sum_{m=1}^{M-1}\frac{e^{-\im(M-m)x}-e^{-\im(M-m)y}}{M-m} \\
    & + \frac{e^{-\im My}}{1-e^{\im (x-y)}}\sum_{m=1}^{M-1}\frac{e^{\im(M-m)y}-e^{\im(M-m)x}}{M-m} \\
    =& \frac{1}{M}\frac{1-e^{\im M(x-y)}}{1-e^{\im (x-y)}}+ \frac{e^{\im Mx}}{1-e^{\im (x-y)}}\sum_{k=1}^{M-1}\frac{e^{-\im kx}-e^{-\im ky}}{k}  + \frac{e^{-\im My}}{1-e^{\im (x-y)}}\sum_{k=1}^{M-1}\frac{e^{\im ky}-e^{\im kx}}{k} \\
    =:& g_1(x,y)+g_2(x,y)+g_3(x,y).
    \end{aligned} \label{eq:expr_g} \end{equation}
We note that for any $x,y\in \R$,
$$|g_1(x,y)|=\frac{1}{M}\left|\frac{\sin (M(x-y)/2)}{\sin((x-y)/2)}\right|\le 1.$$
Next we prove that for any $\delta\in(0,\pi)$, $|g_2(x,y)|$ and $|g_3(x,y)|$ are bounded by $K/\delta^2$ for some absolute constant $K>0$ on $[\delta/2, 2\pi-\delta/2]^2$. Let $z_1=e^{\im x}, z_2=e^{\im y}$ be two distinct points on the arc $\{z\in\C \tq |z|=1, \arg(z)\in [\delta/2, 2\pi-\delta/2]\}$. We have
$$|g_2(x,y)|= \left|\frac{1}{z_1-z_2}\int_{[z_1,z_2]} \sum_{k=0}^{M-2} z^k\dd z\right|\le \sup_{z\in[z_1,z_2]}\left|\frac{1-z^{M-1}}{1-z}\right|\le \frac{1}{\sin^2(\delta/4)},$$
where $[z_1,z_2]$ denotes the segment between $z_1$ and $z_2$. Therefore we have $|g_2(x,y)|\lesssim \delta^{-2}$. The same estimation also applies to $g_3$ and we get $|g(x,y)|\lesssim \delta^{-2}$ for $(x,y)\in [\delta/2,2\pi-\delta/2]^2$. Then we have
$$P_1\lesssim \frac{\|f\|_1^2}{\delta^4},$$
where the implicit constant is absolute.

For $P_2$, because $|1-e^{\im (x-y)}|^2\ge 4\sin^2(\delta/4)$ when $x\in (\delta,2\pi-\delta)$ and $y\in (-\delta/2,\delta/2)$, we have
\begin{equation}
    P_2 \le \frac{\|f\|_{(\theta,\delta)}}{4\sin^2(\frac{\delta}{4})}\int_{\delta}^{2\pi-\delta}\dd x f(x)\int_{-\delta/2}^{\delta/2}|(1-e^{\im(x-y)})g(x,y)|^2\dd y. \label{eq:P_2_contr1}
\end{equation}
We estimate the integral $\int_{-\delta/2}^{\delta/2}|(1-e^{\im(x-y)})g(x,y)|^2\dd y$ for every $x\in (\delta,2\pi-\delta)$. We have
$$\int_{-\delta/2}^{\delta/2}|(1-e^{\im(x-y)})g_1(x,y)|^2\dd y \le \frac{4\delta}{M^2};$$
using the inequality $|a-b|^2\le 2|a|^2+2|b|^2$, we have
$$\begin{aligned}\int_{-\delta/2}^{\delta/2}|(1-e^{\im(x-y)})g_2(x,y)|^2\dd y
   &= \int_{-\delta/2}^{\delta/2}\left|\sum_{k=1}^{M-1}\frac{e^{-\im ky}-e^{-\im kx}}{k}\right|^2\dd y \\
   &\le 2\int_{-\delta/2}^{\delta/2} \left|\sum_{k=1}^{M-1}\frac{e^{-\im ky}}{k}\right|^2\dd y + 2 \delta\left|\sum_{k=1}^{M-1}\frac{e^{-\im kx}}{k}\right|^2 \\
   &\le 2\int_{-\pi}^\pi \left|\sum_{k=1}^{M-1}\frac{e^{-\im ky}}{k}\right|^2\dd y + 2 \delta\left|\sum_{k=1}^{M-1}\frac{e^{-\im kx}}{k}\right|^2 \\
   &= 2\sum_{k=1}^{M-1}\frac{1}{k^2}+2 \delta\left|\sum_{k=1}^{M-1}\frac{e^{-\im kx}}{k}\right|^2.\end{aligned}$$
For $x\in (\delta,2\pi-\delta)$ we have
$$\left|\sum_{k=1}^{M-1}\frac{e^{-\im kx}}{k}\right|=\left|\int_{[0,e^{-\im x}]}\frac{1-z^{M-1}}{1-z}\dd z\right|\le \frac{2}{\sin\delta}$$
where the complex integral is taken along the segment from $0$ to $e^{-\im x}$. Then the following inequality holds with two absolute constants $K_1,K_2$:
$$\int_{-\delta/2}^{\delta/2}|(1-e^{\im(x-y)})g_2(x,y)|^2\dd y \le K_1+K_2/\delta.$$
The integral expression corresponding to $g_3$ is similarly estimated. Taking these into \eqref{eq:P_2_contr1}, we get
$$P_2\le \|f\|_{(\theta,\delta)}\|f\|_1\frac{K_1+K_2\delta+K_3\delta^2/M^2}{\delta^3}\lesssim \frac{\|f\|_{(\theta,\delta)}\|f\|_1}{\delta^3}.$$
The same bound also controls $P_3$.

For $P_4$, from \eqref{eq:g2_density_log} we have
$$P_4\le 8\pi^2\|f\|_{(\theta,\delta)}^2\log M.$$
Summarising the bounds for $P_1,P_2,P_3,P_4$ and dividing $8\pi^2\log M$, the result follows.

To prove \eqref{eq:lim_trQ2}, if $f$ is continuous at $\theta$, then for any $\varepsilon>0$, there is some $\delta>0$ such that
\begin{equation}
    |f(x+\theta)f(y+\theta)-f^2(\theta)|\le \varepsilon, \quad \forall x,y \in (-\delta,\delta). \label{eq:approx_fxfy_f2}
\end{equation}
Note that the sum of the first three integrals in the second line of \eqref{eq:trQ2_integral} is bounded by $P_1+P_2+P_3$, we have
$$\frac{\tr\Qb_M^2(\theta)}{2\log M} = \frac{1}{8\pi^2\log M}\int_{-\delta}^{\delta}\int_{-\delta}^{\delta}f(x+\theta)f(y+\theta)|g(x,y)|^2\dd x\dd y + O\left(\frac{1}{\log M}\right).$$
All the above arguments apply also to $\Rb_M=f^2(\theta)\Ib$, then
$$\frac{\tr\Bb_M^2}{2\log M}f^2(\theta) = \frac{1}{8\pi^2\log M}\int_{-\delta}^{\delta}\int_{-\delta}^{\delta}f^2(\theta)|g(x,y)|^2\dd x\dd y + O\left(\frac{1}{\log M}\right).$$
Take the difference of the last two equations and then take the absolute value, also note \eqref{eq:approx_fxfy_f2} and \eqref{eq:g2_density_log}, we get
$$\left|\frac{\tr\Qb_M^2(\theta)}{2\log M}-\frac{\tr\Bb_M^2}{2\log M}f^2(\theta)\right|\le \varepsilon+O\left(\frac{1}{\log M}\right).$$
Let $M\to\infty$ and $\varepsilon\to 0$, we get the result.
\end{proof}

As a consequence of Proposition~\ref{prop:bound_trQ2}, if $f$ satisfies \ref{ass:spectral_density}, then $\tr \Qb_M^2(\theta)/\log M$ is bounded uniformly in $M$ and in $\theta\in(\delta,2\pi-\delta)$ for any fixed $\delta\in(0,\pi)$.

To summarize what we have obtained, if $f$ satisfies \ref{ass:spectral_density}, \ref{ass:spectral_bound_below} and \ref{ass:spec_den_behav_0}, then
\begin{equation}
    \frac{\tr\Qb_M^2(\theta)}{f^2(\theta)\log M} \label{eq:trQ2_log}
\end{equation}
is bounded uniformly in $M\ge 1$ and
\begin{enumerate}
    \item in $\theta\in [-\frac{A}{M},\frac{A}{M}]$ for any $A>0$, using the global bound \eqref{eq:trQ_bound_trival} and Lemma~\ref{lem:density_property}\ref{item:density_inf}.
    \item in $\theta\in [-\pi,-\delta]\cup [\delta,\pi]$ for any $\delta\in (0,\pi)$, using Proposition~\ref{prop:bound_trQ2}.
\end{enumerate}
Note the difference between \eqref{eq:trQ2_log} and our goal \eqref{eq:trQ2_log2}, using a classic argument, we can find two sequences of positive numbers $1/M \ll \tau_M < \delta_M=o(1)$ such that \eqref{eq:trQ2_log2} is uniformly bounded in $[-\pi,-\delta_M]\cup [-\tau_M,\tau_M]\cup [\delta_M,\pi]$.

In order to achieve the final result, the rest of this section consists of proving the uniform boundedness of \eqref{eq:trQ2_log2} for $|\theta|\in (\tau_M,\delta_M)$.

\begin{lemma} \label{lem:bound_g2} There exists a constant $K>0$ such that for any $-\pi \le x,y \le \pi$ with $xy\ne 0$, we have
$$\left|\frac{1}{x-y}\sum_{k=1}^M\frac{e^{\im kx}-e^{\im ky}}{k}\right|\le \frac{K}{\sqrt{|xy|}}.$$
\end{lemma}
\begin{proof}We first assume that $xy>0$, i.e. $x,y$ have the same sign. Assume further without loss of generality that $0<y<x\le \pi$. Then
$$\begin{aligned}\left|\sum_{k=1}^M\frac{e^{\im kx}-e^{\im ky}}{k}\right| &= \left|\int_y^x \sum_{k=0}^{M-1} e^{\im k s}\dd s\right| \\
  &\le \int_y^x \frac{1}{\sin(s/2)}\dd s \\
  &\le \int_y^x \frac{\pi}{s}\dd s \\
  &= \pi(\log x-\log y).\end{aligned}$$
We prove that $\frac{\log x-\log y}{x-y}\le \frac{1}{\sqrt{|xy|}}$. Let $x=e^u, y=e^v$, it suffices to prove that
\begin{equation}
    e^{\frac{u-v}{2}}-e^{\frac{v-u}{2}}\ge u-v. \label{eq:lem_bound_g2}
\end{equation}
For $u=v$, the equality holds. By differentiating we can see that the function
$$u\mapsto e^{\frac{u-v}{2}}-e^{\frac{v-u}{2}}-u$$
is increasing with $u$. Then for $u>v$, \eqref{eq:lem_bound_g2} holds and we have
$$\left|\frac{1}{x-y}\sum_{k=1}^M\frac{e^{\im kx}-e^{\im ky}}{k}\right|\le \frac{\pi}{\sqrt{|xy|}}.$$

If $-\pi \le y <0< x<\pi$, we can assume without loss of generality that $|y|\le x$. Then
$$\sum_{k=1}^M\frac{e^{\im kx}-e^{\im ky}}{k}=\sum_{k=1}^M\frac{e^{\im kx}-e^{-\im ky}}{k}+ 2\im \sum_{k=1}^M\frac{\sin ky}{k}$$
and by Theorem~1.3 and Remark(b) on Page 183 of \cite{zygmund2002trigonometric}, the series $\sum_{k=1}^M\frac{\sin ky}{k}$ is uniformly bounded (say, by $K$). Then from the first part of the proof, and using the inequality of arithmetic and geometric means, we have
$$\begin{aligned}
\left|\frac{1}{x-y}\sum_{k=1}^M\frac{e^{\im kx}-e^{\im ky}}{k}\right| & \le  \frac{1}{x-(-y)}\left|\sum_{k=1}^M\frac{e^{\im kx}-e^{-\im ky}}{k}\right|+\frac{2K}{x-y} \\
    & \le  \frac{\pi+K}{\sqrt{|xy|}}.
\end{aligned}$$
\end{proof}

From \eqref{eq:trQ2_int}, we have
$$\tr\Qb_M^2(\theta)\le \frac{1}{4\pi^2} \left(\int_{-\pi}^\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{-\pi}^\pi + \int\int_{\frac{\pi}{2}\le |x|,|y|\le \pi}\right)f(x+\theta)f(y+\theta)|g(x,y)|^2\dd x\dd y.$$
From the proof of Proposition~\ref{prop:bound_trQ2}, the function $g(x,y)$ is uniformly bounded on $([-\pi,-\pi/2]\cup [\pi/2,\pi])^2$, thus the third integral is uniformly bounded. We only need to estimate the first integration, because the second one is similar to the first.

When $|x|\le \pi/2, |y|\le \pi$, we have $\frac{|x-y|}{2} < \frac{3\pi}{4}$, then
\begin{equation}
    \frac{|x-y|}{2\left|\sin\frac{x-y}{2}\right|}\le \frac{2\sqrt{2}}{3\pi}. \label{eq:sinc_1_bound}
\end{equation}
We write $g=g_1+g_2+g_3$ as in the proof of Proposition~\ref{prop:traceQ_bound_0}, and we have $|g_1(x,y)|\le 1$.  By Lemma~\ref{lem:bound_g2}, for $|x|\le \pi/2, |y|\le \pi$, we have
$$|g_2(x,y)| \lesssim \frac{1}{\sqrt{|xy|}}.$$
By \eqref{eq:sinc_1_bound}, we have
$$|g_2(x,y)| \lesssim \frac{|x-y|}{2|\sin\frac{x-y}{2}|}\frac{1}{|x-y|}\left|\sum_{k=1}^{M-1}\frac{e^{\im kx}-e^{\im ky}}{k}\right|\lesssim \frac{\log M}{|x-y|}.$$
The above two bounds also controls $|g(x,y)|$ because $|g|\le |g_1|+|g_2|+|g_3|$, where $g_2$ and $g_3$ are similarly bounded, and $|g_1|\le 1$ can be absorbed into the other two bounds. On the other hand $|g(x,y)|$ is obviously bounded by $g(0,0)=2M+1$. In the sequel we will use different bounds in different subsets of $\{(x,y)\in\R^2 \tq |x|\le \pi/2, |y|\le \pi\}$:
\begin{equation}
    |g(x,y)|\lesssim  \begin{cases} M & |x|,|y|<\frac{2}{M} ; \\ \frac{1}{\sqrt{xy}} & |x|,|y|>\frac{1}{M} ; \\ \frac{\log M}{|x-y|} & |x|< \frac{1}{M}, |y|>\frac{2}{M} \text{ or } |y|< \frac{1}{M}, |x|>\frac{2}{M}. \end{cases} \label{eq:3bounds_g}
\end{equation}

Similar to \eqref{eq:trQ2_integral}, we have
\begin{equation}
    \begin{aligned}\int_{-\pi}^\pi\int_{-\frac{\pi}{2}}^\frac{\pi}{2} f(x+\theta)f(y+\theta)|g(x,y)|^2\dd x\dd y \le & \left(\int_{|x|<\frac{2}{M}}\int_{|y|<\frac{2}{M}} + \int_{|x|<\frac{1}{M}}\int_{|y|>\frac{2}{M}} +\int_{|x|>\frac{2}{M}}\int_{|y|<\frac{1}{M}} \right. \\
    & \left. + \int_{|x|>\frac{1}{M}}\int_{|y|>\frac{1}{M}}\right)f(x+\theta)f(y+\theta)|g(x,y)|^2\dd x\dd y \\
    & =: P_1+P_2+P_3+P_4,  \end{aligned}
\end{equation}
where the integrals on the RHS are all taken within the region $(x,y)\in (-\pi/2,\pi/2)\times (-\pi,\pi)$, which are omitted in order to abbreviate the notations.

When $x\in [-2/M,2/M]$ and $|\theta| \in [\tau_M,\delta_M]$, as $M$ is large enough, we have $\theta^{-1}(\theta+x)\in (1-\varepsilon,1+\varepsilon)$ for some $\varepsilon\in (0,1)$. Then from the Uniform Convergence Theorem (UCT, Theorem~1.2.1 in \cite{bingham1989regular}) of slowly varying function, we have
\begin{equation}
    \frac{f(x+\theta)}{f(\theta)}= \frac{|\theta|^a L(|\theta+x|^{-1})}{|\theta+x|^a L(|\theta|^{-1})}\xrightarrow[M\to\infty]{} 1 \label{eq:unif_conv_foverf}
\end{equation}
uniformly in $x$ and $\theta$. Combining with the first clause of \eqref{eq:3bounds_g}, we have
$$P_1 \lesssim M^2\int_{|x|<\frac{2}{M}}\int_{|y|<\frac{2}{M}} f(x+\theta)f(y+\theta)\dd x\dd y = O(f^2(\theta)).$$

For $P_2$ (and similarly $P_3$), by the third clause of \eqref{eq:3bounds_g} and \eqref{eq:unif_conv_foverf}, we have
\begin{equation}
    P_2 \lesssim \log^2 M \int_{|x|<\frac{1}{M}}\int_{|y|>\frac{2}{M}}\frac{f(x+\theta)f(y+\theta)}{|x-y|^2}\dd x\dd y\lesssim f(\theta)M^{-1}\log^2 M \int_{|y|>\frac{2}{M}}\frac{f(y+\theta)}{(|y|-\frac{1}{M})^2}\dd y
\end{equation}
with
\begin{equation}
    \int_{|y|>\frac{2}{M}}\frac{f(y+\theta)}{(|y|-\frac{1}{M})^2}\dd y=\int_{-\pi}^{-\frac{2}{M}}\frac{f(y+\theta)}{(y+\frac{1}{M})^2}\dd y+\int_{\frac{2}{M}}^{\pi}\frac{f(y+\theta)}{(y-\frac{1}{M})^2}\dd y. \label{eq:P_2_part_bound}
\end{equation}
The function $f$ being even, the LHS of the above equality is even for $\theta$. Thus we can assume that $\theta>0$. Then as $M\to\infty$, the following holds uniformly for $y\in (0,\pi)$ and $\theta\in (\tau_M,\delta_M)$:
$$f(y+\theta)\le \sup_{t\ge \theta}f(t)\sim f(\theta).$$
Thus
$$\int_{\frac{2}{M}}^{\pi}\frac{f(y+\theta)}{(y-\frac{1}{M})^2}\dd y = O(f(\theta))\int_{\frac{1}{M}}^{\pi-\frac{1}{M}}\frac{1}{y^2}\dd y = O(f(\theta)M).$$
For the other integral on the RHS of \eqref{eq:P_2_part_bound}, we have
$$\begin{aligned}\int_{-\pi}^{-\frac{2}{M}}\frac{f(y+\theta)}{(y+\frac{1}{M})^2}\dd y &= \int_{-\pi+\theta}^{-\frac{2}{M}+\theta}\frac{f(y)}{(y-\theta+\frac{1}{M})^2}\dd y\\
    &= \int_{|y|>\frac{\theta}{2}, y<\theta-2/M}\frac{f(y)}{(y-\theta+\frac{1}{M})^2}\dd y + \int_{-\frac{\theta}{2}}^{\frac{\theta}{2}}\frac{f(y)}{(y-\theta+\frac{1}{M})^2}\dd y \\
    &\le  \sup_{|t|>\theta/2}f(t)\int_{-\pi+1/M}^{-1/M}\frac{1}{y^2}\dd y+O(\theta^{-2})\int_{-\frac{\theta}{2}}^{\frac{\theta}{2}} f(y)\dd y \\
    &= O(f(\theta/2)M)+O(\theta^{-2}\theta^{1-a}L(\theta^{-1})), \end{aligned}$$
and by the definition of slowly varying function,
$$f(\theta/2)= 2^a \theta^{-a}L(2\theta^{-1})\sim 2^a \theta^{-a}L(\theta^{-1}) = 2^a f(\theta)$$
uniformly for $\theta\in (\tau_M,\delta_M)$, thus
$$\int_{-\pi}^{-\frac{2}{M}}\frac{f(y+\theta)}{(y+\frac{1}{M})^2}\dd y=O(f(\theta)M)+O(\theta^{-1}f(\theta)).$$
Combining the above estimations, we have
$$P_2 \lesssim f(\theta)M^{-1}\log^2 M [f(\theta)M+\theta^{-1}f(\theta)] =(1+\frac{1}{M\theta})f^2(\theta)\log^2 M \sim f^2(\theta)\log^2 M.$$

For $P_4$, using the second clause of \eqref{eq:3bounds_g}, we have
$$\begin{aligned}P_4 & = \int_{|x|>\frac{1}{M}}\int_{|y|>\frac{1}{M}}f(x+\theta)f(y+\theta)|g(x,y)|^2\dd x\dd y \\
    & \lesssim \int_{|x|,|y|>\frac{1}{M}} f(x+\theta)f(y+\theta)\frac{1}{|xy|}\dd x \dd y \\
    & = \left(\int_{\frac{1}{M}}^\frac{\pi}{2} \frac{f(\theta+x)+f(\theta-x)}{x}\dd x\right)\left(\int_{\frac{1}{M}}^\pi \frac{f(\theta+y)+f(\theta-y)}{y}\dd y \right). \end{aligned}$$
Using similar methods as in the estimation of $P_2$, we have
$$P_4\lesssim f^2(\theta)\log^2 M.$$

Summing the above estimations for $P_1, P_2, P_3, P_4$, the result follows.

\section{The other proofs} \label{sec:other_proofs}
\subsection{Proof of Corollary~\ref{corol:conv_1/2power}}\label{subsec:proof_conv_1/2power}
We write
\begin{equation}
    \left\|\hRb_M^{1/2}\Rb_M^{-1/2}-\sqrt{\xi_N}\Ib\right\|\le \|\Rb_M^{1/4}\|\left\|\Rb_M^{-1/4}\hRb_M^{1/2}\Rb_M^{-1/4}-\sqrt{\xi_N}\Ib\right\|\|\Rb_M^{-1/4}\|, \label{eq:corol_1/2conv}
\end{equation}
where $\|\Rb_M^{-1/4}\|$ is bounded, and from Lemma~\ref{lem:norm_toeplitz}, $\|\Rb_M^{1/4}\|=O(M^{1/4+\varepsilon})$ with any $\varepsilon\in (0,1/8)$.

The spectral norm $\left\|\Rb_M^{-1/4}\hRb_M^{1/2}\Rb_M^{-1/4}-\sqrt{\xi_N}\Ib\right\|$ equals to
$$\max\left\{\left|\lmax(\Rb_M^{-1/4}\hRb_M^{1/2}\Rb_M^{-1/4})-\sqrt{\xi_N}\right|, \left|\lmin(\Rb_M^{-1/4}\hRb_M^{1/2}\Rb_M^{-1/4})-\sqrt{\xi_N}\right|\right\}.$$
The positive definite Hermitian matrix $\Rb_M^{-1/4}\hRb_M^{1/2}\Rb_M^{-1/4}$ has the same eigenvalues as $\hRb_M^{1/2}\Rb_M^{-1/2}$, so the latter matrix has $M$ positive eigenvalues. On the other hand, all the eigenvalues of $\hRb_M^{1/2}\Rb_M^{-1/2}$ are between its smallest and largest singular values, that is,
$$\sqrt{\lmin(\Rb_M^{-1/2}\hRb_M\Rb_M^{-1/2})}\le \lmin(\hRb_M^{1/2}\Rb_M^{-1/2})\le \lmax(\hRb_M^{1/2}\Rb_M^{-1/2})\le \sqrt{\lmax(\Rb_M^{-1/2}\hRb_M\Rb_M^{-1/2})}.$$
From Theorem~\ref{th:main_th}, if $\xi_N$ is bounded away from $0$, as $N,M\to\infty$ with $N/M\to c\in\positR$, we have almost surely
$$\left|\sqrt{\lmax(\Rb_M^{-1/2}\hRb_M\Rb_M^{-1/2})}-\sqrt{\xi_N}\right|=O(M^{-1/2+\varepsilon}).$$
The same result also holds for $\sqrt{\lmin(\Rb_M^{-1/2}\hRb_M\Rb_M^{-1/2})}$. Therefore, we have almost surely
$$\left\|\Rb_M^{-1/4}\hRb_M^{1/2}\Rb_M^{-1/4}-\sqrt{\xi_N}\Ib\right\|=O(M^{-1/2+\varepsilon}).$$
Taking the above estimations into \eqref{eq:corol_1/2conv}, the result follows.

\subsection{Proof of Proposition~\ref{prop:inconsis_Rb}} \label{subsec:proof_inconsis_Rb}
Define
$$r_k^b:=\left(1-\frac{|k|}{M}\right)r_k,$$
and
$$\hUpsilon^b_M(\theta):=\sum_{k=-M+1}^{M-1}\hat r_k^b e^{\im k\theta}, \quad \Upsilon^b_M(\theta):=\sum_{k=-M+1}^{M-1} r_k^b e^{\im k\theta}=\sum_{k=-M+1}^{M-1}\left(1-\frac{|k|}{M}\right) r_k e^{\im k\theta}.$$
Note that $\Upsilon^b_M$ is just the Ces\`aro mean of $\Upsilon_M(\theta):=\sum_{k=-M+1}^{M-1} r_k e^{\im k\theta}$, therefore
\begin{equation}
    \Upsilon^b_M(\theta)=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)F_M(\theta-x)\dd x,
\end{equation}
where $F_M(x)=\frac{\sin^2(Mx/2)}{M\sin^2(x/2)}$ is the Fej\'er kernel. Thus for any $\theta\in\R$, we have
\begin{equation}
    \essinf_{t}f(t)\le \Upsilon^b_M(\theta)\le \esssup_{t} f(t). \label{eq:range_upsilonb}
\end{equation}
By the assumptions on the spectral density $f$, $\Upsilon^b_M$ is positive and uniformly lower bounded from $0$.

Following the same idea as Section~\ref{subsec:preliminary_pr_main}, we only need to estimate
\begin{equation}
     \P\left(\sup_{\theta\in[0,2\pi]}\left|\frac{\hUpsilon^b_M(\theta)}{\Upsilon^b_M(\theta)}-1\right|>x\right) \label{eq:to_ctrl_prob_bias}
\end{equation}
for any $x>0$.

We use the same discretization strategy as Section~\ref{sec:proof_main_th}. For any fixed $\theta\in [0,2\pi]$, define
$$d_M(\theta)=\frac{1}{\sqrt{M}}(1,e^{\im \theta},\dots, e^{\im (M-1)\theta})^\tran,\quad \Qb_M^b(\theta):= \Rb_M^{1/2}d_M(\theta)d_M^*(\theta)\Rb_M^{1/2}$$
then by Lemma~3 of \cite{vinogradova2015estimation}, we have
$$\hUpsilon^b_M(\theta)=\frac{1}{N}d_M^*(\theta)\Rb_M^{1/2}\Zb\Cb_N\Zb^*\Rb_M^{1/2}d_M(\theta)=\frac{1}{N}\sum_{n=1}^N c_n\zb_n^*\Qb_M^b(\theta)\zb_n,$$
and by the unitary invariance of $\zb_n$, note also that $\Qb_M^b(\theta)$ is of rank one with a positive eigenvalue $d_M^*(\theta)\Rb_M d_M(\theta)=\Upsilon^b_M(\theta)$, we have
$$\frac{\hUpsilon^b_M(\theta)}{\Upsilon^b_M(\theta)}\eqinlaw \frac{1}{N}\sum_{n=1}^N c_n|Z_{1,n}|^2.$$
Then
$$\P\left(\left|\frac{\hUpsilon^b_M(\theta)}{\Upsilon^b_M(\theta)}-1\right|>x\right)=\P\left(\left|\frac{1}{N}\sum_{n=1}^N c_n(|Z_{1,n}|^2-1)\right|>x\right).$$
Using the same method as Section~\ref{subsec:large_dev_ind_theta}, we get the concentration inequality
$$\P\left(\left|\frac{1}{N}\sum_{n=1}^N c_n(|Z_{1,n}|^2-1)\right|>x\right)\le 2\exp\left(-\frac{KNx^2}{\log^2 M}\right)$$
for some constant $K>0$, for any $x>0$ and $M,N$ large enough, where the $\log^2M$ comes from the upper bound $\max_n\{c_n\}\le \kappa \log M$.

For the discretization step, we use the same method as in Section~\ref{subsec:discret_unbias}, and the proof of Lemma~4, Lemma~6 in \cite{vinogradova2015estimation}, along with the norm bound $\|\Rb_M\|\lesssim M^aL(M)$. Note also that $\Upsilon_M^b(\theta)$ are positive and uniformly lower bounded from $0$. We finally get
$$\P\left(\esssup_{\theta}\left|\frac{\hUpsilon^b_M(\theta)}{\Upsilon^b_M(\theta)}-1\right|>x\right)\le 2M^\beta\exp\left(-\frac{KNx^2}{\log^2 M}\right)$$
for some $\beta>0$, $K>0$ and any $x>0$, large enough $M,N$. This implies that
$$\|(\Rb_M^b)^{-1/2}\hRb^b_M(\Rb_M^b)^{-1/2}-\Ib\|\to 0$$
as $M,N\to\infty$ with $N\gg \log^3 M$.

Next we prove the inconsistency \eqref{eq:inconsistency}. We first prove that for two sequences of invertible matrices $\Rb_{1,M}, \Rb_{2,M}$, a necessary condition for the convergence
\begin{equation}
    \|\Rb_{1,M}^{-1/2}\Rb_{2,M}\Rb_{1,M}^{-1/2}-\Ib\|\xrightarrow[M\to\infty]{} 0\,, \label{eq:assumpt_absurd}
\end{equation}
is that
\begin{equation}
    \lim_{M\to\infty} \frac{\lmax(\Rb_{2,M})}{\lmax(\Rb_{1,M})}=1. \label{eq:max_lim_gamma}
\end{equation}
Take an arbitrary $\varepsilon>0$. Let $u$ be un eigenvector of $\Rb_{1,M}$ associated with $\lmax(\Rb_{1,M})$, then from \eqref{eq:assumpt_absurd}, for large enough $M$,
\begin{equation}
    1-\varepsilon<u^*(\Rb_{1,M})^{-1/2}\Rb_{2,M}(\Rb_{1,M})^{-1/2}u=\frac{u^*\Rb_{2,M} u}{\lmax(\Rb_{1,M})}\le \frac{\lmax(\Rb_{2,M})}{\lmax(\Rb_{1,M})}. \label{eq:1ep_le_lmax}
\end{equation}
Note that $(\Rb_{1,M}^{-1/2}\Rb_{2,M}\Rb_{1,M}^{-1/2})^{-1}$ has the same eigenvalues as $\Rb_{2,M}^{-1/2}\Rb_{1,M}\Rb_{2,M}^{-1/2}$. Recall that for a sequence of Hermitian matrices $\Ab_M$, the convergence $\|\Ab_M-\Ib\|\to 0$ is equivalent to the convergence of its eigenvalues, i.e. $\lmax(\Ab_M)\to 1$, $\lmin(\Ab_M)\to 1$. Therefor \eqref{eq:assumpt_absurd} also implies that
$$\|(\Rb_{2,M})^{-1/2}(\Rb_{1,M})(\Rb_{2,M})^{-1/2}-\Ib\|\xrightarrow[M\to\infty]{} 0\,.$$
Using the same arguments as \eqref{eq:1ep_le_lmax}, we get, for large enough $M$,
\begin{equation}
    1-\varepsilon \le \frac{\lmax(\Rb_{1,M})}{\lmax(\Rb_{2,M})}. \label{eq:lmin_lmax_1ep}
\end{equation}
Combining \eqref{eq:1ep_le_lmax} and \eqref{eq:lmin_lmax_1ep}, we have
\begin{equation}
    \lim_{M\to\infty}\frac{\lmax(\Rb_{2,M})}{\lmax(\Rb_{1,M})}= 1\,, \label{eq:lambda_ratio_1}
\end{equation}
and \eqref{eq:max_lim_gamma} follows.

However, we will prove that almost surely \eqref{eq:max_lim_gamma} cannot be satisfied by $\hRb_M^b$ and $\Rb_M$. Indeed, from \eqref{eq:ratio_consist_hRbRb} we conclude that almost surely
$$\frac{\lmax(\hRb_M^b)}{\lmax(\Rb_M^b)}\to 1.$$
Thus we only need to prove that
\begin{equation}
    \frac{\lmax(\Rb_M^b)}{\lmax(\Rb_M)}\not\to 1. \label{eq:lmax_lim_not1}
\end{equation}
Let $\K$ and $\K^b$ be two integral operators on $L^2(0,1)$ defined by
$$\K(\varphi)(x)=\int_0^1\frac{1}{|x-y|^{1-a}}\varphi(y)\dd y,\quad \K^b(\varphi)(x)=\int_0^1\frac{1-|x-y|}{|x-y|^{1-a}}\varphi(y)\dd y,\ \text{for } \varphi\in L^2(0,1).$$
We first assume that the slowly varying function $L$ in \ref{ass:spec_den_behav_0} equals to $1$. Then
$$r_k\sim \frac{K}{(1+|k|)^{1-a}}$$
as $k\to\infty$ with some constant $K>0$. Using Widom-Shampine's Lemma, which states that for any $M\times M$ matrix $\Ab=(a_{i,j})_{1\le i,j\le M}$, a non-zero number $\lambda$ is an eigenvalue of $\Ab$ if and only if $\lambda/M$ is an eigenvalue of the operator $\K_A$ defined on $L^2(0,1)$ by
$$\K_A(\varphi)(x)=\int_0^1 a_{[x],[y]}\varphi(y)\dd y,\quad \forall \varphi\in L^2(0,1),$$
one can prove that
\begin{equation}
    \frac{\lmax(\Rb_M)}{K M^a}\to \lambda_1(\K), \quad \frac{\lmax(\Rb^b_M)}{K M^a}\to \lambda_1(\K^b), \label{eq:cv_normalized_eigenvalues}
\end{equation}
Some more details can be found in the proof of \cite[Theorem~2.3]{merlevede2019unbounded}. Thus
\begin{equation}
    \frac{\lmax(\Rb^b_M)}{\lmax(\Rb_M)}\to \frac{\lambda_1(\K^b)}{\lambda_1(\K)}. \label{eq:ratio_lmax}
\end{equation}
If the slowly varying function $L$ in the assumption \ref{ass:spec_den_behav_0} is not constant, let
$$\tilde f(\theta)=\frac{1}{|\theta|^a}, \quad \theta \in [-\pi,\pi]$$
and $\tilde \Upsilon^b_M$, $\tilde \Rb_M^b$ be defined with $\tilde f$ in the same way as $\Upsilon^b_M$, $\Rb_M^b$ with $f$. Note that the F\'ejer kernel $F_M$ has the same upper bound as the Dirichlet kernel used in the proof of Theorem~2.3 in \cite{tian2019joint}, that is, for $\theta\in [-3\pi/2, 3\pi/2]$,
$$|F_M(\theta)|\lesssim \min\{M, |\theta|^{-1}\}.$$
Then using the same technique there, one can prove that
$$\sup_{\theta}\left\|\frac{\Upsilon^b_M(\theta)}{M^aL(M)}-\frac{\tilde\Upsilon^b_M(\theta)}{M^a}\right\|\to 0$$
as $M\to\infty$, which implies that
$$\left\|\frac{\Rb_M^b}{M^aL(M)}-\frac{\tilde\Rb_M^b}{M^a}\right\|\to 0.$$
Also note the result of Theorem~2.3 of \cite{tian2019joint}
$$\left\|\frac{\Rb_M}{M^aL(M)}-\frac{\tilde\Rb_M}{M^a}\right\|\to 0,$$
together with \eqref{eq:cv_normalized_eigenvalues}, we have
$$\frac{\lmax(\Rb_M)}{K M^aL(M)}\to \lambda_1(\K), \quad \frac{\lmax(\Rb^b_M)}{K M^aL(M)}\to \lambda_1(\K^b),$$
and thus \eqref{eq:ratio_lmax} still holds in this case.

We then prove that $\lambda_1(\K)>\lambda_1(\K^b)$. Indeed because the two integral kernels are positive, from the mini-max formula for the largest eigenvalue, their eigenfunctions associated with the largest eigenvalue are positive in $[0,1]$. Let $\varphi^b$ be the eigenfunction of $\K^b$ associated with $\lambda_1(\K^b)$, then
$$\lambda_1(\K^b)=\langle \varphi^b, \K^b\varphi^b\rangle = \langle \varphi^b, \K\varphi^b\rangle- \int_0^1\int_0^1|x-y|^a\varphi^b(x)\varphi^b(y)\dd x\dd y < \lambda_1(\K),$$
from which we conclude that
\begin{equation}
    \lim_{M\to\infty}\frac{\lmax(\Rb^b_M)}{\lmax(\Rb_M)} = \frac{\lambda_1(\K^b)}{\lambda_1(\K)}<1\,. \label{eq:lmax_less_1}
\end{equation}
This is the end of the proof of proposition since \eqref{eq:lmax_lim_not1} is proved.

\subsection{Proof of Proposition~\ref{prop:lsd_consistent}}\label{subsec:proof_lsd_consist}
Note that from the proof of Proposition~\ref{prop:inconsis_Rb}, almost surely, as $M,N\to\infty$ with $N\gg \log^3 M$,
$$\sup_{\theta}\left|\frac{\hUpsilon_M^b(\theta)}{\Upsilon_M^b(\theta)}-1\right|\to 0.$$
Then almost surely
$$\int_0^{2\pi}|\hUpsilon_M^b(\theta)-\Upsilon_M^b(\theta)|\dd\theta\le \sup_{\theta}\left|\frac{\hUpsilon_M^b(\theta)}{\Upsilon_M^b(\theta)}-1\right|\int_0^{2\pi} |\Upsilon_M^b(\theta)|\dd\theta \to 0,$$
where $\int_0^{2\pi} |\Upsilon_M^b(\theta)|\dd\theta$ is bounded because $\Upsilon_M^b(\theta)$ is the Ces\`aro mean of the Fourier series of $f$, and it is well known that $\Upsilon_M^b$ converges to $f$ in $L^1(0,2\pi)$. Then we deduce that almost surely
$$\int_0^{2\pi}|\hUpsilon_M^b(\theta)-f(\theta)|\dd\theta\to 0.$$
Also note that $\hUpsilon_M^b/f\ge 0$, then $\hUpsilon_M^b$ and $f$ satisfy the conditions of Lemma~\ref{lem:ratio_lsd_cv_toep}. Therefore the result of Proposition~\ref{prop:lsd_consistent} is a corollary of Lemma~\ref{lem:ratio_lsd_cv_toep}.

It remains to prove Lemma~\ref{lem:ratio_lsd_cv_toep}. If $f_M$ converges in $L^1(0,2\pi)$ to $f$, we denote
$$(f_M-f)_+ = \max\{f_M-f, 0\}, \quad (f_M-f)_- = \max\{ f-f_M, 0\},$$
and
$$\Ab_+ =\left(\frac{1}{2\pi}\int_0^{2\pi}(f_M-f)_+(\theta)e^{\im(i-j)\theta}\dd\theta\right),\quad \Ab_-=\left(\frac{1}{2\pi}\int_0^{2\pi}(f_M-f)_-(\theta)e^{\im(i-j)\theta}\dd\theta\right).$$
Then $\Ab_+,\Ab_-$ are two positive semi-definite Toeplitz matrices satisfying
$$\frac{1}{M}\tr\Ab_\pm=\frac{1}{2\pi}\int_0^{2\pi}(f_M-f)_\pm\dd\theta\to 0.$$
It is easy to prove that there exists a sequence of positive numbers $(\varepsilon_M)_{M\ge 1}$ converging to $0$, such that
$$\frac{\# \{k\tq \lambda_k(\Ab_\pm)>\varepsilon_M\}}{M}\le \varepsilon_M\to 0$$
where "$\# S$" denotes the cardinal of the set $S$. We have already known that $\mu^{\Rb_M}\cvweak \mu^{\Rb}$. We next prove successively that $\mu^{(\Rb_M+\Ab_+)}\cvweak \mu^{\Rb}$ and $\mu^{(\Rb_M+\Ab_+-\Ab_-)}\cvweak \mu^{\Rb}$, which is the first result of the lemma.

Let $\Ab_+=U\diag(\lambda_1,\dots,\lambda_M)U^*$ be a diagonalization of $\Ab_+$ with $\lambda_1,\dots,\lambda_M$ its eigenvalues. Let $\Ab_+^{(1)}=U\diag(\lambda_1\ind_{\lambda_1>\varepsilon_M},\dots,\lambda_M\ind_{\lambda_M>\varepsilon_M})U^*$, and $\Ab_+^{(2)}=U\diag(\lambda_1\ind_{\lambda_1\le\varepsilon_M},\dots,\lambda_M\ind_{\lambda_M\le \varepsilon_M})U^*$. Then the rank of $\Ab_+^{(1)}$ is at most $M\varepsilon_M$, and $\|\Ab_+^{(2)}\|\le \varepsilon_M$. Using Theorem~A.43 and A.45 in \cite{bai2010spectral} successively, we can prove that
$$\mu^{\Rb_M+\Ab_+}\cvweak \mu^{\Rb}.$$
Repeating the same procedure, we have also
$$\mu^{\Rb_M+\Ab_+-\Ab_-}\cvweak \mu^{\Rb}.$$
Thus the first part of the lemma is proved.

Next we prove the second part.
\begin{Rq}
Inspired by the logarithmic potential method used in the non-Hermitian random matrix theory, we would like to prove that for almost every $z\in\C$,
\begin{equation}
    \frac{1}{M}\log \left|\det(\Rb_M^{-1}\Rb^{(M)}_M-z\Ib)\right|\xrightarrow[M\to\infty]{} \log |1-z|. \label{eq:potential_principe}
\end{equation}
Note that the LHS of the above convergence can be written as
\begin{equation}
    \frac{1}{M}\log \left|\det(\Rb^{(M)}_M-z\Rb_M)\right|-\frac{1}{M}\log \left|\det(\Rb_M)\right|, \label{eq:potential_decomp}
\end{equation}
and by Theorem~2 in \cite{capizzano2002test} and the hypothesis that $f$ is positive and bounded away from $0$,  we know that
\begin{equation}
    \lim_{M\to\infty}\frac{1}{M}\log \left|\det(\Rb_M)\right|=\frac{1}{2\pi}\int_0^{2\pi}\log|f(\theta)|\dd\theta. \label{eq:log_det_toep_conv}
\end{equation}
If we could do the same thing on the first term in \eqref{eq:potential_decomp}, we would have
\begin{equation}
    \lim_{M\to\infty}\frac{1}{M}\log \left|\det(\Rb^{(M)}_M-z\Rb_M)\right|=\frac{1}{2\pi}\int_0^{2\pi}\log|f(\theta)-zf(\theta)|\dd\theta. \label{eq:limit_potential_mov}
\end{equation}
Take the difference of the last two equalities, \eqref{eq:potential_principe} would be proved.

However we meet the following difficulties. Firstly, when establishing \eqref{eq:limit_potential_mov}, if $z$ is not a real number, the Toeplitz matrix $\Rb^{(M)}_M-z\Rb_M$ is not Hermitian and do not have a fixed spectral density. Thus neither Theorem~2 of \cite{capizzano2002test}, nor the first part of this lemma can be directly used. Secondly, the logarithmic potential continuity theorems presented in the literature often suppose that the sequence of measures $\mu^{\Rb_M^{-1}\Rb_M^{(M)}}$ are all supported in one compact subset of $\C$, in order to ensure the existence of logarithmic potentials. (See e.g. \cite[Theorem~2.8.3 with Erratum]{tao2012topics}, or the combination of Lower Envelop Theorem \cite[Theorem 6.9, p.73]{saff2013logarithmic} and Unicity Theorem \cite[Corollary 2.2, p.98]{saff2013logarithmic}). This obliges us to assume that $|f_M/f|$ are uniformly bounded, which is often complex to check. Even these two difficulties may be easy to solve by relaxing some conditions of these existing results, we choose to develop a new tool adapted to our case.
\end{Rq}

\begin{lemma}\label{lem:my_potential_th}Suppose that the probability measures $\mu_n,\mu, n=1,2,\dots$ are supported on $[a,+\infty)$. If for any $x<a$,
\begin{equation}
    \lim_{n\to\infty}\int \log |x-t|\dd\mu_n(t)=\int \log |x-t|\dd\mu(t)<\infty, \label{eq:conv_real_potential}
\end{equation}
then $\mu_n$ converges weakly to $\mu$.
\end{lemma}
\begin{proof} From every subsequence of $(\mu_n)$ we can extract a subsequence converging vaguely to a positive measure $\nu$ with total mass less than or equal to $1$. Take an arbitrary $x_0<a$. Then for any $z\in \C\backslash (x_0,+\infty)$, because the function $t\mapsto (z-t)^{-1}$ is continuous on the support of $\mu_n$ and $\nu$, and tends to $0$ as $t\to\infty$, we have
$$s_n(z):=\int\frac{1}{z-t}\dd\mu_n(t) \xrightarrow[n\to\infty]{}\int\frac{1}{z-t}\dd\nu(t)=:s(z).$$
By dominated convergence theorem,
$$\int_{x_0}^z s_n(w)\dd w\xrightarrow[n\to\infty]{} \int_{x_0}^z s(w)\dd w,$$
where the integral is taken along the segment from $x_0$ to $z$. Changing the order of integrals, we get
$$\int(\log(z-t)-\log(x_0-t))\dd\mu_n(t)\xrightarrow[n\to\infty]{}\int_{x_0}^z s(w)\dd w,$$
where $\log z=\log |z|+i\arg z$ with $\arg z\in [0,2\pi)$. When $z=x\in (-\infty,x_0]$, the above convergence and the condition \eqref{eq:conv_real_potential} imply that
$$\int_{x_0}^x s(w)\dd w=\int\log(x-t)\dd\mu(t)-\int\log(x_0-t)\dd\mu(t).$$
This equality can be extended to $z\in \C\backslash (x_0,+\infty)$ by analyticity, then
$$\int_{x_0}^z s(w)\dd w= \int\log(z-t)\dd\mu(t)-\int\log(x_0-t)\dd\mu(t).$$
Differentiating the two sides, we get
$$s(z)= \int\frac{1}{z-t}\dd\mu(t),$$
and thus $\mu=\nu$. Because the vague limit $\mu$ is a probability measure, we actually have the weak convergence $\mu_n\cvweak \mu$.
\end{proof}

Using this result, and assume that $f_M/f > a$ for some $a\in\R$, we only need to prove \eqref{eq:limit_potential_mov} for $z\in (-\infty,a)$. When $z=x<a$, we have $f_M-xf>0$, thus the matrix $\Rb^{(M)}_M-x\Rb_M$ is positive definite. For any $A>1$ and $t>0$, let
$$\ell_A(t):=\min(\log(t), A), h_A(t):=\max(\log(t),A)-A.$$
Then it is easily seen that $\ell_A(t)+h_A(t)=\log(t)$, and $\ell_A(t)\to\log (t)$, $h_A(t)\to 0$ for every $t>0$ when $A\to\infty$. Note that $f_M-xf\to (1-x)f$ in $L^1$, then from the first part of the lemma, we have
\begin{equation}
    \frac{1}{M}\sum_{k=1}^M\ell_A(\lambda_k(\Rb^{(M)}_M-x\Rb_M))\xrightarrow[M\to\infty]{} \frac{1}{2\pi}\int_0^{2\pi} \ell((1-x)f(\theta))\dd \theta. \label{eq:sum1_josf}
\end{equation}
Note that if $A$ is large, $h_A(t)<\sqrt{t}$, thus
\begin{equation}
    \begin{aligned}\frac{1}{M}\sum_{k=1}^M h_A(\lambda_k(\Rb^{(M)}_M-x\Rb_M)) & \le \frac{1}{M}\sum_{k=1}^M (\lambda_k(\Rb^{(M)}_M-x\Rb_M))^{\frac{1}{2}}\ind_{\lambda_k(\Rb^{(M)}_M-x\Rb_M)>A} \\
& \le \frac{1}{M\sqrt{A}}\tr (\Rb^{(M)}_M-x\Rb_M) \\
& = \frac{1}{2\pi\sqrt{A}}\int_0^{2\pi} (f_M-xf)(\theta)\dd\theta \end{aligned} \label{eq:sum2_jost}
\end{equation}
Summing \eqref{eq:sum1_josf} and \eqref{eq:sum2_jost}, and let $A\to+\infty$, we get
$$\lim_{M\to\infty}\frac{1}{M}\log \left|\det(\Rb^{(M)}_M-x\Rb_M)\right|=\frac{1}{2\pi}\int_0^{2\pi}\log|f(\theta)-xf(\theta)|\dd\theta.$$
Take the difference with \eqref{eq:log_det_toep_conv}, we will get
$$\frac{1}{M}\log \left|\det(\Rb_M^{-1}\Rb^{(M)}_M-x\Ib)\right|\xrightarrow[M\to\infty]{} \log |1-x|$$
for $x<a$. From Lemma~\ref{lem:my_potential_th}, we have $\mu^{\Rb_M^{-1}\Rb^{(M)}}\cvweak \delta_1$. The proof of Lemma~\ref{lem:ratio_lsd_cv_toep} is complete.

config: arxiv


TITLE: The forgetful functor $U:T-Alg\rightarrow C$ preserves limits
QUESTION [1 upvotes]: Suppose $F$ is an endofunctor of a complete category $C$.

Let $F-Alg$ be the category that has objects the pairs $(X, \ a:FX\rightarrow X)$ where $X$ is an object in $C$, and that has morphisms $f:(X,a)\rightarrow (Y,b)$ where $f$ is a morphisms in $C$ with the property $fa=b\circ Ff$.

Let $M:I\rightarrow F-Alg$ be a diagram with limit $((\lim M,m),\mu_i)$. Show that the forgetful functor $U:F-Alg\rightarrow C$ preserves this limit.
My attempt.
I need to show that the limit of $UM$ in $C$ is $(\lim M,\mu_i)$. So suppose $X$ is an object together with maps $\nu_i:X\rightarrow object(Mi)$ satisfying $M\alpha\circ \nu_i=\nu_j$ for each $\alpha:i\rightarrow j$ in $I$. My job is now to find a map  $s:X\rightarrow \lim M$ in $C$ with the property $\mu_i\circ s= \nu_i$.
Of course, by the universal property of limits in $F-Alg$, whenever we have an object $(X,a)$ together with morphisms $\nu_i:(X,a)\rightarrow Mi$ satisfying $M\alpha\circ \nu_i=\nu_j$, there is a unique $k:(X,a)\rightarrow (\lim M,m)$ with the property $\mu_i\circ k= \nu_i$. I would like to choose $s=k$. The issue I am having is that we don’t know if the object $X$ in $C$ has a corresponding object $(X,a)$ in $F-Alg$, because it is not guaranteed that a map $a:FX\rightarrow X$ exists in $C$. I tried to raise the map $k$ to $Fk$ and use it is a morphism in $F-Alg$, but with no success...
How do I ensure $X$ comes with a morphism $a:FX\rightarrow X$?
I added a sketch for clarity
￼
I would prefer a hint over an answer.

REPLY [3 votes]: Hint: For simplicity, I'll consider the case of a product of two objects.  Working backwards from the conclusion, if you have two $T$-algebras $(X, f)$ and $(Y, g)$, then the product of these two $T$-algebras will end up being isomorphic to $(X \times Y, h)$ for some $h$.  So, a good place to start would probably be to look for an explicit way to construct a morphism $h : T(X\times Y) \to X\times Y$.  Then, if you can show that this $T$-algebra is a product of $(X, f)$ and $(Y, g)$, you will be done for this type of limit by the uniqueness of limits.  (And then, hopefully, your argument will be straightforward to generalize to more general limits.)


TITLE: Drawing spaces in $\mathbb{R}^3$
QUESTION [0 upvotes]: I want to draw the following spaces:

$D_1=\{(x,y,z)\mid 0\leq x\leq 3, -\sqrt{9-x^2}\leq y\leq \sqrt{9-x^2}, 0\leq z\leq 2\}$.
We can write this set also in the form $\{(x,y,z)\mid 0\leq x\leq 3, x^2+y^2=9, 0\leq z\leq 2\}$.
By $x^2+y^2=9$ we get a circle with center the origin and radius $3$. Since $0\leq x\leq 3$ we consider only the halb of the circle.
Since $0\leq z\leq 2$ we entend the half of the circle upwards with height $z=2$.
So, we get the following:

Is this correct?

$D_2=\{(x,y,z)\mid 0\leq x\leq 1, \ 1-x\leq y\leq 1, \ x\leq z\leq 1\}$
By the inequalities $0\leq x\leq 1, \ 1-x\leq y\leq 1$ we get a triangle on the $xy$-plane with vertices $(0,1)$, $(1,0)$ and $(1,1)$.
So, we have that triangle at the plane $z=0$. Do we shift that triangle to the plane $z=1$ because of the inequality $x\leq z\leq 1$, and so we get all the points between the triangle at $z=0$ till the triangle at $z=1$ ?
$D_3$ is defined by the paraboloid with equation $z=2x^2+y^2$ and the cylinder with the equation $z=8-y^2$.
The paraboloid and the cylinder intersect at $2x^2+y^2=8-y^2 \Rightarrow x^2=4-y^2 \Rightarrow x=\pm \sqrt{4-y^2}$, therefore we get $-\sqrt{4-y^2}\leq x\leq \sqrt{4-y^2}$.
It must hold that $4-y^2\geq 0 \Rightarrow y^2\leq 4 \Rightarrow -2\leq y \leq 2$.
We have that $2x^2+y^2\leq 2\sqrt{4-y^2}^2+y^2=8-y^2 \Rightarrow 2x^2+y^2\leq z\leq 8-y^2$.
Therefore the space $D_3$ is the set $\{(x,y,z)\mid -\sqrt{4-y^2}\leq x\leq \sqrt{4-y^2}, \ -2\leq y \leq 2, \ 2x^2+y^2\leq z\leq 8-y^2\}$, or not?
$D_4$ is defined by the surfaces with equations $z=0$, $x^2+2y^2=2$ and $x+y+2z=2$.
We draw the ellipse $x^2+2y^2=2$ on the $xy$-plane. We extend that ellipse till the plane $x+y+2x=2$.
Is this correct?
 -

REPLY [1 votes]: Correct
Notice that at $z=0$ your only allowed $x=0$ since $x\le z$ and $0\le x$. Then $y=1$. So you have a pyramid.

For this, notice that you already calculated $x^2=4-y^2$. That is an equation of a circle, at origin, of radius $2$. Note however that you need to consider $z$ as well. So you need to draw a parabola $z=8-y^2$ on the side of a cylinder
OK


TITLE: Online math tool for equation visualisation
QUESTION [2 upvotes]: I am looking for a math editor that is able to present equations and relations between them. I am not searching for a traditional Latex / MathML editor (I know about various online services that allow typing in equations in various formats then outputting a pretty picture). I know about online sage notebook, about mathjax and that like. I also know about Lurch and Wolfram Alpha (thanks to tp1 below).
What I'm looking for is something like pearltrees in interface with pearls being equations and edges indicating the source of the concepts (other equations or axioms). The equations should be editable and each may have an additional page that explains the equation, maybe from external source like Wikipedia.
Here is a sample tree with a direct proof directly from Wikipedia.

Consider two even integers x and y. Since they are even, they can be written as x = 2a and y = 2b respectively for integers a and b. Then the sum x + y = 2a + 2b = 2(a + b). From this it is clear x + y has 2 as a factor and therefore is even, so the sum of any two even integers is even.

(can't embed images yet)

REPLY [1 votes]: Trees for some equational proofs
Mathematica (which is Online) recently added some functionality to find proofs for some equational logic theorems and present them in various ways in both text and a tree.
There are a lot of examples/pictures at the reference page for FindEquationalProof. One particularly striking picture is a tree for a proof that one axiom implies a collection of 3 for NAND. If $(ab)$ denotes $a\,\operatorname{NAND}\,b$, then the three axioms

$((aa)(ab))=a$
$(a(ab))=(a(bb))$
$(a(a(bc)))=(b(b(ac)))$

all follow from $(((ab)c)(a((ac)a)))=c$. Mathematica proves this in 250 steps with a tree like:

Trees for many theorems
This doesn't cover equational proofs yet because it's only been done for the first 1000 proofs, but Antony is working on automatically converting metamath proofs to trees at this github repository. For example, this page shows the following tree for the theorem "ccase", which roughly says that if we have $P\land R\to T$ and $Q\land R\to T$ and $P\land S\to T$ and $Q\land S\to T$, then we can derive $(P\lor Q)\land (R\lor S)\to T$:


TITLE: CHSH violation and entanglement of quantum states
QUESTION [12 upvotes]: How is the violation of the usual CHSH inequality by a quantum state related to the entanglement of that quantum state?
Say we know that exist Hermitian and unitary operators $A_{0}$, $A_{1}$, $B_{0}$ and $B_{1}$ such that
$$\mathrm{tr} ( \rho ( A_{0}\otimes B_{0} + A_{0} \otimes B_{1} + A_{1}\otimes B_{0} - A_{1} \otimes B_{1} )) = 2+ c > 2,$$
then we know that the state $\rho$ must be entangled.  But what else do we know?  If we know the form of the operators $A_{j}$ and $B_{j}$, then there is certainly more to be said (see e.g. http://prl.aps.org/abstract/PRL/v87/i23/e230402 ).  However, what if I do not want to assume anything about the measurements performed?
Can the value of $c$ be used to give a rigourous lower bound on any of the familar entanglement measures, such as log-negativity or relative entropy of entanglement?
Clearly, one could argue in a slightly circular fashion and define an entanglement measure as the maximal possible CHSH violation over all possible measurements.  But is there anything else one can say?

REPLY [5 votes]: In a paper C.-E. Bardyn et al., PRA 80(6): 062327 (2009), arXiv:0907.2170, they discuss constrains on the state, given how much the CHSH equality is violated ($S=2+c$), but without putting any assumptions on the operator used.
In general people consider schemes, when operators (for a Bell-type measurement) are random or one or more parties cannot be trusted. One of the key phrases is device-independent and maybe also loophole-free (as even a slight misalignment of operators may change the results dramatically).


TITLE: About the use of Set algebra
QUESTION [1 upvotes]: I'm stocked with the following exercise:  Show that $A\cap(B\setminus C) =(A\cap B)\setminus (A\cap C) $.
Proof (Attempt): $A\cap (B\setminus C)= A\cap(B\cap C^c) = (A\cap B)\cap C^c $ since intersection is associative. But then... Iv'e got no idea on a following step, or even if this is the right way to get the proof. So, any help would be grateful.

REPLY [1 votes]: Another possible solution (not the shortest, probably, but it does the job): now start working with the right-hand side, to have "meet in the middle":
$$(A\cap B)\setminus(A\cap C)=(A\cap B)\cap(A\cap C)^c=(A\cap B)\cap(A^c\cup C^c)=(A\cap B\cap A^c)\cup(A\cap B\cap C^c)=\varnothing\cup(A\cap B\cap C^c),$$
as desired.


TITLE: How does $\frac{1}{i} \cdot \frac{\partial{u}}{\partial{y}}$ convert into $- i \frac{\partial{u}}{\partial{y}}$?
QUESTION [1 upvotes]: A proof I am studying showed the following step:
$\dfrac{1}{i} \cdot \dfrac{\partial{u}}{\partial{y}} + \dfrac{\partial{v}}{\partial{y}} = \dfrac{\partial{v}}{\partial{y}} - i \dfrac{\partial{u}}{\partial{y}}$
I do not understand how $\dfrac{1}{i} \cdot \dfrac{\partial{u}}{\partial{y}}$ converts into $- i \dfrac{\partial{u}}{\partial{y}}$.
I would greatly appreciate it if someone can please take the time to show me the reasoning behind this.

REPLY [4 votes]: This is because
$$\frac{1}{i}=-i$$
Proof:
$$\frac{1}{i}=\frac{1}{i}\times\frac{-i}{-i}=\frac{1\times (-i)}{i\times (-i)}=\frac{-i}{-i^2}=\frac{-i}{1}=-i$$


TITLE: Why is the downward transition of electrons in a three level pumping scheme of laser non radiative?
QUESTION [3 upvotes]: Laser produces coherant monochromatic waves by stimulated emission of radiation. In the three level pumping scheme of laser the downward transition of electrons from E3 to E2 is spontaneous and from E2 to E1 is stimulated emission. However electrons after being pumped to E3 undergo downward transition to measurable level E2 through nonradiative transitions. How can these transitions not radiate photons? How does this not violate the law of conservation of energy? Moreover the atoms are trapped at the level as spontaneous transition from level E 2 to E1 is forbidden. How is it forbidden?

REPLY [2 votes]: Although Stefan Bischof's answer is correct, I would like to add a few things.
Your main question is, how can an excited electron relax in a non-radiative way, without emitting photons.
The answer to the question is phonons. In non-radiative relaxation, the energy is dissipated in the form of phonons, that is, lattice vibrations of a solid (or liquid).
Phonons are collective excitations in a periodic, elastic arrengament of atoms or molecules in a condensed matter. A phonon is a QM description of an elementary vibrational motion in which a lattice of atoms or molecules uniformly oscillates at a single frequency.
So the answer to your question, is that the electrons relax by releasing phonons, that is, vibrational motions in the lattice, and this is a form of energy and heat dissipation.


TITLE: Do $|\mathbb{Z}^{+}| < |\mathcal{P}(\mathbb{Z}^{+})|$ and $|\mathcal{P}(\mathbb{Z}^{+})| \leq |\mathbb{R}|$ imply $|\mathbb{Z}^{+}| < |\mathbb{R}|$?
QUESTION [0 upvotes]: I am reading "A First Course in Analysis vol.3" written by Kazuo Matsuzaka.
First, he proves $\lvert X \rvert < \lvert \mathcal{P}(X) \rvert$ for any set $X$.
Next, he proves $\lvert \mathcal{P}(\mathbb{Z}^{+}) \rvert \leq \lvert \mathbb{R} \rvert$.
Then, he concludes that $\lvert \mathbb{Z}^{+} \rvert < \lvert \mathbb{R} \rvert$ because $\lvert \mathbb{Z}^{+} \rvert < \lvert \mathcal{P}(\mathbb{Z}^{+}) \rvert$ and $\lvert \mathcal{P}(\mathbb{Z}^{+}) \rvert \leq \lvert \mathbb{R} \rvert$.
But I cannot understand why $\lvert \mathbb{Z}^{+} \rvert < \lvert \mathbb{R} \rvert$ holds.

REPLY [2 votes]: Generalising slightly: for arbitrary sets $X$, $Y$ and $Z$, to say that $|X| < |Y| \le |Z|$ means that there are injections $f : X \to Y$ and $g : Y \to Z$ but there is no injection $Y \to X$. Let's prove that this implies that $|X| < |Z|$.
To deduce that $|X| \le |Z|$, note that $g \circ f : X \to Z$ is injective.
To finally deduce that $|X| < |Z|$, suppose there were an injection $h : Z \to X$. Then the composite $h \circ g : Y \to X$ would be injective, contradicting the assumption that $|X| < |Y|$. So indeed we have $|X|<|Z|$.
So here, let $X=\mathbb{Z}^+$, $Y=\mathcal{P}(\mathbb{Z}^+)$ and $Z=\mathbb{R}$ and apply the above result.

REPLY [1 votes]: The relations $<$ and $\le$ are both transitive. Thus, if we have, for cardinals $A,B,C$, that $A<B<C$, then we can conclude $A<C$. Similarly, if we have $A\le B\le C$, we can conclude $A\le C$. This is a combination of the two, namely: $A<B\le C$.
Since $A<B\implies A\le B$, then we can plug that in, and obtain $A\le C$. Then, if we can rule out $A=C$, we will have $A<C$. However, $A=C$ is impossible, for we know $B\le C$, and substituting $A$ for $C$ contradicts what we know about $A$ and $B$.
In this particular case, we have $A=|\Bbb Z^+|$, $B=|\mathcal P(\Bbb Z^+)|$, and $C=|\Bbb R|$ Does this answer your question?


TITLE: From elementary equivalence to isomorphism
QUESTION [5 upvotes]: A few years ago, when I took the basic course in Logic, I was very surprised to discover that given a signature $\sigma$ and two structures $M$ and $N$ of $\sigma$ which are elementarily equivalent (that is, they satisfy the same first-order $\sigma$-sentences), they are not necessarily isomorphic, even if they are of the same cardinality.
My question is, are there some "reasonable" conditions that one may pose on either the signature $\sigma$ or the structures $M$ and $N$ that guarantee the implication: "If $M$ and $N$ are elementarily equivalent, then $M \cong N$"?

REPLY [2 votes]: First, you should note that the signature is determined by the structures and any two structures that are elementarily equivalent or isomorphic have to have the same signature.
For your question, I don't think much can be done in the realm of first order logic.  Asaf's answer is correct that you need $Th(M)$ to be categorical for $M \equiv N$ to imply $M \cong N$.  Fixing the language is even worse as, for every signature, there are non-categorical theories in it.
Things get a little more interesting when you move to infinitary languages.  $L_{\omega_1, \omega}$ is the logic formed by allowing countably many conjunctions and disjunctions (as opposed to finitely many in first order), while still requiring that there are only finitely many free variables.  In this logic, every countable structure has a Scott sentence which characterizes its countable isomorphism type.  That is: if $M$ is a countable structure, then there is a sentence $\phi_M \in L_{\omega_1, \omega}$ (of the language of $M$) such that if $N$ is a countable structure so $N \vDash \phi_M$, then $M \cong N$.
This doesn't generalize as nicely to larger infinitary languages as one might like, but I believe that if $cf(\lambda)=\omega$ and $M$ is a $\lambda$ sized model, then there is a $L_{\lambda^+, \omega}$ sentence with the same property.


TITLE: Evaluate $\sum\limits_{n=1}^{\infty} \frac{1}{2^n}\tan\frac{x}{2^n}$
QUESTION [2 upvotes]: Evaluate $\sum\limits_{n=1}^{\infty} \frac{1}{2^n}\tan\frac{x}{2^n}$.

My attempt:
Let $S$ = $\sum\limits_{n=1}^{\infty} \frac{1}{2^n}$ $tan{\frac{x}{2^n}}$
$\int S dx$ = $\sum\limits_{n=1}^{\infty}$ $\int \frac{1}{2^n}$ $tan{\frac{x}{2^n}} dx$
= $\sum\limits_{n=1}^{\infty}$ $\ln \sec {\frac{x}{2^n}}$
= $\ln(\prod_{n=1}^{\infty}\sec {\frac{x}{2^n}})$
not sure how to proceed with this,
Due to the limitations of my school syllabus, I would greatly appreciate a solution involving complex numbers.
Thanks

REPLY [3 votes]: Hint: $\sec x=\dfrac1{\cos x}$ , and $\sin2t=2\sin t\cos t$. Replace the upper limit of your product with N, then multiply the entire expression with $\csc\bigg(\dfrac x{2^N}\bigg)=\dfrac1{\sin\bigg(\dfrac x{2^N}\bigg)}$, and see what happens $\ldots$


TITLE: Are there any random variables so that E[X] and E[Y] exist but E[XY] doesn't?
QUESTION [2 upvotes]: Are there any random variables so that E[X] and E[Y] exist but E[XY] doesn't?

REPLY [0 votes]: Having recently discussed here the difference between "$X$ has expectation" (in the wide sense) and "$X$ is integrable", let us give an example where $X$ and $Y$ are integrable (that is, have finite expectation) but $XY$ does not admit an expectation (that is, ${\rm E}(XY)^+ = {\rm E}(XY)^- = \infty$).
Let $Z$ be any nonnegative random variable satisfying ${\rm E}(Z)<\infty$ and ${\rm E}(Z^2) = \infty$ (for example, $Z=1/\sqrt{U}$, where $U \sim {\rm uniform}(0,1)$; cf. user8268's comment above). Let $R$ be independent of $Z$ with ${\rm P}(R=1) = {\rm P}(R=-1) = 1/2$. Define the random variables $X$ and $Y$ by $X=ZR$ and $Y=Z$. Then, $X$ and $Y$ have finite expectation. Next note that $XY = Z^2$ if $R=1$ and $XY = -Z^2$ if $R=-1$. Hence,
$$
(XY)^ +  := \max \{ XY,0\}  = Z^2 {\mathbf 1}(R = 1)
$$
and
$$
(XY)^ -   :=  -\min \{ XY,0\} = Z^2 {\mathbf 1}(R = -1),
$$
where ${\mathbf 1}$ denotes the indicator function. Since $Z$ and $R$ are independent (and, by assumption, ${\rm E}(Z^2) = \infty$), we get ${\rm E}(XY)^+ = {\rm E}(XY)^- = \infty$, as desired.
In the example (mentioned above) where $Z=1/\sqrt{U}$, $U \sim {\rm uniform}(0,1)$, one can find that $XY$ has density function $f_{XY}$ given by $f_{XY} = 1/(2x^2)$, $|x|>1$. Thus, $\int_0^\infty  {xf_{XY} (x)\,{\rm d}x}  = \int_1^\infty  {(2x)^{ - 1} \,{\rm d}x}  = \infty $ and $\int_{ - \infty }^0 {|x|f_{XY} (x)\,{\rm d}x}  = \int_{ - \infty }^{ - 1} {(2|x|)^{ - 1} \,{\rm d}x}  = \infty $, corresponding to ${\rm E}(XY)^+ = \infty$ and ${\rm E}(XY)^- = \infty$, respectively.