TITLE: Find the moment generating function of $r_n Y_n$
QUESTION [0 upvotes]: Let $(Y_n)_{n\geq 1}$ be a sequence of geometrical random variables
  with parameter $r_n$ such that $r_n\to 0$ as $n\to \infty$.
Find the moment generating function of $r_n Y_n$.
  You can assume that $r_n$ is a function on $\mathbb{R}$ in $C^1 ([0,g])$ for $g\geq 1$

This is what I did
\begin{equation}
\begin{split}
    M_{r_nY_n}(t)&=\sum_{k=0}^{\infty} e^{tk}P(r_nY_n=k) \\
    &=\sum_{k=0}^{\infty} e^{tk}P(Y_n=\frac{k}{r_n})=\sum_{k=0}^{\infty} e^{tk}r_n(1-r_n)^{\frac{k}{r_n}-1}\\
    &=\frac{r_n}{1-r_n} \sum_{k=0}^{\infty} e^{tk}(1-r_n)^{\frac{k}{r_n}}= \frac{r_n(1-r_n)^{\frac{1}{r_n}-1}}{1-r_n} \sum_{k=0}^{\infty}e^{tk}(1-r_n)^k \\
    &=(1-r_n)^{\frac{1}{r_n}-2}r_n \sum_{k=0}^{\infty}[e^{t}(1-r_n)]^k\\
    &= \frac{(1-r_n)^{\frac{1}{r_n}-2}r_ne^t}{1-(1-r_n)e^t}\\
 \notag
\end{split}
\end{equation}
I was wondering if this was the right way to do this?

REPLY [1 votes]: No. $Y_n$ takes integer values, whereas $r_nY_n$ doesn’t generally take integer values, so it makes no sense to sum over probabilities that it’s an integer. You want
$$
M_{r_nY_n}(t)=\sum_{k=0}^\infty\mathrm e^{tr_nk}\mathsf P(Y_n=k)\;.
$$
(Note also that there’s an excess factor $\mathrm e^t$ in your last line.)


text:
\begin{document}
\begin{abstract}
\noindent
In Einstein's theory of relativity, the interaction of two collinearly polarized plane gravitational waves can be described by a Goursat problem for the Euler--Darboux equation in a triangular domain. In this paper, using a representation of the solution in terms of Abel integrals, we give a full asymptotic expansion of the solution near the diagonal of the triangle. The expansion is related to the formation of a curvature singularity of the spacetime. In particular, our framework allows for boundary data with derivatives which are singular at the corners. This level of generality is crucial for the application to gravitational waves.
\end{abstract}

\maketitle

\noindent
{\small{\sc AMS Subject Classification (2010)}: 35Q75, 41A60, 83C35.}

\noindent
{\small{\sc Keywords}: Euler--Darboux equation, gravitational waves, Einstein's theory of relativity, collinear polarization.}

\setcounter{tocdepth}{1}

\section{Introduction}\nequation
The collision of two collinearly polarized plane gravitational waves in Einstein's theory of relativity can be described mathematically \cite{G1991} by a Goursat problem for the Euler--Darboux equation \cite{M1973}
\begin{align}\label{linearernst}
V_{xy} - \frac{V_x + V_y}{2(1-x-y)} = 0, \qquad (x,y) \in D,
\end{align}
in the triangular region $D$ defined by (see Figure \ref{D.pdf})
\begin{align}\label{Ddef}
D = \{(x,y) \in \R^2\, | \, x \geq 0, \; y \geq 0, \; x+y < 1\}.
\end{align}

Besides its importance for collinearly polarized plane waves, equation \eqref{linearernst} can also be viewed as the linear limit of the nonlinear hyperbolic Ernst equation \cite{CF1984}. The hyperbolic Ernst equation is a reduction of the vacuum Einstein field equations which is similar to the elliptic version of the same equation \cite{E1968}. It describes the collision of two (not necessarily collinearly polarized) plane gravitational waves.  While many exact solutions of the hyperbolic Ernst equation have been found (see \cite{ET1989,EGA1988,FI1987b,NH1977}), it seems that the problem of determining the solution from given data has only been treated in the series of papers \cite{HE1989a, HE1989b, HE1990, HE1991} and in \cite{FST1999, AG2004,PM2017}. In \cite{HE1989a, HE1989b, HE1990, HE1991}, the authors relate the Goursat problem to the solution of a homogeneous Hilbert problem and in \cite{FST1999, AG2004, PM2017} the problem is studied by means of inverse scattering. In the case of collinear polarization, both \cite{FST1999} and \cite{HE1989a} provide representations for the solution in terms of Abel integrals.

In the same way that the vacuum Einstein equations in the case of colliding plane gravitational waves can be reduced to the hyperbolic Ernst equation, the Einstein--Maxwell equations in vacuum can be reduced to an integrable system of two coupled nonlinear PDEs in the case of colliding electromagnetic plane waves \cite{E1968b, CX1985}. In the linear limit, both of these coupled equations reduce to the Euler--Darboux equation \eqref{linearernst}.

Of particular interest is the behavior of the solution of \eqref{linearernst} near the triangle's diagonal edge $x + y = 1$. Indeed, this behavior is related to the formation of a curvature singularity of the spacetime by the mutual focusing of the colliding waves. In \cite{S1972}, it was observed that the solution must behave like $\ln(1-x-y)$ as $x+y\to 1$. In this paper, using an Abel integral representation of the solution, we derive an  asymptotic expansion to all orders as $x+y\to 1$.

\begin{figure}
	\bigskip
	\begin{center}
		\begin{overpic}[width=.4\textwidth]{D.pdf}
			\put(29,33){ $D$}
			\put(103,9.7){\small $x$}
			\put(9.5,101){\small $y$}
			\put(82.5,3.5){\small $1$}
			\put(5,82){\small $1$}
			\put(27,3.5){\small $V(x,0) = V_0(x)$}
			\put(2,26){\small \rotatebox{90}{$V(0,y) = V_1(y)$}}

		\end{overpic}
		\begin{figuretext}\label{D.pdf}
			The triangular region $D$ defined in \eqref{Ddef} and the boundary conditions relevant for the Goursat problem.
		\end{figuretext}
	\end{center}
\end{figure}
Our first result (Theorem \ref{linearmainth}) establishes a mathematically precise version of the classical Abel integral representation of the solution of the Goursat problem for \eqref{linearernst} with given boundary data. This type of representation is well known (cf. e.g. \cite{HE1989a, GS2002, FST1999}) and the main purpose of Theorem~\ref{linearmainth} is to provide a formulation that is suitable for our needs. We discuss regularity, the behavior at the boundary of $D$, and uniqueness of the solution under reasonable assumptions on the boundary data. In this context, reasonable means that our results can be applied to the common examples of collinear solutions (cf. e.g. \cite{FI1987,KP1971,S1972}). In particular, we allow for boundary data with singular derivatives at the corners of $D$. As a consequence, our representation of the solution of \eqref{linearernst} contains singular integrands, which is the main challenge in the analysis of the Goursat problem.

Our second result (Theorem \ref{GeneralThmAsymptotics}) describes the asymptotic behavior of the solution near the diagonal of $D$. We show that $V(x,1-x-\epsilon)$ admits an asymptotic expansion to all orders of the form
\begin{align*}
V(x,1-x-\epsilon) =\sum_{j=0}^{J}  f_j(x) \epsilon^j \ln(\epsilon)
\; +\sum_{j=0}^{J} g_j(x)\epsilon^j
+O(\epsilon^{J+1}\ln(\epsilon)), \qquad \epsilon \downarrow 0,
\end{align*}
where the coefficients $f_j,g_j$ are given explicitly in terms of the boundary data $V_0(x)=V(x,0)$ and $V_1(y)=V(0,y)$, and where the error term is uniform on compact subsets of~$(0,1)$. The first few terms of the asymptotic formula are given by
\begin{align}\label{formulanaiveasymptotics}\begin{split}
V(x,1-x-\epsilon) =& \; -\frac{1}{\pi} \left( h_0(x)+ h_1(x) \right) \ln(\epsilon)\\
&+\frac{1}{\pi} \frac{d}{dx} \left(\int_{0}^{x} h_0(k) \ln(4(x-k)) dk - \int_{x}^{1} \ln(4(k-x)) h_1(k)dk\right) \\
& -\frac{1}{2\pi} \left( h_0'(x)+ h_1'(x) \right) \epsilon \ln(\epsilon)\\
&+ \frac{1}{2\pi} \frac{d^2}{dx^2} \left(\int_{0}^{x} h_0(k) \ln(4(x-k)) dk - \int_{x}^{1} \ln(4(k-x)) h_1(k)dk\right) \epsilon \\
&+ O(\epsilon^2 \ln(\epsilon)), \qquad \epsilon \downarrow 0,
\end{split}
\end{align}
where
\begin{align}\label{Defh0h1}
h_0(k) = \sqrt{1-k} \int_0^k \frac{V_{0x}(x')}{\sqrt{k-x'}} dx', \hspace{0.5cm} 	h_1(k) = \sqrt{k} \int_0^{1-k} \frac{V_{1y}(y')}{\sqrt{1-k-y'}} dy'.
\end{align}

We finally point out that representations for solutions of boundary value problems similar to the Goursat problem for \eqref{linearernst} have been subject of recent mathematical research. For instance, such representations have been found for solutions of Protter problems for equations of Keldysh type in $(3+1)$ dimensions \cite{PHNS2017a,PHNS2017b}. In~\cite{PHNS2017a}, the representations have also been applied to derive asymptotic formulas for the solution. The Euler--Darboux equation and integral representations of its solution have also arisen in the context of hierarchies of integrable systems such as the KdV hierarchy~\cite{KMM2010}.

The two main results of the paper are presented in Section \ref{SectionMainresults}, and their proofs are given in Section \ref{SectionProof1} and Section \ref{SectionAsymptotics}, respectively.
In Section \ref{SectionApplication} we apply our results to collinearly polarized colliding gravitational waves and give full asymptotic expansions for the components of the Weyl tensor.

\section{Main results} \label{SectionMainresults}
Let $D$ be defined by \eqref{Ddef}. Since \eqref{linearernst} is a linear equation, we can assume that $V$ is real-valued and that $V(0,0)=0$. We introduce a notion of $C^n$-solution of the Goursat problem for \eqref{linearernst} in $D$ as follows.

\begin{definition}\label{linearsolutiondef}\upshape
	Let $V_0(x)$, $x \in [0, 1)$, and $V_1(y)$, $y \in [0,1)$, be real-valued functions.
	A function $V:D \to \R$ is called a {\it $C^n$-solution of the Goursat problem for \eqref{linearernst} in $D$ with data $\{V_0, V_1\}$} if there exists an $\alpha \in [0,1)$ such that
	\begin{align*}
	\begin{cases}
	V \in C(D) \cap C^n(\Int(D)),
	\\
	\text{$V(x,y)$ satisfies the Euler--Darboux equation \eqref{linearernst} in $\Int(D)$,}
	\\
	\text{$x^\alpha V_x, y^\alpha V_y, x^\alpha y^\alpha V_{xy} \in C(D)$},
	\\
	\text{$V(x,0) = V_0(x)$ for $x \in [0,1)$,}
	\\
	\text{$V(0,y) = V_1(y)$ for $y \in [0,1)$.}
	\end{cases}
	\end{align*}
\end{definition}

The following theorem solves the Goursat problem for \eqref{linearernst} in $D$ by providing a representation for the solution in terms of the boundary data.

\begin{theorem}[Solution of the Euler--Darboux equation in a triangle]\label{linearmainth}
	Let $n \geq 2$ be an integer and suppose $\alpha \in [0,1)$. Let $V_0(x)$, $x \in [0, 1)$, and $V_1(y)$, $y \in [0,1)$, be two real-valued functions such that
	\begin{align}\label{V0V1assumptions}
	\begin{cases}
	V_0, V_1 \in C([0,1)) \cap C^n((0,1)),
	\\
	\text{$x^\alpha V_{0x}, y^\alpha V_{1y}, x^{\alpha+1}V_{0xx}, y^{\alpha+1} V_{1yy} \in C([0,1))$,}
	\\
	V_0(0) = V_1(0) = 0.
	\end{cases}
	\end{align}
	Then
	\begin{align}\begin{split}\label{linearVrepresentation}
	V(x,y) = &\; \frac{1}{\pi} \int_0^x \frac{\sqrt{1-k}}{\sqrt{(1-y-k)(x-k)}} \bigg(\int_0^k \frac{V_{0x}(x')}{\sqrt{k - x'}}dx'\bigg) dk
	\\
	&  + \frac{1}{\pi} \int_{1-y}^1 \frac{\sqrt{k}}{\sqrt{(k - (1-y))(k-x)}} \bigg(\int_0^{1-k} \frac{V_{1y}(y')}{\sqrt{1 - y' - k}} dy'\bigg) dk, \qquad (x,y) \in D,
		\end{split}
	\end{align}
	is a $C^{n-1}$-solution of the Goursat problem for \eqref{linearernst} in $D$ with data $\{V_0, V_1\}$ (and with the same $\alpha$). Furthermore, $V$ is the only solution in the sense of Definition \ref{linearsolutiondef}.
\end{theorem}

\begin{remark} \upshape
	In the case of $n=2$, Theorem \ref{linearmainth} states that $V$ is at least in $C^1$. However, the derivative $V_{xy}$ always exists and is continuous in $\Int D$. Furthermore, $x^\alpha y^\alpha V_{xy}$ is still in $C(D)$ in this case as it is required in Definition \ref{linearsolutiondef}.
\end{remark}

\begin{remark}\upshape
	An alternative integral representation for the solution of the Goursat problem for the Euler--Darboux equation was already derived in \cite{S1972} using Riemann's classical method \cite{CH1962, G1964}. The representation \eqref{linearVrepresentation} relies on Abel integrals and is analogous to formulas derived in \cite{HE1989a}, whereas the formula of \cite{S1972} involves the Legendre function $P_{-1/2}$ of order $-1/2$.
\end{remark}

The representation \eqref{linearVrepresentation} can be used to study the behavior of the solution $V(x,y)$ near the diagonal $x+y=1$, i.e. the behavior of $V(x,1-x-\epsilon)$ as $\epsilon \downarrow 0$. Letting $y=1-x-\epsilon$ in our representation formula \eqref{linearVrepresentation}, we find
\begin{align}\begin{split} \label{splitVasympt}
V(x,1-x-\epsilon) = &\; \frac{1}{\pi} \int_0^x \frac{\sqrt{1-k}}{\sqrt{(x-k+\epsilon)(x-k)}} \bigg(\int_0^k \frac{V_{0x}(x')}{\sqrt{k - x'}}dx'\bigg) dk
\\
&  + \frac{1}{\pi} \int_{x+\epsilon}^1 \frac{\sqrt{k}}{\sqrt{(k - x-\epsilon)(k-x)}} \bigg(\int_0^{1-k} \frac{V_{1y}(y')}{\sqrt{1 - y' - k}} dy'\bigg) dk \\
=:&\; X_1 + X_2.
\end{split}
\end{align}

We define $h_0, h_1$ by \eqref{Defh0h1} and the kernels $K_0$ and $K_1$ by
\begin{align} \label{DefKernels}
K_0(u,t) = \frac{1}{\sqrt{u}\sqrt{u+t}}, \quad K_1(u,t)= \frac{1}{\sqrt{u}\sqrt{u-t}},
\end{align}
for $u>0$ and $-u<t<u$. Furthermore, we define for $j\ge 0$ the constants
\begin{align} \label{Defcj}
	c_j =  \frac{1}{2^j} \left( \prod_{l=0}^{j-1} (2l+1) \right) = (-1)^j \frac{\partial^j}{\partial t^j}K_0(1,0) =\frac{\partial^j}{\partial t^j}K_1(1,0)
\end{align}
and
\begin{align} \label{DefH}
	H_0^j(x,k) = h_0(k) - \sum_{l=0}^{j} \frac{h_0^{(l)}(x) (-1)^l}{l!} (x-k)^l, \quad H_1^j(x,k) = h_1(k) - \sum_{l=0}^{j} \frac{h_1^{(l)}(x)}{l!} (k-x)^l.
\end{align}

The following theorem provides the asymptotics of $V(x,1-x-\epsilon)$ as $\epsilon \downarrow 0$ up to all orders.

\begin{theorem}[Asymptotic expansion to all orders] \label{GeneralThmAsymptotics}
	Let $J\ge 0$ be an integer. Suppose that $V_0,V_1\in C^{J+2}((0,1))$ satisfy the conditions \eqref{V0V1assumptions}. Then the unique solution $V(x,y)$ of the Goursat problem for \eqref{linearernst} with data $\{ V_0,V_1 \}$ enjoys the asymptotic expansion
	\begin{align*}
	V(x,1-x-\epsilon) =&-\frac{1}{\pi}\sum_{j=0}^{J}  \frac{c_j}{(j!)^2}(h_0^{(j)}(x)+h_1^{(j)}(x)) \epsilon^j \ln(\epsilon) \\
	&  +\frac{1}{\pi}\sum_{j=0}^{J} ((-1)^jA_j(x)+B_j(x)) \epsilon^j   \\
	&+O(\epsilon^{J+1}\ln(\epsilon)), \qquad \epsilon \downarrow 0, \quad 0<x<1,
	\end{align*}
	where the error term is uniform with respect to $x$ in compact subsets of $(0,1)$,
	\begin{align*}
	A_j(x) =& \; \frac{c_j}{j!}\left( \int_{0}^{x} \frac{H_0^j(x,k)}{(x-k)^{j+1}}dk +\sum_{l =0}^{j-1} \frac{(-1)^l}{l!(l-j)}h_0^{(l)}(x)x^{l-j} +\frac{(-1)^jh_0^{(j)}(x)}{j!} \ln(x) \right) \\
	&+ \frac{h_0^{(j)}(x)}{j!}\left( \int_0^1\frac{v^j}{\sqrt{v+1}\sqrt{v}}dv+\int_{1}^{\infty} v^j \left( \frac{1}{\sqrt{v+1}\sqrt{v}}-\sum_{l=0}^{j} \frac{1}{l!} (-1)^l c_l  v^{-l-1}\right)dv  \right)\\
	&+ \frac{h_0^{(j)}(x)}{j!}\sum_{l=0}^{j-1} (-1)^l \frac{c_l}{l! (l-j)},
\\
	B_j(x) =&\; \frac{c_j}{j!} \left( \int_{x}^{1} \frac{H_1^j(x,k)}{(k-x)^{j+1}}dk + \sum_{l=0}^{j-1}  \frac{1}{l!(l-j)} h_1^{(l)}(x) (1-x)^{l-j} + \frac{h_1^{(j)}(x)}{j!} \ln(1-x) \right)\\
	&+ \frac{h_1^{(j)}(x)}{j!}\left( \int_{1}^{\infty} v^j \left( \frac{1}{\sqrt{v-1}\sqrt{v}}-\sum_{l=0}^{j} \frac{c_l}{l!} v^{-l-1}\right)dv +\sum_{l=0}^{j-1} \frac{c_l}{l!(l-j)} \right),
	\end{align*}
	and $h_0$, $h_1$, $H_0^j$, $H_1^j$ are defined by \eqref{Defh0h1} and \eqref{DefH}. The first few terms can also be represented as in \eqref{formulanaiveasymptotics} (see Remark \ref{TheoremNaiveAsymptotics}).
\end{theorem}

\section{Proof of Theorem \ref{linearmainth}} \label{SectionProof1}
In order to prove Theorem \ref{linearmainth} we study some properties of Abel integrals.

\begin{definition} \upshape
	Let $\alpha \in [0,1)$ and suppose $k^{\alpha} h \in C([0,1))$. Then the \emph{Abel transform $\mathcal{A} h$ of the function $h$} is defined by
	\[
	\mathcal{A} h(x) = \int_{0}^{x} \frac{h(k)}{\sqrt{x-k}} dk, \qquad x \in [0,1).
	\]
\end{definition}

The following lemma shows that the Abel transform is well defined and gives some basic properties.

\begin{lemma} \label{PropertiesAbel}
	Let $n\ge 0$ and $\alpha \in [0,1)$. Suppose $h \in C^n((0,1))$ and $k^{\alpha+j} h^{(j)} \in C([0,1))$ for $0\le j \le n$. Then the Abel transform $f=\mathcal{A}h$ of $h$, defined by
	\[
	f(x)= \mathcal{A}h(x) = \int_{0}^{x} \frac{h(k)}{\sqrt{x-k}}dk,
	\]
	is in $C^n((0,1))$. Furthermore, it holds that $x^{\alpha-1/2+j} f^{(j)} \in C([0,1))$ for all $0\le j \le n$.
\end{lemma}

\begin{proof}
	First, we show that the function $f$ is well defined. Fix some $x\in (0,1)$. Then we can define the constant $M =\sup_{k\in [0,x]} |h(k)k^\alpha|$ and get
	\[
	|f(x)| \le \int_{0}^{x} \left| \frac{h(k)}{\sqrt{x-k}} \right| dk \le M \int_{0}^{x} \frac{1}{k^\alpha \sqrt{x-k}} dk.
	\]
	The transformation $u=k/x$ gives
	\[
	\int_{0}^{x} \frac{1}{k^\alpha \sqrt{x-k}} dk = \int_{0}^{1} \frac{x}{(ux)^\alpha \sqrt{x-ux}} du = x^{1/2-\alpha} \int_{0}^{1} \frac{1}{u^\alpha \sqrt{1-u}} du = x^{1/2-\alpha} B\left(1-\alpha,\frac{1}{2}\right),
	\]
	where $B$ denotes the Beta function. By the identity $B(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$, $\Gamma$ denoting the Gamma function, this yields
	\begin{align}\label{KeyEstimate}
	\int_{0}^{x} \left| \frac{h(k)}{\sqrt{x-k}} \right| dk \le M \, x^{1/2-\alpha} \sqrt{\pi} \frac{\Gamma(1-\alpha)}{\Gamma(3/2-\alpha)}.
	\end{align}
	Hence $f$ is well defined for all $x\in(0,1)$.

	Next we show that $f$ is continuous on $(0,1)$. We apply the transformation $u=k/x$ and obtain
	\begin{align} \label{Substitution of Abel integral}
	f(x) = \int_{0}^{x} \frac{h(k)}{\sqrt{x-k}}dk = \sqrt{x}\int_{0}^{1} \frac{h(ux)}{\sqrt{1-u}} du.
	\end{align}
	Then it follows easily by Lebegue's dominated convergence theorem that $f$ is in $C((0,1))$.

	In order to show that $x^\gamma f \in C([0,1))$ for $\gamma = \alpha-1/2$ we show that the limit $\lim_{x\to0} x^\gamma f(x)$ exists. In a similar way as \eqref{KeyEstimate}, we obtain the estimate
	\[
	\inf_{k\in [0,x]}(k^\alpha h(k)) x^{1/2-\alpha} \sqrt{\pi} \frac{\Gamma(1-\alpha)}{\Gamma(3/2-\alpha)}
	\le f(x) \le \sup_{k\in [0,x]}(k^\alpha h(k)) x^{1/2-\alpha} \sqrt{\pi} \frac{\Gamma(1-\alpha)}{\Gamma(3/2-\alpha)}
	\]
	for all $x \in (0,1)$. Hence, by multiplying the inequality above with $x^\gamma$, we find
	\[
	\lim_{x\to0} x^{\alpha-1/2} f(x)= \lim_{k\to 0} (k^\alpha h(k))  \sqrt{\pi} \frac{\Gamma(1-\alpha)}{\Gamma(3/2-\alpha)}.
	\]
	In particular, the function $x^\gamma f$ can be continuously extended onto $[0,1)$.

	In order to show differentiability of $f$, we consider formula \eqref{Substitution of Abel integral} again. By applying Lebeugue's dominated convergence theorem to the difference quotient,  and, by using that $k^{1+\alpha}h' \in C([0,1))$, we obtain
	\begin{align} \begin{split}\label{Formula of derivative of Abel integral}
	\frac{d}{dx} \mathcal{A}h(x) &= \frac{1}{2\sqrt{x}} \int_{0}^{1} \frac{h(ux)}{\sqrt{1-u}} du + \sqrt{x} \int_{0}^{1} \frac{h'(ux) u}{\sqrt{1-u}} du \\
	&= \frac{1}{2x} \mathcal{A}h(x)+ \frac{1}{x} \mathcal{A}(kh')(x).
	\end{split}
	\end{align}
	Since $kh' $ also fulfills the assumptions of the Lemma (but with reduced $n$), all remaining statements follow.
\end{proof}

\begin{remark} \label{RemarkHadamardFP}\upshape
	We will also use the following formula for the derivative of the Abel transform, given by
	\begin{align} \label{FormulaDerivative}
	\frac{d}{dx} \left( \int_{0}^{x} \frac{h(k)}{\sqrt{x-k}}dk\right) =\frac{h(\delta)}{\sqrt{x-\delta}}- \frac{1}{2} \int_{0}^{\delta} \frac{h(k)}{(x-k)^{3/2}}dk + \int_{\delta}^{x} \frac{h'(k)}{\sqrt{x-k}}dk
	\end{align}
	for an arbitrary $0<\delta<x$. This is obtained by splitting the integral, integrating by parts, and differentiating.
\end{remark}

\begin{remark} \label{RemarkUpperBoundirr}\upshape
	Note that the proof of Lemma \ref{PropertiesAbel} does not change if we replace the segment $[0,1)$ in the lemma by a segment $[0,b)$, where $0<b<1$.
\end{remark}

The following lemma will be useful in the proof of Theorem \ref{linearmainth}, since the Abel integrals in the integral representation \eqref{linearVrepresentation} depend on two variables.

\begin{lemma} \label{AbelWithParameter}
	Let $h \in C((0,1))$ and $x^{\alpha} h\in C([0,1))$ for some $\alpha<1$. Suppose that $g \in C(D)$ and that there exists a number $\beta \ge 0$ and a constant $C >0$ such that
	\[
	\left| g(y,k) \right| \le \frac{C}{(1-x-y)^{\beta}}, \hspace{0.5cm} \text{for all } (x,y) \in D \text{ and } k \in [0,x].
	\]
	Then the function $f$ defined by
	\[
	f(x,y)=  \int_{0}^{x} g(y,k) \frac{h(k)}{\sqrt{x-k}}dk
	\]
	is in $C(\Int D)$. Furthermore, we have $x^{\alpha-1/2} f \in C(D)$.
\end{lemma}

\begin{proof}
	Let $U$ be an open set in $D\backslash \{ x=0 \}$ such that $\overline{U} \subset D\backslash \{ x=0 \}$, where $\overline{U}$ denotes the closure of $U$. Substituting $u=k/x$ and using the assumptions of the lemma yields
	\[
	\int_{0}^{x} \left| g(y,k) \frac{h(k)}{\sqrt{x-k}} \right| dk=  \sqrt{x}\int_{0}^{1} \left| g(y,ux) \frac{h(ux)}{\sqrt{1-u}} \right| du \le  \frac{C\sqrt{x}}{(1-x-y)^{\beta}} \int_{0}^{1} \frac{|h(ux)|}{\sqrt{1-u}} du.
	\]
	The  integrand $|h(ux)|/\sqrt{1-u}$ is bounded by an $L^1((0,1))$-function independent of $x$ and the term $(1-x-y)^{-\beta}$ is bounded by a constant on $U$. This shows that $f$ is in $C(D \backslash \{ x=0 \})$ by applying Lebegue's dominate convergence theorem.

	In order to prove the last statement of the lemma, we observe
	\[
	\int_{0}^{1}\frac{x^\alpha|h(ux)|}{\sqrt{1-u}} du \le \sup_{k\in [0,x)} (k^\alpha |h(k)|) \int_0^1 \frac{1}{u^\alpha \sqrt{1-u}} du.
	\]
	Hence, if we fix a neighborhood $W$ around a point $(0,y) \in D$ such that $\overline{W}\subset D$, we can bound the expression
	\[
	x^{\alpha-1/2} \sqrt{x}\int_{0}^{1} \left| g(y,ux) \frac{h(ux)}{\sqrt{1-u}} \right| du
	\]
	independently of $x$ and $y$. This shows that $x^{\alpha-1/2} f \in C(D)$.
\end{proof}

The next lemma gives a well-known inversion formula for Abel integrals. This will be needed in order to check the boundary conditions $V(x,0) = V_0(x)$ and $V(0,y)=V_1(y)$.

\begin{lemma} \label{InversionAbel}
	Let $h,f$ be as in Lemma \ref{PropertiesAbel}. Then we have
	\[
	h(k)= \frac{1}{\pi} \frac{d}{dk} \left( \int_{0}^{k}\frac{f(x)}{\sqrt{k-x}} dx \right), \qquad k\in (0,1).
	\]
\end{lemma}

\begin{proof}
	Due to Lemma \ref{PropertiesAbel} the integral
	\[
	\int_{0}^{k}\frac{f(x)}{\sqrt{k-x}} dx
	\]
	exists and is differentiable on $(0,1)$. By the definition of $f$, we have
	\[
	\int_0^k \frac{f(x)}{\sqrt{k-x}} dx = \int_{0}^{k} \int_{0}^{x} \frac{h(\kappa)}{\sqrt{x-\kappa}\sqrt{k-x}} d\kappa dx = \int_{0}^{k} h(\kappa) \int_{\kappa}^{k} \frac{1}{\sqrt{x-\kappa}\sqrt{k-x}} dx d\kappa.
	\]
	We apply the transformation $y=(k-x)/(k-\kappa)$ and get
	\begin{align*}
	\int_0^k \frac{f(x)}{\sqrt{k-x}} dx = \int_0^k h(\kappa) \int_{0}^{1} \frac{dy}{\sqrt{y}\sqrt{1-y}}d\kappa=\pi \int_{0}^{k}h(\kappa)d\kappa.
	\end{align*}
	This implies the desired statement.
\end{proof}

\begin{proof}[Proof of Theorem \ref{linearmainth}]
	It follows immediately by Lemma \ref{AbelWithParameter} with
	 \[
	 g(y,k) = \sqrt{1-k} /\sqrt{1-y-k}
	 \]
	 that $V$ is in $C(D)$ with $V(0,0) = 0$. Furthermore, letting
	\begin{align} \label{DefG}
	G(y,k) =  \frac{\sqrt{1-k}}{\sqrt{1-y-k}} \mathcal{A}V_{0x}(k),
	\end{align}
	we have
	\begin{align}\begin{split}\label{FirstStepDerivativeV}
	V_x(x,y) = &\; \frac{1}{\pi} \frac{\partial}{\partial x}\int_0^x \frac{G(y,k)}{\sqrt{x-k}} dk
	\\
	&  + \frac{1}{\pi} \frac{\partial}{\partial x} \left( \int_{1-y}^1 \frac{\sqrt{k}}{\sqrt{(k - (1-y))(k-x)}} \bigg(\int_0^{1-k} \frac{V_{1y}(y')}{\sqrt{1 - y' - k}} dy'\bigg) dk \right).
	\end{split}
	\end{align}
	Note that $x^{\alpha-1/2}\mathcal{A}V_{0x}(x)$ is in $C([0,1))$ by Lemma \ref{PropertiesAbel} and hence $G$ satisfies all assumptions in Lemma \ref{PropertiesAbel} in the sense of Remark \ref{RemarkUpperBoundirr}. Thus the first integral in \eqref{FirstStepDerivativeV} can be differentiated. For the second integral in \eqref{FirstStepDerivativeV}, it is clear that we can interchange the derivative and the integral sign. Under consideration of \eqref{Formula of derivative of Abel integral} we find
	\begin{align*}
	V_x(x,y) =& \frac{1}{2x\pi}\mathcal{A}\left( G(y,\cdot) \right)(x) + \frac{1}{x\pi} \mathcal{A}\left(  kG_k(y,\cdot) \right)(x)\\ &+\frac{1}{2\pi}\mathcal{A}\left(  \frac{\sqrt{1-(\cdot)}}{(1-x-(\cdot))^{3/2}} \mathcal{A}V_{1y}(\cdot) \right)(y).
	\end{align*}

	Since Lemma \ref{AbelWithParameter} can be applied to all three terms (with different choices of $g$, $h$ and in the third case with $x$ and $y$ interchanged), we obtain that $V_x \in C(\Int D)$ and $x^\alpha V_x \in C(D)$. The cases $V_y$ and $V_{xy}$ are similar and we find $V \in C^{n-1}(\Int D)$ by induction.

	In order to prove that $V$ satisfies equation \eqref{linearernst}, we use the alternative formula \eqref{FormulaDerivative} to represent the derivative.  We only prove that the first term
	\[
	X_1(x,y) =  \frac{1}{\pi} \int_0^x \frac{\sqrt{1-k}}{\sqrt{(1-y-k)(x-k)}} \bigg(\int_0^k \frac{V_{0x}(x')}{\sqrt{k - x'}}dx'\bigg) dk
	\]of \eqref{linearVrepresentation} satisfies the equation, since the other term is similar. We fix some $(x,y) \in \Int D$ and define $G(y,k)$ by \eqref{DefG}.
	By \eqref{FormulaDerivative} it holds for any fixed $\delta \in (0,x)$ that
	\begin{align*}
	\pi X_{1x}(x,y)=&\frac{G(y,\delta)}{\sqrt{x-\delta}}- \frac{1}{2} \int_{0}^{\delta} \frac{G(y,k)}{(x-k)^{3/2}}dk + \int_{\delta}^{x} \frac{G_k(y,k)}{\sqrt{x-k}}dk, \\
	\pi X_{1y}(x,y) =&\int_{0}^{x} \frac{G_y(y,k)}{\sqrt{x-k}}dk,  \\
	\pi X_{1xy}(x,y) =& \,\frac{G_y(y,\delta)}{\sqrt{x-\delta}}- \frac{1}{2} \int_{0}^{\delta} \frac{G_y(y,k)}{(x-k)^{3/2}}dk + \int_{\delta}^{x} \frac{G_{yk}(y,k)}{\sqrt{x-k}}dk.
	\end{align*}
	In particular,
	\begin{align*}
	\pi X_{1x}(x,y) &= \lim_{\epsilon\rightarrow 0} \left( \frac{G(y,x)}{\sqrt{\epsilon}} - \frac{1}{2} \int_{0}^{x-\epsilon} \frac{G(y,k)}{(x-k)^{3/2}}dk\right), \\
	\pi X_{1xy}(x,y) &=\lim_{\epsilon\rightarrow 0} \left( \frac{G_y(y,x)}{\sqrt{\epsilon}} - \frac{1}{2} \int_{0}^{x-\epsilon} \frac{G_y(y,k)}{(x-k)^{3/2}}dk \right).
	\end{align*}
	The identity $G_y(y,k) = (2(1-y-k))^{-1}G(y,k)$ yields
	\begin{align*}
	&\pi X_{1x}(x,y) +\pi X_{1y}(x,y)\\
	=&\lim_{\epsilon\rightarrow 0} \left( \frac{G(y,x)}{\sqrt{\epsilon}} - \frac{1}{2} \int_{0}^{x-\epsilon} \frac{G(y,k)(1-y-k)-G(y,k)(x-k)}{(1-y-k)(x-k)^{3/2}}dk\right)\\
	=&\; 2\pi (1-x-y) X_{1xy}(x,y).
	\end{align*}
	The other term of \eqref{linearVrepresentation} is similar and hence $V$ satisfies \eqref{linearernst}.

	Next we prove that $V$ satisfies the boundary conditions. We have
	\[
	V(x,0) = \frac{1}{\pi} \mathcal{A} (\mathcal{A}V_{0x})(x).
	\]
	Furthermore, Lemma \ref{InversionAbel} (with $h=V_{0x}$ and $f=\mathcal{A}V_{0x}$) gives
	\[
	V_{0x}(x)=\frac{1}{\pi}\frac{d}{dx} \mathcal{A} (\mathcal{A}V_{0x})(x)=V_x(x,0)
	\]
	for all $x \in (0,1)$ and hence $V(x,0) = V_{0}(x)+c$ for a constant $c \in \R$. But $0 = V(0,0) = V_{0}(0)$. Thus we have $c=0$ and $V(x,0) = V_{0}(x)$. The case $V(0,y)$ is completely analogous.

	It remains to show that $V$ is the only solution. For that, it suffices to show that a function $V$ satisfying \eqref{linearernst} with $V(x,0) = 0 = V(0,y)$ must vanish in $D$. Let $D_T = \{ (x,y) \in D: x+y < T \}$ for $0<T<1$. Then the coefficients
	\[
	\frac{1}{2(1-x-y)}
	\]
	of \eqref{linearernst} are bounded on $D_T$. It follows by the method of successive approximations \cite[pp. 136--137]{G1964} that $V$ is unique on $D_T$ and hence vanishes on $D_T$. Since $T$ was arbitrary in $(0,1)$, the solution $V$ is zero on $D$. This completes the proof.
\end{proof}

\section{Proof of Theorem \ref{GeneralThmAsymptotics}} \label{SectionAsymptotics}

We define $h_0$, $h_1$, $K_0$, $K_1$, $H_0^j$, $H_1^j$, $X_1,X_2$, and $c_j$ by \eqref{Defh0h1}, \eqref{splitVasympt}, \eqref{DefKernels}, \eqref{Defcj}, and \eqref{DefH}. Note that, by Lemma \ref{PropertiesAbel}, $h_0$ and $h_1$ have the same order of regularity as $V_{0x}$ and $V_{0y}$, respectively, and that $k^{\alpha-1/2} h_0(k)$ and $k^{\alpha-1/2} h_1(k)$ are in $C([0,1))$. Furthermore, the kernels satisfy $K_i(au,at)=a^{-1} K_i(u,t)$ for $i=0,1$ and $a>0$.

\begin{proof}[Proof of Theorem \ref{GeneralThmAsymptotics}]
	Fix some compact subset $I \subset (0,1)$. Consider $X_1$ first. Then, for fixed $x \in I$, we have
	\[
	X_1 = \frac{1}{\pi}\int_{0}^{x} \frac{H_0^J(x,k) }{\sqrt{x-k}\sqrt{x-k+\epsilon}} dk  + \frac{1}{\pi} \sum_{j=0}^{J}\frac{h_0^{(j)}(x)(-1)^j}{j!}\int_{0}^{x}K_0(x-k,\epsilon)(x-k)^jdk.
	\]
	After the substitution $u=x-k$, the rightmost integral in the expression above can be written as
	\[
	\int_{0}^{x} u^jK_0(u,\epsilon) du = \int_{0}^{\epsilon} u^jK_0(u,\epsilon) du+\int_{\epsilon}^{x} u^jK_0(u,\epsilon) du = \epsilon^j \int_0^1v^jK_0(v,1)dv + \int_{\epsilon}^{x} u^jK_0(u,\epsilon) du,
	\]
	where we applied the transformation $u=\epsilon v$. Denoting
	\[
	R_{K_0,J}(u,\epsilon) = K_0(u,\epsilon)-\sum_{l=0}^{J} \frac{\epsilon^l}{l!} (-1)^l c_l u^{-l-1},
	\]
	we find
	\begin{align} \begin{split} \label{EquationFirstPartAsym}
	\int_{\epsilon}^{x} u^jK_0(u,\epsilon) du =& \int_{\epsilon}^{x} u^j \left( K_0(u,\epsilon)-\sum_{l=0}^{J} \frac{\epsilon^l}{l!} (-1)^l c_l u^{-l-1}\right)du +\sum_{l=0}^{J} \frac{\epsilon^l}{l!} (-1)^l c_l\int_{\epsilon}^{x}u^{j-l-1}du \\
	=& \;  \epsilon^j \int_{1}^{\infty} v^j \left( K_0(v,1)-\sum_{l=0}^{J} \frac{1}{l!} (-1)^l c_l  v^{-l-1}\right)dv - \int_{x}^{\infty} u^j R_{K_0,J}(u,\epsilon)du\\  &+ \sum_{\substack{l=0\\ l \neq j}}^J \frac{\epsilon^l}{l!} (-1)^l c_l\frac{1}{j-l} (x^{j-l}-\epsilon^{j-l}) + \frac{\epsilon^j}{j!} (-1)^j c_j (\ln(x) - \ln(\epsilon)), \end{split}
	\end{align}
	where we again substituted $u=\epsilon v$. We can split the sum $\sum_{l=0}^{J} \frac{1}{l!} (-1)^l c_l  v^{-l-1}$ at $l=j$ and obtain
	\begin{align*}
	\int_{\epsilon}^{x} u^jK_0(u,\epsilon) du =& \; \epsilon^j \int_{1}^{\infty} v^j \left( K_0(v,1)-\sum_{l=0}^{j} \frac{1}{l!} (-1)^l c_l  v^{-l-1}\right)dv - \int_{x}^{\infty} u^j R_{K_0,J}(u,\epsilon)du\\  &+ \mathcal{C}_j(x)+\sum_{l=j+1}^{J} \frac{\epsilon^l}{l!} (-1)^l c_l\frac{1}{j-l} x^{j-l} - \sum_{l=0}^{j-1} \frac{\epsilon^l}{l!} (-1)^l c_l\frac{1}{j-l} \epsilon^{j-l}\\
	&+ \frac{\epsilon^j}{j!} (-1)^j c_j (\ln(x) - \ln(\epsilon)),
	\end{align*}
	where
	\[
	\mathcal{C}_j(x)= \sum_{l=0}^{j-1} \frac{\epsilon^l}{l!} (-1)^l c_l\frac{1}{j-l} x^{j-l}.
	\]
	In the first term of $X_1$ we can expand $K_0$ (since $H_0^J(x,k) \sim (x-k)^{J+1}$) and find
	\begin{align*}
	\int_{0}^{x} \frac{H_0^J(x,k) }{\sqrt{x-k}\sqrt{x-k+\epsilon}} dk &= \sum_{j=0}^{J} \frac{\epsilon^j}{j!} (-1)^j c_j  \int_{0}^{x} H_0^J(x,k) (x-k)^{-j-1} dk \\
	&+ \int_0^x H_0^J(x,k) R_{K_0,J}(x-k,\epsilon) dk.
	\end{align*}
	In the integrals $\int_0^x H_0^J(x,k)(x-k)^{-j-1}dk$, we write
	\[
	H_0^J(x,k) = h_0(k) - \sum_{l=0}^{j}\frac{h_0^{(l)}(x)(-1)^l}{l!}(x-k)^l - \sum_{l=j+1}^{J}\frac{h_0^{(l)}(x)(-1)^l}{l!}(x-k)^l.
	\]
	Then the sums $\sum_{l = j+1}^J (\cdots)$ vanish together with the terms
	\[
	\sum_{j=0}^J \frac{h_0^{(j)}(x)(-1)^j}{j!}\mathcal{C}_j(x).
	\]
	It remains to show that the remainders are in $O(\epsilon^{J+1}\ln(\epsilon))$. We write
	\[
	R_{K_0,J}(u,\epsilon) = \frac{\partial^{J+1}K_0(u,\xi_\epsilon)}{\partial\epsilon^{J+1}} \frac{\epsilon^{J+1}}{(J+1)!} = \epsilon^{J+1}\frac{c_{J+1}}{(J+1)!} \frac{(-1)^{J+1}}{\sqrt{u}(\sqrt{u+\xi_\epsilon})^{2J+3}},
	\]
	for some $\xi_\epsilon \in [0,\epsilon]$. Then
	\[
	|R_{K_0,J}(u,\epsilon)| \le \epsilon^{J+1} \frac{C_J}{u^{J+2}}
	\]
	for some constant $C_J>0$. Hence
	\[
	\left| \int_{x}^{\infty} u^j R_{K_0,J}(u,\epsilon)du \right| \le \epsilon^{J+1} C_J \int_x^\infty u^{j-J-2} du = O(\epsilon^{J+1}), \qquad \epsilon \downarrow 0,
	\]
	uniformly on $I$, since $C_J$ does not depend on $x$. For the second remainder, we denote $m=\min I>0$ and $M=\max I <1$ and we split the integral
	\[
	\int_0^x H_0^J(x,k) R_{K_0,J}(x-k,\epsilon) dk = \int_{0}^{x/2} H_0^J(x,k) R_{K_0,J}(x-k,\epsilon) dk + \int_{x/2}^{x} H_0^J(x,k) R_{K_0,J}(x-k,\epsilon) dk.
	\]
	Since $H_0^J(x,k)k^\gamma$, where $\gamma = \alpha-1/2$, is bounded by a constant independently of $x$ on $[0,M/2]$, there exists a constant $C>0$ such that
	\begin{align*}
	\left|\int_{0}^{x/2} H_0^J(x,k) R_{K_0,J}(x-k,\epsilon) dk\right| &\le C \epsilon^{J+1} \int_{0}^{x/2} \frac{1}{k^{\gamma}(x-k)^{J+2}}dk  \\ &\le C \epsilon^{J+1} \left( \frac{2}{M} \right)^{J+2} \frac{M^{1-\gamma}}{1-\gamma} = O(\epsilon^{J+1}), \qquad \epsilon \downarrow 0,
	\end{align*}
	uniformly on $I$. For the other integral, we observe
	\[
	|H_0^J(x,k)| = \left|\int_{k}^{x} \frac{h_0^{(J+1)}(t)}{J!} (k-t)^{J} dt \right| \le \sup_{t \in [m/2,M]} |h_0^{(J+1)}(t)| \frac{(x-k)^{J+1}}{J!}
	\]
	for $k \in [x/2,x]$ and hence there exist constants $C,C'>0$, independent of $x$, such that
	\begin{align*}
	\left| \int_{x/2}^{x} H_0^J(x,k) R_{K_0,J}(x-k,\epsilon) dk \right| &\le C \int_{x/2}^{x} (x-k)^{J+1}\int_{0}^{\epsilon}\left| \frac{\partial^{J+1}}{\partial t^{J+1}}K_0(x-k,t) \right| (\epsilon-t)^{J} dt dk \\
	&= C'  \int_{x/2}^{x} \int_{0}^{\epsilon} \frac{(x-k)^{J+1}}{\sqrt{x-k}(\sqrt{x-k+t})^{2J+3}} (\epsilon-t)^{J} dt dk \\
	&\le C' \epsilon^{J+1} \int_{x/2}^{x}\int_{0}^{1} \frac{1}{(x-k)+\epsilon s}ds dk \\
	&=C' \epsilon^J \left( (x/2+\epsilon)\ln(x/2+\epsilon)-(x/2) \ln(x/2) - \epsilon \ln (\epsilon) \right).
	\end{align*}
	We observe
	\[
	\frac{ (x/2+\epsilon)\ln(x/2+\epsilon)-(x/2) \ln(x/2) }{\epsilon} = \ln(\xi_\epsilon) + 1
	\]
	for some $\xi_\epsilon \in [x/2,x/2+\epsilon]$ and hence the quotient is bounded on $I$. This implies that
	\[
	\int_{x/2}^{x} H_0^J(x,k) R_{K_0,J}(x-k,\epsilon) dk = \mathcal{O} (\epsilon^{J+1} \ln(\epsilon)), \qquad \epsilon \downarrow 0,
	\]
	uniformly on $I$. A similar procedure for $X_2$ completes the proof.
\end{proof}

\begin{remark} \upshape \label{TheoremNaiveAsymptotics}
	The first few terms of the asymptotic expression found in Theorem \ref{GeneralThmAsymptotics} can also be written as in \eqref{formulanaiveasymptotics}, where $x\in (0,1)$ and where we assume that $V_0, V_1 \in C^3((0,1))$ satisfy the conditions \eqref{V0V1assumptions}. Indeed, consider the case of $X_1$; the case of $X_2$ is similar. By splitting
\[
\pi X_1= \int_{0}^{x} h_0(k)K_0(x-k,\epsilon)dk =  \int_{0}^{x/2} h_0(k)K_0(x-k,\epsilon)dk + \int_{x/2}^{x} h_0(k)K_0(x-k,\epsilon)dk,
\]
we can expand $K_0$ in the first term and in the second term we integrate by parts. This yields
\begin{align*}
\pi X_1= &\;\int_{0}^{x/2} \frac{h_0(k)}{x-k}dk +\left( - h_0(x)\ln(\epsilon) +2h_0(x/2)\ln(\sqrt{x/2}+\sqrt{x/2+\epsilon}) \right) \\
&+2\int_{x/2}^x \ln(\sqrt{x-k}+\sqrt{x-k+\epsilon})h_0'(k)dk + O(\epsilon),
\end{align*}
as $\epsilon \downarrow 0$. Since $h_0'$ is bounded on $[x/2,x]$, this gives
\begin{align*}
\pi X_1= &\int_{0}^{x/2} \frac{h_0(k)}{x-k}dk  - h_0(x)\ln(\epsilon) +h_0(x/2)\ln(2x)  \\
&+\int_{x/2}^x \ln(4(x-k))h_0'(k)dk + O(\ln(\epsilon)\epsilon), \qquad \epsilon \downarrow 0.
\end{align*}
In a similar way as \eqref{FormulaDerivative} we obtain the identity
\[
\frac{d}{dx} \left( \int_{0}^{x} h_0(k) \ln(4(x-k))dk \right) =\int_{0}^{x/2} \frac{h_0(k)}{x-k}dk+h_0(x/2)\ln(2x)+\int_{x/2}^x \ln(4(x-k))h_0'(k)dk.
\]
The following coefficients can be computed similarly by integrating by parts iteratively.
\end{remark}

Alternatively, \eqref{formulanaiveasymptotics} can be obtained directly from Theorem \ref{GeneralThmAsymptotics} by writing
\begin{align*}
\frac{d}{dx} \left( \int_{0}^{x} h_0(k) \ln(4(x-k))dk \right) &= \frac{d}{dx} \left( \int_{0}^{x} H^0_0(x,k) \ln(4(x-k))dk + \int_{0}^{x} h_0(x) \ln(4(x-k))dk \right) \\
&= \int_0^x \frac{H_0^0(x,k)}{x-k} dk - h_0'(x) \int_{0}^{x} \ln(4(x-k))dk \\
& \; \; \; + \frac{d}{dx} \left( h_0(x) \int_0^x \ln(4(x-k))dk \right) \\
&= \int_0^x \frac{H_0^0(x,k)}{x-k} dk + h_0(x) \ln(4) +h_0(x) \ln(x),
\end{align*}
which is equal to $A_0^0(x)$ in Theorem \ref{GeneralThmAsymptotics}. The cases of $h_1$ and the next coefficients are similar.

\begin{example}[Solution of Khan and Penrose] \upshape
	The solution of Khan and Penrose (cf.~\cite{G1991,KP1971}) is given by
	\begin{align*}
	V(x,y) = -\ln \left(\frac{1+\sqrt{x}\sqrt{1-y} + \sqrt{y} \sqrt{1-x}}{1-\sqrt{x}\sqrt{1-y} -\sqrt{y} \sqrt{1-x}}  \right).
	\end{align*}
	If we directly compute the asymptotic behavior of $V$ as $x+y\to1$ we find
	\begin{align}\label{SolutionKhanPenrose}
	V(x,1-x-\epsilon) = 2\ln(\epsilon) - \ln(16(1-x)x) - \frac{1-2x}{2(1-x)x} \epsilon +\frac{3(1-2x+2x^2)}{16(1-x)^2x^2} \epsilon^2 + O(\epsilon^3).
	\end{align}
	If we compute the asymptotics by applying Theorem \ref{GeneralThmAsymptotics}, we have $h_0=-\pi =h_1$ and
	\begin{align*}
	A_0(x) &= -\pi (\ln(x) +\ln(4)), &
	A_1(x) &= \frac{\pi}{2x}, &
	A_2(x) &=\frac{3\pi}{16 x^2},
	\\
	B_0(x) &= -\pi(\ln(1-x) + \ln(4) )&
	B_1(x) &=\frac{\pi}{2(1-x)}, &
	B_2(x) &= \frac{3\pi}{16 (1-x)^2},
	\end{align*}
	which is consistent with \eqref{SolutionKhanPenrose}.
\end{example}

\section{Application to gravitational waves} \label{SectionApplication}
It is shown \cite{G1991} (Eq.~(10.2)) that the colliding gravitational wave problem for collinearly polarized plane waves reduces to solving the equation
\begin{align}
\label{GriffithsEq}V_{fg} + \frac{V_f + V_g}{2(f+g)} = 0
\end{align}
in the triangular region
\[
\left\{(f,g) \in \R^2 \, | \, f \leq \frac{1}{2}, \; g \leq \frac{1}{2}, \; f + g > 0 \right\}.
\]
The change of variables $x = \frac{1}{2} - g$, $y = \frac{1}{2} - f$ transforms \eqref{GriffithsEq} into \eqref{linearernst}. The components of the Weyl tensor (cf. Eq. (10.4) in \cite{G1991}) are given by
\begin{align*}
\Psi_0^\circ & = -  \frac{g'(v)^2}{4}\bigg( 2V_{gg}+\frac{3}{f+g}V_g - (f+g)V_g^3 \bigg)
\\
& = -  \frac{g'(v)^2}{4}\bigg( 2V_{xx}-\frac{3}{1-x-y}V_x +(1-x-y)V_x^3 \bigg),
\\
\Psi_2^\circ & =  f'(u) g'(v)\bigg(V_fV_g-\frac{1}{(f+g)^2}\bigg)
\\
& =  f'(u) g'(v)\bigg( V_xV_y-\frac{1}{(1-x-y)^2} \bigg),
\\
\Psi_4^\circ & = - \frac{f'(u)^2}{4}\bigg(2V_{ff}+\frac{3}{f+g}V_f - (f+g)V_f^3\bigg)
\\
& = - \frac{f'(u)^2}{4}\bigg( 2V_{yy}-\frac{3}{1-x-y}V_y +(1-x-y)V_y^3 \bigg),
\end{align*}
where $u,v \ge 0$ are suitable null coordinates in the interaction region \cite[Chapter 6]{G1991}. Using that $f(u)$ and $g(v)$ have the form (see Eq. (7.6) in \cite{G1991})
\begin{align*}
f(u) & = \frac{1}{2} - (c_1u)^{n_1}, \qquad g(v) = \frac{1}{2} - (c_2v)^{n_2},
\\
f'(u) & = - c_1 n_1 (c_1u)^{n_1-1}
= - \frac{n_1}{u}(\frac{1}{2} - f)
= - \frac{n_1}{u}y=- c_1 n_1y^{1- \frac{1}{n_1}},
\\
 g'(v) &= - c_2 n_2 (c_2v)^{n_2-1}
= - c_2 n_2x^{1- \frac{1}{n_2}},
\end{align*}
with some constants $c_1,c_2,n_1,n_2$, we find
\begin{align} \begin{split} \label{ComponentsWeyl}
\Psi_0^\circ & = \frac{(c_2 n_2x^{1- \frac{1}{n_2}})^2}{4}\bigg( 2V_{xx}-\frac{3}{1-x-y}V_x +(1-x-y)V_x^3 \bigg),
\\
\Psi_2^\circ & = c_1 n_1y^{1- \frac{1}{n_1}}c_2 n_2x^{1- \frac{1}{n_2}}\bigg( V_xV_y-\frac{1}{(1-x-y)^2} \bigg),
\\
\Psi_4^\circ & =\frac{(c_1 n_1y^{1- \frac{1}{n_1}})^2}{4}\bigg( 2V_{yy}-\frac{3}{1-x-y}V_y +(1-x-y)V_y^3 \bigg).
\end{split}
\end{align}
Theorem \ref{GeneralThmAsymptotics} implies
\[
V(x,y) =  \sum_{j=0}^{N}f_j(x)\epsilon^j \ln (\epsilon) +\sum_{j=0}^{N} g_j(x) \epsilon^j + O\big(\epsilon^{N+1}\ln(\epsilon)\big)
\]
as $\epsilon=\epsilon(x,y)=1-x-y \to 0$ uniformly for $x,y$ in some neighborhood of the diagonal away from the corners, where
\[
f_j(x) = \frac{-c_j}{\pi (j!)^2}\left(h_0^{(j)}(x)+h_1^{(j)}(x)\right), \quad g_j(x)=\frac{1}{\pi}\left((-1)^jA_j(x)+B_j(x)\right).
\]
 Due to symmetry of the representation \eqref{linearVrepresentation}, it holds that
\[
V(x,y) =  \sum_{j=0}^{N}f_j(1-y)\epsilon^j \ln (\epsilon) +\sum_{j=0}^{N} \tilde{g}_j(1-y) \epsilon^j + O\big(\epsilon^{N+1}\ln(\epsilon)\big)
\]
as $\epsilon=1-x-y \to 0$, where $\tilde{g}_j$ is defined by interchanging $h_0$ and $h_1$ in the definition of $g_j$. It is easy to see from the proof of Theorem \ref{GeneralThmAsymptotics} that the $y$-derivative of the remainder is $O((1-x-y)^N \ln(1-x-y))$ uniformly. Letting $\epsilon=\epsilon(x,y)=1-x-y$, this gives
\begin{align*}
V_y(x,y) &= -\sum_{j=0}^{N}j f_j(x)\epsilon^{j-1} \ln (\epsilon) - \sum_{j=0}^{N} \gamma_j (x) \epsilon^{j-1}
+ O\big((\epsilon^N \ln(\epsilon)\big), \quad \epsilon \downarrow 0,
\end{align*}
and
\begin{align*}
V_{yy}(x,y) &= \sum_{j=0}^{N}(j-1)jf_j(x)\epsilon^{j-2} \ln (\epsilon)+ \sum_{j=0}^{N} G_j(x)\epsilon^{j-2} + O\big(\epsilon^{N-1}\ln(\epsilon)\big), \quad \epsilon \downarrow 0,
\end{align*}
where $\gamma_j = jg_j +f_j$ and $G_j =(j-1)\gamma_j +jf_j$. In the same way we get
\begin{align*}
V_x(x,y) &= -\sum_{j=0}^{N}jf_j(1-y)\epsilon^{j-1} \ln (\epsilon) - \sum_{j=0}^{N} \tilde{\gamma}_j(1-y)\epsilon^{j-1}+ O\big(\epsilon^{N}\ln(\epsilon)\big), \quad \epsilon \downarrow 0,
\\
V_{xx}(x,y) &= \sum_{j=0}^{N}(j-1)jf_j(1-y)\epsilon^{j-2} \ln (\epsilon)  + \sum_{j=0}^{N} \tilde{G}_j(1-y) \epsilon^{j-2} + O\big(\epsilon^{N-1}\ln(\epsilon)\big), \quad \epsilon \downarrow 0,
\end{align*}
where $\tilde{\gamma}_j = j\tilde{g_j} +f_j$ and $\tilde{G}_j =(j-1)\tilde{\gamma}_j +jf_j$. Together with \eqref{ComponentsWeyl} this leads to full expansions of the components of the Weyl tensor near the diagonal. We have shown the following corollary of Theorem \ref{GeneralThmAsymptotics}.
\begin{corollary}[Asymptotics of the Weyl tensor] Let $N \ge 1$ be and integer. Suppose  $V_0,V_1 \in C^{N+2}((0,1))$ satisfy \eqref{V0V1assumptions}. Then the components of the Weyl tensor associated to the solution $V(x,y)$ of the Goursat problem for \eqref{linearernst} with data $\{ V_0,V_1 \}$ have the following asymptotic behaviour as $\epsilon=\epsilon(x,y)=1-x-y \to 0:$
	\begin{align*}
	\Psi_0^o(x,y)&=\frac{(c_2 n_2x^{1- \frac{1}{n_2}})^2}{4}\Bigg[ \sum_{j=0}^{N} (2\tilde{G}_j(1-y) +3\tilde{\gamma}_j(1-y))\epsilon^{j-2}\\ &\quad+ 	 \sum_{j=0}^{N}(j+2)jf_j(1-y)\epsilon^{j-2} \ln (\epsilon)
	\\
	&\quad-\epsilon\left( \sum_{j=0}^{N}jf_j(1-y)\epsilon^{j-1} \ln (\epsilon) + \sum_{j=0}^{N} \tilde{\gamma}_j(1-y)\epsilon^{j-1} \right)^3+ O\big(\epsilon^{N-1}\ln(\epsilon)^3\big) \Bigg],
	\end{align*}
	\begin{align*}
	\Psi_2^o(x,y)&=c_1 n_1c_2 n_2y^{1- \frac{1}{n_1}}x^{1- \frac{1}{n_2}}\Bigg[ \sum_{j=0}^{N} \left( \sum_{k+l=j} \gamma_k(x) \tilde{\gamma}_l(1-y) \right)\epsilon^{j-2}  - \frac{1}{\epsilon^2}
	\\
	&\quad + \sum_{j=0}^{N}\left( \sum_{k+l=j} kf_k(x) \tilde{\gamma}_l(1-y) + kf_k(1-y) {\gamma}_l(x)  \right)\epsilon^{j-2} \ln (\epsilon)
	\\
	&\quad + \sum_{j=0}^{N} \left( \sum_{k+l=j} klf_k(x) {f}_l(1-y) \right)\epsilon^{j-2} \ln(\epsilon)^2
+ O\big(\epsilon^{N-1}\ln(\epsilon)^2\big) \Bigg],
\\
	\Psi_4^o(x,y)&=\frac{(c_1 n_1y^{1- \frac{1}{n_1}})^2}{4}\Bigg[ \sum_{j=0}^{N} (2{G}_j(x) +3{\gamma}_j(x))\epsilon^{j-2}
	\\
	&\quad + \sum_{j=0}^{N}(j+2)jf_j(x)\epsilon^{j-2} \ln (\epsilon)
	\\
	&\quad-\epsilon\left( \sum_{j=0}^{N}jf_j(x)\epsilon^{j-1} \ln (\epsilon) + \sum_{j=0}^{N} {\gamma}_j(x)\epsilon^{j-1} \right)^3 + O\big(\epsilon^{N-1}\ln(\epsilon)^3\big) \Bigg],
	\end{align*}
where the error terms are uniform for $(x,y) \in D$ bounded away from the corners of $D$.
\end{corollary}

\bigskip
\noindent
{\bf Acknowledgement} {\it The author thanks Jonatan Lenells for bringing this problem to his attention. Support is acknowledged from the European Research Council, Grant Agreement No. 682537.}

\bibliographystyle{plain}

config: arxiv


TITLE: Dual space of $\mathbb{P}^{g-1}$
QUESTION [1 upvotes]: We know that:
If $V$ is a vector space, the collection of all linear functionals on $V$
forms a vector space, this space is the dual space of $V$. We denote
this space by $V^*$.
But I wanted to understand what it means $(\mathbb{P}^{g-1})^* $ the dual projective space of  $\mathbb{P}^{g-1}$.
$\mathbb{P}^{g-1}$ is not a vector space, so how could you understand its dual?

REPLY [2 votes]: If $\mathbb{P}^{g-1}=\mathbb{P}V$ for some vector space $V$ of dimension $g$. Then $(\mathbb{P}^{g-1})^*=(\mathbb{P}V)^*$.
An element of $(\mathbb{P}V)^*$ corresponds to a dimension one subspace of $V^*$ spanned by some linear functional $l$.
There is a one to one correspondence between dimension one subspace of $V^*$ and the corresponding vanishing set of $\mathbb{P}V$:
$\{ [x]\in \mathbb{P}V\ |\ l(x)=0 \}$, which is a codimension one linear subspace of $\mathbb{P}V$. Therefore, you can regard element of $(\mathbb{P}V)^*$ as codimension one linear subspace of $\mathbb{P}V$.


TITLE: Does $v\in L^p$ imply the derivative or integral of $v$ also in $L^p$?
QUESTION [0 upvotes]: And furthermore, will the composition $(voF)$ be also in $L^p$ if $F$ is bijective and $v\in L^p$.

REPLY [1 votes]: Regarding the title:
The answer is no. Consider $(X,\mathcal A, \mu) = ((0,1), \mathcal B \cap (0, 1), \lambda)$ and take $v(x) = \ln(x)$ then the derivative is in no $L^p$, but $v\in L^p$

Regarding the body:
The answer is no again, with the example of $(X,\mathcal A, \mu) = ((1,\infty), \mathcal B \cap (1, \infty), \lambda)$, $v(x) = \frac1{x^2}$ and $F(x) = \sqrt x$ such that
$$v\circ F(x) = \frac1{|x|} \notin L^p$$


TITLE: Why does $\int_{-\infty} ^{\infty}\frac{\sin x}{x}dx$ have a definite integral?
QUESTION [3 upvotes]: My question is kind of misleading but here is the thing. I know that $\int_{-\infty} ^{\infty}\frac{\sin x}{x}dx$ has an integral, as a lot of proofs are available here (and it is easily observable from its plot).
But what I don't get is that it has a removable singularity and hence no residue whatsoever in the upper semicircle (I am assuming a regular upper semicircular contour) and by this, its integral should be 0 [Relating to: removable singularities give integral as 0].
So what am I getting wrong here? I am sure I messed up my fundamentals somewhere. Please help me out.

PS: My inspiration for this question is a physics-based H.W. So less rigorous math would be appreciated. Thanks in advance.

REPLY [2 votes]: You have a singularity 'on' the simple closed contour, which should not be the case to apply Cauchy' s Theorem or Cauchy's Integral Formula!


TITLE: Help me with this equation
QUESTION [1 upvotes]: I'm having difficulty getting the variable $t$ out of this equation
$$\alpha x  t + \beta  y  t + \gamma  z  t + \delta = 0$$

REPLY [1 votes]: \begin{align}\alpha xt+\beta yt+\gamma zt+\delta&=0\\
t(\alpha x+\beta y+\gamma z)+\delta&=0\tag{factorise out $t$}\\
t(\alpha x+\beta y+\gamma z)&=-\delta\tag{subtract $\delta$}\\
t&=-\frac{\delta}{\alpha x+\beta y+\gamma z}\tag{divide by $\alpha x+\beta y+\gamma z$}\end{align}
This is only valid when $\alpha x+\beta y+\gamma z\neq 0$ as otherwise the fraction is undefined


TITLE: Is it possible to split a unit cube in such way?
QUESTION [2 upvotes]: Is it possible to split a three-dimensional unit cube into $42$ tetrahedrons of equal volume with nonoverlapping interiors?  The two-dimesional case is discussed in "On Dividing a Square into Triangles" by Paul Monsky, American Mathematical Monthly 77 :2 (1970), 161-164.
Addition: Is it possible for $44$ tetrahedrons instead of $42$ ones?

REPLY [3 votes]: Yes, I believe so. It is possible to split a cube into six equal-volume tetrahedra (link), generated by three diagonal planes through the cube, and all of those are the full height of the cube. Therefore that dissection also applies to a flattened version of the cube scaled along one major axis (e.g. height), so we can divide the cube into seven slices, each $1\times 1\times \frac{1}{7}$, and six-dissect each slice.


TITLE: Continuity of functions in a metric space.
QUESTION [1 upvotes]: Let $\mathbb N^{\mathbb N}$ be the set of sequences of natural numbers ($0 \notin \mathbb N$) and let for sequences $a=(n_1,n_2,\ldots)$, $b=(m_1,m_2,...)$ be:
$$d(a,b)=\begin{cases}
\frac{1}{\min(i: n_i\neq m_i)},  & \text{if $a \neq b$} \\
0, & \text{if $a=b$}  \\
\end{cases}
$$
Let $f,g \colon \mathbb N^{\mathbb N} \to \mathbb R$ be described as below:
$$f(n_1,n_2,...)=\sum_{i=1}^\infty 2^{-i}\frac{1}{n_i} $$
$$g(n_1,n_2,...)=\sup \left\{ \frac{n_i}{1+n_i}:i=1,2,\ldots \right\}$$
Are the functions $f$, $g$ continuous if we consider $\mathbb N^{\mathbb N}$ with topology induced by metric $d$?

REPLY [1 votes]: Hint. Consider $|f(a)-f(b)|$. If $d(a, b) = \frac 1 k$, how large can $|f(a)-f(b)|$ get?
Hint. Consider that $g$ doesn't care about the ordering of the sequence and $d$ does. Use this to build a counter example.


text:
\begin{document}

\begin{center}
\vskip 1cm{\LARGE\textbf{Combinatorial remarks on the cyclic sum formula for multiple zeta values}}
\vskip 1cm
\large

Shingo Saito\\
Faculty of Mathematics\\
Kyushu University\\
744, Motooka, Nishi-ku, Fukuoka, 819--0395, Japan\\
\href{mailto:ssaito@math.kyushu-u.ac.jp}{\texttt{ssaito@math.kyushu-u.ac.jp}}\\
\ \\
Tatsushi Tanaka\\
Faculty of Mathematics\\
Kyushu University\\
744, Motooka, Nishi-ku, Fukuoka, 819--0395, Japan\\
\href{mailto:t.tanaka@math.kyushu-u.ac.jp}{\texttt{t.tanaka@math.kyushu-u.ac.jp}}\\
\ \\
Noriko Wakabayashi\\
Faculty of Engineering\\
Kyushu Sangyo University\\
3--1, Matsukadai 2-chome, Higashi-ku, Fukuoka, 813--8503, Japan\\
\href{mailto:noriko@ip.kyusan-u.ac.jp}{\texttt{noriko@ip.kyusan-u.ac.jp}}\\
\end{center}

\vskip .2 in

\begin{abstract}
 The multiple zeta values are generalizations of the values of the Riemann zeta function
 at positive integers.
 They are known to satisfy a number of relations,
 among which are the cyclic sum formula.
 The cyclic sum formula can be stratified via linear operators
 defined by the second and third authors.
 We give the number of relations belonging to each stratum
 by combinatorial arguments.
\end{abstract}

\renewcommand{\theequation}{\arabic{equation}}

\section{Introduction}
The Riemann zeta function
\[
 \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}
\]
is one of the most important functions in mathematics,
and its values $\zeta(k)$ at $k\in\Z_{\ge2}$ are among the most interesting real numbers.
The \emph{multiple zeta values} (MZVs for short) are defined as generalizations of the values $\zeta(k)$:
\begin{defn}
 For $k_1,\ldots,k_l\in\Z_{\ge1}$ with $k_1\ge2$,
 the \emph{multiple zeta value} $\zeta(k_1,\ldots,k_l)$ is a real number defined by
 \[
  \zeta(k_1,\ldots,k_l)=\sum_{n_1>\cdots >n_l\ge1}\frac{1}{n_1^{k_1}\cdots n_l^{k_l}}.
 \]
\end{defn}

Little is known about the irrationality or transcendentality of MZVs;
we however know that MZVs abound with relations among themselves,
the simplest example being $\zeta(2,1)=\zeta(3)$.

The \emph{cyclic sum formula} (CSF for short), described in Section~\ref{sec:CSF},
is a class of $\Q$-linear relations among MZVs, established by Hoffman-Ohno~\cite{HO}.
Their proof appealed to partial fraction decomposition,
whereas another proof given by the second and third authors~\cite{TW} of the present article
proceeded by showing that the CSF is included in Kawashima's relation,
which is believed to be rich enough to yield all relations among MZVs.

This present paper is aimed at providing combinatorial arguments to find the ranks
of linear operators defined in \cite{TW}.
In order to facilitate access for both algebraists and combinatorists,
we strive to make the exposition as self-contained as possible.

\subsection*{Sets of multi-indices}
The study of MZVs inevitably requires frequent use of multi-indices.
We here summarize the sets of multi-indices used in this paper.
For $k,l\in\Z_{\ge1}$, put
\begin{align*}
 I_{k,l}^1&=\{(k_1,\ldots,k_l)\in\Z_{\ge1}^l\mid k_1+\cdots+k_l=k\},\\
 I_{k,l}^0&=\{(k_1,\ldots,k_l)\in I_{k,l}^1\mid k_1\ge2\},\\
 \check{I}_{k,l}^1&=\{(k_1,\ldots,k_l)\in I_{k,l}^1\mid\text{not all of $k_1,\ldots,k_l$ are $1$}\};
\end{align*}
for $k\in\Z_{\ge1}$, put
\[
 I_k^1=\bigcup_{l=1}^{\infty}I_{k,l}^1,\qquad
 I_k^0=\bigcup_{l=1}^{\infty}I_{k,l}^0,\qquad
 \check{I}_k^1=\bigcup_{l=1}^{\infty}\check{I}_{k,l}^1;
\]
put
\[
 I^1=\bigcup_{k,l=1}^{\infty}I_{k,l}^1,\qquad
 I^0=\bigcup_{k,l=1}^{\infty}I_{k,l}^0,\qquad
 \check{I}^1=\bigcup_{k,l=1}^{\infty}\check{I}_{k,l}^1.
\]

The elements of $I_{k,l}^1$ are said to have \emph{weight} $k$ and \emph{depth} $l$.
For each $\ve{k}=(k_1,\ldots,k_l)\in I^0$
the MZV $\zeta(\ve{k})=\zeta(k_1,\ldots,k_l)$ is defined.

\section{Hoffman's algebra}
In the discussion of MZVs, it is convenient to use the algebra introduced by Hoffman.
Let $\H=\Q\langle x,y \rangle$ be the noncommutative polynomial algebra over $\Q$
in two indeterminates $x$ and $y$,
and put $\H^0=\Q+x\H y$, which is a subalgebra of $\H$.
Write $z_k=x^{k-1}y\in\H$ for $k\in\Z_{\ge1}$, so that
\[
 \{1\}\cup\{z_{k_1}\cdots z_{k_l}\mid \ve{k}=(k_1,\ldots,k_l)\in I^0\}
\]
is a $\Q$-vector space basis for $\H^0$.
It follows that we may define a $\Q$-linear map $Z\colon\H^0\to\R$ by
setting $Z(1)=1$ and $Z(z_{k_1}\cdots z_{k_l})=\zeta(\ve{k})$
for $\ve{k}=(k_1,\ldots,k_l)\in I^0$.

The MZVs are known to fulfill a number of $\Q$-linear relations,
each of which corresponds to an element of $\Ker Z\subset\H^0$.
Since Goncharov~\cite{Go} conjectured that
the MZVs $\zeta(\ve{k})$ with $\ve{k}$ having different weights
are $\Q$-linearly independent,
we look at the $\Q$-vector subspaces defined by
\begin{align*}
 \H_k^0&=\spa_{\Q}\{z_{k_1}\cdots z_{k_l}\mid \ve{k}=(k_1,\ldots,k_l)\in I_k^0\}\\
 &=\{w\in\H^0\mid\text{$w$ is a homogeneous polynomial of degree $k$}\}\cup\{0\}\subset\H^0,\\
 \ZZ_k&=Z(\H_k^0)=\spa_{\Q}\{\zeta(\ve{k})\mid \ve{k}\in I_k^0\}\subset\R
\end{align*}
for each $k\in\Z_{\ge1}$.

Let $(d_k)_{k\ge1}$ be the Padovan sequence (\seqnum{A000931})
defined by $d_1=0$, $d_2=d_3=1$, and $d_k=d_{k-2}+d_{k-3}$ for $k\ge4$.
Zagier~\cite{Z} conjectured the following:
\begin{conj}[Zagier~\cite{Z}]\label{conj:Z}
 We have $\dim\ZZ_k=d_k$ for all $k\in\Z_{\ge1}$.
\end{conj}

Let $k\in\Z_{\ge2}$.
In light of the fact that $\dim\H_k^0=\#I_k^0=2^{k-2}$,
Conjecture~\ref{conj:Z} means that the MZVs must satisfy plenty of $\Q$-linear relations.
Note that Conjecture~\ref{conj:Z} is equivalent to saying that
the restriction $Z\vert_{\H_k^0}\colon\H_k^0\to\R$ of $Z$ to $\H_k^0$ has rank $d_k$,
which is also equivalent to
\[
 \dim\Ker Z\vert_{\H_k^0}=\dim(\Ker Z\cap\H_k^0)=2^{k-2}-d_k.
\]

\begin{table}[h]
 \caption{Dimensions concerning $Z\vert_{\H_k^0}$}
 \begin{center}
  \begin{tabular}{c|ccccccccc|c}
   $k$&$2$&$3$&$4$&$5$&$6$&$7$&$8$&$9$&$10$&Sequence Number\\\hline
   $2^{k-2}(=\dim\H_k^0)$&$1$&$2$&$4$&$8$&$16$&$32$&$64$&$128$&$256$&\seqnum{A000079}\\
   $d_k(\stackrel{?}{=}\rank Z\vert_{\H_k^0})$&$1$&$1$&$1$&$2$&$2$&$3$&$4$&$5$&$7$&\seqnum{A000931}\\
   $2^{k-2}-d_k(\stackrel{?}{=}\dim\Ker Z\vert_{\H_k^0})$&$0$&$1$&$3$&$6$&$14$&$29$&$60$&$123$&$249$&\seqnum{A038360}
  \end{tabular}
 \end{center}
\end{table}

Goncharov~\cite{G} and Terasoma~\cite{T} partially proved Conjecture~\ref{conj:Z}:
\begin{thm}[Goncharov~\cite{G}, Terasoma~\cite{T}]\label{thm:GT}
 We have $\dim\ZZ_k\le d_k$ for all $k\in\Z_{\ge1}$.
\end{thm}

Since their proofs of Theorem~\ref{thm:GT} resort to algebraic geometry
and fails to give concrete relations among MZVs,
it still lies at the heart of research to find sufficiently many $\Q$-linear relations.
Also, the converse inequality is far from being solved.

\section{Cyclic sum formula}\label{sec:CSF}
Numerous concrete $\Q$-linear relations among MZVs have been obtained so far,
and our focus is on the following \emph{cyclic sum formula} (CSF for short),
first proved by Hoffman-Ohno~\cite{HO}:
\begin{thm}[Cyclic sum formula]\label{thm:CSF}
 If $(k_1,\ldots,k_l)\in\check{I}^1$, then
 \[
  \sum_{j=1}^{l}\sum_{i=1}^{k_j-1}\zeta(k_j-i+1,k_{j+1},\ldots,k_l,k_1,\dots,k_{j-1},i)
  =\sum_{j=1}^{l}\zeta(k_j+1,k_{j+1},\ldots,k_l,k_1,\ldots,k_{j-1}).
 \]
\end{thm}

\begin{ex}
 The cyclic sum formula for $l=1$ and $k_1=2$ gives
 $\zeta(2,1)=\zeta(3)$.
\end{ex}

In dealing with the CSF, it is convenient to extend the indices of $k_j$ to all $j\in\Z$
by declaring $k_j=k_{j'}$ whenever $j\equiv j'\pmod{l}$.
Then the CSF can simply be written as
\[
 \sum_{j=1}^{l}\sum_{i=1}^{k_j-1}\zeta(k_j-i+1,k_{j+1},\ldots,k_{j+l-1},i)
 =\sum_{j=1}^{l}\zeta(k_j+1,k_{j+1},\ldots,k_{j+l-1}).
\]
This convention will be used tacitly throughout the paper.

In order to describe the CSF in terms of Hoffman's algebra, we write
\begin{align*}
 \check{\H}^1&=\spa_{\Q}\{z_{k_1}\cdots z_{k_l}\mid \ve{k}=(k_1,\ldots,k_l)\in\check{I}^1\}\\
 &=\spa_{\Q}\{w\in\H\mid\text{$w$ is a monomial ending with $y$ but not a power of $y$}\}\subset\H
\end{align*}
and define a $\Q$-linear map $\rho\colon\check{\H}^1\to x\H y\subset\H^0$ by setting
\[
 \rho(z_{k_1}\cdots z_{k_l})
 =\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}z_{k_j-i+1}z_{k_{j+1}}\cdots z_{k_{j+l-1}}z_i
 -\sum_{j=1}^{l}z_{k_j+1}z_{k_{j+1}}\cdots z_{k_{j+l-1}}
\]
for $\ve{k}=(k_1,\ldots,k_l)\in\check{I}^1$.
Then the CSF is equivalent to saying that
$\Im\rho\subset\Ker Z$.

For each $k\in\Z_{\ge1}$, if we put
\begin{align*}
 \check{\H}_k^1&=\spa_{\Q}\{z_{k_1}\cdots z_{k_l}\mid(k_1,\ldots,k_l)\in\check{I}_k^1\}\\
 &=\{w\in\check{\H}^1\mid\text{$w$ is a homogeneous polynomial of degree $k$}\}\cup\{0\},
\end{align*}
then $\rho$ satisfies that $\rho(\check{\H}_k^1)\subset\H_{k+1}^0$.
Therefore, it follows from Theorem~\ref{thm:CSF} that
\[
 \rho(\check{\H}_{k-1}^1)\subset\Ker Z\cap\H_k^0
\]
for all $k\in\Z_{\ge2}$.

In view of Conjecture~\ref{conj:Z},
which is equivalent to $\dim(\Ker Z\cap\H_k^0)=2^{k-2}-d_k$ for $k\in\Z_{\ge2}$,
it is natural to ask for $\dim\rho(\check{\H}_{k-1}^1)$
because it can be regarded as the number of relations given by the CSF.
The following theorem is a folklore:
\begin{thm}\label{thm:dim_CSF}
 For $k\in\Z_{\ge2}$, we have
 \[
  \dim\rho(\check{\H}_{k-1}^1)=\frac{1}{k-1}\sum_{m\mid k-1}\varphi\biggl(\frac{k-1}{m}\biggr)2^m-2,
 \]
 where $\varphi$ denotes Euler's totient function.
\end{thm}

\begin{table}[h]
 \caption{Dimensions concerning the CSF}
 \begin{center}
  \begin{tabular}{c|ccccccccc|c}
   $k$&$2$&$3$&$4$&$5$&$6$&$7$&$8$&$9$&$10$&Sequence Number\\\hline
   $2^{k-2}-d_k\bigl(\stackrel{?}{=}\dim(\Ker Z\cap\H_k^0)\bigr)$&
   $0$&$1$&$3$&$6$&$14$&$29$&$60$&$123$&$249$&\seqnum{A038360}\\
   $\dim\rho(\check{\H}_{k-1}^1)$&$0$&$1$&$2$&$4$&$6$&$12$&$18$&$34$&$58$&\seqnum{A052823}
  \end{tabular}
 \end{center}
\end{table}

\section{The operator $\rho_n$ and the statement of our main theorem}
\subsection{The operator $\rho_n$}
The second and third authors~\cite{TW} of the present article
defined linear maps $\rho_n\colon\H\to\H$
with the aim of giving an algebraic proof of the CSF
by reducing it to Kawashima's relation.
We will not elaborate on their proof here,
but focus on the stratification of the CSF provided by $\rho_n$.
Note that our usage of indices is different from that in~\cite{TW}:
what we mean by $\rho_n$ is denoted by $\rho_{n+1}$ in \cite{TW}.

Let $n\in\Z_{\ge0}$ and consider the $(n+2)$nd tensor power $\H^{\otimes(n+2)}$
of $\H$ over $\Q$.
We first make $\H^{\otimes(n+2)}$ an $\H$-bimodule by setting
\[
 a\diamond(w_1\otimes\dots\otimes w_{n+2})\diamond b
 =w_1b\otimes w_2\otimes\dots\otimes w_{n+1}\otimes aw_{n+2}
\]
for $a,b,w_1,\dots,w_{n+2}\in\H$.
Writing $z=x+y$, we define a $\Q$-linear map $\C_n\colon\H\to\H^{\otimes(n+2)}$ by
setting $\C_n(1)=0$, $\C_n(x)=x\otimes z^{\otimes n}\otimes y$,
$\C_n(y)=-x\otimes z^{\otimes n}\otimes y$, and
\[
 \C_n(ww')=\C_n(w)\diamond w'+w\diamond\C_n(w').
\]
We next define a $\Q$-linear map $M_n\colon\H^{\otimes(n+2)}\to\H$ by setting
\[
 M_n(w_1\otimes\cdots\otimes w_{n+2})=w_1\cdots w_{n+2}.
\]
Finally, set $\rho_n=M_n\circ\C_n\colon\H\to\H$.

\begin{rem}
 The recurrence relation in the definition of $\C_n$ shows that
 \[
  \C_n(w_1\cdots w_k)=
  \sum_{j=1}^{k}\bigl(w_1\cdots w_{j-1}\diamond\C_n(w_j)\diamond w_{j+1}\cdots w_k\bigr)
 \]
 for $w_1,\ldots,w_k\in\H$.
\end{rem}

\begin{prop}\label{prop:rho0=rho}
 We have $\rho_0(w)=\rho(w)$ for all $w\in\check{\H}^1$.
\end{prop}

\begin{proof}
 We may assume that $w=z_{k_1}\cdots z_{k_l}$ for some
 $\ve{k}=(k_1,\ldots,k_l)\in\check{I}^1$.
 For $k\in\Z_{\ge1}$, we have
 \begin{align*}
  \C_0(z_k)&=\C_0(x^{k-1}y)
  =\sum_{i=1}^{k-1}\bigl(x^{i-1}\diamond\C_0(x)\diamond x^{k-i-1}y\bigr)+x^{k-1}\diamond\C_0(y)\\
  &=\sum_{i=1}^{k-1}\bigl(x^{i-1}\diamond(x\otimes y)\diamond x^{k-i-1}y\bigr)
    +x^{k-1}\diamond(-x\otimes y)\\
  &=\sum_{i=1}^{k-1}(x^{k-i}y\otimes x^{i-1}y)-x\otimes x^{k-1}y
   =\sum_{i=1}^{k-1}(z_{k-i+1}\otimes z_i)-x\otimes z_k.
 \end{align*}
 It follows that
 \begin{align*}
  \C_0(w)
  &=\C_0(z_{k_1}\cdots z_{k_l})
   =\sum_{j=1}^{l}\bigl(z_{k_1}\cdots z_{k_{j-1}}\diamond\C_0(z_{k_j})\diamond z_{k_{j+1}}\cdots z_{k_l}\bigr)\\
  &=\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}
    \bigl(z_{k_1}\cdots z_{k_{j-1}}\diamond(z_{k_j-i+1}\otimes z_i)\diamond z_{k_{j+1}}\cdots z_{k_l}\bigr)\\
  &\qquad-\sum_{j=1}^{l}
   \bigl(z_{k_1}\cdots z_{k_{j-1}}\diamond(x\otimes z_{k_j})\diamond z_{k_{j+1}}\cdots z_{k_l}\bigr)\\
  &=\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}
    (z_{k_j-i+1}z_{k_{j+1}}\cdots z_{k_l}\otimes z_{k_1}\cdots z_{k_{j-1}}z_{i})
  -\sum_{j=1}^{l}(xz_{k_{j+1}}\cdots z_{k_l}\otimes z_{k_1}\cdots z_{k_j}),
 \end{align*}
 and so
 \begin{align*}
  \rho_0(w)
  &=M_0\bigl(\C_0(w)\bigr)\\
  &=\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}
   z_{k_j-i+1}z_{k_{j+1}}\cdots z_{k_l}z_{k_1}\cdots z_{k_{j-1}}z_i
   -\sum_{j=1}^{l}xz_{k_{j+1}}\cdots z_{k_l}z_{k_1}\cdots z_{k_j}\\
  &=\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}
   z_{k_j-i+1}z_{k_{j+1}}\cdots z_{k_{j+l-1}}z_i
   -\sum_{j=1}^{l}xz_{k_{j+1}}\cdots z_{k_{j+l}},
 \end{align*}
 which is equal to $\rho(w)$ because
 \[
  \sum_{j=1}^{l}xz_{k_{j+1}}\cdots z_{k_{j+l}}
  =\sum_{j=1}^{l}xz_{k_j}\cdots z_{k_{j+l-1}}
  =\sum_{j=1}^{l}z_{k_j+1}z_{k_{j+1}}\cdots z_{k_{j+l-1}}.\qedhere
 \]
\end{proof}

\subsection{Properties of $\rho_n$}
\begin{defn}
 Define $\sgn\colon\{x,y,z\}\to\{1,-1,0\}$ by
 \[
  \sgn(u)=
  \begin{cases}
   1&\text{if $u=x$};\\
   -1&\text{if $u=y$};\\
   0&\text{if $u=z$},
  \end{cases}
 \]
 so that $\C_n(u)=\sgn(u)(x\otimes z^{\otimes n}\otimes y)$
 for $u\in\{x,y,z\}$.
\end{defn}

\begin{lem}\label{lem:rho_n_prod_xyz}
 If $w=u_1\cdots u_k$, where $u_1,\ldots,u_k\in\{x,y,z\}$, then
 \[
  \rho_n(w)=\sum_{j=1}^{k}\sgn(u_j)xu_{j+1}\cdots u_kz^nu_1\cdots u_{j-1}y.
 \]
\end{lem}

\begin{proof}
 We have
 \begin{align*}
  \C_n(w)&=\C_n(u_1\cdots u_k)
  =\sum_{j=1}^{k}\bigl(u_1\cdots u_{j-1}\diamond\C_n(u_j)\diamond u_{j+1}\cdots u_k\bigr)\\
  &=\sum_{j=1}^{k}\sgn(u_j)
    \bigl(u_1\cdots u_{j-1}\diamond(x\otimes z^{\otimes n}\otimes y)\diamond u_{j+1}\cdots u_k\bigr)\\
  &=\sum_{j=1}^{k}\sgn(u_j)
    (xu_{j+1}\cdots u_k\otimes z^{\otimes n}\otimes u_1\cdots u_{j-1}y),
 \end{align*}
 and so
 \[
  \rho_n(w)=M_n\bigl(\C_n(w)\bigr)
  =\sum_{j=1}^{k}\sgn(u_j)xu_{j+1}\cdots u_kz^nu_1\cdots u_{j-1}y.\qedhere
 \]
\end{proof}

\begin{prop}\label{prop:rho_n_n+1}
 We have
 \[
  \rho_{n+1}(w)=\rho_n(zw)
 \]
 for all $w\in\H$.
\end{prop}

\begin{proof}
 We may assume that $w=u_1\cdots u_k$ for some $u_1,\ldots,u_k\in\{x,y\}$.
 Then Lemma~\ref{lem:rho_n_prod_xyz} shows that
 \begin{align*}
  \rho_n(zw)
  &=\sgn(z)xu_1\cdots u_kz^ny+\sum_{j=1}^{k}\sgn(u_j)xu_{j+1}\cdots u_kz^nzu_1\cdots u_{j-1}y\\
  &=\sum_{j=1}^{k}\sgn(u_j)xu_{j+1}\cdots u_kz^{n+1}u_1\cdots u_{j-1}y
   =\rho_{n+1}(w).\qedhere
 \end{align*}
\end{proof}

\begin{cor}\label{cor:stratification}
 We have
 \[
  \{0\}=\rho_{k-2}(\check{\H}_1^1)\subset\rho_{k-3}(\check{\H}_2^1)
  \subset\cdots\subset\rho_0(\check{\H}_{k-1}^1)=\rho(\check{\H}_{k-1}^1).
 \]
 for $k\in\Z_{\ge2}$.
\end{cor}

\begin{proof}
 For each $n\in\{0,\ldots,k-3\}$,
 we have $\rho_{n+1}(\check{\H}_{k-n-2}^1)\subset\rho_n(\check{\H}_{k-n-1}^1)$
 by Proposition~\ref{prop:rho_n_n+1} and by the fact that
 if $w\in\check{\H}_{k-n-2}^1$, then $zw\in\check{\H}_{k-n-1}^1$.
\end{proof}

\subsection{Statement of our main theorem}
Corollary~\ref{cor:stratification} can be interpreted as stratifying
the $\Q$-linear relations provided by the CSF.
Since Theorem~\ref{thm:dim_CSF} tells us the dimension of the whole space,
we may well wish to find the dimensions of the subspaces $\rho_n(\check{\H}_k^1)$ in general.
Our main theorem (Theorem~\ref{thm:main}) provides a complete solution to this problem,
and it uses the following generalization of the Lucas sequence:

\begin{defn}
 For $n\in\Z_{\ge1}$, the \emph{$n$-step Lucas sequence} $(L_m^{n})_{m\ge1}$
 is defined by
 \[
  L_m^{n}=
  \begin{cases}
   2^m-1,&\text{for $m=1,\ldots,n$};\\
   L_{m-1}^{n}+\cdots+L_{m-n}^{n},&\text{for $m\ge n+1$}.
  \end{cases}
 \]
 We adopt the convention that $L_m^0=0$ for all $m\in\Z_{\ge1}$.
\end{defn}

\begin{table}[h]
 \caption{$n$-step Lucas sequences}
 \begin{center}
  \begin{tabular}{c|ccccccc|c}
   $n$&$1$&$2$&$3$&$4$&$5$&$6$&$7$&Sequence Number\\\hline
   $L_m^{1}$&$1$&$1$&$1$&$1$&$1$&$1$&$1$&\seqnum{A000012}\\
   $L_m^{2}$&$1$&$3$&$4$&$7$&$11$&$18$&$29$&\seqnum{A000032}, \seqnum{A000204}\\
   $L_m^{3}$&$1$&$3$&$7$&$11$&$21$&$39$&$71$&\seqnum{A001644}\\
   $L_m^{4}$&$1$&$3$&$7$&$15$&$26$&$51$&$99$&\seqnum{A073817}, \seqnum{A001648}\\
   $L_m^{5}$&$1$&$3$&$7$&$15$&$31$&$57$&$113$&\seqnum{A074048}, \seqnum{A023424}
  \end{tabular}
 \end{center}
\end{table}

\begin{lem}\label{lem:n-step}
 Let $n\in\Z_{\ge0}$.
 \begin{enumerate}
  \item We have $L_{n+1}^{n}=2^{n+1}-n-2$.
  \item We have $L_m^{n}=2L_{m-1}^{n}-L_{m-n-1}^{n}$ for $m\ge n+2$.
 \end{enumerate}
\end{lem}

\begin{proof}
 Easy.
\end{proof}

\begin{thm}[Main Theorem]\label{thm:main}
 If $n\in\Z_{\ge0}$ and $k\in\Z_{\ge1}$, then
 \[
  \dim\rho_n(\check{\H}_{k}^1)
  =\frac{1}{n+k}\sum_{m\mid n+k}\varphi\biggl(\frac{n+k}{m}\biggr)(2^m-L_m^n)-2.
 \]
\end{thm}

\begin{rem}
 We may easily see that this theorem is a generalization of Theorem~\ref{thm:dim_CSF}.
\end{rem}

\section{Proof of our main theorem}
\subsection{Cyclic equivalence and $\rho_0$}
\begin{defn}
 Let $l\in\Z_{\ge1}$.
 Set $X_l=\{y,z\}^l$, and let the cyclic group $\Z/l\Z$ of order $l$
 act on $X_l$ by cyclic shifts, i.e.,
 \[
  j(u_1,\ldots,u_l)=(u_{j+1},\ldots,u_{j+l})
 \]
 for $j\in\Z/l\Z$ and $(u_1,\ldots,u_l)\in X_l$.
 The equivalence relation on $X_l$ induced by the action is called the \emph{cyclic equivalence}
 and denoted by $\sim$.
 Put $Y_l=X_l/{\sim}$.
\end{defn}

\begin{prop}
 For each $l\in\Z_{\ge1}$,
 we may define $\tilde{\rho}_0\colon Y_l\to\H$ by setting
 \[
  \tilde{\rho}_0\bigl([(u_1,\ldots,u_l)]\bigr)=\rho_0(u_1\cdots u_l)
 \]
 for $(u_1,\ldots,u_l)\in X_l$.
\end{prop}

\begin{proof}
 Lemma~\ref{lem:rho_n_prod_xyz} shows that if $(u_1,\ldots,u_l)\in X_l$, then
 \begin{align*}
  \rho_0(u_1\cdots u_l)
  &=\sum_{j=1}^{l}\sgn(u_j)xu_{j+1}\cdots u_lu_1\cdots u_{j-1}y\\
  &=\sum_{\substack{j=1,\ldots,l\\(v_1,\ldots,v_l)=j(u_1,\ldots,u_l)}}\sgn(v_1)xv_2\cdots v_ly.
 \end{align*}
 It follows that $\rho_0(u_1\cdots u_l)$ depends only on
 the equivalence class $[(u_1,\ldots,u_l)]\in Y_l$,
 as required.
\end{proof}

\begin{ex}
 For $l=4$, we have
 \begin{align*}
  \tilde{\rho}_0\bigl([(z,z,z,z)]\bigr)&=0,\\
  \tilde{\rho}_0\bigl([(z,z,z,y)]\bigr)&=-xzzzy,\\
  \tilde{\rho}_0\bigl([(z,z,y,y)]\bigr)&=-x(yzz+zzy)y,\\
  \tilde{\rho}_0\bigl([(z,y,z,y)]\bigr)&=-x(zyz+zyz)y,\\
  \tilde{\rho}_0\bigl([(z,y,y,y)]\bigr)&=-x(yyz+yzy+zyy)y,\\
  \tilde{\rho}_0\bigl([(y,y,y,y)]\bigr)&=-x(yyy+yyy+yyy+yyy)y.
 \end{align*}
\end{ex}

\begin{prop}\label{prop:rho_U_lin_indep}
 The family
 $\bigl\{\tilde{\rho}_0(U)\bigm|U\in Y_l\setminus\{[(\underbrace{z,\ldots,z}_{l})]\}\bigr\}$
 is $\Q$-linearly independent.
\end{prop}

\begin{proof}
 This is because the values $\tilde{\rho}_0(U)$
 for different equivalence classes $U\in Y_l\setminus\{[(z,\ldots,z)]\}$
 are nonzero and consist of different monomials.
\end{proof}

\subsection{Relationship between $\rho_n$ and $\rho_0$}
\begin{defn}
 For $l\in\Z_{\ge1}$ and $n\in\{0,\ldots,l\}$,
 we write $X_{l,n}$ for the subset of $X_l$ consisting of
 all $(u_1,\ldots,u_l)\in X_l$ that contain at least $n$ consecutive $z$'s
 when written cyclically.
\end{defn}

\begin{ex}
 For $l=4$, we have
 \begin{align*}
  X_{4,0}&=\{(z,z,z,z),\\
   &\phantom{{}=\{}(z,z,z,y),(z,z,y,z),(z,y,z,z),(y,z,z,z),\\
   &\phantom{{}=\{}(z,z,y,y),(z,y,y,z),(y,y,z,z),(y,z,z,y),\\
   &\phantom{{}=\{}(z,y,z,y),(y,z,y,z),\\
   &\phantom{{}=\{}(z,y,y,y),(y,y,y,z),(y,y,z,y),(y,z,y,y),\\
   &\phantom{{}=\{}(y,y,y,y)\}\\
   &=[(z,z,z,z)]\cup[(z,z,z,y)]\cup[(z,z,y,y)]\cup[(z,y,z,y)]\cup[(z,y,y,y)]\cup[(y,y,y,y)]\\
   &=X_4,\\
  X_{4,1}&=\{(z,z,z,z),\\
   &\phantom{{}=\{}(z,z,z,y),(z,z,y,z),(z,y,z,z),(y,z,z,z),\\
   &\phantom{{}=\{}(z,z,y,y),(z,y,y,z),(y,y,z,z),(y,z,z,y),\\
   &\phantom{{}=\{}(z,y,z,y),(y,z,y,z),\\
   &\phantom{{}=\{}(z,y,y,y),(y,y,y,z),(y,y,z,y),(y,z,y,y)\}\\
   &=[(z,z,z,z)]\cup[(z,z,z,y)]\cup[(z,z,y,y)]\cup[(z,y,z,y)]\cup[(z,y,y,y)],\\
  X_{4,2}&=\{(z,z,z,z),\\
   &\phantom{{}=\{}(z,z,z,y),(z,z,y,z),(z,y,z,z),(y,z,z,z),\\
   &\phantom{{}=\{}(z,z,y,y),(z,y,y,z),(y,y,z,z),(y,z,z,y)\}\\
   &=[(z,z,z,z)]\cup[(z,z,z,y)]\cup[(z,z,y,y)],\\
  X_{4,3}&=\{(z,z,z,z),\\
   &\phantom{{}=\{}(z,z,z,y),(z,z,y,z),(z,y,z,z),(y,z,z,z)\}\\
   &=[(z,z,z,z)]\cup[(z,z,z,y)],\\
  X_{4,4}&=\{(z,z,z,z)\}\\
   &=[(z,z,z,z)].
 \end{align*}
\end{ex}

\begin{rem}
 Each $X_{l,n}$ is invariant under the action of $\Z/l\Z$,
 which allows us to make the following definition.
\end{rem}

\begin{defn}
 For $l\in\Z_{\ge1}$ and $n\in\{0,\ldots,l\}$, we write
 \[
  Y_{l,n}=X_{l,n}/{\sim}
  =\bigl\{[(\underbrace{z,\ldots,z}_{n},u_{n+1},\ldots,u_l)]\in Y_l
   \bigm|u_{n+1},\ldots,u_l\in\{y,z\}\bigr\}.
 \]
\end{defn}

\begin{prop}\label{prop:dim_and_Y}
 For $n\in\Z_{\ge0}$ and $k\in\Z_{\ge1}$, we have
 \[
  \dim\rho_n(\check{\H}_k^1)=\#Y_{n+k,n}-2.
 \]
\end{prop}

\begin{proof}
 Since
 \begin{align*}
  \check{\H}_k^1
  &=\spa_{\Q}\{z_{k_1}\cdots z_{k_l}\mid(k_1,\ldots,k_l)\in\check{I}_k^1\}\\
  &=\spa_{\Q}\bigl\{u_1\cdots u_{k-1}y\bigm|
    (u_1,\ldots,u_{k-1})\in\{x,y\}^{k-1}\setminus\{(\underbrace{y,\ldots,y}_{k-1})\}\bigr\}\\
  &=\spa_{\Q}\bigl\{u_1\cdots u_{k-1}y-y^k\bigm|
    (u_1,\ldots,u_{k-1})\in\{y,z\}^{k-1}\bigr\},
 \end{align*}
 Proposition~\ref{prop:rho_n_n+1} shows that
 \begin{align*}
  \rho_n(\check{\H}_k^1)
  &=\spa_{\Q}\bigl\{\rho_n(u_1\cdots u_{k-1}y)-\rho_n(y^k)\bigm|
    (u_1,\ldots,u_{k-1})\in\{y,z\}^{k-1}\bigr\}\\
  &=\spa_{\Q}\bigl\{\rho_0(z^nu_1\cdots u_{k-1}y)-\rho_0(z^ny^k)\bigm|
    (u_1,\ldots,u_{k-1})\in\{y,z\}^{k-1}\bigr\}\\
  &=\spa_{\Q}\bigl\{\tilde{\rho}_0(U)-
    \tilde{\rho}_0\bigl([(\underbrace{z,\ldots,z}_{n},\underbrace{y,\ldots,y}_k)]\bigr)\bigm|
    U\in Y_{n+k,n}\setminus\{[(\underbrace{z,\ldots,z}_{n+k})]\}\bigr\},
 \end{align*}
 which implies that $\dim\rho_n(\check{\H}_k^1)=\#Y_{n+k,n}-2$
 because of Proposition~\ref{prop:rho_U_lin_indep}.
\end{proof}

\subsection{Calculation of $\#Y_{l,n}$}
Proposition~\ref{prop:dim_and_Y} reduces our main theorem (Theorem~\ref{thm:main})
to the following proposition:
\begin{prop}\label{prop:Yln}
 For $l\in\Z_{\ge1}$ and $n\in\{0,\ldots,l\}$, we have
 \[
  \#Y_{l,n}
  =\frac{1}{l}\sum_{m\mid l}
   \varphi\biggl(\frac{l}{m}\biggr)(2^m-L_m^n).
 \]
\end{prop}

We first invoke the Cauchy-Frobenius lemma:

\begin{prop}[Cauchy-Frobenius lemma]
 If a finite group $G$ acts on a finite set $X$, then we have
 \[
  \#(X/G)
  =\frac{1}{\#G}\sum_{g\in G}\#\{x\in X\mid gx=x\}.
 \]
\end{prop}

\begin{lem}\label{lem:Yln_1}
 For $l\in\Z_{\ge1}$ and $n\in\{0,\ldots,l\}$, we have
 \[
  \#Y_{l,n}
  =\frac{1}{l}\sum_{m\mid l}
   \varphi\biggl(\frac{l}{m}\biggr)
   \#\{\ve{u}\in X_{l,n}\mid m\ve{u}=\ve{u}\}.
 \]
\end{lem}

\begin{proof}
 Applying the Cauchy-Frobenius lemma with $G=\Z/l\Z$ and $X=X_{l,n}$ gives
 \begin{align*}
  \#Y_{l,n}
  &=\frac{1}{l}\sum_{j\in\Z/l\Z}
    \#\{\ve{u}\in X_{l,n}\mid j\ve{u}=\ve{u}\}\\
  &=\frac{1}{l}\sum_{j\in\Z/l\Z}
    \#\{\ve{u}\in X_{l,n}\mid \gcd(j,l)\ve{u}=\ve{u}\}\\
  &=\frac{1}{l}\sum_{m\mid l}
    \bigl(\#\{j\in\Z/l\Z\mid\gcd(j,l)=m\}\cdot
    \#\{\ve{u}\in X_{l,n}\mid m\ve{u}=\ve{u}\}\bigr)\\
  &=\frac{1}{l}\sum_{m\mid l}
    \varphi\biggl(\frac{l}{m}\biggr)
    \#\{\ve{u}\in X_{l,n}\mid m\ve{u}=\ve{u}\}.\qedhere
 \end{align*}
\end{proof}

\begin{lem}\label{lem:Yln_2}
 If $l\in\Z_{\ge1}$, $n\in\{0,\ldots,l\}$, and $m$ is a positive divisor of $l$,
 then
 \[
  \#\{\ve{u}\in X_{l,n}\mid m\ve{u}=\ve{u}\}=2^m-L_m^n.
 \]
\end{lem}

\begin{proof}
 The map $f_{m,l}\colon X_m\to X_l$ defined by
 \[
  f_{m,l}(\ve{v})=(\underbrace{\ve{v},\ldots,\ve{v}}_{l/m})
 \]
 is injective and has image $\{\ve{u}\in X_l\mid m\ve{u}=\ve{u}\}$.
 Therefore it suffices to show that
 \[
  \#f_{m,l}^{-1}(X_{l,n})=2^m-L_m^n.
 \]
 Observe that for each $(m,n)\in\Z_{\ge1}\times\Z_{\ge0}$,
 the set $f_{m,l}^{-1}(X_{l,n})$ is the same for all multiples $l$ of $m$
 with $l\ge n$.
 Therefore, we may put $Z_{m,n}=X_m\setminus f_{m,l}^{-1}(X_{l,n})=f_{m,l}^{-1}(X_{l,n}^c)$,
 aiming to show that $\#Z_{m,n}=L_m^n$ for all $m\in\Z_{\ge1}$ and $n\in\Z_{\ge0}$.

 If $m\le n$, then
 \[
  \#Z_{m,n}
  =\#\{\ve{v}\in X_m\mid\text{at least one component of $\ve{v}$ is $y$}\}
  =2^m-1=L_m^n.
 \]
 If $m=n+1$, then
 \begin{align*}
  \#Z_{m,n}&=\#\{\ve{v}\in X_m\mid\text{at least two components of $\ve{v}$ are $y$}\}\\
  &=2^m-m-1=2^{n+1}-n-2=L_m^n
 \end{align*}
 by Lemma~\ref{lem:n-step} (1).
 Suppose that $m\ge n+2$.
 If we put
 \[
  Z_{m,n,i}=\{(v_1,\ldots,v_m)\in Z_{m,n}\mid v_1=\cdots=v_{i-1}=z,\;v_i=y\}
 \]
 for $i=1,\ldots,n$, then $\{Z_{m,n,i}\mid i=1,\ldots,n\}$
 is a partition of $Z_{m,n}$.
 Set
 \begin{align*}
  Z_{m,n,i}^y&=\{(v_1,\ldots,v_m)\in Z_{m,n,i}\mid v_{i+1}=y\},\\
  Z_{m,n,i}^z&=\{(v_1,\ldots,v_m)\in Z_{m,n,i}\mid v_{i+1}=z\}.
 \end{align*}
 Then removing the $(i+1)$st component gives a bijection from $Z_{m,n,i}^y$ to $Z_{m-1,n,i}$
 and an injection from $Z_{m,n,i}^z$ to $Z_{m-1,n,i}$ with image $Z_{m-1,n,i}\setminus Z_{m-1,n,i}^{z\cdots zy}$,
 where
 \[
  Z_{m-1,n,i}^{z\cdots zy}=\{(v_1,\ldots,v_{m-1})\in Z_{m-1,n,i}\mid v_{i+1}=\cdots=v_{i+n-1}=z,\;v_{i+n}=y\}.
 \]
 Moreover removing components from the $(i+1)$st to the $(i+n)$th gives a bijection from
 $Z_{m-1,n,i}^{z\cdots zy}$ to $Z_{m-n-1,n,i}$.
 It follows that
 \begin{align*}
  \#Z_{m,n,i}&=\#Z_{m,n,i}^y+\#Z_{m,n,i}^z
  =\#Z_{m-1,n,i}+\#(Z_{m-1,n,i}\setminus Z_{m-1,n,i}^{z\cdots zy})\\
  &=2\#Z_{m-1,n,i}-\#Z_{m-n-1,n,i}.
 \end{align*}
 Summing up for $i=1,\ldots,n$ gives $\#Z_{m,n}=2\#Z_{m-1,n}-\#Z_{m-n-1,n}$.
 Hence the lemma follows from Lemma~\ref{lem:n-step} (2).
\end{proof}

Lemmas~\ref{lem:Yln_1} and \ref{lem:Yln_2} imply Proposition~\ref{prop:Yln},
thereby establishing our main theorem.

\section{Multiple zeta-star values}
\subsection{Multiple zeta-star values and Hoffman's algebra}
The \emph{multiple zeta-star values} (MZSVs for short) are defined as
the MZVs with equality allowed in the index of summation:
\begin{defn}
 For $\ve{k}=(k_1,\ldots,k_l)\in I^0$, the \emph{multiple zeta-star value}
 $\zeta^{\star}(k_1,\ldots,k_l)$ is a real number defined by
 \[
  \zeta^{\star}(\ve{k})=\zeta^{\star}(k_1,\ldots,k_l)
  =\sum_{n_1\ge\cdots\ge n_l\ge1}\frac{1}{n_1^{k_1}\cdots n_l^{k_l}}.
 \]
\end{defn}

\begin{ex}
 We have
 \begin{align*}
  \zeta^{\star}(4,2,1)
  &=\sum_{n_1\ge n_2\ge n_3\ge1}\frac{1}{n_1^4n_2^2n_3}\\
  &=\sum_{n_1>n_2>n_3\ge1}\frac{1}{n_1^4n_2^2n_3}
    +\sum_{n_1=n_2>n_3\ge1}\frac{1}{n_1^4n_2^2n_3}
    +\sum_{n_1>n_2=n_3\ge1}\frac{1}{n_1^4n_2^2n_3}
    +\sum_{n_1=n_2=n_3\ge1}\frac{1}{n_1^4n_2^2n_3}\\
  &=\sum_{n_1>n_2>n_3\ge1}\frac{1}{n_1^4n_2^2n_3}
    +\sum_{n_1>n_3\ge1}\frac{1}{n_1^6n_3}
    +\sum_{n_1>n_2\ge1}\frac{1}{n_1^4n_2^3}
    +\sum_{n_1\ge1}\frac{1}{n_1^7}\\
  &=\zeta(4,2,1)+\zeta(6,1)+\zeta(4,3)+\zeta(7).
 \end{align*}
\end{ex}

As the above example indicates,
each MZSV can be expressed as a $\Z$-linear combination of MZVs.
Hoffman's algebra is useful for describing this relationship between MZVs and MZSVs.

\begin{defn}
 Define a $\Q$-linear map $\bar{Z}\colon\H^0\to\R$ by
 setting $\bar{Z}(1)=1$ and $\bar{Z}(z_{k_1}\cdots z_{k_l})=\zeta^{\star}(\ve{k})$
 for $\ve{k}=(k_1,\ldots,k_l)\in I^0$.
\end{defn}

\begin{defn}
 Let $\gamma$ denote the algebra automorphism on $\H$
 satisfying $\gamma(x)=x$ and $\gamma(y)=z$.
 Define a $\Q$-linear transformation $d$ on $\Q+\H y$ by setting
 $d(1)=1$ and $d(wy)=\gamma(w)y$ for $w\in\H$.
\end{defn}

\begin{prop}
 We have $\bar{Z}=Z\circ d\colon\H^0\to\R$.
\end{prop}

\begin{proof}
 Easy and well known.
\end{proof}

\subsection{Cyclic sum formula for multiple zeta-star values}
Ohno and the third author~\cite{OW} proved the following analog of the CSF for MZSVs:
\begin{thm}[Cyclic sum formula for multiple zeta-star values]\label{thm:CSF_MZSVs}
 If $(k_1,\ldots,k_l)\in\check{I}^1$, then
 \[
  \sum_{j=1}^{l}\sum_{i=1}^{k_j-1}\zeta^{\star}(k_j-i+1,k_{j+1},\ldots,k_{j+l-1},i)
  =(k_1+\cdots+k_l)\zeta(k_1+\cdots+k_l+1).
 \]
\end{thm}

The second and third authors~\cite{TW}
defined operators $\bar{\rho}_n$ for MZSVs as well as $\rho_n$ for MZVs:
\begin{defn}
 Let $n\in\Z_{\ge0}$.
 Define a $\Q$-linear map $\bar{\C}_n\colon\H\to\H^{\otimes(n+2)}$
 by setting $\bar{\C}_n(1)=0$, $\bar{\C}_n(x)=x\otimes y^{\otimes(n+1)}$,
 $\bar{\C}_n(y)=-x\otimes y^{\otimes(n+1)}$, and
 \[
  \bar{\C}_n(ww')=\bar{\C}_n(w)\diamond\gamma^{-1}(w')+\gamma^{-1}(w)\diamond\bar{\C}_n(w').
 \]
 Write $\bar{\rho}_n=M_n\circ\bar{\C}_n$.
\end{defn}

\begin{prop}\label{prop:diagram}
 We have the following commutative diagram:
 \[
  \xymatrix{
   &x\H\otimes\H^{\otimes n}\otimes\H y
    \ar[dd]^{\gamma\otimes\gamma^{\otimes n}\otimes d}_{\cong}
    \ar[r]_-{M_n}&\H^0\ar[dd]_{d}^{\cong}\ar[rd]_{\bar{Z}}&\\
   \H\ar[ru]_{\bar{\C}_n}\ar[rd]^{\C_n}\ar@/^4pc/[rru]^{\bar{\rho}_n}\ar@/_4pc/[rrd]_{\rho_n}&&&\R.\\
   &x\H\otimes\H^{\otimes n}\otimes\H y\ar[r]^-{M_n}&\H^0\ar[ru]^{Z}&
  }
 \]
\end{prop}

\begin{proof}
 Straightforward.
\end{proof}

\begin{prop}\label{prop:tedious}
 If $k_1,\ldots,k_l\in\Z_{\ge1}$, then
 \[
  \sum_{j=1}^{l}(\bar{\rho}_0\circ\alpha\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})
  =\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}z_{k_j-i+1}z_{k_{j+1}}\cdots z_{k_{j+l-1}}z_i
   -(k_1+\cdots+k_l)z_{k_1+\cdots+k_l+1}.
 \]
 Here $\alpha\colon\H y\to\H y$ denotes the $\Q$-linear map
 representing the division by depth, i.e.,
 \[
  \alpha(z_{k_1'}\cdots z_{k_{l'}'})=\frac{1}{l'}z_{k_1'}\cdots z_{k_{l'}'}
 \]
 for all monomials $z_{k_1'}\cdots z_{k_{l'}'}\in\H y$.
\end{prop}

\begin{proof}
 For simplicity, let $A$ and $B$ respectively denote
 the left- and right-hand sides of the desired identity.
 By the injectivity of $d$, it suffices to show that $d(A)=d(B)$;
 setting $A'=d(A)$ and $B'=d(B)$, we have
 \begin{align*}
  A'&=\sum_{j=1}^{l}(d\circ\bar{\rho}_0\circ\alpha\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})
     =\sum_{j=1}^{l}(\rho_0\circ\alpha\circ d)(z_{k_j}\cdots z_{k_{j+l-1}}),\\
  B'&=\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}d(z_{k_j-i+1}z_{k_{j+1}}\cdots z_{k_{j+l-1}}z_i)
   -(k_1+\cdots+k_l)z_{k_1+\cdots+k_l+1}.
 \end{align*}

 For $a\in\Z_{\ge1}$, let $\delta_a\colon\H y\to\H y$
 be the $\Q$-linear map that extracts the depth-$a$ part, i.e.,
 \[
  \delta_a(z_{k_1'}\cdots z_{k_{l'}'})=
  \begin{cases}
   z_{k_1'}\cdots z_{k_{l'}'}&\text{if $l'=a$};\\
   0&\text{otherwise}
  \end{cases}
 \]
 for all monomials $z_{k_1'}\cdots z_{k_{l'}'}\in\H y$.
 Then it is enough to prove that $\delta_a(A')=\delta_a(B')$ for $a=1,\ldots,l+1$.
 Note that
 \begin{align*}
  (\delta_a\circ\rho_0)(w)&=
  \begin{cases}
   (\delta_a\circ\rho_0\circ\delta_a)(w)+(\delta_a\circ\rho_0\circ\delta_{a-1})(w)&\text{if $a\ge2$};\\
   (\delta_a\circ\rho_0\circ\delta_a)(w)&\text{if $a=1$},
  \end{cases}\\
  (\delta_a\circ\alpha)(w)&=\frac{1}{a}\delta_a(w)
 \end{align*}
 for all $w\in\H y$.

 For $a=1$, we have
 \begin{align*}
  \delta_1(A')&=\sum_{j=1}^{l}(\delta_1\circ\rho_0\circ\alpha\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})
  =\sum_{j=1}^{l}(\delta_1\circ\rho_0\circ\delta_1\circ\alpha\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})\\
  &=\sum_{j=1}^{l}(\delta_1\circ\rho_0\circ\delta_1\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})
   =\sum_{j=1}^{l}(\delta_1\circ\rho_0)(z_{k_j+\cdots+k_{j+l-1}})\\
  &=\sum_{j=1}^{l}(\delta_1\circ\rho_0)(z_{k_1+\cdots+k_l})
   =\sum_{j=1}^{l}(-z_{k_1+\cdots+k_l+1})=-lz_{k_1+\cdots+k_l+1},\\
  \delta_1(B')
  &=\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}(\delta_1\circ d)(z_{k_j-i+1}z_{k_{j+1}}\cdots z_{k_{j+l-1}}z_i)
   -(k_1+\cdots+k_l)z_{k_1+\cdots+k_l+1}\\
  &=\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}z_{(k_j-i+1)+k_{j+1}+\cdots+k_{j+l-1}+i}
   -(k_1+\cdots+k_l)z_{k_1+\cdots+k_l+1}\\
  &=\sum_{j=1}^{l}(k_j-1)z_{k_1+\cdots+k_l+1}-(k_1+\cdots+k_l)z_{k_1+\cdots+k_l+1}\\
  &=-lz_{k_1+\cdots+k_l+1},
 \end{align*}
 as required.

 For $a=l+1$, we have
 \begin{align*}
  \delta_{l+1}(A')&=\sum_{j=1}^{l}(\delta_{l+1}\circ\rho_0\circ\alpha\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})\\
  &=\sum_{j=1}^{l}(\delta_{l+1}\circ\rho_0\circ\delta_{l+1}\circ\alpha\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})
   +\sum_{j=1}^{l}(\delta_{l+1}\circ\rho_0\circ\delta_l\circ\alpha\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})\\
  &=\frac{1}{l+1}\sum_{j=1}^{l}(\delta_{l+1}\circ\rho_0\circ\delta_{l+1}\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})
   +\frac{1}{l}\sum_{j=1}^{l}(\delta_{l+1}\circ\rho_0\circ\delta_l\circ d)(z_{k_j}\cdots z_{k_{j+l-1}})\\
  &=\frac{1}{l}\sum_{j=1}^{l}(\delta_{l+1}\circ\rho_0)(z_{k_j}\cdots z_{k_{j+l-1}})
  =\frac{1}{l}\sum_{j=1}^{l}\sum_{j'=1}^{l}\sum_{i=1}^{k_{j+j'-1}-1}
   z_{k_{j+j'-1}-i+1}z_{k_{j+j'}}\cdots z_{k_{j+j'+l-2}}z_i\\
  &=\sum_{j=1}^{l}\sum_{i=1}^{k_j-1}z_{k_j-i+1}z_{k_{j+1}}\cdots z_{k_{j+l-1}}z_i
   =\delta_{l+1}(B'),
 \end{align*}
 as required.

 Now let $2\le a\le l$.
 We first compute $\delta_a(A')$.
 It is easy to see that $\delta_a(A')=P+Q$, where
 \begin{align*}
  P&=\frac{1}{a}\sum_{j=1}^{l}
     (\delta_a\circ\rho_0\circ\delta_a\circ d)(z_{k_j}\cdots z_{k_{j+l-1}}),\\
  Q&=\frac{1}{a-1}\sum_{j=1}^{l}
     (\delta_a\circ\rho_0\circ\delta_{a-1}\circ d)(z_{k_j}\cdots z_{k_{j+l-1}}).
 \end{align*}
 For $b\in\Z_{\ge1}$, set
 \[
  M_b=I_{l,b}^1=\{\ve{m}=(m_1,\ldots,m_b)\in\Z_{\ge1}^b\mid m_1+\cdots+m_b=l\}.
 \]
 For each $\ve{m}=(m_1,\ldots,m_b)\in M_b$,
 we extend the indices of $m_p$ to all $p\in\Z$ by declaring $m_p=m_{p'}$
 whenever $p\equiv p'\pmod{b}$.
 For $j=1,\ldots,l$, $\ve{m}=(m_1,\ldots,m_b)\in M_b$, and $p=1,\ldots,b$, write
 \[
  k_{j,\ve{m},p}=\sum_{i=j+m_1+\cdots+m_{p-1}}^{j+m_1+\cdots+m_p-1}k_i
  =k_{j+m_1+\cdots+m_{p-1}}+\cdots+k_{j+m_1+\cdots+m_p-1}.
 \]
 Then we have
 \begin{align*}
  P&=\frac{1}{a}\sum_{j=1}^{l}(\delta_a\circ\rho_0)\Biggl(
     \sum_{\ve{m}\in M_a}z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a}}\Biggr)\\
   &=-\frac{1}{a}\sum_{\substack{j=1,\ldots,l\\\ve{m}\in M_a}}\sum_{p=1}^{a}
      z_{k_{j,\ve{m},p}+1}z_{k_{j,\ve{m},p+1}}\cdots z_{k_{j,\ve{m},p+a-1}}.
 \end{align*}
 Note here that
 \[
  \sum_{\substack{j=1,\ldots,l\\\ve{m}\in M_a}}
  z_{k_{j,\ve{m},p}+1}z_{k_{j,\ve{m},p+1}}\cdots z_{k_{j,\ve{m},p+a-1}}
 \]
 does not depend on $p=1,\ldots,a$,
 because the bijection from $\{1,\ldots,l\}\times M_a$ to itself defined by
 \[
  (j,\ve{m})=\bigl(j,(m_1,\ldots,m_a)\bigr)
  \mapsto(j',\ve{m}')=\bigl(j+m_1+\cdots+m_{p-1},(m_p,\ldots,m_{p+a-1})\bigr)
 \]
 has the property that
 \[
  z_{k_{j,\ve{m},p}+1}z_{k_{j,\ve{m},p+1}}\cdots z_{k_{j,\ve{m},p+a-1}}
  =z_{k_{j',\ve{m}',1}+1}z_{k_{j',\ve{m}',2}}\cdots z_{k_{j',\ve{m}',a}}.
 \]
 It follows that
 \[
  P=-\sum_{\substack{j=1,\ldots,l\\\ve{m}\in M_a}}
     z_{k_{j,\ve{m},a}+1}z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a-1}}.
 \]
 Similar reasoning shows that
 \begin{align*}
  Q&=\frac{1}{a-1}\sum_{j=1}^{l}(\delta_a\circ\rho_0)\Biggl(\sum_{\ve{m}\in M_{a-1}}
     z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a-1}}\Biggr)\\
   &=\frac{1}{a-1}\sum_{\substack{j=1,\ldots,l\\\ve{m}\in M_{a-1}}}
     \sum_{p=1}^{a-1}\sum_{i=1}^{k_{j,\ve{m},p}-1}
     z_{k_{j,\ve{m},p}-i+1}z_{k_{j,\ve{m},p+1}}\cdots z_{k_{j,\ve{m},p+a-2}}z_i\\
   &=\sum_{\substack{j=1,\ldots,l\\\ve{m}\in M_{a-1}}}\sum_{i=1}^{k_{j,\ve{m},a-1}-1}
      z_{k_{j,\ve{m},a-1}-i+1}z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a-2}}z_i.
 \end{align*}

 We next compute $\delta_a(B')$.
 Observe that $\delta_a(B')=R+S$, where
 \begin{align*}
  R&=\sum_{\substack{j=1,\ldots,l\\\ve{m}\in M_{a-1}}}\sum_{i=1}^{k_j-1}
  z_{k_{j,\ve{m},1}-i+1}z_{k_{j,\ve{m},2}}\cdots z_{k_{j,\ve{m},a-1}}z_i,\\
  S&=\sum_{\substack{j=1,\ldots,l\\\ve{m}\in M_a}}\sum_{i=1}^{k_j-1}
  z_{k_{j,\ve{m},1}-i+1}z_{k_{j,\ve{m},2}}\cdots z_{k_{j,\ve{m},a-1}}
  z_{k_{j,\ve{m},a}+i}.
 \end{align*}
 Since the bijection from $\{1,\ldots,l\}\times M_{a-1}$ to itself defined by
 \[
  (j,\ve{m})=\bigl(j,(m_1,\ldots,m_{a-1})\bigr)
  \mapsto(j',\ve{m}')=\bigl(j+m_1,(m_2,\ldots,m_{a-1},m_1)\bigr)
 \]
 has the property that
 \[
  \sum_{i=1}^{k_j-1}
  z_{k_{j,\ve{m},1}-i+1}z_{k_{j,\ve{m},2}}\cdots z_{k_{j,\ve{m},a-1}}z_i
  =\sum_{i=1}^{k_{j'+m_1'+\cdots+m_{a-2}'}-1}
  z_{k_{j',\ve{m}',a-1}-i+1}z_{k_{j',\ve{m}',1}}
  \cdots z_{k_{j',\ve{m}',a-2}}z_i,
 \]
 we have
 \[
  R=\sum_{\substack{j=1,\ldots,l\\\ve{m}=(m_1,\ldots,m_{a-1})\in M_{a-1}}}
  \sum_{i=1}^{k_{j+m_1+\cdots+m_{a-2}}-1}
  z_{k_{j,\ve{m},a-1}-i+1}z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a-2}}z_i.
 \]
 Similar reasoning shows that
 \[
  S=\sum_{\substack{j=1,\ldots,l\\\ve{m}=(m_1,\ldots,m_a)\in M_a}}
  \sum_{i=1}^{k_{j+m_1+\cdots+m_{a-1}}-1}
  z_{k_{j,\ve{m},a}-i+1}z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a-2}}
  z_{k_{j,\ve{m},a-1}+i}.
 \]

 What needs to be shown is that $P+Q=R+S$.
 Note that
 \begin{align*}
  S-P
  &=\sum_{\substack{j=1,\ldots,l\\\ve{m}=(m_1,\ldots,m_a)\in M_a}}
  \sum_{i=0}^{k_{j+m_1+\cdots+m_{a-1}}-1}
  z_{k_{j,\ve{m},a}-i+1}z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a-2}}
  z_{k_{j,\ve{m},a-1}+i}\\
  &=\sum_{\substack{j=1,\ldots,l\\\ve{m}=(m_1,\ldots,m_a)\in M_a}}
  \sum_{i=k_{j,\ve{m},a-1}}^{k_{j,\ve{m},a-1}+k_{j+m_1+\cdots+m_{a-1}}-1}
  z_{k_{j,\ve{m},a-1}+k_{j,\ve{m},a}-i+1}
  z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a-2}}z_i.
 \end{align*}
 Fix $j=1,\ldots,l$ and consider the map $\psi\colon M_a\to M_{a-1}$ defined by
 \[
  \psi(m_1,\ldots,m_a)=(m_1,\ldots,m_{a-2},m_{a-1}+m_a).
 \]
 If $\psi(\ve{m})=\ve{m}'$, then
 $k_{j,\ve{m},a-1}+k_{j,\ve{m},a}=k_{j,\ve{m}',a-1}$ and
 $k_{j,\ve{m},p}=k_{j,\ve{m}',p}$ for $p=1,\ldots,a-2$.
 Moreover, for each $\ve{m}'=(m_1',\ldots,m_{a-1}')\in M_{a-1}$, the sets
 \[
  \{i\in\Z\mid k_{j,\ve{m},a-1}\le i\le k_{j,\ve{m},a-1}+k_{j+m_1+\cdots+m_{a-1}}-1\}
 \]
 for $\ve{m}=(m_1,\ldots,m_a)\in\psi^{-1}(\ve{m}')$
 are disjoint with union
 \[
  \{i\in\Z\mid k_{j+m_1'+\cdots+m_{a-2}'}\le i\le k_{j,\ve{m}',a-1}-1\}.
 \]
 It follows that
 \[
  S-P=\sum_{\substack{j=1,\ldots,l\\\ve{m}'=(m_1',\ldots,m_{a-1}')\in M_{a-1}}}
  \sum_{i=k_{j+m_1'+\cdots+m_{a-2}'}}^{k_{j,\ve{m}',a-1}-1}
  z_{k_{j,\ve{m}',a-1}-i+1}z_{k_{j,\ve{m}',1}}\cdots z_{k_{j,\ve{m}',a-2}}z_i.
 \]
 We therefore conclude that
 \[
  S-P+R=\sum_{\substack{j=1,\ldots,l\\\ve{m}\in M_{a-1}}}
  \sum_{i=1}^{k_{j,\ve{m},a-1}-1}
  z_{k_{j,\ve{m},a-1}-i+1}z_{k_{j,\ve{m},1}}\cdots z_{k_{j,\ve{m},a-2}}z_i
  =Q,
 \]
 as required.
\end{proof}

\begin{prop}\label{prop:rho-bar-ker}
 We have $\bar{\rho}_0(\check{\H}^1)\subset\Ker\bar{Z}$
 and $\bar{\rho}_0(\check{\H}_{k-1}^1)\subset\Ker\bar{Z}\cap\H_k^0$ for $k\in\Z_{\ge2}$.
\end{prop}

\begin{proof}
 If $w\in\check{\H}^1$, then Proposition~\ref{prop:diagram},
 Proposition~\ref{prop:rho0=rho}, and Theorem~\ref{thm:CSF} show that
 \[
  \bar{Z}\bigl(\bar{\rho}_0(w)\bigr)
  =Z\bigl(\rho_0(w)\bigr)
  =Z\bigl(\rho(w)\bigr)
  =0,
 \]
 from which the first assertion follows.
 The second assertion is now obvious.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:CSF_MZSVs}]
 Immediate from Propositions~\ref{prop:tedious} and \ref{prop:rho-bar-ker}.
\end{proof}

\begin{prop}
 We have
 \[
  \{0\}=\bar{\rho}_{k-2}(\check{\H}_1^1)\subset\bar{\rho}_{k-3}(\check{\H}_2^1)
  \subset\cdots\subset\bar{\rho}_0(\check{\H}_{k-1}^1)\subset\Ker\bar{Z}\cap\H_k^0
 \]
 for all $k\in\Z_{\ge2}$.
\end{prop}

\begin{proof}
 Observe that $\bar{\rho}_{n+1}(w)=\bar{\rho}_n(zw)$ for all $n\in\Z_{\ge0}$ and $w\in\H$;
 indeed, we have
 \[
  \bar{\rho}_{n+1}(w)=d^{-1}\bigl(\rho_{n+1}(w)\bigr)=d^{-1}\bigl(\rho_n(zw)\bigr)=\bar{\rho}_n(zw).
 \]
 This finishes the proof.
\end{proof}

\begin{thm}
 If $n\in\Z_{\ge0}$ and $k\in\Z_{\ge1}$, then
 \[
  \dim\bar{\rho}_n(\check{\H}_k^1)
  =\dim\rho_n(\check{\H}_k^1)
  =\frac{1}{n+k}\sum_{m\mid n+k}\varphi\biggl(\frac{n+k}{m}\biggr)(2^m-L_m^n)-2.
 \]
\end{thm}

\begin{proof}
 The $\Q$-vector spaces $\bar{\rho}_n(\check{\H}_k^1)$ and $\rho_n(\check{\H}_k^1)$
 are isomorphic as given by $d$,
 and the theorem follows from Theorem~\ref{thm:main}.
\end{proof}

\section{Acknowledgements}
The authors are grateful to Professor Yasuo Ohno for valuable comments that inspired
this work, and to Mr.~Hiroki Kondo for carefully reading the manuscript.