text:
\section{Introduction}\label{int}

Let $\mu$ be any probability measure on $\mathbb{R}$. Denote by $(S_n)$ a random walk with step distribution
$\mu$, such that $S_0=0$, a.s. Define the first ladder times associated to $(S_n)$ by
\[\tau_-=\inf\{n\ge1:S_n<0\}\,,\;\;\;\tau_+=\inf\{n\ge1:S_n\ge0\}\,.\]
Then the Wiener-Hopf factorization of the characteristic function $\varphi(t)=\int e^{itx}\mu(dx)$ of $\mu$ can be written
as,
\begin{equation}\label{6283}
1-s\varphi(t)=[1-\chi_-(s,it)][1-\chi_+(s,it)]\,,\;\;\;|s|\le1,\,t\in\mathbb{R}\,,
\end{equation}
where $\chi_-$ and $\chi_+$ are the downward and upward space-time Wiener-Hopf factors,
\[\chi_-(s,it)=\mathbb{E}(s^{\tau_-}e^{itS_{\tau_-}}1_{\{\tau_-<\infty\}})\;\;\;\mbox{and}\;\;\;\chi_+(s,it)=
\mathbb{E}(s^{\tau_+}e^{itS_{\tau_+}}1_{\{\tau_+<\infty\}})\,.\]
To paraphrase W. Feller \cite{fe}, XVIII.3, {\it the remarkable feature of the factorization $(\ref{6283})$ is that it represents
an arbitrary characteristic function $\varphi$ in terms of two (possibly defective) distributions, one being concentrated on the
half line $(-\infty,0)$ and the other one on the half line  $[0,\infty)$}. However, this feature only exploits identity (\ref{6283}) for
fixed $s\neq0$ and reflects the fact that $\mu$ is determined by the knowledge of the distributions of both $S_{\tau_-}$ and
$S_{\tau_+}$. But one may wonder about the extra information brought by the joint distributions $(\tau_-,S_{\tau_-})$ and
$(\tau_+,S_{\tau_+})$. In particular, is it true in general that $\mu$ is determined by only one of these joint distributions?
or equivalently, is it true that $\varphi$ is determined by only one of its space-time Wiener-Hopf factors?\\

The aim of this paper is an attempt to answer the latter question. We will actually show that $\mu$ is determined by $\chi_+(s,it)$
in some quite large classes of distributions including the case where $\mu$ has some positive exponential moments, or when
$t\mapsto\mu(t,\infty)$ is completely monotone on $(a,\infty)$, for some $a\ge0$, or satisfies some property which is slightly stronger
than analyticity. Obviously all these assumptions can be verified from the sole data of $\chi_+(s,it)$. These different cases cover a
sufficiently large range of distributions for us to allow ourselves to raise the following conjecture. Let $\mathcal{M}_1$ be
the set of probability measures on $\mathbb{R}$.\\

\noindent {\bf Conjecture C.} {\it  Any distribution $\mu\in\mathcal{M}_1$ whose support is not included in $(-\infty,0)$ is
determined by its upward space-time Wiener-Hopf factor $\chi_+(s,it)$, $|s|<1$, $t\in\mathbb{R}$.}\\

\noindent A crucial step in the proof of (\ref{6283}) is the following development of the factor $\chi_+(s,t)$, for $|s|<1$ and
$t\in\mathbb{R}$,
\begin{equation}\label{1476}
\log\frac1{1-\chi_+(s,it)}=\sum_{n=1}^\infty\frac{s^n}n\int_{[0,\infty)}e^{itx}\mu^{*n}(dx)\,,
\end{equation}
see \cite{fe}, XVIII.3, where $\mu^{*n}$ is the nth fold convolution product of $\mu$ by itself. We will actually refer to
$\mu^{*n}$, $n\ge0$ as the convolution powers of $\mu$. This identity shows that the
data of $\chi_+$ is equivalent to the knowledge of the measures $\mu^{*n}$, $n\ge1$ on $[0,\infty)$ and leads to the
following equivalent conjecture.\\

\noindent {\bf Conjecture C'.} {\it Any distribution  $\mu\in\mathcal{M}_1$  whose support is not included in $(-\infty,0)$ is
determined by its convolution powers $\mu^{*n}$, $n\ge1$ restricted to $[0,\infty)$.}\\

Each of the next sections corresponds to a class of probability distributions for which Conjecture C holds. For the first one in
Section \ref{FourierLaplace}, we prove that distributions having some particular exponential moments
satisfy conjecture C. Then in Section \ref{5673} we consider three other classes for which a much
stronger result than Conjectures C and C' is true. We will see that there are actually many distributions which are determined by
the sole data of $\mu$ and $\mu^{*2}$ on $[0,\infty)$. This is the case when the function $t\mapsto\mu(t,\infty)$ is smooth
enough. In Subsection \ref{monotone} we will consider the case where the function $t\mapsto\mu(t,\infty)$ is completely
monotone on $(a,\infty)$, for some $a\ge0$ and in Subsection \ref{analytic} we will make a slightly stronger assumption than
analyticity on this function. We will also present the discrete counterpart of the completely monotone case in Subsection \ref{6288}.
Finally in Section \ref{2482} we will consider Conjecture C in the restricted class of infinitely divisible distributions and show that if
the upper tail of the L\'evy measure is completely monotone, then $\mu$ is determined by its upper Wiener-Hopf factor. Then we
end this paper in Section \ref{2450} with some important remarks on the possibility of extending the classes of distributions
studied.

Throughout this paper, we will denote by  $\mathscr{C}$ the set of distributions satisfying Conjecture C. Let us give a proper
definition of this set.

\begin{definition}
Let $\mathscr{C}$ be the set of distributions $\mu\in\mathcal{M}_1$ which are
determined by their upward Wiener-Hopf factor $\chi_+(s,it)$, for $|s|\le1$ and $t\in\mathbb{R}$ or equivalently by the data of
their convolution powers $\mu^{*n}$, $n\ge1$ restricted to $[0,\infty)$. More formally,
\[\mathscr{C}=\{\mu\in\mathcal{M}_1:\mbox{if $\mu_1\in\mathcal{M}_1$
satifies $\mu^{*n}=\mu_1^{*n}$, on $[0,\infty)$, for all $n\ge0$, then $\mu=\mu_1$}\}\,.\]
\end{definition}

The problem we investigate here originates from a result in V.~Vigon's PhD thesis \cite{vi}, see Section 4.5 therein, where a question
equivalent to Conjecture C is raised in the setting of  L\'evy processes. Our question is actually more general since it concerns distributions
which are not necessarily infinitely divisible. In particular, a positive answer to Conjecture C would imply that the law of any L\'evy process
$(X_t,t\ge0)$  is determined by one of its space-time Wiener-Hopf factors or equivalently by the marginals of the process $(X_t^+,t\ge0)$,
see Section \ref{2482}.

\section{When $\mu$ admits exponential moments.}\label{FourierLaplace}

\subsection{Recovering the characteristic function and the moment generating function.}
In this paper, we will always assume that the support of the measure $\mu$ is not included in $(-\infty,0)$.
Let us observe that from the data of the upward Wiener-Hopf factor $\chi_+(s,it)$, for $|s|\le1$ and $t\in\mathbb{R}$ or
equivalently from the data of the measures $\mu^{*n}$, $n\ge1$ restricted to $[0,\infty)$, we know the sequences, $\mathbb{P}(S_n<0)$
and $\mathbb{P}(S_n\ge0)$, $n\ge0$, as well as the distributions of both $\tau_-$ and $\tau_+$. In particular we know whether $(S_n)$
oscillates, drifts to $-\infty$, or drifts to $\infty$.  The next result shows that provided $n\mapsto\mathbb{P}(S_n<0)$ tends to 0 sufficiently
fast along some subsequence, it is possible to recover the characteristic function $\varphi$ of $\mu$ on some interval containing
0, from the measures $\mu^{*n}$ restricted to $[0,\infty)$.

\begin{lemma}\label{2361}
Assume that there is $\alpha>0$ such that, at least for a subsequence,
\begin{equation}\label{0236}
\mathbb{P}(S_{n}<0)\le e^{-\alpha n}\,.
\end{equation}
Then for all $t$ such that $|\varphi(t)|>e^{-\alpha}$, along this subsequence,
\[\lim_{n\rightarrow+\infty}\mathbb{E}(e^{itS_{n}}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})^{1/{n}}=\varphi(t)\,.\]
In particular, if $(\ref{0236})$ holds then $\varphi$ can be determined on some neighborhood of $0$.
\end{lemma}
\begin{proof} Recall that $\varphi(t)$ tends to 1 as $t$ tends to 0. Then let $t$ be sufficiently small so that $\varphi(t)\neq0$ and
let us write,
\begin{eqnarray}
\mathbb{E}(e^{it S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})^{1/n}&=&\left[\varphi(t)^n-\mathbb{E}(e^{it S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n<0\}})\right]^{1/n}\nonumber\\
&=&\varphi(t)[1-\varphi(t)^{-n}\mathbb{E}(e^{it S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n<0\}})]^{1/n}\,.\label{5454}
\end{eqnarray}
From the assumption, whenever is $t$ such that $|\varphi(t)|>e^{-\alpha}$ and for all $n$ such that $\mathbb{P}(S_n<0)<e^{-\alpha n}$,
\begin{eqnarray*}
 \left|\varphi(t)^{-n}\mathbb{E}(e^{it S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n<0\}})\right|&\le&| \varphi(t)^{-n}\mathbb{P}(S_n<0)|\,,\\
 &\le&\left(|\varphi(t)|e^{\alpha}\right)^{-n}\,.
 \end{eqnarray*}
Therefore, the left hand side of the above inequality tends to 0 along a subsequence and this yields the result thanks to
equation (\ref{5454}).
\end{proof}
\noindent Since, for all $n\ge0$,
\[\mathbb{P}(S_1<0,S_2-S_1<0,\dots,S_n-S_{n-1}<0)=\mathbb{P}(S_1<0)^n\le\mathbb{P}(S_n<0)\,,\]
$(\ref{0236})$ cannot hold for all $\alpha>0$, unless $\mathbb{P}(S_1\ge0)=1$. Note also that if $(\ref{0236})$ holds then the random
walk $(S_n)$ cannot drift to $-\infty$. Moreover, if it holds for all suffciently large $n$, then $(S_n)$ necessarily drifts to $\infty$
thanks to Spitzer's criterion which asserts that this happens if and only if $\sum n^{-1}\mathbb{P}(S_n<0)<\infty$.\\

\begin{lemma}\label{3478}
Let $\mu_1,\mu_2\in\mathcal{M}_1$. Denote by $\varphi_1$ and $\varphi_2$ their characteristic functions
and by $\phi_1$ and $\phi_2$ their moment generating functions, that is $\varphi_j(u):=\int_{\mathbb{R}}e^{iu x}\,\mu_j(dx)$
and $\phi_j(v):=\int_{\mathbb{R}}e^{v x}\,\mu_j(dx)$, $u,v\in\mathbb{R}$, $j=1,2$. Assume that there
exists $\lambda>0$ such that $\phi_1(\lambda)<\infty$ and $\phi_2(\lambda)<\infty$ and that there is an open interval $I$
such that $\varphi_1(u)=\varphi_2(u)$, for all $u\in I$. Then $\mu_1=\mu_2$.
\end{lemma}
\begin{proof}
Let $D=\{z=u+iv\in\mathbb{C}:u\in\mathbb{R},\,-\lambda<v<0\}$. From the assumptions, the function $f:=\varphi_1-\varphi_2$ admits
an analytic continuation in the open domain $D$. Then let $O_+=\{z=u+iv\in\mathbb{C}:u\in I,\,0<v<\lambda\}$ and
$O_-=\{z=u+iv\in\mathbb{C}:u\in I,\,-\lambda<v<0\}$. From Schwarz reflection principle, $f$ admits an analytic continuation in the open
domain $O_+\cup I \cup O_-$. From the principle of isolated zeroes, $f$ vanishes in $O_+\cup I \cup O_-$ and from the same principle,
it vanishes in $D$. By continuity, $f(u)=0$ for all $u\in\mathbb{R}$ and the result follows from injectivity of the Fourier
transform.\\
\end{proof}

\noindent  We will say that a distribution $\mu\in\mathcal{M}_1$ admits an exponential moment if there is
$\lambda\in\mathbb{R}\setminus\{0\}$ such that $\phi(\lambda):=\int_{\mathbb{R}}e^{\lambda x}\,\mu(dx)<\infty$.\\

\begin{lemma}\label{3024}
The characteristic function of a distribution having an exponential moment cannot vanish identically in an interval.
\end{lemma}
\begin{proof}
Let $\varphi$ be the characteristic function of $\mu\in\mathcal{M}_1$. If $\varphi$ vanishes in an interval and if  there is
$\lambda\in\mathbb{R}\setminus\{0\}$ such that $\phi(\lambda)<\infty$, then we conclude
from the same arguments as in the proof of Lemma \ref{3478} (replacing $f$ by $\varphi$) that $\varphi(u)=0$ for all
$u\in\mathbb{R}$, which is absurd since $\varphi$ is a characteristic function.
\end{proof}

\noindent Lemma \ref{3024} was noticed in \cite{sm} for nonnegative random variables.\\

\begin{theorem}\label{4572}
Assume that we know the measures $\mu^{*n}$ restricted to $[0,\infty)$. Then we can determine if $(\ref{0236})$ holds.
Moreover if $(\ref{0236})$ holds, then $\mu$ belongs to the class $\mathscr{C}$.
\end{theorem}
\begin{proof}
The first assertion is trivial since $1-\mu^{*n}[0,\infty)=\mathbb{P}(S_{n}<0)$. Now if $(\ref{0236})$ holds, then from Lemma \ref {2361},
the characteristic function $\varphi$ of $\mu$ can be determined on some neighborhood of 0. Since the function
$t\mapsto \chi_+(1,it)$ is known, from (\ref{6283}), it means that the function $t\mapsto \chi_-(1,it)$ can be determined on the
same neighborhood. But $t\mapsto\chi_-(1,it)/\chi_-(1,0)$ is the characteristic function of the non positive random variable
$S_{\tau_-}$ under $\mathbb{P}(\,\cdot\,|\,\tau_-<\infty)$. Since $\mathbb{E}(e^{\lambda S_{\tau_-}}\,|\,\tau_-<\infty)\le1$, for all $\lambda\ge0$,
we derive from Lemma \ref{3478}  that $\chi_-(1,it)$ is determined for all $t\in\mathbb{R}$. Therefore, from (\ref{6283}),
$\varphi(t)$ is determined for all $t\in\mathbb{R}$ and the result follows.\\
\end{proof}

\begin{remark}\label{2699}
Distributions having negative exponential moments provide examples for which $(\ref{0236})$ is satisfied.
More specifically, assume that there is $\lambda<0$ such that $\phi(\lambda)<1$, then
\[\mathbb{P}(S_n<0)\le \mathbb{E}(e^{\lambda S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n<0\}})\le\phi(\lambda)^n\,,\]
which implies $(\ref{0236})$. The question of finding an example of a distribution with no exponential moment which satisfies
$(\ref{0236})$ remains open.\\
\end{remark}

\noindent Recall that $\phi$ is a convex function on the interval $\{\alpha:\phi(\alpha)<\infty\}$. Moreover, since the support of $\mu$ is not included in $(-\infty,0)$,  $\phi(\alpha)$ is nondecreasing for $\alpha$ large enough. If
$\lambda\in\mathbb{R}$ is such that $\lambda=\inf\{\alpha:\phi(\alpha)<\infty\}$, then $\phi'(\lambda)$ will be understood as the right
derivative of $\phi$ at $\lambda$. Similarly,  if  $\lambda=\sup\{\alpha:\phi(\alpha)<\infty\}$, then $\phi'(\lambda)$ will be the left
derivative of $\phi$ at $\lambda$.

\begin{lemma}\label{7371}
For all $\lambda\in\mathbb{R}$ such that $\phi(\lambda)<\infty$ and $\phi'(\lambda)>0$,
\[\lim_{n\rightarrow+\infty}\mathbb{E}(e^{\lambda S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})^{1/n}=\phi(\lambda)\,.\]
\end{lemma}
\begin{proof}
Let $(S^{(\lambda)}_n)$ be a random walk with step distribution
$\displaystyle\mu_\lambda(dx):=\frac{e^{\lambda x}}{\phi(\lambda)}\mu(dx)$.
Since
\[\mathbb{E}(S^{(\lambda)}_1)=\int_{\mathbb{R}}\frac{xe^{\lambda x}}{\phi(\lambda)}\,\mu(dx)=\frac{\phi'(\lambda)}{\phi(\lambda)}>0\,,\]
the random walk $(S^{(\lambda)}_n)$ drifts to $\infty$, so that $\lim_{n\rightarrow\infty}\mathbb{P}(S^{(\lambda)}_n\ge0)=1$. Then the
result follows from the identity
\[\mathbb{P}(S^{(\lambda)}_n\ge0)=\frac{\mathbb{E}(e^{\lambda S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})}{\phi(\lambda)^n}\,.\]
\end{proof}

The following theorem shows that distributions having some negative exponential moments less than 1 or some positive exponential
moments bigger than 1 belong to class~$\mathscr{C}$.

\begin{theorem}\label{2258} The knowledge of the measures $\mu^{*n}$, $n\ge1$ restricted to $[0,\infty)$ allows us
to determine if $\phi$ satisfies one of the two following conditions:
\begin{itemize}
\item[$(a)$] There exists $\lambda<0$ such that $\phi(\lambda)<1$.
\item[$(b)$] There exists  $\lambda>0$ such that $\phi(\lambda)\in(1,\infty)$.
\end{itemize}
When at least one of these two conditions holds, the measure $\mu$
belongs to the class $\mathscr{C}$.
\end{theorem}
\begin{proof}  From Remark \ref{2699}, if $(\ref{0236})$ does not hold, then $(a)$ is not satisfied. Assume that $(\ref{0236})$ holds.
From Theorem \ref{4572}, the measure $\mu$ belongs to class $\mathscr{C}$ so that we can determine if $(a)$ holds.

From our data, for all $\lambda>0$ and $n\ge1$, the expression $\mathbb{E}(e^{\lambda S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})$ is known.
Assume that there is $\lambda>0$ such that $\lim_{n\rightarrow+\infty}\mathbb{E}(e^{\lambda S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})^{1/n}>1$.
Since $\mathbb{E}(e^{\lambda S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})^{1/n}\le \phi(\lambda)$, we have actually $\phi(\lambda)>1$. Moreover,
our data clearly allows us to know if $\phi(\lambda)<\infty$. Then since $\phi(0)=1$, by convexity of the function
$\lambda\mapsto\phi(\lambda)$, it is clear that $\phi'(\lambda)>0$, so that from Lemma \ref{7371},
$\lim_{n\rightarrow+\infty}\mathbb{E}(e^{\lambda S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})^{1/n}=\phi(\lambda)$. This means that from our data, we can
determine if there is $\lambda>0$ such that $\phi(\lambda)\in(1,\infty)$ and in this case
$\lim_{n\rightarrow+\infty}\mathbb{E}(e^{\lambda S_n}\mbox{\rm 1\hspace{-0.04in}I}_{\{S_n\ge0\}})^{1/n}=\phi(\lambda)$, so that $\phi(\lambda)$  is known.
Moreover, from the continuity of $\phi$
on the set $\{x:\phi(x)<\infty\}$, there is an interval $I$ containing $\lambda$ such that for all $x\in I$, $\phi(x)\in(1,\infty)$ and
from the same reasoning as above, $\phi(x)$ is known for all $x\in I$, so that the measure $\mu$ is determined and we conclude
that it belongs to the class $\mathscr{C}$.
\end{proof}
\noindent  We can easily check that condition $(b)$ of Theorem \ref{2258} is satisfied in the two following situations:
\begin{itemize}
\item[$(b_1)$] $\phi(\lambda)<\infty$ for all $\lambda>0$.\\
\item[$(b_2)$] $\mu$ is absolutely continuous in $[0,\infty)$ and its density $f$
satisfies $\ln(f(x))\sim -\lambda_0 x$, as $x\rightarrow\infty$, for some $\lambda_0\in(0,\infty)$.
\end{itemize}
Indeed, in case $(b_1)$, if  $\mu\neq\delta_0$ then since $\phi$ is a nondecreasing convex function such that $\phi(0)=1$
and since the support of $\mu$ is not included in $(-\infty,0)$, $\lim_{\lambda\rightarrow\infty}\phi(\lambda)=\infty$.
In case $(b_2)$ it is clear that $\lim_{\lambda\rightarrow\lambda_0-}\phi(\lambda)=\infty$.

\subsection{Skip free distributions.}\label{skipfree}

A distribution $\mu$ whose support is included in $\mathbb{Z}$ is said to
be downward (resp. upward) skip free if $\mu(n)=0$ for all $n\le-2$ (resp. for all $n\ge2$).
Clearly skip free distributions possess exponential moments.  Moreover, upward skip free distributions belong to class
$\mathscr{C}$ from Theorem \ref{2258} $(b)$ and the note following its proof. Then in this subsection we shall see that the case of
downward skip free distributions allows us to go a little beyond the cases encompassed by Theorems \ref{4572}
and \ref{2258}. We first need to make sure that our data allows us determine if the support of a distribution is included in $\mathbb{Z}$.

\begin{lemma}\label{4261}
The support of the measure $\mu$ is included in $\mathbb{Z}$ if and only if the support of the measures
$\mu^{*n}$, $n\ge1$ restricted to $[0,\infty)$ is included in $\mathbb{Z}_+$.
\end{lemma}
\begin{proof}   The direct implication is obvious. Then assume that the support of $\mu^{*n}$, $n\ge1$  restricted to
$[0,\infty)$ is included in $\mathbb{Z}_+$, whereas the support of $\mu$ restricted to $(-\infty,0]$ is not included in
$\mathbb{Z}_-$.
Then there is an interval $I\subset(-\infty,0]\setminus\mathbb{Z}_-$ such that $\mu(I)>0$.
Let $n\in\mathbb{Z}_+\setminus\{0\}$ such that $\mu(\{n\})>0$ and $h\in\mathbb{Z}_+\setminus\{0\}$ such that $hn+\inf I>0$.
Then
\[0< \mu(I)\mu(\{n\})^h=\mathbb{P}(S_1\in I,S_{i+1}-S_i=n,i=1,\dots,h)\le\mathbb{P}(S_{n+1}\in hn+I)\,.\]
This implies that $\mu^{*(n+1)}(hn+I)>0$, where $hn+I\subset[0,\infty)\setminus\mathbb{Z}_+$, which contradicts the
assumption.
\end{proof}

\begin{theorem}\label{8733}
Downward skip free distributions belong to class $\mathscr{C}$.
\end{theorem}
\begin{proof} Let $\mu\in \mathcal{M}_1$ whose convolution powers $\mu^{*n}$, $n\ge1$ restricted to $[0,\infty)$ are known.
Then from Lemma \ref{4261} we can determine if the support of $\mu$ is included in $\mathbb{Z}$ or not. Let us assume
that it is the case.

As already noticed at the beginning of this section, we can determine if $(S_n)$ drifts to $\infty$ or not.
Assume first that $(S_n)$ drifts to $\infty$.

Let us observe that under this assumption, if $\mu$ is a downward skip free distribution, then there exists $\lambda<0$ such
that $\phi(\lambda)<1$. Indeed in this case, it is clear that $\phi(\lambda)<\infty$, for all $\lambda\le0$. Moreover $\mathbb{E}(S_1)$
exists and is positive. Since $\mathbb{E}(S_1)=\lim_{\lambda\uparrow0}(1-\phi(\lambda))/\lambda$, there is necessarily $\lambda<0$
such that $\phi(\lambda)<1$.

From Theorem \ref{2258} $(a)$, we can determine if there exists $\lambda<0$ such
that $\phi(\lambda)<1$.  If this is not the case, then $\mu$ cannot be downward skip free.
On the contrary, if there exists $\lambda<0$ such that $\phi(\lambda)<1$, then from Theorem \ref{2258}, $\mu$ can be determined.
In particular, we know if $\mu$ is downward skip free or not.

Now assume that $(S_n)$ does not drift to $\infty$ and write for $n\ge0$,
\begin{eqnarray}
\mathbb{P}(S_{\tau_+}>n,\tau_+<\infty)&=&\sum_{k\ge1}\mathbb{P}(S_1<0,\dots,S_{k-1}<0,S_k-S_{k-1}>n-S_{k-1})\nonumber\\
&=&\sum_{k\ge1}\sum_{r\le-1}\mathbb{P}(S_1<0,\dots,S_{k-2}<0,S_{k-1}=r)\mathbb{P}(S_1>n-r)\nonumber\\
&=&\sum_{r\ge1} v(r)\mathbb{P}(S_1>n+r)\,,\label{5202}
\end{eqnarray}
where $v(r)=\sum_{k\ge1}\mathbb{P}(S_1\le0,\dots,S_{k-1}\le0,S_{k}=-r)$ is the renewal measure on $\{1,2,\dots\}$ of the (strict)
downward ladder height process of $(S_n)$, see Chap. XII.2 in \cite{fe}.  In particular, this renewal measure
satisfies $v(r)\le1$ for all $r\ge1$. Moreover, $(S_n)$ is downward skip free and does not drift to $\infty$, if and only if $v(r)=1$,
for all $r\ge1$. Since it is the only unknown in Equation (\ref{5202}), we can determine if it is the case or not.  Finally, knowing that
$\mu$ is downward skip free, we immediately determine this distribution on $\mathbb{R}$ from its knowledge on $[0,\infty)$.
\end{proof}

\noindent It appears in the proof of Theorem \ref{8733} that downward skip
free distributions which drift to $\infty$ actually satisfy condition $(a)$ of Theorem \ref{2258}. Hence the only additional case in
this subsection is that of downward skip free distributions which do not drift to~$\infty$.\\

We will denote by $\mathscr{E}$ the set of measures $\mu$ satisfying the assumptions of  of Theorem \ref{2258} or those of
Theorem \ref{8733}, that is the set of measures satisfying $(a)$ or $(b)$ or downward skip free distributions.
It will be called the {\it exponential} class. From Theorems \ref{2258} and \ref{8733}, we have $\mathscr{E}\subset\mathscr{C}$.
Note that from Theorem \ref{4572},  we have determined a subclass of $\mathscr{C}$ which is presumably bigger than
$\mathscr{E}$.

\section{When $\mu$ is characterized by $\mu$ and $\mu^{*2}$ on $[0,\infty)$.}\label{5673}

\subsection{Preliminary results} We will show that in many cases, the sole data of $\mu$ and $\mu*\mu$ on $[0,\infty)$
actually suffices to determine $\mu$. In this subsection, we give a theoretical condition for this to hold.\\

\noindent We first observe that in Conjectures C and C', there is no loss of generality in assuming that $\mu$ is absolutely
continuous on $[0,\infty)$.\\

\noindent {\bf Conjecture C''.} {\it Any absolutely continuous distribution $\mu\in\mathcal{M}_1$ whose support is not included
in $(-\infty,0)$ is determined by its convolution powers $\mu^{*n}$, $n\ge1$ restricted to $[0,\infty)$.}

\begin{lemma}\label{3459}
Conjectures {\rm C}, {\rm C'} and {\rm C''} are equivalent.
\end{lemma}
\begin{proof} We already know from Section \ref{int} that Conjectures C and C' are equivalent. Then clearly, it suffices to prove
that if Conjecture C'' is true, then Conjecture C' is true.

Let $\mu,\mu_1\in\mathcal{M}_1$ be any two distributions such that the measures $\mu_1^{*n}$ and $\mu^{*n}$ agree on $[0,\infty)$,
for  all $n\ge1$. Let $g$ be any probability density function on $[0,\infty)$, i.e.$\int_{0}^\infty g(x)\,{\rm d}x=1$ and let
$\bar{\mu},\bar{\mu}_1\in\mathcal{M}_1$ be the absolutely continuous measures whose respective densities are $h(x)=\int_{\mathbb{R}}
g(y-x)\,\mu(dy)$ and $h_1(x)=\int_{\mathbb{R}} g(y-x)\,\mu_1(dy)$, $x\in\mathbb{R}$.

Denoting by $g^{*n}$ the $n$-th convolution product of the function $g$ by itself, it is plain that for all $x\ge0$,
\[\bar{\mu}^{*n}[x,\infty)=\int_0^\infty\mu^{*n}[x+y,\infty)g^{*n}(y)\,dy\;\;\mbox{and}\;\;\bar{\mu}_1^{*n}[x,\infty)=
\int_0^\infty\mu_1^{*n}[x+y,\infty)g^{*n}(y)\,dy\,.\]
Therefore since the measures $\mu^{*n}$ and $\mu_1^{*n}$ agree on $[0,\infty)$, for  all $n\ge1$, the measures  $\bar{\mu}^{*n}$
and $\bar{\mu}_1^{*n}$ also agree on $[0,\infty)$, for  all $n\ge1$ and from the assumption that conjecture C'' is true, we conclude that
the measures $\bar{\mu}$ and $\bar{\mu}_1$ agree on $\mathbb{R}$. Then we can identify both characteristic functions:
\begin{eqnarray*}\int_{\mathbb{R}}e^{itx}\,\bar{\mu}(dx)&=&\int_{\mathbb{R}}e^{itx}\,\mu(dx)
\int_{0}^\infty e^{-itx}g(x)\,dx\;\;\:\mbox{and}\\
\int_{\mathbb{R}}e^{itx}\,\bar{\mu}_1(dx)&=&\int_{\mathbb{R}}e^{itx}\,\mu_1(dx)\int_{0}^\infty e^{-itx}g(x)\,dx\,.
\end{eqnarray*}
But from Lemma \ref{3024}, the characteristic function of $g$ cannot vanish identically in an interval.
This implies that $\int_{\mathbb{R}}e^{itx}\,\mu(dx)=\int_{\mathbb{R}}e^{itx}\,\mu_1(dx)$, for all $t\in\mathbb{R}$
by continuity of characteristic functions. Then we conclude that $\mu=\mu_1$ on $\mathbb{R}$, from injectivity of the
Fourier transform.
\end{proof}

\noindent We derive from Lemma \ref{3459} that there is no loss of generality in assuming that $\mu$ is absolutely continuous on
$\mathbb{R}$. We will sometimes make this assumption and denote the density of $\mu$ by $f$.

\begin{lemma}\label{2363} For any probability density function, $f$ on $\mathbb{R}$ and for all $t\ge0$,
\begin{equation}\label{2925}
\int_0^\infty f(t+s)\bar{f}(s)\,ds=\frac12\left(f*f(t)-\int_0^tf(t-s)f(s)\,ds\right)\,,
\end{equation}
where $\bar{f}(s)=f(-s)$.
\end{lemma}
\begin{proof} It suffices to decompose $f*f$ as
\begin{eqnarray*}
f*f(t)&=&\int_{\mathbb{R}}f(t-s)f(s)\,ds\\
&=&\int_{-\infty}^0f(t-s)f(s)\,ds+\int_0^tf(t-s)f(s)\,ds+\int_t^\infty f(t-s)f(s)\,ds\,.
\end{eqnarray*}
Then from a change of variables, we obtain,
$\int_t^\infty f(t-s)f(s)\,ds=\int_{-\infty}^0f(t-s)f(s)\,ds=\int_0^\infty f(t+s)\bar{f}(s)\,ds$, which proves our identity.
\end{proof}

The main idea of this section is to exploit identity (\ref{2925}) in order to characterize the function $\bar{f}$ on $[0,\infty)$
(or equivalently $f $ on $(-\infty,0]$) from the sole data of $f$ and $f*f$ on
$[0,\infty)$. More specifically, assume that $f$ restricted to $[0,\infty)$ fulfills the following property:
for any two nonnegative Borel functions $g_1$ and $g_2$ defined on
$[0,\infty)$ such that $\int_0^\infty f(t+s)g_1(s)\,ds<\infty$, $\int_0^\infty f(t+s)g_2(s)\,ds<\infty$, for all $t\ge0$, the following implication is
satisfied,
\begin{equation}\label{7326}
\int_{[0,\infty)} f(t+s)g_1(s)\,ds=\int_{[0,\infty)} f(t+s)g_2(s)\,ds\,,\;\;\mbox{for all $t\ge0$}\;\;\Rightarrow \;\;g_1\equiv g_2\,,\;\;\mbox{a.e.}
\end{equation}
Then clearly, the map $t\mapsto\int_{[0,\infty)} f(t+s)\bar{f}(s)\,ds$ characterizes $\bar{f}$ on $[0,\infty)$ and therefore from (\ref{2925}),
 $\mu$ is determined on $\mathbb{R}$ by the sole data of $\mu$ and $\mu^{*2}$ on $[0,\infty)$.

\begin{remark}\label{6922}
There are density functions $f$ which do not satisfy $(\ref{7326})$. For instance
with $f(s)=\frac12e^{-s}$, the operator  $t\mapsto\int_{0}^\infty f(t+s)\bar{f}(s)\,ds=\frac12e^{-t}\int_{0}^\infty e^{-s}\bar{f}(s)\,ds$ provides
a very poor information on $\bar{f}$ and certainly cannot characterize this function on $[0,\infty)$. Also if $\mu$ has a bounded
support in $[0,\infty)$, then clearly the operator $t\mapsto\int_{0}^\infty f(t+s)\bar{f}(s)\,ds$ cannot characterize $\bar{f}$
outside this support.

Note also that from $(\ref{2925})$, the knowledge of  $t\mapsto\int_{0}^\infty f(t+s)\bar{f}(s)\,ds$ and $f(t)$, for $t\ge0$ is equivalent
to this of the functions $f$ and $f^{*2}$ on $[0,\infty)$. Therefore, in the above examples, $f$ is not even determined
by the data of $f$ and $f^{*2}$ on $[0,\infty)$.
\end{remark}

\noindent The following proposition gives a sufficient condition for $(\ref{7326})$ to hold.

\begin{proposition}\label{6412} Assume that $\mu$ is absolutely continuous on $\mathbb{R}$ with density $f$.
Let us introduce the following set of functions defined on $[0,\infty)$,
\[\mathcal{H}:=\left\{\sum_{k=1}^n\alpha_kf(t_k+\cdot):n\ge1,\alpha_k\in\mathbb{R},\,t_k\ge0\right\}\,.\]
If the restriction of $f$ to $[0,\infty)$ belongs to $L^\infty([0,\infty))$ and if
$\mathcal{H}$ is dense in $L^\infty([0,\infty))$, then for any $g\in L^1([0,\infty))$, the following implication is satisfied,
\begin{equation}\label{7325}
\int_{0}^\infty f(t+s)g(s)\,ds=0,\;\;\mbox{for all $t\ge0$}\;\;\Rightarrow \;\;g\equiv0,\;\;\mbox{a.e.}
\end{equation}
When $(\ref{7325})$ holds, the measure $\mu$ is determined on $\mathbb{R}$ by $\mu$ and $\mu^{*2}$ on $[0,\infty)$.
\end{proposition}
\begin{proof} If $\int_{0}^\infty f(t+s)g(s)\,ds=0$, for all $t\ge0$, then clearly, since $\mathcal{H}$ is dense in $L^\infty([0,\infty))$,
$\int_{0}^\infty h(s)g(s)\,ds=0$ for all $h\in L^\infty([0,\infty))$ and this implies that $g\equiv 0$, a.e.

Assume now that the restriction of the measures $\mu$ and $\mu^{*2}$ are known on $[0,\infty)$. Recall the notation
$\bar{f}$ from Lemma \ref{2363} and observe that the right hand side of identity (\ref{2925}) is known for all $t\ge0$.
From (\ref{7325}) this determines $\bar{f}$ on $[0,\infty)$ and the measure $\mu$ is determined.
\end{proof}
\noindent Unfortunately we do not know any example of function satisfying the condition of Proposition~$\ref{6412}$ and finding a
simple criterion on $f$ for it to satisfy this condition remains an open problem. More specifically, we may wonder if the converse of
Proposition~$\ref{6412}$ holds, that is if assertion $(\ref{7325})$ implies that $\mathcal{H}$ is dense in $L^\infty([0,\infty))$.
The latter problem can be compared with Wiener's approximation theorem which asserts that for a function $f$ in
$L^1(\mathbb{R})$ the set $\mathcal{H}$ $($thought as a set of functions defined on $\mathbb{R}$$)$ is dense in
$L^1(\mathbb{R})$ if and only if the Fourier transform of $f$ does not vanish, see \cite{ru}.\\

\noindent In Subsection \ref{7235}, we give a class of density functions such that (\ref{7326}) holds and in Subsection \ref{3468},
 we give a class of density functions which are bounded on $[0,\infty)$ and such that (\ref{7325}) holds.

\subsection{The completely monotone class}\label{monotone}\label{7235}

In this subsection, we assume that $\mu$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}$ and we
denote by $f$ its density.\\

\noindent We will show that if $f$ restricted to $(a,\infty)$, for some $a\ge0$ is a completely monotone function satisfying some mild
additional assumption, then $\mu$ is characterized from $\mu$ and $\mu^{*2}$ on $[0,\infty)$. Let us first recall that from Bernstein
Theorem, the function $f$ is completely monotone on $(a,\infty)$, for $a\ge0$, if and only if there is a positive Borel measure $\nu$
on $(0,\infty)$ such that for all $t>a$,
\begin{equation}\label{1842}
f(t)=\int_0^\infty e^{-ut}\nu(du)\,.
\end{equation}

\begin{theorem}\label{9524} Assume that there is $a\ge0$ such that the restriction of $f$ to $(a,\infty)$ is completely monotone.
Assume moreover  that the support of the measure $\nu$ in $(\ref{1842})$ contains an increasing sequence $(a_n)$ such that
$\sum_n a_n^{-1}=+\infty$. Then $(\ref{7326})$ holds and the measure $\mu$ is characterized by the restriction of $\mu$ and
$\mu^{*2}$ to $[0,\infty)$. In particular, $\mu$ belongs to class $\mathscr{C}$.
\end{theorem}
\begin{proof} Let $g$ be any nonnegative Borel function defined on $[0,\infty)$ such that $\int_0^{\infty}f(t+s)g(s)\,ds<\infty$, for all
$t>a$. Then from Fubini's Theorem, for all $t>a$,
\begin{eqnarray*}
\int_0^{\infty}f(t+s)g(s)\,ds&=&\int_0^{\infty}\int_0^\infty e^{-u(t+s)}\nu(du)g(s)\,ds\\
&=&\int_0^{\infty}e^{- ut}\int_0^\infty e^{-us}g(s)\,ds\,\nu(du)\,.
\end{eqnarray*}
This expression is the Laplace transform of the measure $\theta(du):=\int_0^\infty e^{-us}g(s)\,ds\,\nu(du)$.
The knowledge of this Laplace transform for all $t>a$ characterizes the measure $\theta(du)$ so that a version of the density
function $u\mapsto \int_0^\infty e^{-us}g(s)\,ds$ is known on a Borel set $B\subset (0,\infty)$ such that $\nu(B^c)=0$. Since this
density function is continuous, it is known on $\overline{B}$ and hence it is known everywhere on the support of $\nu$. Therefore,
from the assumption on $\nu$, we can find an increasing sequence $(a_n)$ such that $\sum_n a_n^{-1}=+\infty$ and such that the
Laplace transform $\int_0^\infty e^{-a_ns}g(s)\,ds$ of the function $g$ at $a_n$ is known for each $n$.
From a result in \cite{fe1}, this is enough to determine the function $g$ and (\ref{7326}) holds, see also \cite{fe}, p.430.

Then we derive from (\ref{7326}) and Lemma \ref{2363} that the function $\bar{f}$ is determined on $[0,\infty)$ from the restriction
on $[0,\infty)$ of $\mu$ and $\mu^{*2}$. Therefore the measure $\mu$ is determined.
\end{proof}

\noindent We will denote by $\mathscr{M}$ the set of absolutely continuous measures $\mu$ whose density $f$ satisfies the
assumption of Theorem \ref{9524}. This class will be called the completely monotone class. Theorem \ref{9524} shows that
$\mathscr{M}\subset\mathscr{C}$.

\begin{remark}
Note that since $f$ is a density function, the measure $\nu$ in $(\ref{1842})$ should
also satisfy
\begin{equation}\label{6759}
\int_0^\infty f(t)\,dt=\int_0^\infty \frac{e^{-au}}u\nu(du)\le 1\,.
\end{equation}
\end{remark}

\begin{remark}\label{8644} Clearly class $\mathscr{E}$ is not included in class $\mathscr{M}$. Moreover, it is easy to find an
example of a measure in class $\mathscr{M}$ which does not belong to class $\mathscr{E}$.
Indeed, we readily check that whenever the support $S$ of $\nu$ is such that $S\cap(0,\varepsilon)\neq\emptyset$,
for all $\varepsilon>0$, then $\mu$ has no positive exponential moments, so that it cannot belong to class
$\mathscr{E}$. Let us take for instance the measure $\nu(du)$ on $[0,\infty)$ with density
$u^{2}\mbox{\rm 1\hspace{-0.04in}I}_{\{u\in[0,1]\}}+\frac14u^{-1}\mbox{\rm 1\hspace{-0.04in}I}_{\{u\in(1,\infty)\}}$.
This measure satisfies the assumption of Theorem $\ref{9524}$,
for $a=0$. \\
\end{remark}

\begin{remark}
Theorem $\ref{9524}$ excludes completely monotone functions of the type $f(t)=\sum_{k=1}^n\alpha_ke^{-\beta_k t}$, $t>a$, for
some $\alpha_k,\beta_k>0$ and some finite $n$ since in this case the measure $\nu(du)=\sum_{k=1}^n\alpha_k\delta_{\beta_k}(du)$
does not satisfy the condition required by this theorem. It also excludes functions $f$ whose support is bounded since completely
monotone functions on $(a,\infty)$ are analytic on $(a,\infty)$.   This remark is consistent with Remark $\ref{6922}$.
\end{remark}

\subsection{The analytic class}\label{analytic}\label{3468}

Let us assume again that $\mu$ has density $f$ on $\mathbb{R}$. We will now exploit the same kind of arguments as in the previous
subsection by assuming that $f$ is the Fourier transform of some complex valued function.

\begin{theorem}\label{3720} Assume that there is a complex valued function $k$ such
that for all $t\ge0$,
\begin{equation}\label{8179}
f(t)=\int_{\mathbb{R}}e^{iut}k(u)\,du\,.
\end{equation}
Assume moreover that
\begin{itemize}
\item[1.] the absolute moments $M_n=\int_{\mathbb{R}}|u|^n|k(u)|\,du$, $n\ge0$ are finite and satisfy
\[\limsup_n\frac{M_n}{n!}<\infty\,,\]
\item[2.] the function $k$ does not vanish on any interval of $\mathbb{R}$.
\end{itemize}
Then $f$ is bounded on $[0,\infty)$ and $(\ref{7325})$ holds. In particular, the measure $\mu$ is characterized by
the restrictions of $\mu$ and $\mu^{*2}$ to $[0,\infty)$ and $\mu$ belongs to class $\mathscr{C}$. Moreover $f$ admits an analytic
continuation on $\mathbb{R}$.
\end{theorem}
\begin{proof} The fact that $f$ is bounded follows directly from (\ref{8179}) and $1.$ Let $g\in L^1([0,\infty))$,
then from (\ref{8179}) and Fubini's theorem, we can write for all $t\ge0$,
\begin{eqnarray*}
\int_{0}^\infty f(t+s)g(s)\,ds&=&\int_0^{\infty}\int_{\mathbb{R}}e^{iu(t+s)}k(u)\,du\,
g(s)\,ds\\
&=&\int_{\mathbb{R}}e^{iut}\int_0^{\infty}e^{ius}g(s)\,ds\,k(u)\,du\,.
\end{eqnarray*}
Assume that this expression vanishes for all $t\ge0$ and set $\varphi(u)=\int_0^{\infty}e^{ius}g(s)\,ds$.
This means that the Fourier transform
\[\Psi(t):=\int_{\mathbb{R}}e^{iut}\varphi(u)\,k(u)\,du\,,\]
of the function $u\mapsto \varphi(u)k(u)$, $u\in\mathbb{R}$ vanishes for all $t\ge0$. Then let us show that under our assumptions,
the function $\Psi$ is analytic on the whole real axis. First note that since $|\varphi(u)|\le\|g\|_{L^1}$, $u\in\mathbb{R}$ and since all the
moments $M_n$ are finite, then $\Psi$ is infinitely differentiable on $\mathbb{R}$ and
\begin{equation}\label{4168}
\Psi^{(n)}(t)=\int_{\mathbb{R}}(iu)^ne^{iut}\varphi(u)\,k(u)\,du\,,\;\;\;t\in\mathbb{R}\,.
\end{equation}
Then notice that for all $t,u,x\in\mathbb{R}$,
\[\left|e^{iux}\left(e^{itu}-1+\frac{itu}{1!}-\dots-\frac{(itu)^{n-1}}{(n-1)!}\right)\right|\le \frac{|tu|^n}{n!}\,.\]
We derive from this inequality and (\ref{4168}) that
\begin{equation}\label{4199}
\left|\Psi(x+t)-\Psi(x)-\frac t{1!}\Psi'(x)-\dots-\frac{t^{n-1}}{(n-1)!}\Psi^{(n-1)}(x)\right|\le
\frac{M_n}{n!}|t|^n\,.
\end{equation}
Set $c=\limsup_nM_n/n!$, then from Stirling's formula, for $|t|<1/(3c)$, the righthand side of (\ref{4199})
tends to 0, as $n\rightarrow+\infty$, hence the Taylor series of $\Psi$ converges in some interval around $x$,
for all $x\in\mathbb{R}$. It follows that $\Psi$ is analytic on $\mathbb{R}$. As a consequence,
$\Psi$  is determined by its expression on the positive half line. Hence the Fourier transform of the continuous function
$u\mapsto\varphi(u)k(u)$ vanishes on $\mathbb{R}$, which means that this function vanishes a.e. on $\mathbb{R}$.
Since $k$ does not vanish on any interval of $\mathbb{R}$ and $\varphi$ is continuous, it implies that $\varphi(u)=0$,
for all $u\in\mathbb{R}$ and we conclude that $g(t)=0$, for almost every $t\in[0,\infty)$. We have proved that
$(\ref{7325})$ holds and from the second part of Proposition \ref{6412}, $\mu$ is characterized by the restriction of $\mu$
and $\mu^{*2}$ on $[0,\infty)$.

We have proved above that $\Psi(t)=\int_{\mathbb{R}}e^{iut}\varphi(u)\,k(u)\,du$ is analytic on $\mathbb{R}$.
It follows from the same arguments that the continuation of $f$ on $\mathbb{R}$ which is defined in a natural way by
$f(t)=\int_{\mathbb{R}}e^{iut}k(u)\,du$, $t\in\mathbb{R}$ is analytic, which proves the last assertion of the theorem.
\end{proof}

We will denote by $\mathscr{A}$ the class of distributions which satisfy the assumptions of Theorem \ref{3720}. It will be
called the analytic class. From Theorem \ref{3720}, $\mathscr{A}\subset\mathscr{C}$.\\

It is very easy to construct examples of distributions in class $\mathscr{A}$, simply by choosing any symmetric function
$k$ which satisfies assumptions $1.$ and $2.$ in Theorem \ref{3720}. Let us consider for instance $k(u)=e^{-|u|}$,  for
$u\in\mathbb{R}$. Then
\[f(t)=\int_{\mathbb{R}}e^{iut}e^{-|u|}\,du=\frac{1}{2(1+t^2)}\,,\;\;\;t\ge0\,.\]
Any extension on $\mathbb{R}$ of this function into a density function determines a distribution of $\mathscr{A}$ which does
not belong to classes $\mathscr{E}$ and $\mathscr{M}$. Conversely, none of the classes $\mathscr{E}$ and $\mathscr{M}$
is included in $\mathscr{A}$. It is straightforward for $\mathscr{E}$.
Then let us consider
\[f(t)=\frac14\left(\int_0^1 e^{-ut}\,u^{1/2}du+\int_1^\infty e^{-ut}\,u^{-1}du\right)\,,\;\;\;t>0\,.\] The measure
$\nu(du)=\left(u^{1/2}\mbox{\rm 1\hspace{-0.04in}I}_{[0,1]}(u)+u^{-1}\mbox{\rm 1\hspace{-0.04in}I}_{[1,\infty)}(u)\right)\,du$ satisfies the
condition of Theorem \ref{9524} so that $\mu\in\mathscr{M}$ (here we  choose $a=0$). However,
$\lim_{t\rightarrow0+}f(t)=\infty$, hence it does not admit an analytic continuation on $\mathbb{R}$, so that $f$ does not belong
to class $\mathscr{A}$, from Theorem \ref{3720}. (Note also that since the support of $\nu$ intersects any interval
$(0,\varepsilon)$, $\varepsilon>0$, the measure $\mu$ has no positive exponential moments, see Remark \ref{8644}.)\\

Here is a consequence of Theorem \ref{3720} on stable distributions.

\begin{corollary}\label{2885}
Let $S$ be a stable distribution on $\mathbb{R}$ with index $\alpha\in[1,2]$, whose support is not
included in $(-\infty,0)$. If a measure $\mu$ satisfies $\mu=S$ and $\mu*\mu=S*S$ on $[0,\infty)$ then $\mu=S$ on
$\mathbb{R}$.
\end{corollary}
\begin{proof} Let $f$ be the density of $\mu$ in $[0,\infty)$. From the expression of the characteristic exponent of stable
distributions and Fourier inverse transform, for all $t\ge0$,
\[f(t)=\frac1{2\pi}\int_{\mathbb{R}}e^{-itu}e^{-c|u|^\alpha(1-i\beta\mbox{\rm\tiny sgn}(u)\tan(\pi\alpha/2))}\,du\,,\]
where $\beta\in[-1,1]$ and $c$ is some positive constant. Then we can easily check that if $\alpha\in[1,2]$, $f$ satisifes the
conditions of Theorem \ref{3720}  and the result follows.
\end{proof}

\noindent Corollary \ref{2885} should be compared with a result from Rossberg and Jesiak \cite{rj} which asserts that if
$F_1$ and $F_2$ are the distribution functions of two stable distributions and if $F_1(x)=F_2(x)$ for $x$ belonging to
a set which contains at least three accumulation points, $l_1$, $l_2$ and $l_3$ such that $F_i(l_j)\neq 0,1$ for $i=1,2$
and $j=1,2,3$, then $F_1\equiv F_2$, on $\mathbb{R}$. We stress that in Corollary \ref{2885} it is not even known a priori
that $\mu$ is an infinitely divisible distribution. It is actually conjectured in \cite{rj} that if $F$ is the distribution
function of an infinitely divisible distribution and satisfies $F(x)=S(x)$, for all $x\ge0$, then $F\equiv S$ on $\mathbb{R}$.
This problem is solved (and proved to be true) only for $\alpha=2$.

\subsection{A class of discrete distributions}\label{6288}

In this section, we present the discrete counterpart of class $\mathscr{M}$. More specifically, we will
consider distributions whose support is included in $\mathbb{Z}$.\\

First we need the following equivalent of Lemma \ref{2363} for discrete distributions. Its proof is straightforward
so we omit it.

\begin{lemma}\label{9742}  Let $(q_n)_{n\in\mathbb{Z}}$ be any probability on $\mathbb{Z}$. Define
$q^{*2}_n=\sum_{k\in\mathbb{Z}}q_{n-k}q_k$.
Then for all $n\ge1$
\begin{equation}\label{5821}
\sum_{k=-\infty}^0q_{n-k}q_k=\frac12\left(q^{*2}_n-\sum_{k=1}^{n-1}q_{n-k}q_k\right)\,,
\end{equation}
where we set $\sum_{k=1}^{n-1}q_{n-k}q_k=0$ if $n=1$.
\end{lemma}

A sequence $(a_k)_{k\ge0}$ of nonnegative real numbers is called completely monotone if for all $k\ge0$ and $n\ge1$,
\[\Delta^na_k:=\Delta^{n-1}a_k-\Delta^{n-1}a_{k+1}\ge0\,,\]
where $\Delta^0a=a$. A result from Hausdorff asserts that $(a_k)_{k\ge0}$ is completely monotone if and only if there is a
finite measure $\nu$ on $[0,1]$ such that for all $k\ge0$,
\begin{equation}\label{3936}
a_k=\int_0^1t^k\,\nu(dt)\,.
\end{equation}

\noindent Let us set $\mu(\{n\})=\mu_n$, for $n\in\mathbb{Z}$. Then by assuming that $(\mu_n)_{n\ge0}$ is completely
monotone, we obtain a new class of distributions satisfying conjecture $\mathscr{C}$ as shows the following theorem.

\begin{theorem}\label{9756}  Assume that $(\mu_n)_{n\ge0}$ is completely monotone and that the support of the measure
$\nu$ in the representation $(\ref{3936})$ contains a decreasing sequence $(c_n)$ such that $\sum (-\ln c_n)^{-1}=\infty$.
Then the data of $\mu$ and $\mu^{*2}$ restricted to $[0,\infty)$ allows us to determine if the support of $\mu$ is included in
$\mathbb{Z}$. Assume that this is the case, then the measure $\mu$ is determined by the restriction of $\mu$ and $\mu^{*2}$
to $[0,\infty)$. In particular $\mu$ belongs to class $\mathscr{C}$.
\end{theorem}
\begin{proof} First let us observe that we can derive from the data of
of $\mu$ and $\mu^{*2}$ restricted to $[0,\infty)$ that the support of $\mu$ is included in $\mathbb{Z}$. Indeed, assume
that  the support of $\mu$ restricted to $(-\infty,0]$ is not included in $\mathbb{Z}_-$. Then there is an interval
$I\subset(-\infty,0]\setminus\mathbb{Z}_-$ such that $\mu(I)>0$. From (\ref{3936}), $\mu(\{n\})>0$ for all $n\ge1$.
Let $n\in\mathbb{Z}_+\setminus\{0\}$ such that $n+\inf I>0$, then
\[0< \mu(I)\mu(\{n\})=\mathbb{P}(S_1\in I,S_{2}-S_1=n)\le\mathbb{P}(S_{2}\in n+I)\,.\]
This implies that $\mu^{*2}(n+I)>0$, where $n+I\subset[0,\infty)\setminus\mathbb{Z}_+$, which contradicts the assumption.

From the Haussdorff representation recalled above, there is a unique finite measure $\nu$ on [0,1] such that
$\mu_k=\int_0^1t^k\,\nu(dt)$. Using this representation and Fubini's theorem, we can write for all $n\ge0$,
\[\sum_{k=0}^\infty \mu_{-k}\mu_{n+k}=\int_0^1t^n\left(\sum_{k=0}^\infty \mu_{-k}t^k\right)\,\nu(dt)\,.\]
From (\ref{5821}) in Lemma \ref{9742}
applied to $(\mu_k)$, we derive that this expression is determined from the knowledge of $\mu$ and $\mu^{*2}$
on  $[0,\infty)$.  This means that we know the moments of the measure $\left(\sum_{k=0}^\infty \mu_{-k}t^k\right)\cdot\nu(dt)$.
This measure is finite and its support is included in $[0,1]$, hence it is determined by its moments. Then we know
the generating function $t\mapsto \sum_{k=0}^\infty \mu_{-k}t^k$ of the sequence $(\mu_{-k})_{k\ge0}$ on the support of $\nu$,
since this function is continuous. From the assumption, we know this generating function on a sequence $(c_n)$ such that
$\sum (-\ln c_n)^{-1}=\infty$. This is enough to determine the sequence $(\mu_{-k})_{k\ge0}$, from \cite{fe1}.

We conclude that the measures $\mu$ and $\mu^{*2}$ restricted to $[0,\infty)$ allow us to determine $\mu$ on $\mathbb{Z}$.
\end{proof}

The set of measures satisfying the assumptions  of Theorem \ref{4261} will be called the discrete monotone class and will be denoted by
$\mathscr{M}_d$. Theorem \ref{4261} shows that $\mathscr{M}_d\subset \mathscr{C}$. Moreover, it is clear that none of the classes
$\mathscr{E}$, $\mathscr{M}$ and  $\mathscr{A}$ is included in $\mathscr{M}_d$ and that these classes do not contain  $\mathscr{M}_d$.

\section{When $\mu$ is infinitely divisible}\label{2482}

The aim of this section is to present a problem equivalent to conjecture  $\mathscr{C}$ in the framework of infinitely divisible distributions.
When $\mu$ is infinitely divisible, the Wiener-Hopf factorization  can be understood in two different ways: we can either factorize the
characteristic function $\varphi$ as in (\ref{6283}), or we can factorize the characteristic exponent $\psi$, which is defined by
\[\varphi(t)=e^{\psi(t)}\,,\;\;\;t\in\mathbb{R}\,.\]
Then let us recall the Wiener-Hopf factorization in the latter context.
Let $(X_t,\,t\ge0)$ be a real L\'evy process issued from 0 under the probability $\mathbb{P}$ and such that $X_1$ has law $\mu$
under this probability, that is $\mathbb{E}(e^{iuX_t})=e^{-t\psi(u)}$, for all $t\ge0$. The  characteristic exponent of $\mu$ is
given explicitly according to the L\'evy-Khintchine formula by
\[\psi(u)=iau+\frac{\sigma^2}2u^2+\int_{\mathbb{R}\setminus\{0\}}(1-e^{iux}+iux1_{\{|x|\le1\}})\,\Pi(dx)\,,\]
where $a\in\mathbb{R}$, $\sigma\ge0$ and $\Pi$ is a measure on $\mathbb{R}\setminus\{0\}$, such that
$\int(x^2\wedge 1)\,\Pi(dx)<\infty$. Then the Wiener-Hopf factorization
of $\psi$ has the following form:
\begin{equation}\label{6228}
s+\psi(u)=\kappa_+(s,-iu)\kappa_-(s,iu)\,,\;\;\;u\in\mathbb{R}\,,\;\;s\ge0\,,
\end{equation}
where $\kappa_+$ and $\kappa_-$ are the Laplace exponents of the upward and downward ladder processes
$(\tau^+,H^+)$ and $(\tau^-,H^-)$ of $X$, that is $\mathbb{E}(e^{-\alpha\tau_t^{+/-}-i\beta H_t^{+/-}})=e^{-t\kappa_{+/-}(\alpha,\beta)}$. These
exponents are given explicitly for $\alpha,\beta\ge0$ by the identities,
\begin{eqnarray}
\kappa_-(\alpha,\beta)&=&k_-\exp\left(\int_0^\infty\int_{(-\infty,0)}(e^{-t}-e^{-\alpha t-\beta x})\frac1t\mathbb{P}(X_t\in dx)\,dt\right)\\
\kappa_+(\alpha,\beta)&=&k_+\exp\left(\int_0^\infty\int_{[0,\infty)}(e^{-t}-e^{-\alpha t-\beta x})\frac1t\mathbb{P}(X_t\in dx)\,dt\right)\,,\label{2324}
\end{eqnarray}
where $k_-$ and $k_+$ are positive constants depending on the normalization of the local times at the infimum and at the supremum of $X$.
The joint law of $(\tau_1^{+},H_1^{+})$ is the continuous time counterpart of the joint law $(\tau_{+},S_{\tau_{+}})$
defined in Section \ref{int}, in the setup of random walks. We refer to Chap. VI of \cite{be}, Chap. IV of \cite{ky} or Chap. of \cite{do} for
complete definitions of these notions. Note that our formulation of the Wiener-Hopf factorization (\ref{6228}) includes compound Poisson
processes since expression (\ref{2324}) takes account of a possible mass at 0 for the measure $\mathbb{P}(X_t\in dx)$. This slight extension can be
derived from p. 24 and 25 of \cite{vi}, see also the end of Section 6.4, p.183 in \cite{ky}.

Set $\overline{\Pi}(t)=\Pi(t,\infty)$, $t>0$ and let $\mu_t$ be the law of $X_t$.

\begin{lemma}\label{4237}
The data of $\kappa_+$ is equivalent to that of $\mu_t$ on $[0,\infty)$, for all $t\ge0$.
Moreover, the knowledge of $\kappa_+$ allows us to determine the function $\overline{\Pi}(t)$, $t>0$.
\end{lemma}
\begin{proof}
From identity
\begin{equation}\label{1263}
\frac1t\mathbb{P}(X_t\in dx)\,dt=\int_0^\infty\mathbb{P}(\tau_u\in dt,H_u\in dx)\,\frac{du}u\,,\;\;x\ge0,t>0\,,
\end{equation}
which can be found in Section 5.2 of \cite{do}, we see that the law of $X_t$ in $[0,\infty)$, for all $t\ge0$ is determined
by the law of $(\tau,H)$ and hence by $\kappa^+$. (Note that equation (\ref{1263}) is also valid for compound Poisson processes.)
Conversely, it follows directly from formula (\ref{2324}) that $\kappa^+$ is determined by the data of the measure $\mu_t$ on $[0,\infty)$,
for all $t\ge0$.

The second assertion is a consequence of the first one and Exercise~1 of chap.I in \cite{be}, which asserts that the
family of measures $\frac1t\mathbb{P}(X_t\in dx)$ converges vaguely toward $\Pi$, as $t\rightarrow0$.
\end{proof}

\noindent The above lemma enables us to make the connection between the two Wiener-Hopf factorizations
(\ref{6283}) and (\ref{6228}). Let us state it more specifically in the following proposition.

\begin{proposition}\label{3783}
The Wiener-Hopf factor $\kappa_+$ allows us to determine the Wiener-Hopf factor $\chi_+$.
\end{proposition}
\begin{proof}
The result is straightforward from Lemma \ref{4237}. Indeed, knowing $\kappa_+$ we can determine $\mu_n=\mu^{*n}$ restricted
to $[0,\infty)$, for all $n\ge1$ and from Section \ref{int} that this data is equivalent to that of $\chi_+$.
\end{proof}

\begin{definition}
We will denote by $\mathscr{C}_i$ the class of infinitely divisible distributions $\mu$ which are determined by the data of their
upward Wiener-Hopf factor $\kappa_+(s,t)$, for $s,t\ge0$ or equivalently by the data of the measures $\mu_t$,
$t>0$ restricted to $[0,\infty)$.
\end{definition}

\noindent Let us denote by $\mathscr{I}$ the set of infinitely divisible distributions. Then it is straightforward
that $\mathscr{I}\cap\mathscr{C}\subset\mathscr{C}_i$. In particular if Conjecture C is true, then $\mathscr{C}_i=\mathscr{I}$.
It was proved in Chapter 4 of \cite{vi} that infinitely divisible distributions having some exponential moments belong to class $\mathscr{C}_i$,
which is a consequence of our results. The latter work uses a different technique based on the analytical continuation of the Wiener-Hopf
factors $\kappa_+$ and $\kappa_-$. \\

Let $k_-,\delta_-,\gamma_-$ and $k_+,\delta_+,\gamma_+$ be the killing rate, the drift and the L\'evy measure of
the subordinators $H_-$ and $H_+$, respectively and let us set
$\bar{\gamma}_+(x)=\gamma_+(x,\infty)$ and $\bar{\gamma}_-(x)=\gamma_-(x,\infty)$. Let also
 $U_-$ be the renewal measure of the  downward ladder height process $H^-$, that is $U_-(dx)=\int_0^\infty\mathbb{P}(H_t^-\in dx)$.

\begin{theorem}\label{3920}  Assume that the function $t\in(a,\infty)\mapsto \overline{\Pi}(t)$ is completely monotone, for some $a\ge0$,
that is there exists a Borel measure $\nu$ on $(0,\infty)$ such that for all $t>a$,
\begin{equation}\label{9422}
\overline{\Pi}(t)=\int_0^\infty e^{-ut}\,\nu(du)\,.
\end{equation}
Assume moreover that the support of $\nu$ contains an increasing sequence $(a_n)$ such that $\sum_{n}a_n^{-1}=+\infty$.
Then the measure $\mu$ belongs to the class $\mathscr{C}_i$.
\end{theorem}
\begin{proof}
The proof relies on Vigon's \'equation amicale invers\'ee, see \cite{vi}, p.71, or (5.3.4) p.44 in \cite{do} which can be written as
\begin{equation}\label{4581}
\bar{\gamma}_+(x)=\int_{[0,\infty)}U_-(dy)\overline{\Pi}(x+y)\,,\;\;\;x>0\,.
\end{equation}
Note that (\ref{4581}) is analoguous to (\ref{5202}). From Lemma \ref{4237}, given $\kappa_+$, we know both $\bar{\gamma}_+(x)$,
for $x>0$ and $\overline{\Pi}(t)$, for $t>0$. Then we will show that under our assumption, equation (\ref{4581}) allows us to determine
the renewal measure $U_-(dy)$, so that the law of $X$ will be entirely determined, thanks to the relation:
\begin{eqnarray}
\hat{U}_-(z)&=&\int_\mathbb{R_+}e^{-yz}U_-(dy)\nonumber\\
&=&\frac1{\kappa_-(0,z)}\label{3471}\,,\;\;\;z>0\,,
\end{eqnarray}
and the Wiener-Hopf factorization (\ref{6228}).

From (\ref{9422}), (\ref{4581}) and Fubini's Theorem, we can write for all $x>0$,
\begin{eqnarray}
\bar{\gamma}_+(x)&=&\int_{[0,\infty)}U_-(dy)\int_0^\infty e^{-(x+y)z}\,\nu(dz)\nonumber\\
&=&\int_0^\infty e^{-xz}\hat{U}_-(z)\,\nu(dz)\,.\label{4925}
\end{eqnarray}
Then the left hand side of equation (\ref{4925}) determines the measure $\hat{U}_-(z)\,\nu(dz)$.
Since $z\mapsto\hat{U}_-(z)$ is a continuous function, then it is determined on the support of $\nu$. From our assumption
on this support and \cite{fe1} we derive that $\hat{U}_-$ (and hence $\kappa^-(0,z)=-\log \mathbb{E}(e^{-z H^-_1})$)
for $z>0$, is determined.
\end{proof}

Note that an analogous result to Theorem \ref{3920}  holds when $\mu$ has support in $\mathbb{Z}$ and the
sequence $\Pi(n)$, $n\ge1$
satisfies the same assumptions as $(\mu_n)_{n\ge0}$ in Theorem \ref{9756}. One may also wonder if an assumption such as
(\ref{8179}) for $\overline{\Pi}(t)$ would lead to a similar result to Theorem \ref{3720}. However, in order to use the same
argument as in the proof of this theorem together with equation (\ref{4581}), we need $\hat{U}_-(z)$ to be bounded, which is
not the case in general.\\

\begin{remark}
As already observed above, the class $\mathscr{C}_i$ contains at least all probability measures in the set
$\mathscr{I}\cap(\mathscr{E}\cup\mathscr{M}\cup\mathscr{A}\cup\mathscr{M}_d)$ but Theorem $\ref{3920}$ shows that there
are other distributions in $\mathscr{C}_i$. Indeed, it is easy to construct an example of a compound Poisson process $(X_t,\,t\ge0)$
with intensity $1$, whose L\'evy measure $\Pi$ satisfies conditions of Theorem $\ref{3920}$ but such that the law
$\mu(dx)=e^{-1}\sum_{n\ge0}\Pi^{*n}(dx)/{n!}$ of $X_1$ does not belong to any of the classes
$\mathscr{E}$, $\mathscr{M}$, $\mathscr{A}$ and $\mathscr{M}_d$.
\end{remark}

An infinitely divisible distribution is said to be downward skip free (respectively upward skip free) if the support of the measure $\Pi$ is
included in $(-\infty,0]$ (respectively in $[0,\infty)$). Upward skip free distributions clearly belong to class $\mathscr{C}_i$ from
the Wiener-Hopf factorization (\ref{6228}). Then here is a counterpart of Theorem \ref{8733}.

\begin{theorem}\label{2455}
Downward skip free infinitely divisible distributions belong to the class $\mathscr{C}_i$.
\end{theorem}
\begin{proof} The proof relies on Vigon's \'equation amicale, p.71 in \cite{vi}. See also equation (5.3.3), p.44 in \cite{do}.
If $\delta_->0$, then from \cite{vi}, the L\'evy measure $\gamma_+$ is absolutely continuous and we will denote by $\gamma_+(x)$ its
density. Then Vigon's \'equation amicale can be written as
\[\overline{\Pi}(x)=\int_0^\infty\gamma_+(x+du)\bar{\gamma}_-(u)+\delta_-\gamma_+(x)+k_-\bar{\gamma}_+(x)\,,\;\;\;x>0\,.\]
It is plain that in the right hand side, the term $\int_0^\infty\gamma_+(x+du)\bar{\gamma}_-(u)$ is identically 0 if and only if the L\'evy
process $X$ is spectrally positive, that is $\mu$ is downward skip free. Moreover, (\ref{6228}) for $u=0$ entails that the knowledge of
$\kappa_+$ implies that of $\kappa_-(s,0)$, for all $s\ge0$. In particular, we know the killing rate of the subordinator $(\tau_t^-,t\ge0)$,
and this killing rate is the same as this of $(H_t^-,t\ge0)$, that is $k_-$.
Then we conclude that $X$ is spectrally positive if and only if there is a constant $\delta_-$ such that
\[\overline{\Pi}(x)=\delta_-\gamma_+(x)+k_-\bar{\gamma}_+(x)\,,\;\;\;x>0\,,\]
and this can be determined, since from our data, we know $k_-$, $\overline{\Pi}(x)$ and $\bar{\gamma}_+(x)$, for $x>0$.
\end{proof}

\section{More classes of distributions}\label{2450}

In the previous sections, we have highlighted the subclasses $\mathscr{E}$, $\mathscr{M}$, $\mathscr{M}_d$ and $\mathscr{A}$ of
$\mathscr{C}$ and proved that these sets of distributions are distinct  from each other. More specifically, none of them is included into
another one. Then the aim of this section is to show that some of these classes can be substantially enlarged through simple
arguments.\\

Actually for most of the subclasses investigated in this paper, we imposed conditions bearing only on $\mu$ restricted to $[0,\infty)$,
but one is also allowed to make assumptions on $\mu^{*n}$ restricted to $[0,\infty)$. In order to move in this direction, let us mention
the following straightforward extension of results of Section \ref{5673}.

\begin{proposition}\label{3555} Let $\mu\in\mathcal{M}_1$ be absolutely continuous with density $f$.
If there is $n\ge1$ such that the density function $f^{*n}$ satisfies the same conditions as $f$ in Theorems $\ref{9524}$ or in
Theorem $\ref{3720}$, then $\mu$ is determined by $\mu^{*n}$ and $\mu^{*2n}$ restricted to $[0,\infty)$. In particular $\mu$
belongs to class $\mathscr{C}$.
\end{proposition}
\noindent It is plain that an analogous extension of Theorem \ref{9756} is satisfied.
Then here is a more powerful result  allowing us to extend our classes of distributions.

\begin{theorem}\label{8455}
Let $\mu\in\mathcal{M}_1$. If there is $\nu\in\mathcal{M}_1$ whose support is included in $(-\infty,0]$ and such that
$\mu*\nu\in\mathscr{C}$, then $\mu\in\mathscr{C}$.
\end{theorem}
\begin{proof} Let $\mu,\mu_1\in\mathcal{M}_1$ such that for each $n\ge1$, the measures
$\mu^{*n}$ and $\mu_1^{*n}$ restricted to $[0,\infty)$ coincide. Set $\bar{\mu}=\mu*\nu$ and $\bar{\mu}_1=\mu_1*\nu$. Then
from commutativity of the convolution product, $\bar{\mu}^{*n}=\mu^{*n}*\nu^{*n}$ and $\bar{\mu}^{*n}_1=\mu^{*n}_1*\nu^{*n}$.
Since the support of $\nu^{*n}$ is included in $(-\infty,0]$ and $\mu^{*n}$ and $\mu_1^{*n}$ restricted to $[0,\infty)$ are known
and coincide, the measures $\bar{\mu}^{*n}$ and $\bar{\mu}^{*n}_1$ restricted to $[0,\infty)$ and known
coincide. Since $\bar{\mu}\in\mathscr{C}$, the measures $\bar{\mu}$ and $\bar{\mu}_1$ are equal. Finally, from Lemma \ref{3024},
the characteristic function of $\nu$ does not vanish on any interval of $\mathbb{R}$ and the identity $\mu=\mu_1$ follows
from continuity and injectivity of the Fourier transform.
\end{proof}
\noindent Theorem \ref{8455} entails in particular that  Conjecture C' is equivalent to the following one: {\it Any distribution
$\mu\in\mathcal{M}_1$  whose support is not included in $(-\infty,0)$ is determined by its convolution powers $\mu^{*n}$, $n\ge1$
restricted to $[a,\infty)$, for some $a\ge0$.} Indeed, it suffices to choose $\nu=\delta_{-a}$ in Theorem \ref{8455}. Finding more
general examples illustrating this result is an open problem. In order to do so, one needs for instance to find the characteristic function
$\varphi$ of a random variable which belongs to class $\mathscr{C}$ and the characteristic function $\varphi_Y$ of a nonnegative
random variable $Y$ such that the ratio $\varphi(t)/\varphi_Y(-t)$ is the characteristic function of some random variable $X$. Then
since the law of $X-Y$ belongs to class $\mathscr{C}$, so does the law of $X$ from Theorem \ref{8455}.\\

Note that neither Proposition \ref{3555} nor Theorem \ref{8455} allows us to enlarge class $\mathscr{E}$. In order to do so in the
same spirit as in Theorem \ref{8455}, one needs to find an invertible transformation $T(\mu)\in\mathcal{M}_1$ of a distribution
$\mu\in\mathcal{M}_1\setminus\mathscr{E}$, such that $T(\mu)^{*n}$, $n\ge1$ restricted to $[0,\infty)$ would be known and
such that $T(\mu)$ belongs to class $\mathscr{E}$.\\

Let us end this paper with an example of a distribution which satisfies conjecture C, although it does not belong to any of the
classes studied here. Assume that the support of $\mu$ is included in $\mathbb{Z}$ and recall that according to Lemma \ref{4261},
this assumption can be checked from the data of the measures $\mu^{*n}$, $n\ge0$ restricted to $[0,\infty)$. Assume moreover that
there are positive integers $a$ and $b$ such that
\begin{eqnarray}\label{8164}
\qquad\left\{\begin{array}{l}
\mbox{$\mu(n)>0$, for all $n\ge a+b$ and $\mu(n)=0$, for all $n=0,\dots,a+b-1$,}\\
\mbox{$\mu^{*2}(n)=0$, for all $n=0,\dots,a$.}
\end{array}\right.
\end{eqnarray}
Then we  can determine $\mu$ on $\mathbb{Z}_-$, so that $\mu\in\mathscr{C}$. Let us first show that $\mu(n)=0$, for all $n\le-b$.
Assume that there is $n\le-b$ such that $\mu(n)>0$. Then let $k=0,\dots,a$ such that $k-n\ge a+b$. By definition of the
convolution product $0\le \mu(k-n)\mu(n) \le \mu^{*2}(k)$, but from our assumptions $\mu(k-n)\mu(n)>0$ and  $\mu^{*2}(k)=0$, which is
contradictory, hence $\mu(n)=0$, for all $n\le-b$.  On the other hand, assumptions  (\ref{8164}) entail that for all $k=a+1,\dots,a+b-1$,
\[\mu^{*2}(k)=\sum_{i=a+b}^{k+b-1}\mu(k-i)\mu(i)\,,\]
that is $\mu^{*2}(a+1)=\mu(-b+1)\mu(a+b)$, $\mu^{*2}(a+2)=\mu(-b+2)\mu(a+b)+\mu(-b+1)\mu(a+b+1)$,...
Therefore, this system allows us to determine $\mu(n)$, for $n=-b+1,-b+2,\dots,-1$ and the conclusion follows.\\

Let us consider for instance $a=1$, $b=3$ and
\[\mu(-2)=\mu(-1)=\frac{1-c}2\;\;\;\mbox{and}\;\;\;\mu(n)=\frac1{n^3}\,,\;\;\;n\ge 4\,,\]
where $c=\sum_{n\ge 4}n^{-3}$. Clearly, such a distribution does not belong to any of the classes $\mathscr{A}$, $\mathscr{M}$ or
$\mathscr{M}_d$. Then let us check that it does not belong to class $\mathscr{E}$. The mean of $\mu$ satisfies
\[\sum_{k\ge-2}k\mu(k)=-\frac32(1-c)+\sum_{n\ge 4}\frac1{n^2}<0\,,\]
so that (\ref{0236}) does not hold. Moreover $\mu$ has no positive exponential moments. Therefore conditions of Theorems \ref{4572}
and \ref{2258} are not satisfied and since $\mu$ is not downward skip free, we obtain the conclusion. Finally it cannot be proved that $\mu$
belong class $\mathscr{C}$ by applying Proposition \ref{3555} or Theorem \ref{8455}. However, from the above arguments, $\mu$ does
belong to class $\mathscr{C}$.

\vspace*{.5in}

\noindent {\bf Acknowledgement} We would like to thank Rodolphe Garbit, Philippe Jaming and Jean-Jacques Loeb for
motivating discussions on this subject.

\vspace*{1in}

\newpage

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October is fast approaching and for Black History Month this year I want to do something different.
Despite the many brilliant sheroes who feature on the blog, not enough of them are women of colour. If women's histories are neglected, then black women's histories are neglected even more so. I want to correct the imbalance so I'm putting out a call to #TellHerStory during Black History month this year.
I'd like to share the stories of as many black sheroes as possible during Black History Month, but I can only do that with your help!
Check they're not already featured on the blog!


The Vashikaran specialist in Guwahati has said that the word Vashikaran is not a new word in society. It is one of the ancient sacred arts to create those things which are in control. There are many situations where a person got stuck and failed to get out of the same. The word Vashikaran is once a spiritual art of making such a mind which is not under self control but in control, but to win the wrong. Not only in the country of India but throughout the world, is the holy mantra of Vashikaran. Here we present you with them, where you will get the full solution of your daily life. He is India's gold medalist astrologer. He provides the best service in all of Guwahati. He has said that people of Guwahati believe or faith in the field of Vashikaran mantra, tantra and Yantra. He has given different advice and other services are how to sort out these problems as the love of lovers' problem, inter caste marriage of couples, and job or business problem etc.
Here we bring a situation where you can understand better than others. This means that we make a relationship of love and trust that is one of the partners, here are our specialists who are spelled with love, who possess full knowledge of Vashikaran mantras, which is included in this way, as the first Vashikaran mantra for Wife, the second is Vashikaran mantra for husband, the third one is Vashikaran mantra for the former partner, etc and many more. The Vashikaran specialist in Guwahati has said that there are separate Vashikaran spells applied under different circumstances. He said that Vashikaran is a kind of magic that is used to control someone and attracts love for someone. He said that the Vashikaran is a very powerful ritual between other rituals. It can solve all kinds of problems that we face in our lives. Vashikaran logic is used primarily as a girl friend and boy friend in the case of love for most lovers as girl friend and boy friend.


ComboFrame Wall Clip | VKF Renzel USA Corp.
Our Wall Clips let you mount a ComboFrame to the wall. Simply snap the Wall Clips onto the ComboFrame and attach the clips to the wall with screws. Then you can change the art in your ComboFrame without having to take it off the wall. Just slide your art into the slot on the side of the frame.
You will need at least two Wall Clips to hang a ComboFrame. The clips are made of nearly unbreakable plastic and will fit any of our ComboFrame sizes.


text:
Town of Simmesport
Together We Stand
Gas Department
Town Events
Atchafalaya River Festival
Simmesport Christmas Parade
Joe Simon Boulevard
SHS Reunion
Town Legal Matters
Kirk LaCour Law Firm
Leo Ehrhardt Central Park
Simmesport Police Department
Joseph A. "Tony Whitmore
Wiltz "Prof" Gremillion
Gerald Ortego
Lifelong Simmesport resident Joseph "Tony" Whitmore was born on December 20, 1940. He graduated from Simmesport High School in 1958, but did not join the Army until March of 1959. He was sent to basic training at Fort Chafee, Arkansas and then to Fort McLellan in Alabama. After a stay at Fort Ord in Monterey, California, Tony was assigned to the 557th Quartermaster Company stationed in Evreau, France. While in France, Whitmore joined the Evreux-Fauville Sky-Diving club since he was a certified paratrooper and rigger. Joseph was stationed there during the tense times of the Berlin Crisis, when the Russians erected the wall around East Berlin. Tony recounted stories of many (then secret) missions air dropping supplies by parachute into communist-occupied East Berlin. He would also often recount the frightening tank standoff at Checkpoint Charlie in October of 1961.
Joseph was honorably discharged from service in March of 1962 having attained the rank of Sergeant E-5. He returned home to Simmesport and was nearly called up again for active service in October of 1963. He married Juanita Moulard, his wife of over 40 years, and together they raised five children in Simmesport, James, Rene, Eric, Mark and Michelle. His son Mark and grandson Jacob also served in the U.S. Army. Tony, as people in Simmesport knew him, was a nickname given to him by his father, Wiley Whitmore.
In the early 1970's he was hired by the Louisiana Department of public safety working at the State Police barracks. His knack for finding escaped inmates led Joe, as he was called in State Police circles, to be asked to join the first Louisiana State Police intelligence unit. There he spent over 30 years as an investigator working many high-profile cases such as organized crime's infiltration into Louisiana casino gambling, church burnings, serial killers and political crimes among others. Tony also spent time working with the FBI on anti-terrorism efforts and became deputized by the Bureau gaining one of the FBI's highest security clearances. He became known throughout the State Police and around the country for his meticulous attention to detail and prolific report writing. Mr. Joe, as most everyone in the State Police called him, became a legend in the Louisiana State Police. He worked with the City of New York's organized crime unit extensively in the early 1990's.
After retiring from the State Police, Tony took a job as an investigator with the Avoyelles Parish Sheriff's Department, a position he held until he died in 2014. Tony Whitmore never wanted to live anywhere else but in Simmesport and often claimed that the Army taught him everything he needed to prepare for life. The Town of Simmesport would like to pay tribute to this soldier's story as the first in a series of brave men and women who served their country with honor.
(l to r) Joseph "Tony" Whitmore, Jacob Whitmore (grandson), Mark Whitmore (son). Three generations of Airborn as a proud Tony was able to pin the jump wings on his grandson at Fort Benning.
372 Mission Dr
Simmesport LA 71369

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Nico Niquo (aka Nico Callaghan) is releasing album Timeless via Melbourne's Daisart on September 26, with a preview of the title-track premiering via AQNB today. The Australian jazz aficionado presents the LP's hypperreal sonic landscapes on the new label founded by ju ca, alongside a book called Snakes in the Grass exploring the evolution of and ultimate corporate intrusion on ambient music in a contemporary context.
Nico Niquo is releasing album Timeless via Melbourne's Daisart on September 26, 2018.


Our annual Your State Parks Day – Friends biggest volunteer day of the year – is Saturday, September 30th. FREE admission for all volunteers. Come help us Celebrate Georgia and all her beautiful historical and cultural beauty! Each of your parks and historic sites have carefully chosen special projects that are most needed to address various preservation and beautification efforts of these sites. We welcome you to join Friends volunteers and partner with site staff as we Celebrate Georgia Saturday, September 30th!
For more information, visit gastateparks.org.
Posted in State Parks.	This entry was tagged Georgia State Parks, health, state park, volunteer. Bookmark the permalink.


It's Summertime…and the Networking is Easy?
One thought that may have crossed your mind could be 'surely using social media is an effective tool, if so… why spend money on a summer vacation to network when I can do it from home?' The answer is, despite social media's powerful presence and ever-growing popularity, there is no substitute for meeting people 'face to face'.
So maybe it's time to reconsider staying at home and muster up the courage to meet someone in person. Why not try networking this summer holiday? It may mean making the extra effort, but this extra work may be very fruitful.
What Networking Tools should I Pack in My suitcase before going on Holiday?
Stop right there! Before you book anything, have you looked on your contact list? Many people have contacts they didn't realize they had, even living in foreign countries. Katharina Cavano, writing for Business 2 Community, recommends "taking a look at your contact list and filter or search by current location to see if there's anyone living in where you'll be vacationing". If you are interested in re-contacting them, send them a note on Facebook or tweet them. How about asking them for recommendations for places of interest to visit or to see?
One other thing is to employ the tools you have at your disposal. These are personal attributes that everybody has. Some people are great at building relationships. Some may be more curious, and others may possess a spirit of adventure. All these personal traits are useful when attempting to broaden your network.
Why not try these tools out? You may find that people in the airport lounge, hotel lobby or casually sitting on the beach tend to be more relaxed and easier to talk to. This is where you have to muster up the courage and go talk to them. Go on, you can do it!
Summer is usually the time for relaxing, but surely time was meant to be used productively. Today, one of the ways where we can enrich our personal and working lives is networking. As human beings we are naturally social animals so what better time to choose to network than when going on vacation? Meeting people in person has a distinct advantage; namely the fact we are able assess whether a particular person will enrich our network or not. The other essential thing to make networking work in our favor is to be open to new experiences. By employing this broadmindedness, we increase the opportunities of meeting new and interesting people.
This entry was posted on July 19, 2017 by merrainebrain and tagged business, business skills, doctor, healthcare, healthcare administration, Healthcare Leaders, healthcare leadership, healthcare recruiters, healthcare recruiting, healthcare recruitment, healthcare search firm, healthcare teams, hospital, hospital administration, hospital recruiting, hospital recruitment, job search, leadership tips, networking, nurse, nurse recruitment, nursing, nursing recruitment, patient centered care, recruiting, success, summer, summertime, vacation.
We staff the nations hospitals!


This comprehensive inspection took place on 13 November 2018 and was announced. We gave the service 24 hours' notice of the inspection visit because the location provides a small care home for younger adults who are often out during the day. We needed to be sure that they would be in. The inspection was carried out by one inspector.
We requested information from the local authority commissioning and quality teams and safeguarding team, to aid us with our planning.We observed interactions between people and staff and observed the support offered to people. We spoke with two people living in the home. We looked at two people's care plans.
During the inspection we spoke with the manager and two support workers. We also reviewed a range of relevant documents relating to how the home was run including training records, audits and quality assurance survey 2017.


text:
Pavonia aurantia är en malvaväxtart som beskrevs av Brother Alain. Pavonia aurantia ingår i släktet påfågelsmalvor, och familjen malvaväxter. Inga underarter finns listade i Catalogue of Life.

Källor

Påfågelsmalvor
aurantia

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An emulator adapter provides a DIP socket on the top and plugs into a PLCC production socket or solders down to a QFP land pattern. We offer both generic 1:1 wired adapters and device specific adapters.
This page includes the most popular emulator adapters, use the searches above to find adapters not listed. Click on Adapters in the menu above to search by device manufacturer. Click on the blue part numbers below for photos and data sheets or to order our adapters.
Most of our emulator adapters offer a choice of Aries DIP ZIF socket or open-frame socket. Our PLCC plugs are supplied in a standard height of 21.6mm. They are available in 29.2mm by request. If clearance is still an issue let us know and we'll work with you.
DIP / QFP Emulator Adapters - user configurable (shown with optional ZIF socket).


In celebration of the 50th Anniversary of the Stonewall Riots Bradbury-Sullivan LGBT Community Center will be looking at the artists and societal movements that surrounded and empowered LGBTQ people leading up to and after the Stonewall Rebellion in 1969, when everything about the Queer community began to change. Queer artists of that period were part of the second wave of Abstract Expressionism, Pop, Op, Photo Realism, and the beginnings of Performance Art. Their decisions to be out and proud fueled and were fueled by the Civil Rights, the Anti-War, and Women's Movements, the Sexual Revolution, and the Stonewall Rebellion. This class will weave it all together and there will be great themed snacks.


Good morning everyone. My name is Brianna Payne. I am 20 years old, originally from New Bedford and I recently relocated to Springfield. I am speaking to you all today to share my journey and my experience at YWCA YouthBuild Springfield.
I was supposed to graduate high school in 2014. Due to certain circumstances, I was unable to graduate. I then gave birth to my two children in December of 2014. I struggled for about a year, deciding how I was going to accomplish my goals of getting my diploma and attending college. Becoming homeless with my two newborn babies was the most challenging obstacle I had to overcome.
I relocated to Springfield where I was given the greatest opportunity I've ever received in my life. It was a new beginning that gave me hope for a brighter and promising future. The YWCA where I currently live had referred me to the YWCA YouthBuild program.
I was a little skeptical at first because I had wanted a high school diploma and I had felt that I was lowering my standards. However, YouthBuild is one of the most extraordinary programs I've had the privilege of being a part of. There are tremendous benefits, opportunities and advantages that YouthBuild has provided me within just a short amount of time.
We have amazing staff and support counselors that work to help each of us with our own individual needs. The YouthBuild program is preparing me to be self sufficient and giving me the ability to further my education and move on to college.
I plan on attending Holyoke Community College in the spring of 2017 for Psychology and Criminal Justice thanks to the support, encouragement and motivation of YouthBuild. I am honestly so appreciative, grateful and privileged to say that I've made it here today through my personal perseverance, hard work and dedication but also the support from the director and staff at the YWCA of YouthBuild Springfield. Thank you all so much.
The YWCA of Western Massachusetts YouthBuild Springfield program recently held an Open House at their site that attracted an overflow crowd of community supporters. One focal point of the event featured the Tiny Home that students have been building. This project is part of a statewide effort of the Massachusetts YouthBuild Coalition and will be used to provide housing for a homeless veteran. This project was funded in part by a grant from the Corporation for National and Community Service, as one of only five grants awarded nationally for 911 Remembrance. This is an excellent teaching project because it involves all of the skills needed to construct a full size house, but can be completed more quickly due to the size. Another part of the project was the education of the students about the events of 911 and the impact that it had on people all over the world because of the events of that fateful day, an event that happened when they were small children.
Mayor Domenic Sarno offered words of encouragement to the students and was also very pleased to see the work that the students performed to help out a fellow citizen in need. He was especially taken by the words of the YouthBuild students. During the event, the students welcomed their guests and gave them a guided tour of the facility. They spoke in glowing terms about the opportunity that YouthBuild provides them. They were happy to talk about how much they had learned since enrolling in the program.
Listening to them talk about the work on the tiny house revealed an amazing sense of pride in their accomplishment. They readily described the various features of the structure and the role that they played in the construction. The satisfaction and achievement that the students took from this experience told the story. The smiles on their faces needed no explanation.


In this Tantric Mysticism experience, we focus on ecstatic union with the Cosmos and the primal forces of the elements. The techniques and states of mind required to do this and several practices you can do to deepen this connection every day.
Beginning with an overview, energy flow exercises, and a cosmic Alchemy experience through direct transmission. You will actually see energy as light, colors, multidimensional fluid reality, and other metaphysical phenomena.
Then moving into the three core techniques, and how to apply them in various ways to achieve deeper union with these primal elements. Then actually practicing with each element.


text:
Q: An elegant proof for a claim about orthogonal and positive-definite matrices Let $P$ be a real $n \times n$ symmetric positive-semidefinite matrix.
Suppose that $\langle O,P \rangle = \langle I,P \rangle$ for some orthogonal matrix $O$. Here $\langle , \rangle$ is the Euclidean (Frobenius) inner product. The assumption is equivalent to $\text{tr}(O^TP)=\text{tr}(P)$.
Then $OP=P$ holds. Since we can replace $O$ with $O^T$ the claim is equivalent to the seemingly surprising assertion $\text{tr}(OP)=\text{tr}(P) \implies OP=P$.

I have a proof, and I wonder whether there are other shorter proofs.
  Is there a proof which does not require diagonalizing $P$?

Here is my proof: By orthogonally diagonalizing $P$, we can reduce to the case where $P=\Sigma$ is diagonal, with non-negative entries $\sigma_i$.
The assumption implies $\sum_i \sigma_i O_{ii}=\sum_i \sigma_i$ where the sums run over all the indices $i$ where $\sigma_i \neq 0$. Since $|O_{ii}| \le 1$ this forces $O_{ii}=1$ for all these $i$. A direct check then implies that $O\Sigma=\Sigma$:
Indeed, $O\Sigma=\Sigma$ is equivalent to $O_{ij}\sigma_j=\Sigma_{ij}$.
If  $\sigma_j=0$, then $\Sigma_{ij}=0$ so equality holds. If $\sigma_j \neq 0$, then $O_{jj}=1$ so equality holds for $i=j$. For $i \neq j$,  $\Sigma_{ij}=0$ (since $\Sigma$ is diagonal) and , and $O_{ij}=0$, since whenever $O_{jj}=1$, all the rest of the entries in the $j$-th column of $O$ are zero. (Since $O$ is orthogonal).

A: The following proof does diagonalize $P$, but not in matrix form.
Let $\{v_1,\ldots,v_n\}$ be an orthonormal eigenbasis of $P$ and $Pv_i=\lambda_iv_i$ for each $i$. Then
$$
\langle P,I\rangle
= \langle P,O\rangle = \sum_i\langle Pv_i,Ov_i\rangle
\le \sum_i\|Pv_i\|\|Ov_i\|
= \sum_i\lambda_i
= \operatorname{tr}(P)
= \langle P,I\rangle.
$$
Therefore $\langle Pv_i,Ov_i\rangle=\|Pv_i\|\|Ov_i\|$ for each $i$. Since $Pv_i=\lambda v_i$ and $Ov_i$ is a unit vector, we have $\langle v_i,Ov_i\rangle=1$ and in turn $Ov_i=v_i$ whenever $\lambda_i>0$. Hence $OP$ and $P$ agree on $\{v_1,\ldots,v_n\}$, meaning that $OP=P$.

A: Trace is invariant under cyclic permutations, so $$\mathrm{tr}(O^TP)=\mathrm{tr}(P^{1/2}O^TP^{1/2})=\langle OP^{1/2},P^{1/2}\rangle.$$
Letting $\|\cdot\|$ denote the Frobenius norm, one form of the Cauchy-Schwarz inequality says that
$$\langle O,P\rangle=\langle OP^{1/2},P^{1/2}\rangle\leq \|OP^{1/2}\|\cdot\|P^{1/2}\|=\mathrm{tr}(P)^{1/2}\mathrm{tr}(P)^{1/2}=\langle I,P\rangle$$
with equality only when
$$OP^{1/2}=P^{1/2}\frac{\|OP^{1/2}\|}{\|P^{1/2}\|}=P^{1/2}.$$
Multiplying on the right by $P^{1/2},$ this condition implies
$$OP=P.$$

redpajama_set_name: RedPajamaStackExchange


The Bible has a lot to say about money. The word is used over 130 times; 16 of 38 parables refer to the handling of money; and there are over 2,300 references to money, wealth, or property. You are involved with money in some form every day of your life. You live in a house or an apartment that costs money. Your food, clothes and transportation all cost money.
Jesus said, "Where your treasure is, there will your heart be also" (Matthew 6:21).
There is a tendency to think about "my money." It is reasoned that since I earned it, it is mine and I will decide how to spend it, save it, or give it. But we need to recognize that God owns everything: "the earth is the Lord's and the fulness thereof; the world, and they that dwell therein" (Psalm 24:1).
Acknowledging God's ownership of all things changes our perspective.
We come to recognize that rather than this being "my money" that I can do with as I please, it is God's money and I am to be a good steward of it.
While money may be received through an inheritance, the usual method by which we secure the funds which supply our needs is through work. Throughout the Scriptures slothfulness and laziness is condemned and diligent work is encouraged. In fact Christians are admonished to work with the thought they are not just pleasing their superiors but are serving unto the Lord. As Moses declared to Israel, it is God who gives the power to get wealth and—while man must work—God gives the wisdom and strength to perform the task.
The Israelites were warned that when they had been blessed to see their flocks and their gold multiplied they should be careful not to forget the Lord. They were exhorted not to say, "My power and the might of mine hand hath gotten me this wealth" (Deuteronomy 8:17). How easy it is to take pride in what you have accomplished and fail to acknowledge that what you have is a blessing from the Lord.
Most advertising is designed to make you discontent with what you have and therefore desire and purchase the product being promoted. The use of credit cards makes impulse buying easy and as a result millions of Americans today find themselves facing an indebtedness they may never be able to pay. Credit cards can be a convenient way of keeping a record of what you spend as long as the bill is paid when it is due, but when a balance is carried and 18% interest is paid this is failing to be a good steward of the money God has put in your care.
Paul puts things in perspective for us when he writers "godliness with contentment is great gain…and having food and raiment let us be therewith content"(1 Timothy 6:6, 8). In his letter to the church at Philippi the apostle said he had learned to be content whether he was abased or abounding. He indicated this was something he had to learn. Our flesh is prone to always be seeking more of the things of this world, but we must learn to pursue God rather than money and find contentment in Him.
Jesus taught that it is impossible to serve God and riches (Matthew 6:24).
"They that will be rich fall into temptation and a snare, and into many foolish and hurtful lusts, which drown men in destruction and perdition. For the love of money is the root of all evil: which while some coveted after, they have erred from the faith, and pierced themselves through with many sorrows" (1 Timothy 6:9-10).
Money is not of itself evil but the love of it is. Whether it is a love for money or the things it can buy, we see that affection will lead one down the wrong road to the point they may depart from the faith.
So while the love of money can lead to destruction, godliness with contentment is great gain. Being thankful for what God has blessed you to receive and using these resources to the honor of His name brings joy to the believer's heart.
One aspect of stewardship is giving as God has prospered. God cares about your giving. In Mark 12:41 we read that "Jesus sat over against the treasury, and beheld how the people cast money into the treasury…"Jesus was pleased with the widow who gave but two mites, because she gave all she had.
It was not then the amount which was of prime importance but the spirit of sacrifice that influenced this giving.
Just as Jesus observed those casting in their money in that day, our heavenly Father sees our giving and knows the motives of our heart.
The key to our giving is revealed as Paul describes the generosity of the churches of Macedonia. He says they "first gave their own selves to the Lord" (2 Corinthians 8:5). If we recognize we are not our own but have been bought with a price and then give ourselves to the Lord, giving will never be a burden but always a joy. "Every man according as he purposeth in his heart, so let him give; not grudgingly, or of necessity: for God loveth a cheerful giver" (2 Corinthians 9:7).
"The administration of this service not only supplies the want of the saints, but is abundant also by many thanksgivings unto God; whiles by the experiment of this ministration they glorify God for your professed subjection unto the gospel of Christ, and for your liberal distribution unto them, and unto all men" (2 Corinthians 9:12-13).
What a blessing when one has developed the right view of money!
Remember it is God who owns all things. We are to be good stewards of what God puts in our care. We must not worry for He has promised to supply our needs. We are to learn to be content with what we have. We then find joy in giving. When God's servants are supported by our giving, that support is a blessing in their life. When there is giving for the circulation of the gospel, those who receive the message are blessed. When the needy are cared for, it is a blessing to them.
The giver gives with joy, the recipient receives the blessing with joy and all are praising God for his abundant mercies.
What a blessing when the spirit of giving prevails in a church! A greedy, covetous spirit will cause a church to grow cold, but the spirit of giving brings joy and permeates the whole atmosphere of the body of Christ with love and joy.
For more on money: listen to the two-part radio broadcast "Money", purchase the three-volume series: Money, and check out our other resources for personal finances.


text:
7 Tips For Parents Teaching Teens How To Drive
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Where to Go for Your Adult Driver's Ed Class in Dallas
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