text:
\begin{document}

\maketitle

\abstract{\noindent
    We establish a moderate deviation principle for processes with independent increments under certain growth conditions for the characteristics of the process. Using this moderate deviation principle, we give a new proof for Strassen's functional law of the iterated logarithm. In particular, we show that any square-integrable L\'evy process satisfies Strassen's law.
\par\medskip

\noindent\emph{Keywords:} moderate deviations; additive processes; Strassen's law; functional limit theorem. \par \medskip

\noindent\emph{2010 Mathematics Subject Classification:} Primary: 60F10. Secondary: 60F17, 60G51, 60G17.
}

\section{Introduction} \label{s-intro}
Let $(X_t)_{t \geq 0}$ be a L\'evy process such that $\mathbb{E}X_t=0$, and assume that the weak Cram\'er condition holds, i.\,e.\ $\mbb{E}e^{\lambda |X_1|}<\infty$ for some $\lambda>0$. It is known (see e.\,g.\ Mogulskii \cite{mog93} or Feng--Kurtz \cite{feng}) that the family of scaled L\'evy process $(X(t \lcdot)/S(t))_{t>0}$ obeys a moderate deviation principle in the space of \cadlag functions $(D[0,1],\|\cdot\|_{\infty})$ with good rate function \begin{equation*}
		I(f) := \begin{cases} \displaystyle\frac{1}{2 \var X_1} \int_0^1 f'(s)^2 \, ds, & f :[0,1] \to \mbb{R}\,\, \text{absolutely continuous}, f(0)=0, \\ \displaystyle\infty, & \text{otherwise}, \end{cases}
	\end{equation*}
	and speed $S(t)^2/t$ if the scaling function $S$ satisfies \begin{equation*}
		\frac{S(t)}{\sqrt{t}} \xrightarrow[]{t \to \infty} \infty \quad \text{and} \quad \frac{S(t)}{t} \xrightarrow[]{t \to \infty} 0.
	\end{equation*}
	The assumptions on the moments of $X_1$ have been substantially weakened by Gao \cite{gao05}. In this paper, we consider the corresponding moderate deviation principle for additive processes, see Section~\ref{s-def}. Large deviation results for additive processes have been obtained by Puhalskii \cite{puc94} and Liptser--Puhalskii \cite{lip-puc} for the scaling $S(t)=t$ under rather abstract conditions on the characteristics and the stochastic exponential, respectively.
	We state sufficient conditions in terms of the growth of the characteristics and give a direct proof of the moderate deviation principle using the well-known G\"artner--Ellis theorem. In particular, we obtain two representations for the good rate function $I$. \par
	As an application, we study Strassen's law for additive processes. Strassen \cite{stra64} proved that for a (one-dimensional) Brownian motion $(B_t)_{t \geq 0}$ the set \begin{equation*}
		\left\{ f \in C[0,1];\; \exists t>0: f(s)=\frac{B(t s)}{\sqrt{2t \log \log (t \vee e^e)}} \, \, \text{for all} \, s \in [0,1] \right\}
	\end{equation*}
	is almost surely relatively compact in $(C[0,1],\|\cdot\|_{\infty})$ and its limit points are given by \begin{equation*}
		\left\{f: [0,1] \to \mathbb{R};\; f(0)=0, f \,\, \text{absolutely continuous}, \int_0^1 f'(s)^2 \, ds  \leq 1 \right\};
	\end{equation*}
	this result is called Strassen's law (or functional law of the iterated logarithm). There have been several generalizations since then; Wichura \cite{wic73} and Wang \cite{wang} studied Strassen's law for additive processes and Maller \cite{mal08} obtained a small-time version of Strassen's law for L\'evy processes. More recently, Gao \cite{gao09} proved a Strassen law for a subclass of locally square-integrable martingales. Strassen's law has a variety of applications, e.\,g.\ functional limit theorems such as the law of the iterated logarithm; see e.\,g.\ Strassen \cite{stra64}, Wichura \cite{wic73} and Buchmann et al. \cite{bms08}. \par
	We show that, whenever a certain moderate deviation principle holds for an additive processes, the process satisfies a functional law of the iterated logarithm. Combining this result with the moderate deviation principle proved in the first part of this paper, this covers the corresponding results in Wang \cite{wang}, cf.\ Corollary~\ref{add-n05}, and Wichura \cite{wic73}. As a special case, we find that any square-integrable L\'evy process satisfies Strassen's law.
	Let us emphasize that we do not intend to formulate the results in the most general form but to present an alternative proof for Strassen's law. The results discussed here are generalizations of these obtained by K\"uhn \cite{fk14}. \par
	The paper is organized as follows. In Section~\ref{s-def}, we introduce basic definitions and notation. The main results are stated in Section~\ref{s-main} and proved in Section~\ref{s-proof}. Finally, in Section~\ref{s-rem}, we sketch some generalizations.

\section{Basic definitions and notation} \label{s-def}
	Let $(X_t)_{t \geq 0}$ be a (real-valued) stochastic process on a complete probability space $(\Omega,\mc{A},\mbb{P})$. We call $(X_t)_{t \geq 0}$ an \emph{additive process} if $(X_t)_{t \geq 0}$ has \cadlag sample paths, independent increments and $X_0=0$. If, additionally, $(X_t)_{t \geq 0}$ has stationary increments, i.\,e.\ $X_t-X_s \sim X_{t-s}-X_0$, $s \leq t$, then $(X_t)_{t \geq0}$ is a \emph{L\'evy process}. Any mean-zero square-integrable additive process $(X_t)_{t \geq 0}$ admits a \emph{L\'evy--It\^o decomposition} of the form
\begin{equation*}
		X(t) = X^c(t) + \int_0^t \!\! \! \int z \, (N(dz,dr)-\nu(dz,dr)), \qquad t \geq 0,
	\end{equation*}
	where $X^c$ denotes the continuous martingale part, $N$ the jump measure of $(X_t)_{t \geq 0}$ and $\nu$ its compensator. Moreover, there exist increasing deterministic functions $A,C$ and a family of $\sigma$-finite measures $K_r$ on $(\mbb{R},\mc{B}(\mbb{R}))$ such that $C_t \geq 0$, $\int (z^2 \wedge 1) K_r(dz) \leq 1$, and \begin{equation}
		\komp{X^c}_t = C_t, \qquad  \qquad \nu(dz,dr) = K_r(dz) \, dA_r. \label{add-neq05}
	\end{equation}
	If $(X_t)_{t \geq 0}$ is a L\'evy process, then $A$ and $C$ are linear, and $K_r$ does not depend on $r$. We denote by  \begin{equation*}
		[\nu]_{s,t} := \inf\left\{\varrho>0;\; \nu(B(0,\varrho)^c \times [s,t])=0\right\}, \qquad (\inf \emptyset := \infty)
	\end{equation*}
	the maximal jump height during the time interval $[s,t]$. The process $(X_t)_{t \geq 0}$ has no \emph{fixed jump discontinuities} if $\nu(\mbb{R} \times \{t\} ) =0$ for any $t>0$. Our standard reference for additive processes is the monograph by Jacod--Shiryaev \cite{jac}, we use Sato \cite{sato} for L\'evy processes. \par
	For simplicity, we assume that the mapping $(\Omega,\mc{A}) \ni \omega \mapsto X(\lcdot,\omega) \in (D[0,1],\|\cdot\|_{\infty})$ is measurable. Here $D[0,1]$ denotes the space of \cadlag (i.\,e.\ right-continuous with left-hand limits) functions $f:[0,1] \to \mbb{R}$ endowed with the uniform norm $\|\cdot\|_{\infty}$.  By $\ac$ we denote the set of absolutely continuous functions $f:[0,1] \to \mbb{R}$. \par
	Recall that a family $(X^t)_{t>0}$ of stochastic processes with values in a metric space $(M,d)$ satisfies a  \emph{large deviation principle in $(M,d)$ with good rate function $I:M \to [0,\infty]$ and speed $(a_t)_{t>0} \subseteq (0,\infty)$} if $I$ has compact sublevel sets $\Phi(r):=\{f \in M; I(f) \leq r\}$, $a_t \to \infty$ as $t \to \infty$, and \begin{equation*}
		- \inf_{f \in U} I(f) \leq \liminf_{t \to \infty} \frac{1}{a_t} \log \mbb{P}(X^t \in U), \qquad
		\limsup_{t \to \infty} \frac{1}{a_t} \log \mbb{P}(X^t \in F) \leq - \inf_{f \in F} I(f)
	\end{equation*}
	holds for any open set $U \subseteq M$ and closed set $F \subseteq M$, respectively. Let $(X_t)_{t \geq 0}$ be a square-integrable stochastic process with \cadlag sample paths and $S:(0,\infty) \to (0,\infty)$ such that $S(t)/\var X_t \to 0$ as $t \to \infty$. We say that $(X(t \lcdot)/S(t))_{t>0}$ satisfies a \emph{moderate deviation principle with good rate function $I$ and speed $(a_t)_{t>0}$} if the family satisfies a large deviation principle in $(D[0,1],\|\cdot\|_{\infty})$ with good rate function $I$ and speed $(a_t)_{t>0}$.
	For a detailed discussion of large deviation theory, we refer the reader to the monographs by Dembo--Zeitouni \cite{dz} and Feng--Kurtz \cite{feng}.

\section{Main results} \label{s-main}

We are now in the position to state the main results.

\begin{thm} \label{add-n01}
	Let $(X_t)_{t \geq 0}$ be an additive process without fixed jump discontinuities such that $\mathbb{E}X_t = 0$ and \begin{equation}
		\lim_{t \to \infty} \frac{\var X_t}{t^\gamma} =: \sigma^2>0\label{add-neq11}
	\end{equation}
	exists for some $\gamma >0$. Let $S:(0,\infty) \to (0,\infty)$ be such that \begin{equation}
			\frac{S(t)}{t^{\gamma/2}} \xrightarrow[]{t \to \infty} \infty, \qquad \qquad \frac{S(t)}{t^{\gamma}} \xrightarrow[]{t \to \infty} 0, \label{add-neq13}
		\end{equation}
	and \begin{equation}
		 \frac{S(t)}{S(\floor{t})} \xrightarrow[]{t \to \infty} 1. \label{add-neq15}
	\end{equation}
	Suppose that one of the following conditions holds.
	\begin{enumerate}[label*=\upshape(C\arabic*),ref=\upshape(C\arabic*)]
		\item\label{C2}  There exists a measure $G$ on $(\mbb{R},\mc{B}(\mbb{R}))$ such that $\int_{|z|>1} e^{2\lambda_0 |z|} \, G(dz)<\infty$ for some $\lambda_0>0$ and \begin{equation*}
			K_t(B(0,r)^c) \leq G(B(0,r)^c) \fa  t \geq 0,\: r \geq 1.
		\end{equation*}
		Furthermore,
		\begin{equation}
				\lim_{t \to \infty} |A_t| \frac{S(t)}{t^{2\gamma}}= 0. \label{add-neq19}
		\end{equation}
		\item\label{C1} \strut \vspace{-\baselineskip} \begin{equation}
				\lim_{t\to \infty} [\nu]_{0,t} \frac{S(t)}{t^\gamma} =0 \label{add-neq17}
		\end{equation}
		(In this case, we implicitly set $\lambda_0 := \infty$.)
	\end{enumerate}
	Then $(X(t\lcdot)/S(t))_{t > 0}$ satisfies a moderate deviation principle in $(D[0,1],\|\cdot\|_{\infty})$ with speed $S(t)^2/t^\gamma$ and good rate function \begin{equation}
		I(f) := \begin{cases} \displaystyle\frac{1}{2\sigma^2 \gamma} \int_0^1 \frac{f'(s)^2}{s^{\gamma-1}} \, ds, & f \in \ac,\: f(0)=0, \\ \displaystyle\infty, & \text{otherwise}. \end{cases} \label{add-neq21}
	\end{equation}
\end{thm}

In particular, if $(X_t)_{t \geq 0}$ is a mean-zero L\'evy process such that $\mbb{E}e^{\lambda |X_1|}<\infty$ for some $\lambda>0$, then condition \ref{C2} holds with $G=\nu$ where $\nu$ is the L\'evy measure of $(X_t)_{t \geq 0}$, cf.\ \cite[Theorem 25.17]{sato}.

\begin{thm}[Strassen's law] \label{add-n03}
	Let $(X_t)_{t \geq 0}$ be an additive process and $S(t)= \sqrt{2t^{\gamma} \log \log (t \vee e^e)}$ for some $\gamma>0$. If $(X(t \lcdot)/S(t))_{t \geq 0}$ satisfies a moderate deviation principle in $(D[0,1],\|\cdot\|_{\infty})$  with speed $S(t)^2/t^{\gamma}$ and good rate function \begin{equation}
		I(f) = \begin{cases} \displaystyle\frac{1}{2 \sigma^2 \gamma} \int_0^1 \frac{f'(s)^2}{s^{\gamma-1}} \, ds, & f \in \ac,\: f(0)=0, \\ \displaystyle\infty, & \text{otherwise}, \end{cases} \label{add-neq23}
	\end{equation}
	for some $\sigma>0$, then \begin{equation*}
		\left\{ \frac{X(t \lcdot)}{\sqrt{2 t^{\gamma}\log \log (t \vee e^e)}}; t > 0\right\}
	\end{equation*}
	is a.\,s.\ relatively compact in $(D[0,1],\|\cdot\|_{\infty})$ as $t \to \infty$, and the set of limit points $\mc{L}(\omega)$ (as $t \to \infty$) is for almost all $\omega \in \Omega$ given by the sublevel set $\Phi(\frac{1}{2})=\{f \in D[0,1]; I(f) \leq \frac{1}{2}\}$ of the good rate function $I$.
\end{thm}

By a standard argument we can replace $t^{\gamma}$ by a regulary varying function $h(t)$ of index $\gamma>0$, cf.\ Section~\ref{s-rem} at the end of the paper. \par
Note that the \emph{law of the iterated logarithm} is a simple consequence of Strassen's law: \begin{equation*}
	\limsup_{t \to \infty} \frac{X_t}{\sqrt{2 t^{\gamma} \log \log t}} = \sigma \als
\end{equation*}
Moreover, if $(X_t)_{t \geq 0}$ is an additive process such that $\var X_t/t^{\gamma} \xrightarrow[]{t \to \infty} \sigma^2>0$ and one of the growth conditions \ref{C1} or \ref{C2} holds, Theorem~\ref{add-n01} shows that Theorem~\ref{add-n03} is applicable.

\begin{kor}\label{add-n05}
	Let $(X_t)_{t \geq 0}$ be an additive process such that $\var X_t/t \xrightarrow[]{t \to \infty} \sigma^2>0$. Suppose that there exists a finite measure $G$ on $(\mbb{R},\mc{B}(\mbb{R}))$ such that \begin{equation*}
		K_t(B(0,r)^c) \leq G(B(0,r)^c) \fa r \geq 1, t \geq 0 \quad \text{and} \quad \int_{\mbb{R}} z^2 \, G(dz)< \infty.
	\end{equation*}
	Then $(X_t)_{t \geq 0}$ satisfies Strassen's law (with $\gamma=1$). In particular, Strassen's law holds for any square-integrable L\'evy process $(X_t)_{t \geq 0}$.
\end{kor}

We close this section with two examples which we state for the more general case of regulary varying functions, see Section~\ref{s-rem} and the remark before Corollary~\ref{add-n05}.

\begin{bsp} \label{add-n065}
	Let $(L_t)_{t \geq 0}$ be a pure-jump L\'evy process,  \begin{equation*}
		L_t = \int_0^t \!\!\! \int z \, (N(dz,ds)-\nu_L(dz) \, ds),
	\end{equation*}
	such that $\var L_1=\sigma^2<\infty$. For a non-decreasing function $\alpha:[0,\infty) \to [0,\infty)$, $\alpha(t) \xrightarrow[]{t \to \infty} \infty$, we define a truncated L\'evy process $(X_t)_{t \geq 0}$ by \begin{equation*}
		X_t := \int_0^t \!\!\! \int z \I_{\{|z| \leq \alpha(s)\}} \, (N(dz,ds)-\nu_L(dz) \, ds).
	\end{equation*}
	From \begin{equation*}
		0 \leq \var L_t-\var X_t = \int_0^t \!\!\! \int z^2 \I_{\{|z| >\alpha(s)\}} \, \nu_L(dz) \, ds
	\end{equation*}
	it follows easily that \begin{equation*}
		\lim_{t \to \infty} \frac{\var X_t}{t} = \lim_{t \to \infty} \frac{\var L_t}{t} = \sigma^2.
	\end{equation*}

	By Corollary~\ref{add-n05}, $(X_t)_{t \geq 0}$ satisfies Strassen's law (with $\gamma=1$). Let us remark that truncated processes of this form are used in \cite{wang} to prove Corollary~\ref{add-n05}.
\end{bsp}

\begin{bsp} \label{add-n06}
	Let $(L_t)_{t \geq 0}$ be a L\'evy process with bounded jumps such that $\mathbb{E}L_1^2 = \sigma_L^2 < \infty$ and $\mathbb{E}L_1=0$. Denote by $\psi$ its characteristic exponent and $\nu_L$ its L\'evy measure. For a regulary varying function $\alpha:[0,\infty) \to [0,\infty)$ of index $\gamma>0$, the additive process $X_t := \int_0^t \alpha(s) \, dL_s$ satisfies a moderate deviation principle with speed $S(t)^2/(t \alpha(t)^2)$ and good rate function \begin{equation*}
		I(f) = \begin{cases}\displaystyle \frac{2\gamma+1}{2\gamma \sigma_L^2} \int_0^1 \frac{f'(s)^2}{s^2 \alpha(s)^2} \,ds, & f \in \ac, f(0)= 0, \\ \infty, & \text{otherwise}, \end{cases}
	\end{equation*}
	for any scaling function $S:(0,\infty) \to (0,\infty)$ such that \eqref{add-neq15} as well as the growth conditions $S(t)/\sqrt{t\alpha(t)} \xrightarrow[]{t \to \infty} \infty$ and $S(t)/(t \alpha(t)) \xrightarrow[]{t \to \infty}0$ hold. Moreover, the set
	\begin{equation*}
		\left\{ \frac{X(t \lcdot)}{\sqrt{2 t \alpha(t)^2 \log \log (t \vee e^e)}}; t > 0\right\}
	\end{equation*}
	is a.\,s.\ relatively compact in $(D[0,1],\|\cdot\|_{\infty})$ as $t \to \infty$, and the set of limit points (as $t \to \infty$) is for almost all $\omega \in \Omega$ given by the sublevel set $\Phi(\frac{1}{2})$. \emph{Indeed:} Using that $(L_t)_{t \geq 0}$ is a martingale with independent increments, it is not difficult to see that \begin{equation*}
		\var X_t = \sigma_L^2 \int_0^t \alpha(s)^2 \, ds \quad \Longrightarrow \quad \frac{\var X_t}{t \alpha(t)^2} \xrightarrow[]{t \to \infty} \frac{\sigma_L^2}{2\gamma+1}.
	\end{equation*}
	The approximation \begin{equation*}
		X_t = \int_0^t \alpha(s) \, dL_s \approx \sum_{j=1}^n \alpha(s_j) (L_{s_j}-L_{s_{j-1}})
	\end{equation*}
	shows that the characteristic function of $X_t$ equals $\exp\left(-\int_0^t \psi(\alpha(s) \xi) \, ds \right)$. In particular, \begin{equation*}
		\nu(dz,ds) \stackrel{\eqref{add-neq05}}{=} K_s(dz) \, dA_s = \nu_L\left(\frac{1}{\alpha(s)} dz \right) \, ds.
	\end{equation*}
	Note that due to the boundedness of the jumps of $(L_t)_{t \geq 0}$, condition \ref{C1} is satisfied. Therefore, the claim follows from Theorem~\ref{add-n01} and Theorem~\ref{add-n03}.
\end{bsp}

\section{Proofs} \label{s-proof}
	We start with the proof of the moderate deviation principle, Theorem \ref{add-n01}, and split the proof into several steps: \begin{enumerate}
		\item The sequence of discretizations $(Z_n/S(n))_{n \in \mbb{N}}$ defined by\begin{equation*}
			\frac{Z_n(s,\omega)}{S(n)}
			:= \frac{1}{S(n)} X(\floor{n s},\omega)
			= \frac{1}{S(n)} \left(\sum_{j=0}^{n-1} X(j,\omega) \I_{[j/n,(j+1)/n)}(s) +  X(n,\omega) \I_{\{1\}}(s) \right)
		\end{equation*}
		is exponentially tight in $(D[0,1],\|\cdot\|_{\infty})$, cf.\ Lemma~\ref{add-n09}.
		\item $(Z_n/S(n))_{n \in \mbb{N}}$ satisfies a moderate deviation principle in $(D[0,1],\|\cdot\|_{\infty})$ with good rate function $J$, \begin{equation}
			J(f) := \sup_{\alpha \in \bv \cap D[0,1]} \left( \int_0^1 f \, d\alpha - \frac{\gamma \sigma^2}{2} \int_0^1 s^{\gamma-1} (\alpha(1)-\alpha(s))^2 \, ds \right), \label{add-neq31}
		\end{equation}
		and speed $a_n = S(n)^2/n^{\gamma}$, cf.\ Theorem~\ref{add-n11}; as usual, $\bv$ denotes the set of functions $\alpha:[0,1] \to \mbb{R}$ of bounded variation.
		\item $(Z_{\floor{t}}/S(\floor{t}))_{t > 0}$ and $(X(t\lcdot)/S(t))_{t >0}$ are exponentially equivalent, cf.\ Lemma~\ref{add-n13}.
		\item The good rate function $J$ equals $I$ defined in \eqref{add-neq21}, cf.\ Theorem~\ref{add-n15}.
	\end{enumerate}
	We essentially follow the lines of de Acosta \cite{acosta-lp}. For the readers' convenience, we include the proofs of (i)-(iv). The next lemma provides an estimate for the exponential moments of $X_t-X_s$.

\begin{lem} \label{add-n07}
	Let $(X_t)_{t \geq 0}$ be as in Theorem~\ref{add-n01}. Then \begin{equation}
		\mbb{E}e^{\lambda (X_t-X_s)} \leq \exp \left( \frac{1}{2} \lambda^2 \var(X_t-X_s) + |\lambda|^3 E_{s,t}\right) \label{add-neq33}
	\end{equation}
	for any $|\lambda| \leq \lambda_0$ and $s \leq t$ where  \begin{equation*}
		E_{s,t} := E_{s,t}(\lambda):=\begin{cases} \frac{1}{6} [\nu]_{s,t} \var(X_t-X_s) e^{|\lambda| [\nu]_{s,t}} , & \text{if \ref{C1} holds}, \\ \frac{1}{6}  |A_t-A_s| \cdot \left(e^{\lambda_0}+\int_{|z|>1} e^{\lambda_0 |z|} |z|^3 \, G(dz) \right), & \text{if \ref{C2} holds}. \end{cases}
	\end{equation*}
	In particular, \begin{equation}
		\lim_{t \to \infty} \frac{S(t)}{t^{2\gamma}} E_{0,t} \left( r \frac{S(t)}{t^\gamma} \right) = 0 \fa r \geq 0. \label{add-neq35}
	\end{equation}
\end{lem}

\begin{proof}
	It follows from the conditions \ref{C1} or \ref{C2}, respectively, that $\mathbb{E}e^{\lambda (X_t-X_s)}<\infty$ for any $s \leq t$, $|\lambda| \leq \lambda_0$, and that  \begin{equation*}
		\mbb{E}e^{\lambda (X_t-X_s)}
		= \exp \left( \frac{1}{2} (C_t-C_s) \lambda^2 + \int_s^t \!\!\! \int (e^{\lambda z}-1-\lambda z)\, \nu(dz,dr)\right),
	\end{equation*}
	cf.\ Fujiwara \cite{fuji}. Since \begin{equation*}
		\var(X_t-X_s) = (C_t-C_s)+ \int_s^t \!\!\! \int z^2 \, \nu(dz,dr)
	\end{equation*}
	Taylor's formula yields \begin{equation*}
		\mbb{E}e^{\lambda (X_t-X_s)}
		\leq \exp \left( \frac{\lambda^2}{2} \var(X_t-X_s) + \frac{|\lambda|^3}{6} \int_s^t \!\!\! \int |z|^3 e^{\lambda \xi} \, \nu(dz,dr) \right)
	\end{equation*}
	for some intermediate value $\xi=\xi(z) \in (0,z)$. From the definition of $[\nu]_{s,t}$ we get \begin{align*}
		\int_s^t \!\!\! \int |z|^3 e^{\lambda \xi} \, \nu(dz,dr)
		&\leq [\nu]_{s,t} \cdot e^{|\lambda| [\nu]_{s,t}} \int_s^t \!\!\! \int z^2 \, \nu(dz,dr)
		\leq [\nu]_{s,t} \var(X_t-X_s) e^{|\lambda| [\nu]_{s,t}} .
	\end{align*}
	This proves \eqref{add-neq33} if \ref{C1} holds. If \ref{C2} is satisfied, the claim follows from the estimate \begin{align*}
		\int_s^t \!\!\! \int |z|^3 e^{\lambda \xi} \, \nu(dz,dr)
		&\leq \int_s^t \!\!\! \int |z|^3 e^{|\lambda| |z|} \, K_r(dz) \, dA_r \\
		&\leq \left(e^{\lambda_0}+ \int_{|z| > 1} |z|^3 e^{\lambda_0 |z|} \, G(dz) \right) |A_t-A_s|.
	\end{align*}
	\eqref{add-neq35} is a direct consequence of the definition of $E_{0,t}$ and the assumptions in \ref{C1} and \ref{C2}, respectively.
\end{proof}

In order to show that the approximations $(Z_n)_{n \in \mbb{N}}$ satisfy a moderate deviation principle, we need the following lemma.

\begin{lem} \label{add-n09}
	For each $n \in \mbb{N}$, $Z_n/S(n)$ is tight in $(D[0,1],\|\cdot\|_{\infty})$. Morover, $(Z_n/S(n))_{n \in \mbb{N}}$ is exponentially tight, i.\,e.\ for any $R \geq 0$ there exists a compact set $K \subseteq D[0,1]$ such that \begin{equation*}
		\limsup_{n \to \infty} \frac{n^{\gamma}}{S(n)^2} \log \mbb{P}\left( \frac{Z_n}{S(n)} \notin K \right) \leq -R.
	\end{equation*} \end{lem}

\begin{proof}
	Since the mapping \begin{equation*}
		(\mbb{R}^n,\|\cdot\|) \ni x \mapsto (T_n x)(t) := \sum_{j=1}^{n-1} x_j \I_{[j/n,(j+1)/n)}(t) + x_n \I_{\{1\}}(t) \in (D[0,1],\|\cdot\|_{\infty})
	\end{equation*}
	is continuous, it follows that $T_n(K)$ is compact for any compact set $K \subseteq \mbb{R}^n$. For $K \subseteq \mbb{R}$ compact and $K^n = K \times \ldots \times K \subseteq \mbb{R}^n$, we have \begin{equation*}
		\mbb{P} \left( \frac{Z_n}{S(n)} \notin T_n(K^n) \right) \leq \sum_{j=1}^n \mbb{P} \left( \frac{X_j}{S(n)} \notin K \right).
	\end{equation*}
	Since $X_j/S(n)$ is tight for $j=1,\ldots,n$, we conclude that $Z_n/S(n)$ is tight in $(D[0,1],\|\cdot\|_{\infty})$.  It remains to prove exponential tightness. To this end, we show that the assumptions of \cite[Lemma 3.3]{feng} are satisfied. Fix $r > 0$ and $\eps>0$. For $K \subseteq \mbb{R}$ and $n \geq m$, we have \begin{align}
		&\mbb{P} \left( d \left( \frac{Z_n}{S(n)}, T_m(K^m) \right) > \eps \right) \notag \\
		&\quad \leq \mbb{P}\left( \frac{Z_n}{S(n)} \notin T_n(K^n) \right) + \mbb{P} \left( \frac{Z_n}{S(n)} \in T_n(K^n), d \left( \frac{Z_n}{S(n)}, T_m(K^m) \right)>\eps \right)
		=: I_1+I_2 \label{add-neq37}
	\end{align}
	with $d(f,A) := \inf_{g \in A} \|f-g\|_{\infty}$, $A \subseteq D[0,1]$. We choose $K:=[-r,r]$ and estimate the terms separately. Applying Etemadi's inequality, Markov's inequality and Lemma~\ref{add-n07} yields \begin{align*}
		I_1
		= \mbb{P} \left( \max_{1 \leq j \leq n} \left|\frac{X_j}{S(n)}\right|>r \right)
		&\leq 3 \max_{1 \leq j \leq n} \mbb{P} \left( |X_j| > \frac{S(n) r}{3} \right)  \\
		&\leq 3 \exp \left(- \frac{S(n) \lambda r}{3} \right) \max_{1 \leq j \leq n} \bigg( \mbb{E}e^{\lambda X_j}+\mbb{E}e^{-\lambda X_j} \bigg) \\
		&\leq 6 \exp \left(- \frac{S(n) \lambda r}{3} \right) \exp \left( \frac{\lambda^2}{2} \var X_n + \lambda^3 E_{0,n}(\lambda) \right)
	\end{align*}
	for any $0 \leq \lambda \leq \lambda_0$. For $\lambda := S(n)/n^\gamma$, we have $\lambda \leq \lambda_0$ for $n$ sufficiently large, and we obtain \begin{equation*}
		I_1 \leq 6 \exp \left(- \frac{S(n)^2}{n^\gamma} \cdot \frac{r-3\sigma^2}{3} \right)
	\end{equation*}
	since \begin{equation*}
		\lim_{n \to \infty} \left(\frac{1}{2}\frac{\var X_n}{n^\gamma} +\frac{S(n)}{n^{2\gamma}} E_{0,n} \left( \frac{S(n)}{n^\gamma} \right) \right) = \frac{\sigma^2}{2},
	\end{equation*}
	cf.\ \eqref{add-neq11} and \eqref{add-neq35}. In order to estimate $I_2$ we observe that for $f_m := f(\floor{m \lcdot}/m)$ it holds that\begin{equation}
		d(f,T_m(K^m))
		\leq \|f-f_m\|_{\infty} \label{add-neq41} \fa f \in T_n(K^n).
	\end{equation}
	In abuse of notation, we write $x+y \wedge z := (x+y) \wedge z$. Then, \begin{align}
		\|f-f_m\|_{\infty}

		&= \adjustlimits\max_{0 \leq i \leq m-1} \sup_{t \in [i/m,(i+1)/m)} \left| f \left( \frac{\floor{n t}}{n} \right)- f \left( \frac{\floor{m t}}{m} \right) \right| \notag \\
		&\leq \adjustlimits\max_{0 \leq i \leq m-1} \max_{1 \leq j \leq \floor{n/m}+1} \left| f \left( \frac{\floor{n \frac{i}{m}}}{n}+\frac{j}{n} \wedge 1 \right)- f \left( \frac{\floor{n \frac{i}{m}}}{n} \right) \right|. \label{add-neq43}
	\end{align}
	For the last line we used that \begin{equation*}
		f \left( \frac{\floor{n \frac{i}{m}}}{n} \right) = f \left( \frac{i}{m} \right)= f \left( \frac{\floor{m t}}{m} \right) \fa t \in \bigg[ \frac{i}{m},\frac{i+1}{m} \bigg)
	\end{equation*}
	as $f \in T_n(K^n)$. Combining \eqref{add-neq41} and \eqref{add-neq43}, we get \begin{align*}
		I_2
		&\leq \mbb{P} \left\{ \adjustlimits\max_{0 \leq i \leq m-1} \max_{1 \leq j \leq \floor{n/m}+1} \left| Z_n \left( \frac{\floor{n \frac{i}{m}}}{n}+\frac{j}{n} \wedge 1 \right)- Z_n \left( \frac{\floor{n \frac{i}{m}}}{n} \right) \right| >\eps  S(n)\right\} \\

		&\leq 3\sum_{i=0}^{m-1} \max_{1 \leq j \leq \floor{n/m}+1} \mbb{P} \left\{ \left| X \left(\floor{n \frac{i}{m}}+j \wedge n \right)-X \left(\floor{n \frac{i}{m}}\right) \right| > \frac{\eps S(n)}{3} \right\} .
	\end{align*}
	By Lemma~\ref{add-n07}, \begin{align*}
		I_2
		& \leq 6 \exp \left(-\frac{S(n) \lambda \eps}{3}\right) \sum_{i=0}^{m-1} \exp \left(\frac{\lambda^2}{2}  \left[\var X\left(\floor{\frac{n i}{m}} + \floor{\frac{n}{m}}+1 \right)- \var X \left( \floor{\frac{n i}{m}} \right) \right]+ \lambda^3 E_{0,n}(\lambda) \right)
	\end{align*}
	for any $0 \leq \lambda \leq \lambda_0$. Writing \begin{equation*}
		\var X_s-\var X_r = s^\gamma \left( \left[\frac{\var X_s}{s^\gamma}-\sigma^2 \right]- \left[\frac{\var X_r}{r^\gamma}-\sigma^2 \right] \right) + \frac{\var X_r}{r^{\gamma}} (s^{\gamma}-r^{\gamma})
	\end{equation*}
	it is not difficult to see that \begin{align*}
		\frac{1}{n^\gamma} \left[\var X\left(\floor{\frac{n i}{m}} + \floor{\frac{n }{m}}+1 \right)- \var X \left( \floor{\frac{n i}{m}} \right) \right]
		&\leq c \sup_{k \geq \floor{\frac{n}{m}}} \left| \frac{\var X_k}{k^\gamma}-\sigma^2 \right| + c \left( \frac{1}{m}+ \frac{1}{n} \right)
		=: \delta(n)
	\end{align*}
	for some constant $c=c(\gamma)$. Note that the first term on the right-hand side converges to $0$ as $n \to \infty$. For $\lambda := r S(n)/n^{\gamma}$, $r \geq 1$, we find \begin{align}
		I_2
		&\leq 6m \exp \left[ \frac{S(n)^2}{n^\gamma} \left(- \frac{r \eps}{3} + \frac{r^2}{2} \delta(n) +  r^3 \frac{S(n)}{n^{2\gamma}} E_{0,n}\left( r\frac{S(n)}{n^\gamma} \right)\right) \right]. \label{add-neq45}
	\end{align}
	Consequently, by \eqref{add-neq35}, \eqref{add-neq37} and \eqref{add-neq45}, \begin{align*}
		\limsup_{n \to \infty} \frac{n^{\gamma}}{S(n)^2} \log \mbb{P} \left( d \left( \frac{Z_n}{S(n)}, T_m(K^m) \right) > \eps \right)
		&\leq \max \left\{\sigma^2-\frac{r}{3}, - \frac{r \eps}{3} + \frac{c  r^2}{2m} \right\} \\
		&\xrightarrow[]{r,m \to \infty} -\infty.
	\end{align*}
	By \cite[Lemma 3.3]{feng}, $(Z_n/S(n))_{n \in \mbb{N}}$ is exponentially tight in $(D[0,1],\|\cdot\|_{\infty})$.
\end{proof}

Now we are ready to prove that $(Z_n/S(n))_{n \in \mbb{N}}$ satisfies a moderate deviation principle.

\begin{thm} \label{add-n11}
	$(Z_n/S(n))_{n \in \mbb{N}}$ satisfies a moderate deviation principle in $(D[0,1],\|\cdot\|_{\infty})$ with speed $S(n)^2/n^{\gamma}$ and good rate function $J$ defined in \eqref{add-neq31}.
\end{thm}

\begin{proof}
	For $\alpha \in \bv \cap D[0,1]$ we set \begin{equation*}
		\Lambda_n(\alpha):=\frac{n^\gamma}{S(n)^2} \log \mbb{E}\exp \left( \frac{S(n)}{n^\gamma} \int_0^1 Z_n(s) \, d\alpha(s) \right).
	\end{equation*}
	By definition, \begin{equation*}
		Z_n(s)
		= \sum_{j=0}^{n-1} X_j \I_{[j/n,(j+1)/n)}(s) + X_n \I_{\{1\}}(s)
		= \sum_{j=1}^n (X_j-X_{j-1}) \I_{[j/n,1]}(s).
	\end{equation*}
	It follows from the independence of the increments that \begin{align*}
		\Lambda_n(\alpha)
		&= \frac{n^\gamma}{S(n)^2} \log \mbb{E} \exp \left( \frac{S(n)}{n^\gamma} \sum_{j=1}^n (\alpha(1)-\alpha(j/n)) \, (X_j-X_{j-1}) \right) \\
		&= \frac{n^\gamma}{S(n)^2}  \sum_{j=1}^n \log \mbb{E} \exp \left( \frac{S(n)}{n^{\gamma}} (\alpha(1)-\alpha(j/n)) \, (X_j-X_{j-1}) \right) \\
		&= \frac{1}{2n^{\gamma}} \sum_{j=1}^n (C_j-C_{j-1}) (\alpha(1)-\alpha(j/n))^2 \\
		& + \frac{n^\gamma}{S(n)^2} \sum_{j=1}^n \int_{j-1}^j \!\! \int \bigg[ \exp \left( \frac{S(n)}{n^\gamma} (\alpha(1)-\alpha(j/n))z \right)-1-\frac{S(n)}{n^\gamma} (\alpha(1)-\alpha(j/n)) z \bigg] \, \nu(dz,dr).
	\end{align*}
	Applying Taylor's formula and using that $\var X_j = C_j + \int_0^j \!\! \int z^2 \, \nu(dz,dr)$, we get \begin{align*}
		\Lambda_n(\alpha)
		&= \frac{1}{2n^\gamma} \sum_{j=1}^n (\var X_j-\var X_{j-1}) (\alpha(1)-\alpha(j/n))^2 \\
		& \quad+ \frac{1}{6} \frac{S(n)}{n^{2\gamma}} \sum_{j=1}^n (\alpha(1)-\alpha(j/n))^3 \int_{j-1}^j \!\! \int \exp \left(\frac{S(n)}{n^\gamma}  (\alpha(1)-\alpha(j/n))\xi_j\right) z^3 \, \nu(dz,dr) \\
		&=: I_1(n)+I_2(n)
	\end{align*}
	for some intermediate value $\xi_j$ between $0$ and $z$. It follows from Abel's summation formula and \eqref{add-neq11} that \begin{align*}
		I_1(n)
		&= \frac{1}{2n^\gamma} \sum_{j=1}^n \var X_j \left( (\alpha(1)-\alpha(j/n))^2  - (\alpha(1)-\alpha(j+1/n))^2 \right) \\
		&= -\frac{1}{2} \sum_{j=1}^n \frac{\var X_j}{j^\gamma} \left(\frac{j}{n}\right)^{\gamma} \left( (\alpha(1)-\alpha((j+1)/n))^2  - (\alpha(1)-\alpha(j/n))^2 \right)  \\
		&\xrightarrow[]{n \to \infty} - \frac{\sigma^2}{2} \int_0^1 s^\gamma d\left( (\alpha(1)-\alpha(s))^2 \right).
	\end{align*}
	Note that this integral is well-defined as $\alpha \in \bv$ (hence $\alpha^2 \in \bv$). Applying integration by parts, we obtain \begin{equation*}
		\lim_{n \to \infty} I_1(n) = \frac{\gamma \sigma^2}{2} \int_0^1 s^{\gamma-1} (\alpha(1)-\alpha(s))^2 \, ds.
	\end{equation*}

	On the other hand, it is not difficult to see that $I_2(n) \xrightarrow[]{n \to \infty} 0$. Consequently,\begin{align*}
		\Lambda(\alpha) := \lim_{n \to \infty} \Lambda_n(\alpha) = \frac{\gamma \sigma^2}{2} \int_0^1 s^{\gamma-1} (\alpha(1)-\alpha(s))^2 \, ds, \qquad \alpha \in \bv \cap D[0,1].
	\end{align*}
	Obviously, $\Lambda$ is G\^ateaux differentiable (with G\^ateaux derivative in $C[0,1]$). Therefore, the claim follows from the G\"artner--Ellis theorem, see e.\,g.\ \cite[Theorem 2.1,Theorem 2.4]{acosta-lp}.
\end{proof}

In order to carry over the moderate deviation principle from $(Z_n/S(n))_{n \in \mbb{N}}$ to $(X(t \lcdot)/S(t))_{t>0}$, we need the following auxiliary result.

\begin{lem} \label{add-n13}
	$(Z_{\floor{t}}/S(\floor{t}))_{t>0}$ and $(X(t\lcdot)/S(t))_{t>0}$ are exponentially equivalent, i.\,e.\ \begin{equation*}
		\limsup_{t \to \infty} \frac{t^{\gamma}}{S(t)^2} \log \mbb{P} \left( \left\| \frac{Z_{\floor{t}}(\lcdot)}{S(\floor{t})} - \frac{X(t\lcdot)}{S(t)} \right\|_{\infty} > \eps \right)=-\infty \fa \eps>0.
	\end{equation*}
\end{lem}

\begin{proof}
	Let $\eps>0$ and $\alpha>0$. Obviously, \begin{equation}
		\left\| \frac{Z_{\floor{t}}}{S(\floor{t})} - \frac{X(t\lcdot)}{S(t)} \right\|_{\infty}
		\leq \left|\frac{S(t)}{S(\floor{t})}- 1\right| \cdot \frac{\|Z_{\floor{t}}\|_{\infty}}{S(t)} + \left\| \frac{Z_{\floor{t}}}{S(t)}-\frac{X(t\lcdot)}{S(t)} \right\|_{\infty} =: A_t+B_t. \label{add-neq51}
	\end{equation}
	We estimate $\mbb{P}(A_t>\eps)$ and $\mbb{P}(B_t>\eps)$ separately. As in the proof of Lemma~\ref{add-n09} we find \begin{align*}
		\mbb{P}(A_t>\eps)
		\leq \mbb{P} \left( \max_{0 \leq k \leq \floor{t}} \frac{|X_k|}{S(t)} > \frac{\eps}{\delta(t)} \right)
		\leq 6 \exp \left[- \frac{S(t)^2}{t^{\gamma}} \left( \frac{\eps}{3\delta(t)} - \sigma^2 \right) \right]
	\end{align*}
	for $t$ sufficiently large where $\delta(t) := \left| \frac{S(t)}{S(\floor{t})} -1 \right| \to 0$ as $t \to \infty$, cf.\ \eqref{add-neq15}. In order to estimate $B_t$ we note that \begin{align*}
			\sup_{s \in [0,1]} |Z_{\floor{t}}(s)-X(t s)|

			&\leq \sup_{\substack{u,v \leq t \\ |u-v| \leq 2}} |X_{u}-X_v|
			\leq 3 \adjustlimits\max_{j \leq \floor{\alpha^{-1}}+1} \sup_{\substack{0 \leq u \leq \alpha t \\  j\alpha t+ u \leq t}} |X_{j \alpha t+u}-X_{j \alpha t}|
	\end{align*}
	for $t \geq t_0(\alpha)$ sufficiently large. Applying again Etemadi's inequality yields \begin{align*}
		\mbb{P}(B_t>\eps)
		&\leq 3 \sum_{j=0}^{\floor{\alpha^{-1}}+1} \sup_{\substack{0 \leq u \leq \alpha t \\ j\alpha t +u \leq t}}  \mbb{P} \left( |X_{j \alpha t+u}-X_{j\alpha t}|> \frac{S(t) \eps}{9} \right).
	\end{align*}
	By Markov's inequality and Lemma~\ref{add-n07}, \begin{align}
		\mbb{P}(B_t>\eps)
		& \leq 6\exp \left(-\frac{S(t) \lambda \eps}{9} \right)  \sum_{j=0}^{\floor{\alpha^{-1}}+1} \exp \left( \frac{\lambda^2}{2} (\var X_{(j+1)\alpha t}-\var X_{j \alpha t}) + \lambda^3 E_{0,t}(\lambda) \right). \label{add-neq53}
	\end{align}
	A similar calculation as in the second part of the proof of Lemma~\ref{add-n09} shows \begin{align*}

		\frac{1}{t^{\gamma}} (\var X_{(j+1)\alpha t}-\var X_{j\alpha t})
		&\leq 2\alpha c+ c \sup_{k \geq 1} \left| \frac{\var X_{k\alpha t}}{(k \alpha t)^\gamma}-\sigma^2 \right|=:\delta(\alpha,t)
	\end{align*}
	for some constant $c=c(\gamma)$. In particular, $\delta(\alpha,t) \xrightarrow[]{t \to \infty} 2 \alpha c$. Setting $\lambda := r S(t)/t^\gamma$, $r \geq 1$, we get \begin{align*}
		\limsup_{t \to \infty} \frac{t^\gamma}{S(t)^2} \log \mbb{P}(B_t>\eps)
		&\leq - \frac{r \eps}{9}+ \lim_{t \to \infty} \left( \frac{r^2}{2} \delta(\alpha,t) + r^3 \frac{S(t)}{t^{2\gamma}} E_{0,t} \left( r \frac{S(t)}{t^\gamma} \right) \right)
		= - \frac{r \eps}{9} +c \alpha r^2.
	\end{align*}
	Finally, we conclude \begin{align*}
		\limsup_{t \to \infty} \frac{t^\gamma}{S(t)^2} \log \mbb{P} \left(\left\| \frac{Z_{\floor{t}}}{S(\floor{t})} - \frac{X(t\lcdot)}{S(t)} \right\|_{\infty} > 2\eps \right)
		&\leq  - \frac{r \eps}{9}+c \alpha r^2  \xrightarrow[]{\alpha \to 0,r \to \infty} -\infty. \qedhere
	\end{align*}
\end{proof}

Combining Lemma~\ref{add-n11} and Lemma~\ref{add-n13}, we find that $(X(t \lcdot)/S(t))_{t>0}$ satisfies a moderate deviation principle with good rate function $J$ and speed $S(t)^2/t^{\gamma}$, cf.\ \cite[Theorem 4.2.13]{dz}. It remains to identify the good rate function.

\begin{thm} \label{add-n15}
	The good rate function $J$, \begin{equation*}
		J(f) = \sup_{\alpha \in \bv \cap D[0,1]} \left( \int f \, d\alpha - \frac{\gamma \sigma^2}{2} \int_0^1 s^{\gamma-1} (\alpha(1)-\alpha(s))^2 \, ds \right),
	\end{equation*}
	equals \begin{equation*}
		I(f) = \begin{cases} \displaystyle\frac{1}{2\gamma \sigma^2} \int_0^1 \frac{f'(s)^2}{s^{\gamma-1}} \, ds, & f \in \ac, f(0)=0, \\ \displaystyle\infty, & \text{otherwise}. \end{cases}
	\end{equation*}
	In particular, $\dom I = \dom J \subseteq \ac$.
\end{thm}

\begin{proof}
	We only consider $\gamma \geq 1$; the case $0<\gamma<1$ is proved similarly. First, we show that $J(f)< \infty$ implies $f \in \ac$ and $f(0)=0$. If so, then \begin{equation}
		f(t) = \int_0^t f'(s) \, ds, \qquad t \in [0,1]. \label{add-neq55}
	\end{equation}
	To this end, let $0<s_1<t_1 < \ldots < s_n < t_n \leq 1$ and define \begin{equation*}
		\alpha(t) := \sum_{j=1}^n c_j \I_{[s_j,t_j)}(t), \qquad t \in [0,1]
	\end{equation*}
	for $c=(c_1,\ldots,c_n) \in \mbb{R}^n$. Obviously, $\alpha \in \bv \cap D[0,1]$ and \begin{equation}
		\int_0^1 f \, d\alpha = \sum_{j=1}^n c_j (f(s_j)-f(t_j)). \label{add-neq57}
	\end{equation}
	By definition of $J$ we have \begin{align*}
		\sum_{j=1}^n c_j (f(s_j)-f(t_j))
		&= \int_0^1 f \, d\alpha
		\leq J(f) + \frac{\gamma \sigma^2}{2} \int_0^1 s^{\gamma-1} \alpha(s)^2 \, ds
		\leq J(f)+ \frac{\gamma \sigma^2 }{2} \sum_{j=1}^n c_j^2 |t_j-s_j|.
	\end{align*}

	If we choose $c_j := r \sgn(f(s_j)-f(t_j))$, $r \geq 1$, we get \begin{equation*}
		\sum_{j=1}^n |f(s_j)-f(t_j)| \leq \frac{J(f)}{r} + \frac{r \gamma \sigma^2 }{2} \sum_{j=1}^n |t_j-s_j|.
	\end{equation*}
	This proves that $f$ is absolutely continuous. A similar calculation shows \begin{equation*}
		|f(t)| \leq \frac{J(f)}{r}+t^{\gamma} \frac{r \sigma^2}{2}.
	\end{equation*}
	Letting $t \to 0$ and $r \to \infty$ we obtain $f(0)=0$. This proves \eqref{add-neq55}.
	Now let $f$ be given by \eqref{add-neq55}. If we set $\varphi(x) := x^2/2$, then we see from the integration by parts formula that \begin{align*}
		\int_0^1 f \, d\alpha - \frac{\gamma \sigma^2}{2} \int_0^1 s^{\gamma-1}& (\alpha(1)-\alpha(s))^2 \, ds \\
		&= \int_0^1  \left[ f'(s) (\alpha(1)-\alpha(s)) - \frac{\gamma \sigma^2}{2} s^{\gamma-1} (\alpha(1)-\alpha(s))^2 \right] \, ds \\
		&= \int_0^1 \left[f'(s) (\alpha(1)-\alpha(s)) - \varphi \left( \sqrt{\gamma \sigma^2 s^{\gamma-1}}  (\alpha(1)-\alpha(s))\right) \right] \, ds \\
		&\leq \int_0^1 \varphi^{\ast} \left( \frac{f'(s)}{\sqrt{\gamma \sigma^2  s^{\gamma-1}}} \right) \, ds \\
		&=\frac{1}{2 \gamma \sigma^2 } \int_0^1 \frac{f'(s)^2}{s^{\gamma-1}} \, ds = I(f)
	\end{align*}
	where \begin{equation*}
		\varphi^{\ast}(\beta) := \sup_{x \in \mbb{R}}\left(\beta x - \varphi(x)\right) = \frac{1}{2} \beta^2, \qquad \beta \in \mbb{R},
	\end{equation*}
	denotes the Legendre transform of $\varphi$. Hence, $J(f) \leq I(f)$. On the other hand, for $f \in \ac$,\begin{equation*}
		\sum_{j=0}^{n-1} \frac{f \left( \frac{j+1}{n} \right)- f \left( \frac{j}{n} \right)}{\frac{1}{n}} \left( \frac{n}{j+1} \right)^{(\gamma-1)/2} \I_{\big[\frac{j}{n},\frac{j+1}{n} \big)}(s) \xrightarrow[]{n \to \infty} \frac{f'(s)}{s^{(\gamma-1)/2}} \qquad \text{for almost all $s \in [0,1]$}.
	\end{equation*}

	Therefore, Fatou's lemma and \eqref{add-neq57} imply \begin{align*}
		I(f)
		&\leq \frac{1}{2\sigma^2 \gamma} \liminf_{n \to \infty} \left( n \sum_{j=0}^{n-1} \left[f \left( \frac{j+1}{n} \right)- f \left( \frac{j}{n} \right) \right]^2 \left( \frac{n}{j+1} \right)^{\gamma-1} \right) \\
		&= \liminf_{n \to \infty} \bigg( \int_0^1 f \, d\alpha^{[n]} - \frac{\sigma^2 \gamma}{2} \sum_{j=0}^{n-1} \int_{j/n}^{(j+1)/n} (\alpha_j^{[n]})^2 \left( \frac{j+1}{n} \right)^{\gamma-1} \,ds \bigg)
	\end{align*}
	for \begin{align*}
		\alpha_j^{[n]} &:= - \frac{n}{\sigma^2 \gamma}  \left( \frac{n}{j+1} \right)^{\gamma-1} \left[f \left( \frac{j+1}{n} \right)- f \left( \frac{j}{n} \right) \right] \\
		\alpha^{[n]}(t) &:= \sum_{j=0}^{n-1} \alpha_j^{[n]} \I_{[j/n,(j+1)/n)}(t).
	\end{align*}
	Using that $\left(\frac{j+1}{n} \right)^{\gamma-1} \geq s^{\gamma-1}$ for any $s \in [j/n,(j+1)/n]$, we get  \begin{equation*}
		I(f)
		\leq \liminf_{n \to \infty} \left( \int_0^1 f \, d\alpha^{[n]} - \frac{\sigma^2 \gamma}{2} \int_0^1  \alpha^{[n]}(s)^2 s^{\gamma-1} \, ds \right)
		\leq J(f). \qedhere
	\end{equation*}
\end{proof}

Next, we prove Strassen's law, Theorem~\ref{add-n03}. The proof is inspired by \cite[Chapter 13]{bm2} where the result is shown for Brownian motion. Without loss of generality, we assume throughout the proof of Theorem~\ref{add-n03} that $\sigma^2=1$. We need the following lemma.

\begin{lem} \label{add-n17}
Let $I$ be the good rate function defined in \eqref{add-neq23}. For $c>0$ set \begin{align*}
	A &:= \left\{f \in D[0,1]; \adjustlimits\sup_{q^{-1} \leq t \leq 1} \sup_{0 \leq s \leq 1} |f(st)-f(s)| \geq c \right\}, & B := \left\{f \in D[0,1]; \sup_{s\in [0,1]} |f(s)| \geq c \right\}.
\end{align*}
Then $A$ and $B$ are closed  in $(D[0,1],\|\cdot\|_{\infty})$ and \begin{align*}
	\inf_{f \in A} I(f) \geq \frac{c^2}{2} \frac{q^{\gamma}}{q^{\gamma}-1}, \qquad \quad \inf_{f \in B} I(f) \geq \frac{c^2}{2}.
\end{align*}
\end{lem}

\begin{proof}
	By the Cauchy--Schwarz inequality, \begin{align*}
		c^2
		\leq \adjustlimits\sup_{q^{-1} \leq t \leq 1} \sup_{0 \leq s \leq 1} |f(st)-f(s)|^2
		&= \adjustlimits\sup_{q^{-1} \leq t \leq 1} \sup_{0 \leq s \leq 1} \left| \int_{st}^{s} f'(r)\, dr \right|^2 \\
		&\leq \adjustlimits\sup_{q^{-1} \leq t \leq 1} \sup_{0 \leq s \leq 1} \left[ \left( \int_0^1 \frac{f'(r)^2}{r^{\gamma-1}} \, dr \right) \left(\int_{st}^s r^{\gamma-1} \,dr \right) \right] \\

		&= 2 (1-q^{-\gamma}) I(f)
	\end{align*}
	for any $f \in A \cap \dom I \subseteq \ac$. Similarly, we find from the Cauchy--Schwarz inequality\begin{equation*}
		c^2

		\leq \sup_{0 \leq s \leq 1} \left| \int_0^s f'(r) \, dr \right|^2
		\leq 2\gamma I(f) \int_0^1 r^{\gamma-1} \, dr = 2 I(f) \fa f \in B. \qedhere
	\end{equation*}
\end{proof}

\begin{lem} \label{add-n19}
	The set of limit points $\mc{L}(\omega)$ satisfies $\mc{L}(\omega) \subseteq \Phi(\frac{1}{2})$ for almost all $\omega \in \Omega$.
\end{lem}

\begin{proof}
	Since $\Phi(\frac{1}{2}) = \bigcap_{r>0} \Phi(\frac{1}{2}+r)$, it suffices to show $\mc{L}(\omega) \subseteq \Phi(\frac{1}{2}+r)$ for any $r>0$. Set $Z_t := X(t \lcdot)/S(t)$, and fix $q>1$, $\delta>0$. Applying the large deviation upper bound (cf.\ \cite[Theorem 3.3.3]{fw}) gives \begin{align*}
		\mbb{P} \left( d\big(Z_{q^n},\Phi(\tfrac{1}{2}+r)\big)>\delta \right)
		&\leq \exp \left(-2 \left( \frac{1}{2}+\frac{r}{2} \right) \log \log q^n \right)
	\end{align*}
	for $n$ sufficiently large. Thus, by the Borel--Cantelli lemma, \begin{equation*}
		d\big(Z_{q^n}(\lcdot,\omega),\Phi(\tfrac{1}{2}+r)\big) \leq \delta \end{equation*}
	for $n \geq n_0(q,\omega)$. It remains to fill the gaps in the sequence $(q^n)_{n \in \mbb{N}}$. Note that \begin{align*}
		\sup_{q^{n-1}\leq t\leq q^n} \|Z_t - Z_{q^n}\|_\infty
		&= \adjustlimits\sup_{q^{n-1}\leq t\leq q^n}\sup_{0\leq s\leq 1}\left|\frac{X(st)}{S(t)} - \frac{X(sq^n)}{S(q^n)}\right|\\
		&\leq \adjustlimits\sup_{q^{n-1}\leq t\leq q^n}\sup_{0\leq s\leq 1} \frac{|X(st)-X(sq^n)|}{S(q^n)}
		+ \adjustlimits\sup_{q^{n-1}\leq t\leq q^n}\sup_{0\leq s\leq 1}  \frac{|X(st)|}{S(q^n)}\left|\frac{S(q^n)}{S(t)}-1\right| \\
		&=: A_n+B_n.
	\end{align*}
	As \begin{equation*}
		\mathbb{P}\left(B_n \geq \frac{\delta}{2} \right)
		= \mathbb{P} \left( \sup_{0 \leq t \leq 1} \left| \frac{X(t q^n)}{S(q^n)} \right| \cdot \left| \frac{S(q^n)}{S(q^{n-1})}-1 \right| \geq \frac{\delta}{2} \right),
	\end{equation*}
	it follows easily from $S(q^n)/S(q^{n-1}) \xrightarrow[]{n \to \infty} q^{\gamma/2}$ and the large deviation upper bound that $\sum_{n \in \mbb{N}} \mathbb{P}(B_n \geq \delta/2)<\infty$ if we choose $q>1$ close to $1$. Hence, by the Borel--Cantelli lemma, $B_n \leq \frac{\delta}{2}$ for $n \geq n_1(q,\omega)$ sufficiently large. In order to estimate $A_n$ we note that\begin{align*}
		\mbb{P}\left( A_n \geq \frac{\delta}{2} \right)
		&= \mbb{P}\left( \adjustlimits\sup_{q^{-1} \leq t \leq 1} \sup_{0 \leq s \leq 1} \frac{|X(s q^n t)-X( sq^n)|}{S(q^n)} \geq \frac{\delta}{2} \right)
		\leq \mbb{P} \left( \frac{X(q^n \lcdot)}{S(q^n)} \in A \right)
	\end{align*}
	for \begin{equation*}
		A := \left\{f \in D[0,1]; \adjustlimits\sup_{q^{-1} \leq t \leq 1} \sup_{0 \leq s \leq 1} |f(st)-f(s)| \geq \frac{\delta}{2} \right\}.
	\end{equation*}
	Therefore, by the large deviation upper bound and Lemma~\ref{add-n17}, \begin{equation*}
		\mbb{P}\left( A_n\geq \frac{\delta}{2} \right) \leq \exp \left(- \frac{\delta^2}{8} \frac{q^\gamma}{q^\gamma-1}  \log \log q^n \right)
	\end{equation*}
	for  $n \geq n_2(q)$ sufficiently large.
	If $q>1$ is close to $1$, this implies $\sum_{n \in \mbb{N}} \mbb{P}(A_n\geq\delta/2)<\infty$. By the Borel--Cantelli lemma, we conclude \begin{equation*}
		\sup_{q^{n-1} \leq t \leq q^n} \|Z_t-Z_{q^n}\|_{\infty} \leq \delta
	\end{equation*}
	for $n \geq n_3(q,\omega)$ sufficiently large. Finally,\begin{align*}
		d\big(Z_s(\lcdot,\omega),\Phi(\tfrac 12+r)\big)
		\leq\|Z_s(\lcdot,\omega)-Z_{q^n}(\lcdot,\omega)\|_\infty + d\big(Z_{q^n}(\lcdot,\omega),\Phi(\tfrac 12+r)\big)
		\leq 2\delta
	\end{align*}
	for $s$ sufficiently large. Since $\Phi(\frac{1}{2}+r)$ is closed, this proves the claim.
\end{proof}

	\begin{lem} \label{add-n21}
		$\mc{L}(\omega) \supseteq \Phi(\frac{1}{2})$ for almost all $\omega \in \Omega$.
	\end{lem}

	\begin{proof}
		Since the sublevel sets are compact, we have $\overline{\bigcup_{r<1/2} \Phi(r)} \subseteq \Phi(\frac{1}{2})$. On the other hand, any $f \in \Phi(\frac{1}{2})$ can be approximated by $(1-\eps) f \in \Phi(\frac{(1-\eps)^2}{2})$. Therefore, $\overline{\bigcup_{r<1/2} \Phi(r)} = \Phi(\frac{1}{2})$. Consequently, it suffices to show that for any $r<\frac{1}{2}$, $\eps>0$, $f \in \Phi(r)$ there is a.\,s.\ a sequence $s_n=s_n(\omega) \to \infty$ such that
		\begin{equation*}
			\limsup_{n \to \infty} \|Z_{s_n}(\omega)-f\|_{\infty} \leq \eps.
		\end{equation*}
		Pick $q>1$. Obviously, \begin{align*}
		 	\|Z_{q^n}-f\|_{\infty}
		 	&\leq \sup_{q^{-1} \leq t \leq 1} \left| \frac{X(t q^n)-X(q^{n-1})}{S(q^n)} - f(t) \right| + \left|\frac{X(q^{n-1})}{S(q^n)}\right| + \sup_{t \leq q^{-1}} |f(t)| + \sup_{t \leq q^{-1}} \left|\frac{X(t q^n)}{S(q^n)}\right|.
		 \end{align*}
		We estimate the terms separately. Setting \begin{align*}
			A := \left\{g \in D[0,1]; \sup_{q^{-1} \leq t \leq 1} |g(t)-g(q^{-1})-f(t)| < \frac{\eps}{4} \right\}
		\end{align*}
		we have \begin{align*}
			\mbb{P}(A_n) := \mbb{P} \left( \sup_{q^{-1} \leq t \leq 1} \left| \frac{X(t q^n)-X(q^{n-1})}{S(q^n)} - f(t) \right| < \frac{\eps}{4} \right)
			&= \mbb{P} \left( \frac{X(q^n \lcdot)}{S(q^n)} \in A \right).
		\end{align*}
		If we choose $q>1$ sufficiently large such that $|f(q^{-1})|<\frac{\eps}{4}$, then $f \in A$. By assumption, $I(f) < \frac{1}{2}$ and therefore we conclude from the large deviation lower bound that $\sum_{n \in \mbb{N}} \mbb{P}(A_n)=\infty$. Taking a subsequence, if necessary, we obtain by applying the Borel--Cantelli lemma \begin{equation*}
			\adjustlimits\limsup_{n \to \infty} \sup_{q^{-1} \leq t \leq 1} \left| \frac{X(t q^n)-X(q^{n-1})}{S(q^n)} - f(t) \right| \leq \frac{\eps}{4}.
		\end{equation*}
		By H\"older's inequality, \begin{equation*}
			\sup_{t \leq q^{-1}} |f(t)|^2
			= \sup_{t \leq q^{-1}}\left|\int_0^{t} f'(s) \, ds\right|^2
			\leq \left( \int_0^1 \frac{f'(s)^2}{s^{\gamma-1}} \, ds \right) \left( \int_0^{q^{-1}} s^{\gamma-1} \, ds \right)
			\leq \frac{1}{q^{\gamma}}.
		\end{equation*}
		Moreover, \begin{align*}
			\mbb{P} \left( \left| \frac{X(q^{n-1})}{S(q^n)} \right| \geq \frac{\eps}{4} \right)+\mbb{P} \left( \sup_{0 \leq t \leq q^{-1}} \left| \frac{X(t q^n)}{S(q^n)} \right| \geq \frac{\eps}{4} \right)
			&\leq 2 \mathbb{P} \left( \sup_{0 \leq t \leq 1} \left|\frac{X(q^{n-1}t)}{S(q^{n-1})} \right| \frac{S(q^{n-1})}{S(q^{n})} \geq \frac{\eps}{4} \right).
			\end{align*}

		By Lemma~\ref{add-n17}, it is not difficult to see that we may apply again the Borel--Cantelli lemma if we choose $q>1$ sufficiently large. Hence,
		\begin{equation*}
			\limsup_{n \to \infty} \left(\left|\frac{X(q^{n-1})}{S(q^n)}\right| + \sup_{t \leq q^{-1}} |f(t)| + \sup_{t \leq q^{-1}} \left|\frac{X(t q^n)}{S(q^n)}\right| \right) \leq \frac{3}{4} \eps.
		\end{equation*}
		This finishes the proof.
	\end{proof}

	Combining  Lemma~\ref{add-n19} and Lemma~\ref{add-n21} yields Theorem~\ref{add-n03}. Finally, it remains to prove Corollary~\ref{add-n05}.

	\begin{proof}[Proof of Corollary~\ref{add-n05}]
		It follows from the assumptions that there exists an additive process $(Y_t)_{t \geq 0}$ satisfying the assumptions of Theorem~\ref{add-n01} such that $((X(t \lcdot)-Y(t \lcdot))/S(t))_{t > 0}$ converges uniformly in $D[0,1]$ to $0$ as $t \to \infty$. Therefore, the claim follows from Theorem~\ref{add-n03}. For more details, we refer the reader to \cite[Proposition 3.1,3.2]{wang}.
	\end{proof}

\section{Concluding remarks} \label{s-rem} \begin{enumerate}
	\item Theorem~\ref{add-n01} still holds if $(X_t)_{t \geq 0}$ has also fixed jump discontinuities. Taking a close look at the proof reveals that we simply have to modify the estimate of the moment generating function in Lemma~\ref{add-n07} appropriately. The corresponding estimate follows from the explicit formula for the exponential moments, cf.\ \cite{fuji}, and well-known elementary inequalities.
	\item Let $(X_t)_{t \geq 0}$ be an additive process such that $\mbb{E}X_t=0$ and \begin{equation*}
		\lim_{t \to \infty} \frac{\var X_t}{h(t)} =\sigma^2>0
	\end{equation*}
	for a regulary varying function $h:(0,\infty) \to (0,\infty)$ of index $\gamma >0$ (see e.\,g.\ \cite{bing}  for the definition). If we replace $t^{\gamma}$ in Theorem~\ref{add-n01} and Theorem~\ref{add-n03} by $h(t)$, then both theorems remain valid.
\end{enumerate}

\begin{ack}
	Financial support through the \emph{Deutsche Forschungsgemeinschaft}, DFG, grant SCHI 419/5--2 is gratefully acknowledged.
\end{ack}


TITLE: How to find $\mu(X\setminus A)?$
QUESTION [0 upvotes]: This question is  highly motivated from this
I was  searching the example .After that this example given below came in my mind
Given that $f_n :X \to [0,\infty]$ is measurable for each positive integer  $n$
let $f_n = \chi_{[n,n+1]}$ where  $A=[n,n+1]$
now $\int_{X}\chi_A d\mu=\mu(A)=n+1-n=1$.Then $\int_X{\chi_{X\setminus A}}=\mu(X\setminus A)=?$
My question is that
How to find $\mu(X\setminus A)?$

REPLY [0 votes]: $X$ supposed to be $\mathbb{R}$
since  $A \subseteq X$, So $X$ is the disjoint union of $A$ and $X \setminus A$. Hence,
$\mu(X) = \mu(A) + \mu(X\setminus A)$.
$\mu(X\setminus A)=\mu(X) - \mu(A)=\infty-1=\infty$


TITLE: Show that $f(x,y)=x^2y^4+x^4y^2-3x^2y^2+1≥0 \ \forall (x,y)$
QUESTION [0 upvotes]: Show that $$f(x,y)=x^2y^4+x^4y^2-3x^2y^2+1≥0 \ \forall (x,y)$$
I tried finding the critical points but $\delta = (f_{12})^2-f_{11}f_{22}$
Which is the test whether the point is saddle or extremum came out to be
$\delta = 0$ which doesn't let us make any conclusions.
Any suggestions on what we should do here?

REPLY [0 votes]: If you want another way, then
$$
\begin{gathered}
  x^{\,2} y^{\,2} \left( {x^{\,2}  + y^{\,2}  - 3} \right) + 1 \geqslant 0 \hfill \\
  r^{\,2} \left( {\cos ^{\,2} \theta \sin ^{\,2} \theta } \right)\left( {r^{\,2}  - 3} \right) + 1 \geqslant 0 \hfill \\
  \frac{{r^{\,2} }}
{4}\left( {\sin ^{\,2} \left( {2\theta } \right)} \right)\left( {r^{\,2}  - 3} \right) + 1 \geqslant 0 \hfill \\
\end{gathered}
$$
and
$$
\min \frac{{x(x - 3)}}
{4} = \left. { - \frac{9}
{{16}}\;} \right|_{\,x = 3/2}
$$


TITLE: Kernel of ring homomorphism is an ideal
QUESTION [1 upvotes]: I am asked to show that if f is a ring homomorphism from R to R' then kernel of f is an ideal of R.
According to definition of ideal : A non empty subset of R is an ideal for any two elements of ideal their substraction must be in that ideal and
the product of any element of R and an element of ideal must be in ideal,
I am not able to prove second condition. Please help

REPLY [2 votes]: Let $f: R\rightarrow R'$ be a ring homomorphism. We assume that $R$ and $R'$ both have an identity. Since $0 \in\ker(f)=\{x\in R\mid f(x)=0\}, \ker(f)$ is non-empty.
Let $u,v \in \ker(f), r \in R$, then $f(ru)=f(r)f(u)=0,\ f(ur)=f(u)f(r)=0,\ f(u-v)=f(u)-f(v)=0$.


TITLE: Primes $p$ such that $5$ is primitive root $\bmod{p}$ , where $p$ is a $321$ prime
QUESTION [4 upvotes]: How to prove following statement :

Let $~p~$ be Thabit $(321)$ prime of the form : $p=3\cdot 2 ^n-1$
and let $~n~$ be an odd number then :
$~5~$ is a primitive root modulo $~p~$ iff $~n \equiv 3,7 \pmod 8 $

I have checked statement for following consecutive odd exponents :
$n \in \{1,3,7,11,43,55,103,143 \}$
I guess that one should prove that :
$ord_p(5)=2\cdot(3\cdot 2^{n-1}-1)$

Added :
Note that if : $n \equiv 3,7 \pmod 8$ then : $p \equiv 3 \pmod 5$
Le't  assume that :
$ord_p(5)=3\cdot 2^{n-1}-1$ , and  $~3\cdot 2^{n-1}-1$ is a prime number ,so:
$$5^{3\cdot 2^{n-1}-1} = 5^{\frac{p-1}{2}} \equiv 1 \pmod p ~~\text {and}~~\left(\frac{5}{p}\right) \equiv 5^{\frac{p-1}{2}} \pmod p \Rightarrow$$
$$\Rightarrow \left(\frac{5}{p}\right) = 1 \Rightarrow p \equiv 1,4 \pmod 5$$
which is contradiction since : $p \equiv 3 \pmod 5$ , therefore :
$ord_p(5) = 2 \cdot(3\cdot 2^{n-1}-1)$ , since it cannot be $2$ , so :
$5$ is primitive root modulo $p$ if both $~3\cdot 2^{n}-1 ~~\text{and}~~ 3\cdot 2^{n-1}-1$
are prime numbers .

However , I didn't manage to prove case when $~3\cdot 2^{n-1}-1~$ is a composite number .
In case when $3 \cdot 2^{n-1}-1$ is a composite number one should prove that :
$(5^m)^2 \not \equiv 1 \pmod p$
where $m$ is a proper factor of  $3 \cdot 2^{n-1}-1$
Note that if $n \equiv 3,7 \pmod 8$ then $p \equiv 3 \pmod 4$

Added :
$x^2 \equiv a \pmod n$ where $n$ is a prime number has no solution if : $\left(\frac{a}{n}\right)=-1$
So let's assume  that $\left(\frac{a}{n}\right) = 1$, if $n \equiv 3 \pmod 4$, Lagrange found that the solutions are given by :
$x \equiv \pm a^{(n+1)/4} \pmod n$
So :
If $~\left(\frac{1}{p}\right) = 1~$ then :
$5^m \equiv 1 \pmod p ~~\text{and}~~ 5^m \equiv -1 \pmod p$
But if $~5^m \equiv 1 \pmod p~$ then $~5^{\frac{p-1}{2}} \equiv 1 \pmod p ~$ since $m$ is a proper factor of $~\frac{p-1}{2}$ .
$$ 5^{\frac{p-1}{2}} \equiv 1 \pmod p ~~\text {and}~~\left(\frac{5}{p}\right) \equiv 5^{\frac{p-1}{2}} \pmod p \Rightarrow$$
$$\Rightarrow \left(\frac{5}{p}\right) = 1 \Rightarrow p \equiv 1,4 \pmod 5$$
which is contradiction since : $p \equiv 3 \pmod 5$ , therefore :
$5^m \not \equiv 1 \pmod p$
Now , I don't know how to prove or disprove that :
$5^m \not \equiv -1 \pmod p$
Since $~p\equiv 3 \pmod 5~$ question whether $~5^m \not \equiv -1 \pmod p$ ?
is equivalent to the question whether $\frac{5^m+1}{p} \not \equiv 2 \pmod {10}$ ?

REPLY [1 votes]: Checking the first few cases with sage, I found
$n=3,4,7,11,43,55,64,76,103,143,216,324$,
suggesting the condition $n\equiv0,3,4,7\pmod8$,
or excluding $n$ even, $n\equiv3\pmod4$.
Your hypothesis should therefore be for $n>1$ odd
(since for $n=1$, $5$ is not a primitive root mod $p=5$).
# http://sage.csuohio.edu/home/pub/87/
def is_primitive_root(a,m):
    if gcd(a,m)>1:
        raise ValueError('a must be relatively prime to m')
    phi=euler_phi(m)
    powers = [ phi//q for q in prime_factors(phi) ]
    return all(pow(a,k,m)!=1 for k in powers)
# our test:
for n in range(1,513):
    p = 3 * 2^n - 1
    if not is_prime(p):
        continue
    k = 1 if gcd(5,p)==1 and is_primitive_root(5,p) else 0
    print 'k=%d n=%3d p=%d' % (k, n, p)

k=0 n=  1 p=5
k=0 n=  2 p=11
k=1 n=  3 p=23
k=1 n=  4 p=47
k=0 n=  6 p=191
k=1 n=  7 p=383
k=1 n= 11 p=6143
k=0 n= 18 p=786431
k=0 n= 34 p=51539607551
k=0 n= 38 p=824633720831
k=1 n= 43 p=26388279066623
k=1 n= 55 p=108086391056891903
k=1 n= 64 p=55340232221128654847
k=1 n= 76 p=226673591177742970257407
k=0 n= 94 p=59421121885698253195157962751
k=1 n=103 p=30423614405477505635920876929023
k=1 n=143 p=33451117797795934712303577408972542258970623
k=0 n=206 p=308532104497726132904056721729503219684262974806296224377864191
k=1 n=216 p=315936875005671560093754083051011296956685286201647333762932932607
k=0 n=306 p=391110907456221328563541572174600606921881931583859760122138966276041209554560647587212296191
k=1 n=324 p=102527377724203683954961041896138501500929817073119332957457997175466546837470746401102180172955647


TITLE: How would this rapid depressurization of water really behave?
QUESTION [1 upvotes]: An in-development game(Starbound) recently released a demo of their water physics. At about 39 seconds into the video, a volume of water in a container is released when the cap on the container is destroyed.
Then the water expands to about 10 times its initial volume in 3 seconds or so.
What would actually happen here? I imagine that compressing water to 1/10th its volume at STP would take an obscene amount of pressure, but I'm not sure about what would actually happen with this rapid decompression.
Would the water freeze from the drop in temperature as it expanded? How fast would the decompression actually take? Is the entire question moot because the pressures involved are too high?

REPLY [2 votes]: There's a lot of info about water phase and density here.
Compressing water to a tenth of it's STP density would give a molar volume of 1.8 cm$^3$. The highest density mentioned in the article is 8 cm$^3$mol$^{-1}$ for ice VIII at about a terapascal, so a density of 1.8 cm$^3$mol$^{-1}$ while technically possible seems unlikely from a pratical point of view.
If water were compressed to anything like this density it would form ice of some form. On release of the pressure the ice would expand to it's normal density at approximately the speed of sound (in the ice) and probably shatter into bits in the process. The ice would them melt relatively slowly and without any further spectacle.
Later: it occurs to me that there would be so much elastic energy stored in the compressed ice that the expansion would actually be more like an explosion. The compressed ice would blow itself to bits along with anything else in the vicinity. Incidentally the density of matter in white dwarf stars is allegedly around $10^6$ g/cm$^3$, so compressing ice by a factor of ten is certainly possible.


TITLE: Inequality involving Moore-Penrose pseudoinverse
QUESTION [0 upvotes]: Let $A=\begin{bmatrix}A_1\\A_2\end{bmatrix}\in \mathbb{F}^{(n+m)\times n}$, with $A_1\in \mathbb{F}^{n\times n}$ and non singular, and $\mathbb{F}=\mathbb{C}$ or $\mathbb{R}$. Show that
$$\lVert A^\dagger\rVert_2 \leq \lVert A_1^{-1}\rVert_2$$
How should I approach the problem?

REPLY [1 votes]: $$
\|A^\dagger\|_2^{-1}
=\min_{\|x\|_2=1}\|Ax\|_2
=\min_{\|x\|_2=1}\left\|\begin{bmatrix}A_1x\\A_2x\end{bmatrix}\right\|_2
\geq\min_{\|x\|_2=1}\left\|A_1x\right\|_2
=\|A_1^{-1}\|_2^{-1}
$$

Start with the definition:
$$
\|A^\dagger\|_2=\max_{x\neq 0}\frac{\|A^\dagger x\|_2}{\|x\|_2}.
$$
Any such $x$ can be written in the form $x=y+z$, where
$$y\in\mathcal{R}(A) \quad\text{and}\quad z\in\mathcal{R}(A)^\perp=\mathcal{N}(A^*)=\mathcal{N}(A^\dagger).$$
We have that $A^\dagger x=A^\dagger y+A^\dagger z=A^\dagger y$ and since $y\perp z$, $\|x\|_2^2=\|y\|_2^2+\|z\|_2^2$. The maximum is therefore attained when $z=0$. We can write $y=Aw$ for some $w$. All this given then
$$
\|A^\dagger\|_2=\max_{w\neq 0}\frac{\|A^\dagger Aw\|_2}{\|Aw\|_2}.
$$
Since the rank of $A$ is $n$, $A^\dagger A=I$, and
$$
\|A^\dagger\|_2=\max_{w\neq 0}\frac{\|w\|_2}{\|Aw\|_2}=\left(\min_{w\neq 0}\frac{\|Aw\|_2}{\|w\|_2}\right)^{-1}.
$$


TITLE: The radial part of a normal distribution
QUESTION [4 upvotes]: I am reading a paper that asks me to sample $s_i$ from a distribution like this:
$s_i \sim (2\pi)^{-\frac{d}{2}}A^{-1}_{d-1}r^{d-1}e^{-\frac{r^2}{2}}$
"Here the normalization constant $A_{d−1}$ denotes the surface volume of the d-dimensional unit ball.
Thus $s_i$ matches the radial part of a normal distribution"
What is the radial part of a normal distribution?
The distribution above is related to the inverse of the hyper-surface of the unit ball, but the radius part is not inverted. Can i just sample from a normal distribution and scale? I am trying to get a sample from this distribution in C++.
Thanks in advance for any help in understanding the distribution above.

REPLY [4 votes]: I gauss you are working on the fast-food method. As it is pointed out in the paper the distribution matches the radial part of a normal distribution. For $d = 2$ it is equal to Rayleigh distribution so you can use the standard implementations to sample $s_i$. For higher value of d you can sample from a $d$ dimensional zero mean gaussian distribution and then compute the length of the random vector. This method works well in terms of correctness. However, generating n samples using this method needs $nd$ time which is much higher than the overall time complexity of the fast-food algorithm. Thus, it is fine just for testing the correctness of the algorithm.
Update: I recently realized that $s_i$ has chi distribution. Equivalently, you can sample from chi-squared distribution and compute the square root of the samples. I succsesfully tested it in the Matlab and following you can find the source code:
function [ ws ] = FF_samples( ndim, nsamples, sigm)
    ws = [];
    d = 2 ^ ceil(log2(ndim));
    x = ceil(nsamples/d);
    for bi = 1:x
        I = eye(d);

        B = diag(2*randi(2, d, 1) - 3);
        H = hadamard(d);
        P = I(randperm(d), :);
        G = diag(normrnd(0,1, d, 1));
        S = diag( chi2rnd(d,d,1) .^ .5) * sum(diag(G) .^ 2).^-.5;
        %%
        nws = (1/sigm / d)^.5 * S * H * G * P * H * B;
        ws = [ws; nws];
    end
    ws = ws(1:nsamples, 1:ndim);
end

P.S.: It still needs some improvement since I don't use the fast Fourier transform like algorithm for multiplication of the Hamard matrix to a vector.


text:
\begin{document}

\author{Artem N. Shevlyakov \footnote{The author was supported by RFII grant 14-01-00068}}
\title{On disjunctions of algebraic sets in completely simple semigroups}

\maketitle

\abstract{A semigroup $S$ is called an equational domain if any finite union of algebraic sets over $S$ is algebraic. We give some necessary and sufficient conditions for a completely simple semigroup to be an equational domain.}

\section{Introduction}

G. Baumslag, A. Miasnikov and V.~Remeslennikov~\cite{AG_over_groupsI,AG_over_groupsII} laid some basics of algebraic geometry over groups. Recall that an equation over a group $G$ is an equality $w(X)=1$, where $w(X)$ is an element of the free product $G\ast F(X)$. An algebraic set over $G$ is the set of all solutions of a system of equations over $G$.

Finite unions of algebraic sets are not necessarily algebraic over $G$. However, there exist groups (which were completely described in the paper written by E.~Daniyarova, A.~Miasnikov and V.~Remeslennikov~\cite{uniTh_IV}) where any finite union of algebraic sets is algebraic. Following~\cite{uniTh_IV}, the groups with such property are called equational domains.

One can similarly pose the problem of existence of equational domains in semigroups. To solve this problem it is natural to search equational domains in the varieties of semigroups which are close to groups. One such ``group-like'' class is that of completely simple semigroups, since any completely simple semigroup is a disjoint union of copies of a group.

Any completely simple semigroup admits operation of the inversion ${}^{-1}$, hence we shall consider completely simple semigroups as algebraic structures in the language $\LL=\{\cdot,{}^{-1}\}$. The use of the language $\LL$ instead of $\{\cdot\}$ is quite convenient, since the class of completely simple semigroups forms a variety in the language $\LL$.

In the current paper we prove the necessary and sufficient conditions for a completely simple semigroup $S$ to be an equational domain (Theorem~\ref{th:criterion}). From the obtained criterion it follows that  free completely simple semigroups of rank $n\geq 2$ and any free product of arbitrary completely simple semigroups are equational domains (similar results hold in the variety of groups). Moreover, in Section~\ref{sec:generalization} we extend Theorem~\ref{th:criterion} to other semigroup languages.

\section{Notions of semigroup theory}

A subset $I\subseteq S$ is called a \textit{left (right) ideal} if $sa\in I$ (respectively, $as\in I$) for any $s\in S$, $a\in I$. If $I$ is a right and left ideal simultaneously, we say that $I$ is a {\it two-sided ideal} (or {\it an ideal} for shortness).

A semigroup $S$ with a unique ideal $I=S$ is called \textit{simple}. A simple semigroup with minimal left and right ideal is \textit{completely simple (c.s.)}.

The next classic theorem completely describes c.s. semigroups.

\begin{theorem}
\label{th:rees}
For any c.s. semigroup $S$ there exists a group $G$ and sets $I,\Lambda$ such that  $S$ is isomorphic to the set of triples $(\lambda,g,i)$, $g\in G$, $\lambda\in\Lambda$, $i\in I$ with multiplication
\[
(\lambda,g,i)(\mu,h,j)=(\lambda,gp_{i\mu}h,j),
\]
where $p_{i\mu}\in G$ is an element of a matrix $\P$ such that
\begin{enumerate}
\item $\P$ consists of  $|I|$ rows and $|\Lambda|$ columns;
\item the matrix $\P$ is \textit{normalised}, i.e.
\[p_{1i}=p_{\lambda 1}=1\in G\mbox{ for all }\lambda\in\Lambda,\;i\in I.\]
\end{enumerate}
\end{theorem}

Following Theorem~\ref{th:rees}, we denote any c.s. semigroup $S$ by $S=(G,\P,\Lambda,I)$. The group $G$ and the matrix $\P$ are called \textit{the structural group} and~\textit{sandwich-matrix}, respectively. The elements $\lambda,i$ occurring in a triple $(\lambda,g,i)\in S$ are \textit{the first and the second indexes}, respectively.

By Theorem~\ref{th:rees}, any c.s. semigroup $S$ is a disjoint union of copies of the structural group  $G$. Clearly, the identity elements of the maximal subgroups are $(\lambda,p_{i\lambda}\1,i)$, $\lambda\in\Lambda$, $i\in I$. The inversion ${}\1$ in the subgroup defined by the indexes $\lambda,i$ is given by
\[
(\lambda,g,i)\1=(\lambda,p_{i\lambda}\1g\1p_{i\lambda}\1,i).
\]

The class of c.s. semigroups is a variety in the language $\{\cdot,{ }\1\}$, since it is defined by the identities:
\[
xx\1 x=x,\; xx\1=x\1 x,\; (x\1)\1=x,\; (xyx)\1(xyx)=x\1 x.
\]

The structure of free c.s. semigroups is described by the next theorem.

\begin{theorem}\textup{\cite{clifford_free,rasin}}
\label{th:free_CSS_structure}
Let $X=\{x_1,x_2,\ldots,x_n\}$, $Y=\{y_{i\lambda}|i\in I,\lambda\in\Lambda\}$ be finite sets of letters, and $I=\Lambda=\{1,2,\ldots, n\}$. Let $F(X\cup Y)$ denote the free group generated by the set $X\cup Y$. Then, the free c.s. semigroup $F_{css}(X)$ generated by $X$ is defined by  $F_{css}=(F(X\cup Y),\P,I,\Lambda)$, where $\P=(y_{i\lambda})$ and the generators $x_i$ correspond to the triples $(i,x_i,i)\in F_{css}(X)$.
\end{theorem}

It is easy to check that the free c.s. semigroup of rank $1$ is isomorphic to the infinite cyclic semigroup of the same rank.

One can define the free product in the variety of c.s. semigroups. The representation of the free product of two c.s. semigroups is given in the next theorem.

\begin{theorem}\textup{\cite{jones}}
\label{th:free_product_CSS_structure}
Let $S=S_1\ast S_2$ be the free product of two c.s. semigroups $S_1=(G_1,\P_1,\Lambda_1,I_1)$, $S_2=(G_2,\P_2,\Lambda_2,I_2)$. Then, $S=(G,\P,\Lambda,I)$, where
\begin{enumerate}
\item $\Lambda=\Lambda_1\sqcup\Lambda_2$, $I=I_1\sqcup I_2$;
\item the elements of the sandwich-matrix $\P=(p_{i\lambda})$ are
\[
p_{i\lambda}=
\begin{cases}
y_{i\lambda},\mbox{ åñëè }i\in I_1, \lambda\in\Lambda_2\mbox{ or }i\in I_2, \lambda\in\Lambda_1\\
p_{i\lambda}^{(1)},\mbox{ åñëè }i\in I_1, \lambda\in\Lambda_1\\
p_{i\lambda}^{(2)},\mbox{ åñëè }i\in I_2, \lambda\in\Lambda_2\\
\end{cases}
\]
where $p_{i\lambda}^{(j)}$ is the element of $\P_j$ with the indexes $i,\lambda$.
\item
\[
G=G_1\ast G_2\ast F(Y),
\]
where $F(Y)$ is the free group generated by the set
\[
Y=\{y_{i\lambda}|i\in I_1, \lambda\in\Lambda_2\mbox{ or }i\in I_2, \lambda\in\Lambda_1\}.
\]
\end{enumerate}

\end{theorem}

\section{Algebraic geometry over semigroups}

All definitions below are derived from the general notions of~\cite{uniTh_I}, where the notions of algebraic geometry were formulated for an arbitrary algebraic structure in the language with no predicates.

Since the inversion ${ }\1$ is an algebraic operation in any c.s. semigroup, we consider the language $\LL_0=\{\cdot,\1\}$. For a given c.s. semigroup $S$ one can extend $\LL_0$ by new constants $\{s|s\in S\}$ which correspond to the elements of $S$. The obtained language is denoted by $\LL_S$ and in the sequel all semigroups are considered in this language.

Let $X$ be a finite set of variables $x_1,x_2,\ldots,x_n$. \textit{A term} of a language $\LL_S$ ($\LL_S$-term) in variables $X$ is one of the following expressions:
\begin{enumerate}
\item variable $x_i$;
\item constant $s$;
\item a product of two terms;
\item $(t(X))\1$, where $t(X)$ is a term.
\end{enumerate}

For example, the expressions $xs(y^2x)\1$, $(xs_1\1 y)\1 s_2x^2$ are $\LL_S$-terms.

{\it An equation} over $\LL_S$ is an equality of two $\LL_S$-terms $t(X)=s(X)$. {\it A system of equations} over $\LL_S$ ({\it a system} for shortness) is an arbitrary set of equations over $\LL_S$.

A point $P=(p_1,p_2,\ldots,p_n)\in S^n$ is a \textit{solution} of a system $\Ss$ in variables $x_1,x_2,\ldots,x_n$, if the substitution $x_i=p_i$ reduces each equation of $\Ss$ to a true equality in the semigroup $S$. The set of all solutions of a system $\Ss$ in the semigroup $S$ is denoted by $\V_S(\Ss)$. A set $Y\subseteq S^n$ is called  {\it algebraic} over the language $\LL_S$ if there exists a system over $\LL_S$ in variables $x_1,x_2,\ldots,x_n$ with the solution set $Y$.

Two systems are called \textit{equivalent} over a c.s. semigroup $S$ if they have the same solution in $S$.

Following~\cite{uniTh_IV}, let us give the main definition of our paper.

A c.s. semigroup $S$ is an {\it equational domain} ({\it e.d.} for shortness) in the language $\LL_S$ if for any finite set of algebraic sets $Y_1,Y_2,\ldots,Y_n$ the union $Y=Y_1\cup Y_2\cup\ldots\cup Y_n$ is algebraic.

The next theorem contains necessary and sufficient conditions for a semigroup to be an e.d.

\begin{theorem}\textup{\cite{uniTh_IV}}
\label{th:about_M}
A semigroup $S$ in the language $\LL_S$ is an e.d. iff the set
\[

\M_{sem}=\{(x_1,x_2,x_3,x_4)|x_1=x_2\mbox{ or }x_3=x_4\}\subseteq S^4
\]
is algebraic, i.e. there exists a system $\Ss$ in the variables $x_1,x_2,x_3,x_4$ with the solution set $\M_{sem}$.
\end{theorem}

Below we study equations over groups, therefore we should give some definitions of algebraic geometry over groups. Any group $G$ below will be considered in the language $\LL_G=\{\cdot,^{-1},1\}\cup\{g|g\in G\}$ where constants of $\LL_G$ correspond to elements of $G$. \textit{An $\LL_G$-term}  in the variables  $X=\{x_1,x_2,\ldots,x_n\}$ is defined in the same way as it is the definition of a term over c.s. semigroup. However, in groups the identity $(xy)\1=x\1 y\1$ takes place, hence any $\LL_G$-term over a group $G$ is equivalent to a finite product that consists of the variables in integer powers and constants $g\in G$. In other words, an $\LL_G$-term is an element of the free product $F(X)\ast G$, where $F(X)$ is a free group generated by the set $X$.

The definitions of equations, algebraic sets and equational domains over groups are similar to the corresponding definitions in the semigroup case.

For the groups of the language $\LL_G$ we have the following result.

\begin{theorem}\textup{\cite{uniTh_IV}}
A group $G$ in the language $\LL_G$ is an e.d. iff the set
\label{th:criterion_for_groups}
\begin{equation*}
\M_{gr}=\{(x_1,x_2)|x_1=1\mbox{ or }x_2=1\}\subseteq G^2
\end{equation*}
is algebraic, i.e. there exists a system $\Ss$ in variables $x_1,x_2$ with the solution set $\M_{gr}$.
\end{theorem}

One can reformulate Theorem~\ref{th:criterion_for_groups} in a simpler  form using the following definition. An element $x\neq 1$ of a group $G$ is a \textit{zero-divisor} if there exists $1\neq y\in G$ such that for any $g\in G$ it holds $[x,y^g]=1$ (here $y^g=gyg^{-1}$, $[a,b]=a^{-1}b^{-1}ab$).

\begin{theorem}\textup{\cite{uniTh_IV}}
\label{th:zero_divisors}
A group $G$ in the language $\LL_G$ is an e.d. iff it does not contain zero-divisors.
\end{theorem}

Using Theorem~\ref{th:zero_divisors}, one can obtain the next properties of free objects in the variety of groups.

\begin{corollary}\textup{\cite{uniTh_IV}}
\label{cor:free_group_is_ED}
Every non-abelian free group $G$ is an e.d. in the group language $\LL_G$ (this was initially proved by G.~Gurevich, see the proof in~\cite{makanin}).
\end{corollary}

\begin{corollary}\textup{\cite{uniTh_IV}}
\label{cor:free_product_of_groups_is_ED}
Every free product $G=G_1\ast G_2$, except for $G=\mathbb{Z}_2\ast\mathbb{Z}_2$ ($\mathbb{Z}_2$ is the cyclic group of the order $2$), is an e.d. in the language $\LL_G$.
\end{corollary}

\section{Main results}

A sandwich-matrix $\P$ is said to be \textit{non-singular} if it does not contain two equal rows or columns.

Let $s_1,s_2$ be two distinct elements of a c.s. semigroup $S$. We say that an $\LL_S$-term $t(x)$ \textit{separates} $s_1,s_2$ if $t(s_1)\neq t(s_2)$.

\begin{lemma}\textup{(Lemma 4.2 of \cite{ED_I})}
\label{l:exists_dist_term}
If a c.s. semigroup $S=(G,\P,\Lambda,I)$ has a non-singular sandwich-matrix $\P$, then for any pair of distinct elements $s_1,s_2\in S$ there exists an $\LL_S$-term
\begin{equation}
t(x)=(1,1,i)x(\lambda,1,1),
\label{eq:dist_term}
\end{equation}
separating $s_1,s_2$ for some $\lambda\in\Lambda$, $i\in I$.
\end{lemma}

Let $t(x)$ be an $\LL_S$-term. By $[t](x)$ we denote the $\LL_S$-term which is obtained from $t(x)$ by omitting all occurrences of the inversion. For example, if $t(x)=((xsy\1)\1zx\1)\1$ then $[t](x)=xsyzx$.

\begin{lemma}
\label{l:exists_2_non_dist_elems}
Suppose the sandwich-matrix $\P$ of a c.s. semigroup $S=(G,\P,\Lambda,I)$ has equal rows (columns) with numbers $i,j$ (respectively, $\lambda,\mu$). Then for the elements $s_1=(1,1,i)$, $s_2=(1,1,j)$ (resp. $s_1=(\lambda,1,1)$, $s_2=(\mu,1,1)$) and an arbitrary $\LL_S$-term $t(x)$ one of the following holds:
\begin{enumerate}
\item $t(s_1)=t(s_2)$;
\item $t(s_1)=(\nu,g,i)$, $t(s_2)=(\nu,g,j)$ for some $g\in G$, $\nu\in\Lambda$, if $[t](x)$ ends with the variable $x$ (resp. $t(s_1)=(\lambda,g,k)$, $t(s_2)=(\mu,g,k)$ for some $g\in G$, $k\in I$, if  $[t](x)$ begins with the variable $x$). In other words, the elements $t(s_1),t(s_2)$ are distinguished only by the second (respectively, first) index.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume that the $\lambda$-th and $\mu$-th columns of $\P$ are equal.(similarly, one can consider $\P$ with equal rows).

We prove this lemma by induction on the construction of a term $t(x)$. If $t(x)$ is a either constant or a variable, lemma obviously holds.

Suppose now $t(x)=(t^\pr(x))\1$, and the lemma holds for $t^\pr(x)$. If $t^\pr(s_1)=t^\pr(s_2)$, then $t(s_1)=t(s_2)$. If $[t(x)]$ begins with the variable $x$, we have
\[
t(s_1)=(\lambda,g,k)\1=(\lambda,p_{k\lambda}\1g\1p_{k\lambda}\1,k),
\]
\[
t(s_2)=(\mu,g,k)\1=(\mu,p_{k\mu}\1g\1p_{k\mu}\1,k).
\]
By the singularity of the matrix $\P$, we have $p_{k\lambda}=p_{k\mu}$, therefore the elements $t(s_1),t(s_2)$ are distinguished only by the first index.

Consider now an $\LL_S$-term $t(x)=t^\pr(x)t^{\pr\pr}(x)$. We have exactly four cases.

\begin{enumerate}
\item Neither of the terms $[t^{\pr}](x)$, $[t^{\pr\pr}](x)$ begins with the variable $x$. By the induction, we have $t^\pr(s_1)=t^\pr(s_2)$, $t^{\pr\pr}(s_1)=t^{\pr\pr}(s_2)$, and, finally, $t(s_1)=t(s_2)$.

\item Only $[t^{\pr\pr}](x)$ begins with the variable $x$. Here we have $t^\pr(s_1)=t^\pr(s_2)=(\nu,h,l)$, $t^{\pr\pr}(s_1)=(\lambda,g,k)$, $t^{\pr\pr}(s_2)=(\mu,g,k)$ for some $g,h\in G, \nu\in\Lambda$, $k,l\in I$.

Thus the values of $t(x)$ equal:
\[
t(s_1)=(\nu,h,l)(\lambda,g,k)=(\nu,hp_{l\lambda}g,k),
\]
\[
t(s_2)=(\nu,h,l)(\mu,g,k)=(\nu,hp_{l\mu}g,k).
\]
Since $p_{l\lambda}=p_{l\mu}$, we have $t(s_1)=t(s_2)$.

\item Only $[t^{\pr}](x)$ begins with the variable $x$. Therefore, $t^{\pr}(s_1)=(\lambda,g,k)$, $t^{\pr}(s_2)=(\mu,g,k)$, $t^{\pr\pr}(s_1)=t^{\pr\pr}(s_2)=(\nu,h,l)$ for some $g,h\in G, \nu\in\Lambda$, $k,l\in I$. We have
\[
t(s_1)=(\lambda,g,k)(\nu,h,l)=(\lambda,gp_{k\nu}h,l),
\]
\[
t(s_2)=(\mu,g,k)(\nu,h,l)=(\mu,gp_{k\nu}h,l),
\]
and the elements $t(s_1),t(s_2)$ are distinguished only by the first index.

\item Both terms $[t^\pr](x)$, $[t^{\pr\pr}](x)$ begin with the variable $x$. By the induction, $t^\pr(s_1)=(\lambda,g,k)$, $t^\pr(s_2)=(\mu,g,k)$, $t^{\pr\pr}(s_1)=(\lambda,h,l)$, $t^{\pr\pr}(s_2)=(\mu,h,l)$ for some $g,h\in G$, $k,l\in I$. Therefore,
\[
t(s_1)=(\lambda,g,k)(\lambda,h,l)=(\lambda,gp_{k\lambda}h,l),
\]
\[
t(s_2)=(\mu,g,k)(\mu,h,l)=(\mu,g_{k\mu}h,l).
\]
Since $p_{k\lambda}=p_{k\mu}$, the elements $t(s_1),t(s_2)$ are distinguished only by the first index.
\end{enumerate}
\end{proof}

\begin{lemma}
\label{l:singular->not_ED}
If the sandwich-matrix $\P$ is singular, a c.s. semigroup $S=(G,\P,\Lambda,I)$ is not an e.d. in the language $\LL_S$.
\end{lemma}
\begin{proof}
Assume the semigroup $S$ is an e.d. with a singular sandwich-matrix $\P$, and columns with numbers $\lambda,\mu$ of $\P$ are equal (similarly, one can consider a matrix $\P$ with equal rows).

Consider a system of equations $\Ss(x,y)$ with the solution set $\M=\{(x,y)|x=(\lambda,1,1)\mbox{ or }y=(\lambda,1,1)\}$. Let $t(x,y)=s(x,y)\in \Ss$ be an equation that is not satisfied by the point $((\mu,1,1),(\mu,1,1))\notin\M$.

Suppose neither of the terms $[t](x,(\mu,1,1))$, $[s](x,(\mu,1,1))$ begins with the variable $x$. Therefore, by Lemma~\ref{l:exists_2_non_dist_elems}, we have
\[
t((\lambda,1,1),(\mu,1,1))=t((\mu,1,1),(\mu,1,1)),\; s((\lambda,1,1),(\mu,1,1))=s((\mu,1,1),(\mu,1,1)).
\]
Since
\[
t((\lambda,1,1),(\mu,1,1))=s((\lambda,1,1),(\mu,1,1)),
\]
we obtain
\[
t(\mu,1,1),(\mu,1,1))=s((\mu,1,1),(\mu,1,1)),
\]
which contradicts the choice of the equation $t(x,y)=s(x,y)$.

Assume now both terms $[t](x,(\mu,1,1))$, $[s](x,(\mu,1,1))$ begin with $x$. By Lemma~\ref{l:exists_2_non_dist_elems}, one can obtain
\[
t((\lambda,1,1),(\mu,1,1))=(\lambda,g,k),\;t((\mu,1,1),(\mu,1,1))=(\mu,g,k).
\]
Since $((\lambda,1,1),(\mu,1,1))\in \M$, then $s((\lambda,1,1),(\mu,1,1)=(\lambda,g,k)$, and, according Lemma~\ref{l:exists_2_non_dist_elems}, we have
\[
s((\mu,1,1),(\mu,1,1))=(\mu,g,k),
\]
It follows that
\[
t((\mu,1,1),(\mu,1,1))=s((\mu,1,1),(\mu,1,1)),
\]
which contradicts the choice of $t(x,y)=s(x,y)$.

Thus, we have the last case: the term $[t](x,(\mu,1,1))$ begins with $x$, and $[s](x,(\mu,1,1))$ begins with a constant $\c$. Therefore one of the following holds:
\begin{enumerate}
\item the constant $\c$ was obtained by the substitution of $(\mu,1,1)$ for the variable $y$, i.e. the term $[s](x,y)$ begins with $y$;
\item $[s](x,y)$ begins with $\c$.
\end{enumerate}
Let us show that both options above are impossible.
\begin{enumerate}
\item Indeed, the equation $t(x,y)=s(x,y)$ is not satisfied by the point $((\lambda,1,1),(\mu,1,1))\in\M$, since the element $t((\lambda,1,1),(\mu,1,1))$ has the first index $\lambda$, whereas  the first index of $s((\lambda,1,1),(\mu,1,1))$ is $\mu$.
\item Suppose $\c=(\nu,g,k)$, so either $t((\lambda,1,1),(\lambda,1,1))\neq s((\lambda,1,1),(\lambda,1,1))$ (if $\nu\neq\lambda$) or $t((\mu,1,1),(\lambda,1,1))\neq s((\mu,1,1),(\lambda,1,1))$ (if $\nu\neq\mu$), since the values of the terms $t(x,y), s(x,y)$ have different first indexes.
\end{enumerate}
\end{proof}

It is directly checked that the subset
\[
\Gamma=\{(1,g,1)|g\in G\}
\]
of a c.s. semigroup $S=(G,\P,\Lambda,I)$ is a group isomorphic to $G$. Since the sandwich-matrix  $\P$ is normalised, $(1,1,1)$ is the identity of $\Gamma$.

Suppose $x,y\in\Gamma$. Then we have the following identities:
\begin{eqnarray}
\label{eq:leech1}
(\lambda,g,i)x=(\lambda,g,1)x=(\lambda,g,1)x(1,1,1),\\
\label{eq:leech2}
x(\lambda,g,i)=x(1,g,i)=(1,1,1)x(1,g,i)\\
\label{eq:leech3}
((\lambda,g,1)x(1,h,j))\1=(\lambda,p_{j\lambda}\1 h\1,1)x\1(1,g\1 p_{j\lambda}\1,j)\\
\label{eq:leech4}
((\lambda,g,i)x)\1=(\lambda,1,1)x\1(1,g\1,1),\\
\label{eq:leech5}
(x(\lambda,g,i))\1=(1,g\1,1)x\1 (1,1,i),\\
\label{eq:leech6}
x(\lambda,g,i)y=x(1,g,1)y,\\
\label{eq:leech7}
(x(\lambda,g,i)y)\1 =y\1 (1,g\1,1)x\1.
\end{eqnarray}

The proof of the equalities~(\ref{eq:leech1},~\ref{eq:leech2}) is straightforward.

Let us prove~(\ref{eq:leech3}). Suppose $x=(1,g_x,1)$, then
\begin{multline*}
((\lambda,g,1)(1,g_x,1)(1,h,j))\1=(\lambda,gg_xh,j)\1=(\lambda,p_{j\lambda}\1 (gg_xh)\1 p_{j\lambda}\1,j)\\
=(\lambda,p_{j\lambda}\1 h\1 g_x\1 g\1 p_{j\lambda}\1,j).
\end{multline*}
On the other hand,
\begin{multline*}
(\lambda,p_{j\lambda}\1 h\1,1)(1,g_x,1)\1(1,g\1 p_{j\lambda}\1,1)=(\lambda,p_{j\lambda}\1 h\1,1)(1,g_x\1,1)(1,g\1 p_{j\lambda}\1,1)\\
=(\lambda,p_{j\lambda}\1 h\1 g_x\1 g\1 p_{j\lambda}\1,j).
\end{multline*}

The equalities~(\ref{eq:leech4},~\ref{eq:leech5}) immediately follows from ~(\ref{eq:leech1},~\ref{eq:leech2},~\ref{eq:leech3}). Let us prove~(\ref{eq:leech6},~\ref{eq:leech7}). Suppose $x=(1,g_x,1)$, $y=(1,g_y,1)$, then
\[
x(\lambda,g,i)y=(1,g_x,1)(\lambda,g,i)(1,g_y,1)=(1,g_xgg_y,1),
\]
\[
(x(\lambda,g,i)y)\1=(1,g_xgg_y,1)\1=(1,(g_xgg_y)\1,1).
\]
Conversely,
\[
x(1,g,1)y=(1,g_x,1)(1,g,1)(1,g_y,1)=(1,g_xgg_y,1),
\]
\[
y\1(1,g\1,1)x\1 =(1,g_y\1,1)(1,g\1,1)(1,g_x\1,1)=(1,g_y\1 g\1 g_x\1,1)=(1,(g_xgg_y)\1,1),
\]
which proves~(\ref{eq:leech6},~\ref{eq:leech7}).

\begin{lemma}
\label{l:S-eq_dom->G-eq_dom}
If a c.s. semigroup $S=(G,\P,\Lambda,I)$ is an e.d. in the language $\LL_S$, then the group $G$ is an e.d. in the language $\LL_G$.
\end{lemma}
\begin{proof}
Since $S$ is an e.d., the set $\M=\{(x,y)|x=(1,1,1)\mbox{ or }y=(1,1,1)\}\subseteq S^2$ is algebraic. In other words, there exists a system $\Ss$ with $\V_S(\Ss)=\M$.

Below we define a system $\Ss^\pr$ which is equivalent to $\Ss$ over the group $\Gamma$, and all constants of $\Ss^\pr$ belong to $\Gamma$.

Applying~(\ref{eq:leech3},~\ref{eq:leech4},~\ref{eq:leech5},~\ref{eq:leech7}) to the system $\Ss$, one can obtain a system $\Ss_1$ such that:
\begin{enumerate}
\item $\Ss_1$ is equivalent to $\Ss$ over $\Gamma$;
\item the inversion ${}^{-1}$ in any equation from $\Ss_1$ is applied only to variables: $x\1, y\1$.
\end{enumerate}

By~(\ref{eq:leech6}) the system $\Ss_1$ is reduced to $\Ss_2$ such that any constant $\c$ in any $t(x,y)=s(x,y)\in\Ss_2$ belongs to $\Gamma$ if $\c$ is neither first nor last symbol in the terms $t(x,y),s(x,y)$. Obviously, $\Ss_2$ is equivalent to $\Ss$ over the group $\Gamma$.

Consider an equation $(\lambda,g,i)t^\pr(x,y)=s(x,y)\in\Ss_2$, where the left part of this equation begins with a constant $\c=(\lambda,g,i)\notin\Gamma$ (similarly, one can consider an equation that has a right part ending by a constant $\c\notin\Gamma$). According to~(\ref{eq:leech1}), one can put $i=1$. Since the system $\Ss_2$ is consistent over $\Gamma$, the term $s(x,y)$ begins with either a constant $(\lambda,h,j)$ or a variable with $\lambda=1$.

Suppose $s(x,y)=(\lambda,h,j)s^\pr(x,y)$ (similarly, one can consider $s(x,y)$ which begins with a variable). It is directly checked that the equation
\[
(\lambda,g,1)t^\pr(x,y)=(\lambda,h,j)s^\pr(x,y)
\]
is equivalent to
\[
(1,g,1)t^\pr(x,y)=(1,h,1)s^\pr(x,y).
\]
Thus, any equation of $\Ss_2$ is equivalent to a system $\Ss^\pr$ whose constants belong to the subgroup  $\Gamma$. Therefore, $\V_\Gamma(\Ss^\pr)=\{(x,y)|x=(1,1,1)\mbox{ or }y=(1,1,1)\}\subseteq \Gamma^2$, and, by Theorem~\ref{th:criterion_for_groups}, the group $\Gamma$ is an e.d. in the language $\LL_\Gamma$. Finally, the isomorphism between the groups $\Gamma,G$ proves the lemma.
\end{proof}

Below the element $(1,1,1)\in\Gamma$ is denoted by $\one$ for shortness.

\begin{lemma}
\label{l:reduce_x=y}
Suppose the sandwich-matrix $\P$ of a c.s. semigroup $S=(G,\P,\Lambda,I)$ is nonsingular. Then the equation $x=y$ is equivalent to a system of the form $\Ss_{=}(x,y)=\{t_i(x,y)=\one |i\in\mathcal{I}\}$.
\end{lemma}
\begin{proof}
Denote by $\M_{=}$ the solution set of $x=y$, i.e. $\M_{=}=\{(x,x)|x\in S\}$.

By Lemma~\ref{l:exists_dist_term}, for any $(s,s^\pr)\notin\M_{=}$ there exists a term $t_{(s,s^\pr)}(x)$ separating $s,s^\pr$.

Put
\[\Ss_{=}(x,y)=\{\one x(\one y\one )\1=\one \}\bigcup_{(s,s^\pr)\notin\M_{=}}\{t_{(s,s^\pr)}(x)(t_{(s,s^\pr)}(y))\1=\one \}.
\]

Let us show that $\M_=$ is the solution set of $\Ss_{=}(x,y)$, and $P=((\mu,g,j),(\mu,g,j))\in\M_=$.

Since
\[\one (\mu,g,j)(\one (\mu,g,j)\one )\1=(1,g,j)(1,g\1,1)=\one ,
\]
the point $P$ satisfies the first equation of $\Ss_=$.
For the other equations of $\Ss_=$ we have (apply the  formula~(\ref{eq:dist_term}) to the term $t_{(s,s^\pr)}(x)$)
\begin{multline*}
t_{(s,s^\pr)}((\mu,g,j))(t_{(s,s^\pr)}((\mu,g,j)))\1=(1,1,i)(\mu,g,j)(\lambda,1,1)((1,1,i)(\mu,g,j)(\lambda,1,1))\1=\\
(1,p_{i\mu}gp_{j\lambda},1)((1,p_{i\mu}gp_{j\lambda},1))\1=(1,p_{i\mu}gp_{j\lambda},1)(1,(p_{i\mu}gp_{j\lambda})\1,1)=\one .
\end{multline*}

Thus, any point of $\M_=$ belongs to the solution set of $\Ss_=$. Conversely, if $(s,s^\pr)\notin\M_=$, Let $(1,h,1)$, $(1,h^\pr,1)$ denote the values of $t_{(s,s^\pr)}(x)$ at $s,s^\pr$, respectively. We obtain
\[
t_{(s,s^\pr)}(s)(t_{(s,s^\pr)}(s^\pr))\1=
(1,h,1)((1,h^\pr,1))\1=(1,h(h^\pr)\1,1)\neq\one ,
\]
since $h\neq h^\pr$. Thus, the equation $t_{(s,s^\pr)}(x)(t_{(s,s^\pr)}(y))\1=\one \in\Ss_=$ is not satisfied by the point $(s,s^\pr)$.

Finally, we obtain $\V_S(\Ss_=)=\M_=$, therefore the system $\Ss_=(x,y)$ is equivalent to the equation $x=y$.
\end{proof}

\begin{lemma}
\label{l:x=1_y=1_implies_domain}
If the set $\M=\{(x,y)|x=(1,1,1)\mbox{ or } y=(1,1,1)\}$ is algebraic over a c.s. semigroup  $S=(G,\P,\Lambda,I)$ with a nonsingular $\P$, then $S$ is an e.d. in the language $\LL_S$.
\end{lemma}
\begin{proof}
By Theorem~\ref{th:about_M} it suffices to prove the existence of a system  $\Ss(x_1,x_2,x_3,x_4)$ with the solution set $\M_{sem}$.

By Lemma~\ref{l:reduce_x=y}, the equations $x_1=x_2$, $x_3=x_4$ are respectively equivalent to the systems $\Ss_=(x_1,x_2)=\{t_i(x_1,x_2)=\one |i\in\mathcal{I}\}$, $\Ss_=(x_3,x_4)=\{t_i(x_3,x_4)=\one |i\in\mathcal{I}\}$.

Applying the distributivity law for algebraic sets, we obtain
\[
\M_{sem}=\bigcap_{i,j\in\mathcal{I}}\V_S(t_i(x_1,x_2)=\one) \cup \V_S( t_j(x_3,x_4)=\one ).
\]

By the condition of the lemma, there exists a system $\Ss^\pr(x,y)$ with the solution $\M$. Hence, the set $\V_S(t_i(x_1,x_2)=\one \cup\V_S(t_j(x_3,x_4)=\one)$ coincides with the solution set of the system $\Ss^\pr(t_i(x_1,x_2),t_j(x_3,x_4))$, and
\[\M_{sem}=\bigcap_{i,j\in\mathcal{I}}\V_S(\Ss^\pr(t_i(x_1,x_2),t_j(x_3,x_4))).\]

Finally, the set $\M_{sem}$ equals to the solution set of the system
\[\Ss(x_1,x_2,x_3,x_4)=\bigcup_{i,j\in\mathcal{I}}\Ss^\pr(t_i(x_1,x_2),t_j(x_3,x_4)).\]
Thus, $\M_{sem}$ is algebraic.
\end{proof}

\begin{lemma}
\label{l:x=1_y=1_is_algebraic}
Suppose the sandwich-matrix $\P$ of a c.s. semigroup $S=(G,\P,\Lambda,I)$ is nonsingular, the group $G$ is an e.d. in the group language $\LL_G$. Then the set $\M=\{(x,y)|x=(1,1,1)\mbox{ or } y=(1,1,1)\}$ is algebraic over $S$ in the language $\LL_S$.
\end{lemma}
\begin{proof}
Let $\one \neq s\in S$. By Lemma~\ref{l:exists_dist_term}, there exists a term of the form~(\ref{eq:dist_term}) separating $\one ,s$. This term is denoted below by $t_s(x)$.

Put
\begin{multline*}
\Ss^\pr(x,y)=\bigcup_{g\in G}\{\one x(1,g\1,1)y(1,g,1)(\one x\one )\1(1,g\1,1)(\one y\one )\1(1,g,1)=\one \}\\
\bigcup_{(s,s^\pr)\notin\M,g\in G}\{t_s(x)(1,g\1,1)t_{s^\pr}(y)(1,g,1)(t_s(x))\1(1,g\1,1)(t_{s^\pr}(y))\1(1,g,1)=\one \}.
\end{multline*}

Let us show that $\V_S(\Ss^\pr(x,y))=\M$.

Firstly, we prove $\M\subseteq\V_S(\Ss^\pr)$.

Let $P=((\mu,h,j),\one )\in\M$ (similarly, one can consider the point $(\one ,(\mu,h,j))\in\M$). We have
\begin{multline*}
\one (\mu,h,j)(1,g\1,1)\one (1,g,1)(\one (\mu,h,j)\one )\1(1,g\1,1)(\one \one \one )\1(1,g,1)=\\
\one (\mu,h,j)\one ((1,h\1,1))(1,g\1,1)(\one )(1,g,1)=\\
(1,h,1)(1,h\1,1)(1,g\1,1)(1,g,1)=\one .
\end{multline*}
In other words, the point $P$ satisfies the first group of equations of $\Ss^\pr$.

Since the terms $t_s(x),t_{s^\pr}(x)$ are of the form~(\ref{eq:dist_term}), we have $t_s(\one )=t_{s^\pr}(\one )=\one $. Let $t_s((\mu,h,j))=(1,f,1)$, $t_{s^\pr}((\mu,h,j))=(1,f^\pr,1)$.

Computing
\begin{multline*}
t_s((\mu,h,j))(1,g\1,1)t_{s^\pr}(\one )(1,g,1)(t_s((\mu,h,j)))\1(1,g\1,1)(t_{s^\pr}(\one ))\1(1,g,1)=\\
(1,f,1)(1,g\1,1)\one (1,g,1)(1,f\1,1)(1,g\1,1)\one (1,g,1)=\one
\end{multline*}
we see that the point $P$ satisfies the second group of equations of $\Ss^\pr$. Finally, we obtained $P\in\V_S(\Ss^\pr)$.

Second, we prove $\M\supseteq\V_S(\Ss^\pr)$.

Let $Q=(s,s^\pr)\notin\M$. Assume that $Q$ satisfies all equations
\[
t_s(x)(1,g\1,1)t_{s^\pr}(y)(1,g,1)(t_s(x))\1(1,g\1,1)(t_{s^\pr}(y))\1(1,g,1)=\one
\]
for all $g\in G$.

By the choice of the terms $t_s(x),t_{s^\pr}(x)$ there exist elements $1\neq f,f^\pr\in G$ with $t_s(s)=(1,f,1)$, $t_{s^\pr}(s^\pr)=(1,f^\pr,1)$.

Therefore, for all $g\in G$ we have
\[
t_s(s)(1,g\1,1)t_{s^\pr}(s^\pr)(1,g,1)(t_s(s)\1(1,g\1,1)(t_{s^\pr}(s^\pr))\1(1,g,1)=\one ,
\]
\[
(1,f,1)(1,g\1,1)(1,f^\pr,1)(1,g,1)(1,f\1,1)(1,g\1,1)(1,(f^\pr)\1,1)(1,g,1)=\one,
\]
\[
(1,fg\1 f^\pr gf\1 g\1(f^\pr)\1 g,1)=\one ,
\]
\[fg\1 f^\pr gf\1 g\1(f^\pr)\1 g=1\Leftrightarrow[f,(f^\pr)^g]=1,
\]
where $[f,(f^\pr)^g]$ is the commutator of $f$ and $f^\pr$ conjugated by $g$ in the group $G$.

Since the last equality holds for all $g\in G$, the elements $f,f^\pr$ are zero-divisors in the group $G$. Thus, by Theorem~\ref{th:zero_divisors}, the group $G$ is not an e.d. that contradicts with the condition of the lemma.

\end{proof}

\begin{theorem}
\label{th:criterion}
A c.s. semigroup $S=(G,\P,\Lambda,I)$ is an e.d. in the language $\LL_S=\{\cdot,{ }\1\}\cup\{s|s\in S\}$ iff the following hold:
\begin{enumerate}
\item the sandwich-matrix $\P$ is non-singular;
\item the structural group $G$ is an e.d. in the group language $\LL_G$.
\end{enumerate}
\end{theorem}
\begin{proof}
The ``only if '' part of the statement immediately follows from Lemmas~\ref{l:singular->not_ED},~\ref{l:S-eq_dom->G-eq_dom}. Let us prove the converse.

First, by Lemma~\ref{l:x=1_y=1_is_algebraic}, we obtain that the set $\M=\{(x,y)|x=(1,1,1)\mbox{ or } y=(1,1,1)\}$ is algebraic over $S$ over the language $\LL_S$. Second, one can apply Lemma~\ref{l:x=1_y=1_implies_domain}, which proves the theorem.
\end{proof}

Theorems~\ref{th:free_CSS_structure},~\ref{th:free_product_CSS_structure}, Corollaries~\ref{cor:free_group_is_ED},~\ref{cor:free_product_of_groups_is_ED} and the criterion provided in Theorem~\ref{th:criterion} allow us to prove easily the following statements.

\begin{corollary}
Any free c.s. semigroup $S=F_{css}(X)$ of the rank $n\geq 2$ is an e.d. in the language $\LL_S$.
\end{corollary}

\begin{corollary}
The free product $S=S_1\ast S_2$ of arbitrary c.s. semigroups $S_1,S_2$ is an e.d. in the language $\LL_S$.
\end{corollary}

The next statement follows from Lemma~\ref{l:reduce_x=y}.

\begin{corollary}
Any algebraic set $Y\subseteq S^n$ over a c.s. semigroup $S=(G,\P,\lambda,I)$ with a non-singular sandwich-matrix $\P$ is defined by a system $\Ss=\{t_i(X)=\one |i \in \mathcal{I}\}$.
\end{corollary}
\begin{proof}
Suppose an algebraic set $Y$ is the solution set of a system $\Ss^\pr=\{t_i(X)=s_i(X)|i \in \mathcal{I}\}$.

According to Lemma~\ref{l:reduce_x=y}, the equation $x=y$ is equivalent to the system $\Ss_=(x,y)$ of the form $\{r_j(x,y)=\one |j\in\mathcal{J}\}$. Therefore every equation $t_i(X)=s_i(X)\in\Ss^\pr$ is equivalent to the system $\Ss_=(t_i(X),s_i(X))$. Thus, $\Ss^\pr$ is equivalent to
\[
\Ss=\bigcup_{i\in\mathcal{I}}\Ss_=(t_i(X),s_i(X))=
\bigcup_{i\in\mathcal{I},\\ j\in\mathcal{J}}r_j(t_i(X),s_i(X)).
\]
\end{proof}

\section{Generalisations to other semigroup languages}
\label{sec:generalization}
A c.s. semigroup $S$ can be considered in any language $\LL_T=\{\cdot,{}\1\}\cup\{s|s\in T\}$, where $T$ is a subsemigroup of $S$. An $\LL_T$-term is an $\LL_S$-term whose constants belong to the subsemigroup $T$.

Similarly, any group $G$ can be considered in a language $\LL_H=\{\cdot,{}\1\}\cup\{g|g\in H\}$, where $H$ is a subgroup of $G$.

The notions of an equation, algebraic set, and equational domain in the language $\LL_T$ ($\LL_H$) are similar to the corresponding definitions in the language $\LL_S$ (respectively, $\LL_G$).

Since $\LL_T\subseteq\LL_S$ ($\LL_H\subseteq\LL_G$), any algebraic set in the language $\LL_T$ (respectively, $\LL_H$) is algebraic in $\LL_S$ ($\LL_G$). Therefore, any equational domain in the language $\LL_T$ ($\LL_H$) is an e.d. in $\LL_S$ (respectively, $\LL_G$). It follows that equational domains in the language $\LL_T$ should satisfy stronger conditions than those in Theorem~\ref{th:criterion}.

Similarly, one can show that the analogue of Theorem~\ref{th:zero_divisors} for equational domains in the language $\LL_H$ should contain stronger conditions. Actually, in~\cite{uniTh_IV} the definition of a zero-divisor with respect to a subgroup $H\leq G$ was given: an element $x\neq 1$ of a group $G$ is an \textit{$H$-zero-divisor} if there exists $1\neq y\in G$ such that $[x,y^h]=1$ for all $h\in H$.

The next theorem describes equational domains in the group language $\LL_H$.

\begin{theorem}\textup{\cite{uniTh_IV}}
\label{th:zero_divisors_for_H}
A group $G$ is an equational domain in the language $\LL_H$ iff it does not contain $H$-zero-divisors.
\end{theorem}

Since the class of c.s. semigroups is a variety, a subsemigroup $T$ of a c.s. semigroup  $S=(G,\P,\Lambda,I)$ has the representation $T=(H,\P^\pr,\Lambda^\pr,I^\pr)$, where $H\leq G$, $\Lambda^\pr\subseteq\Lambda$, $I^\pr\subseteq I$ and $\P^\pr$ is a submatrix of $\P$, $\P^\pr=(p_{i\lambda}|i\in I^\pr,\lambda\in\Lambda^\pr)$.

For the language $\LL_T$ we obtain the next criterion.

\begin{theorem}
A c.s. semigroup $S=(G,\P,\Lambda,I)$ is an e.d. in the language $\LL_T=\{\cdot,{ }\1\}\cup\{s|s\in T\}$ iff the following conditions hold:
\begin{enumerate}
\item the sandwich-matrix $\P$ is non-singular;
\item the structural group $G$ is an e.d. in the group language $\LL_{H}$, where $H$ is the structural group of $T$.
\end{enumerate}
\end{theorem}
\begin{proof}
The ``only if'' part of the statement follows from Theorem~\ref{th:criterion}, since any e.d. in the language $\LL_T$ is an e.d. in $\LL_S$.

To prove the converse statement one can apply Lemmas~\ref{l:reduce_x=y},~\ref{l:x=1_y=1_implies_domain},~\ref{l:x=1_y=1_is_algebraic} with the subgroup $H$ instead of $G$.
\end{proof}

config: arxiv


TITLE: Closed function and the interior
QUESTION [1 upvotes]: I have some troubles with the next problem:

Let $f:X\rightarrow Y$ be a closed function. Prove that for all $y\in Y$ and for all $U\subseteq X$ open set such that $f^{-1}[\{y\}]\subseteq U$ holds that $y\in\text{int}_{Y}\left(f[\text{cl}_X(U)]\right)$ where $\text{int}$ denotes the interior of a set and $\text{cl}$ denotes the closure of a set.

My attempt:
Let $y\in Y$ and $U\subseteq X$ an open set. Then
$f^{-1}[\{y\}]\subseteq U\subseteq\text{cl}_{X}(U)$. In this way, $f^{-1}[\{y\}]\subseteq \text{cl}_{X}(U)$. Then, $\{y\}\subseteq f[\text{cl}_{X}(U)]$, furthermore, $y\in\{y\}\subseteq f[\text{cl}_{X}(U)]$. Then, $y\in f[\text{cl}_{X}(U)]$ and like $f$ is closed, $f[\text{cl}_{X}(U)]$ is closed, but from here, I don't know how to continue my proof.
Another idea is take $X\setminus U$ (is closed because $U$ is open), then, like $f$ is closed, $f[X\setminus U]$ is closed. Furthermore, like $f^{-1}[\{y\}]\subseteq U$ then $f^{-1}[\{y\}]\not\subseteq X\setminus U$ and so, $y\notin X\setminus U$. Then, $y\in U$, but, from here, again, I don't know what can I do.
I really appreciate any help you can provide.

REPLY [1 votes]: I believe the result you are after is not true in general.
Consider the map $f:\mathbb{R}\rightarrow\mathbb{R}$ where $f(x)=0\ \forall x\in\mathbb{R}$, equip $\mathbb{R}$  with the standard topology.
As shown here, the zero function on $\mathbb{R}$ is a closed map.
Take the open set $(0,1)\subseteq\mathbb{R}$. It is clear that $\emptyset=f^{-1}(\{1\})\subseteq (0,1)$, but $1\not\in$int$f($cl$(0,1))=$int$\{0\}=\emptyset$ as the only open set contained in the singleton is the empty set.
In order for your required result to be true, we additionally need to know that the map $f$ is a bijection. Then it follows as seen here that $f$ will be a homeomorphism, then it is also necessarily an open map.


TITLE: Logarithmic Integration
QUESTION [0 upvotes]: Show that
$$\int_{\ln\frac{3}{2}}^{\ln2}\frac{\exp(2x)+1}{\exp(2x)-1}\mathrm dx=\ln\frac{9}{5}$$

REPLY [0 votes]: $$\int_{\ln \frac{3}{2}}^{\ln2} \frac{e^{2x}+1}{e^{2x}-1}dx$$
$$=\int_{\ln \frac{3}{2}}^{\ln2} \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}dx$$
$$=\int_{\ln \frac{3}{2}}^{\ln2} \frac{d(e^{x}-e^{-x})}{(e^{x}-e^{-x})}$$
$$=[\ln |e^{x}-e^{-x}|]_{\ln \frac{3}{2}}^{\ln2}$$
$$=\ln |e^{\ln2}-e^{-\ln 2}|-\ln |e^{\ln \frac{3}{2}}-e^{-\ln \frac{3}{2}}|$$
$$=\ln |\frac{2-\frac{1}{2}}{\frac{3}{2}-\frac{2}{3}}|$$
$$=\ln \frac{\frac{3}{2}}{\frac{5}{6}}$$
$$=\ln \frac{9}{5}$$
$$$$
$$$$