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\begin{document}

\title[A new inequality for superdiffusions]{A new inequality for 
superdiffusions and its  applications to nonlinear differential equations}

\author{E. B. Dynkin}
\address{Department of Mathematics, Cornell University,
Ithaca, NY 14853}
\email{ebd1@cornell.edu}
\thanks{Partially supported by the
National Science Foundation Grant DMS-0204237}
\subjclass[2000]{Primary 60H30; Secondary 35J60, 60J60}
\keywords{Positive solutions of semilinear elliptic PDEs, 
superdiffusions, conditional diffusions, $\mathbb{N}$-measures}

\date{April  23, 2004}

\commby{Mark Freidlin}

\begin{abstract}
Our motivation is the following problem: to  describe all  positive 
solutions of  a 
semilinear elliptic equation $L  u=u^\alpha$ with $\alpha>1$ in a bounded 
smooth  domain $E\subset  \mathbb{R}^d$.  
In 1998  Dynkin and Kuznetsov  solved 
this problem for a class of solutions which they called $\sigma$-moderate. 
The question if all solutions belong to this class remained open. In 2002 
Mselati 
proved that this is true for the equation $\Delta u=u^2$ in a domain of 
class $C^4$. His principal tool---the Brownian snake---is not applicable 
to the case $\alpha\neq 2$. In 2003  Dynkin and Kuznetsov modified most of  
Mselati's arguments by using superdiffusions instead of the snake. 
However a critical  gap remained.  A new inequality established in the 
present paper allows us to close this gap.
\end{abstract}

\maketitle


\section{Introduction}
\subsection{Diffusions and superdiffusions} 
We denote by $\M(S)$  the set of all finite measures, and by $\myP(S)$ 
the set of all probability measures on a measurable space $S$. $\B(E)$ 
stands for the set of all positive Borel functions on $E$. We use 
notation  $\1$.]

\section{Tools}
\subsection{$h$-transform and conditional diffusion}
Suppose $\xi$ is a diffusion in a domain $E$ with the transition function
 $p_t(x,dy)$, and let $h\in\myH(E)$.  Then 
\begin{equation}
\label{A1.2}
p_t^h(x,dy)=\frac {1}{h(x)} p_t(x,dy)h(y)
\end{equation}
is the transition function of a continuous  Markov process $(\xi_t,\hat \Pi_x^h)$  in $E$ called the $h$-transform of $\xi$. We prefer to deal with  measures  
$\Pi_x^h=h(x)\hat \Pi_x^h$ which depend linearly on $h$.
Put  $\Pi_x^\nu=\Pi_x^{h_\nu}$ and  $\hat\Pi_x^y=\hat\Pi_x^{\dd_y}$  
where $\dd_y$ is the unit mass at a point $y$. The process $(\xi_t,\hat
\Pi_x^y)$ can be interpreted as a diffusion starting from $x\in E$ and conditioned to exit from $E$ at $y$.

The following lemma is proved, for instance, in \cite{Dy02}, page 103:
\begin{lem}
\label{Lemma B2.1}
 For every stopping time $\tau$ and every pre-$\tau$ positive $Y$,
\begin{equation}
\label{B2.1}
\Pi_x^hY1_{\tau<\tau_E}=\Pi_xYh(\xi_{\tau})
1_{\tau<\tau_E}.
\end{equation}
\end{lem}

\subsection{Measures $\NN_x$}
 Denote by $\Z_x$  the class of all functions of the form
\begin{equation}
\label{A1.5a}
Z=\sum_1^n \0,
\]
then
\begin{equation}
\label{5.1}
\NN_x\{\RR_E\cap\GG=\emptyset,1-e^{-Z}\}
=-\log P_x\{e^{-Z}\mid \RR_E\cap\GG=\emptyset\}.
\end{equation}

By applying \eqref{B5.1} to $\myll Z$ and passing to the limit as $\myll\to +\infty$, we get
\begin{multline}
\label{5.1a}
-\log P_x\{\RR_E\cap\GG=\emptyset,Z=0\}
=\NN_x\{\RR_E\cap\GG\neq\emptyset\}
+\NN_x\{\RR_E\cap\GG=\emptyset,Z\neq 0\}.
\end{multline}

By Proposition 1.1 in \cite{DK04},
\begin{equation}
\label{5.1d}
\NN_x Z_\nu=P_x Z_\nu \qt{if } P_xZ_\nu<\infty.
 \end{equation}
On the other hand, for every $f\in\B(\bar D)$,
\begin{equation}
\label{5.1e}
P_x\0$ and, for all $Z',Z''\in\Z_x$,
\begin{multline}
\label{nnn}
\NN_x\{\A,(e^{-Z'}-e^{-Z''})^2\}\\
=-2\log P_x\{e^{-Z'-Z''}\mid\A\}
+\log P_x\{e^{-2Z'}\mid\A\}+\log P_x\{e^{-2Z''}\mid\A\}.
\end{multline}
If  $Z'=Z''$ $P_x$-a.s. on $\A$ and if $P_x\{\A,Z'<\infty\}>0$, then  $Z'=Z''$ $\NN_x$-a.s. on $\A$.
\end{prop}
\begin{proof}
First, $P_x{\A}>0$ because $P_x{\A}=e^{-w_\LL(x)}$. 
Next
\[
(e^{-Z'}-e^{-Z''})^2=2(1-e^{-Z'-Z''})-(1-e^{-2Z'})-(1-e^{-2Z''}).
\]
Therefore \eqref{nnn} follows from  \eqref{5.1}.
The second part of the proposition is an obvious implication of \eqref{nnn}.
 \end{proof}
\subsection{Properties of  superdiffusions}
The following properties are often used in the theory of superdiffusions.
[They are  a part of the definition of branching exit Markov systems, and superdiffusions are a special case of such systems (see \cite{Dy02}, Chapters 3 and 4).]

\begin{asm} 
\item\label{2.1.A} (Markov property)
If  $Y\ge 0$ is measurable with respect to the $\myss$-algebra generated by $X_{D'}, D'\subset D$ and $Z\ge 0$ is measurable with respect to the $\myss$-algebra generated by $X_{D''}, D''\supset D$, then
\begin{equation}
\label{Markov}
P_\mu(YZ)= P_\mu(YP_{X_D}Z).
\end{equation} 

\item\label{2.1.B}
If  $\mu(E)=0$, then $P_\mu\{X_E=\mu\}=1$.
\end{asm} 

We use \ref{2.1.A}, \ref{2.1.B} and  Proposition \ref{Proposition B2.1} to prove the next proposition.

\begin{prop}
\label{Proposition 5.1}
Let $D\subset E$ be two open sets. Then, for every $x\in D$, $X_{D}$ and $X_{E}$ coincide  $P_x$-a.s. and $\NN_x$-a.s. on the set  $\A=\{\RR_{D}\subset D^*\}$.
\end{prop}

[Note that 
\begin{equation}
\label{LL}
D^*=\{x\in\bar D: d(x,\LL)>0\}
\end{equation}
where $\LL=\pD\cap E$.]


\section{Relations between superdiffusions and conditional 
diffusions \protect\linebreak in two open sets}
\subsection{}
Now we consider two bounded smooth open sets $D\subset E$. We denote by $\tilde Z_\nu$  the stochastic boundary value of  $\tilde h_\nu(x)=\int_{\pD}  k_D(x,y)\nu(dy)$ in $D$;  $\tilde
\Pi_x^y$ refers to the diffusion in  $D$ conditioned to exit  at $y\in\partial D$. 

\begin{thm}
\label{Theorem 3.1}
Put $\A=\{\RR_D\subset D^*\}$.
For every $x\in D$,
\begin{equation}
\label{CC1}
\RR_E= \RR_D  \qt{$ P_x$-a.s. and\quad  $\NN_x$-a.s.}
\end{equation}
and
\begin{equation}
\label{CC2}
Z_\nu=\tilde Z_\nu  \qt{$ P_x$-a.s. and\quad  $\NN_x$-a.s. on } \A
\end{equation}
for all $\nu\in\N_1^E$ concentrated on $\pD\cap\pE$.
\end{thm}
\begin{proof}
1\grd.\ 
First, we prove \eqref{CC1}.
Clearly, $\RR_D\subset \RR_E$ $P_x$-a.s. and $\NN_x$-a.s. for all $x\in D$. We get \eqref{CC1} if we show that, if $O$ is an open subset  of $E$, then, for every $x\in D$,  $X_O=X_{O\cap D}$  $P_x$-a.s. on $\A$  and, for every $x\in O\cap D$, $X_O=X_{O\cap D}$   $\NN_x$-a.s. on $\A$. For $x\in O\cap D$ this follows from  Proposition \ref{Proposition 5.1} applied  to $O\cap D\subset O$ because $\{\RR_D\subset D^*\}\subset\{\RR_{O\cap D}\subset(O\cap D)^*\}$. For $x\in D\setminus O$, 
$P_x\{X_O=X_{D\cap O}=\dd_x\}=1$.

2\grd.\ Put
\begin{equation}
\label{D*m}
D^*_m=\{x\in\bar D: d(x,E\setminus D)>1/m\}.
\end{equation}
To prove \eqref{CC2}, it is sufficient to prove that it holds  on 
$\A_m=\{\RR_D\subset D^*_m\}$  for all sufficiently large $m$. First we prove that, for all $x\in D$,
\begin{equation}
\label{CC4}
Z_\nu=\tilde Z_\nu\qt{$ P_x$-a.s. on $\A_m$}.
\end{equation}
We get \eqref{CC4} by proving that both $Z_\nu$ and  $\tilde Z_\nu$ coincide $P_x$-a.s. on $\A_m$ with the stochastic boundary value $Z^*$ of $h_\nu$ in $D$.

Let 
\[
E_n=\{x\in E: d(x,\pE)>1/n\},\quad D_n=\{x\in D: d(x,\pD)>1/n\}.
\]
If $n>m$, then 
\[
\A_m \subset \A_n\subset\{\RR_D\subset D_n^*\}
\subset\{\RR_{D_n}\subset D_n^*\}.
\]
We apply Proposition \ref{Proposition 5.1}  to $D_n\subset E_n$ and we get 
that, $P_x$-a.s.  on $\{\RR_{D_n}\subset D_n^*\}\supset \A_m$,  
$X_{D_n}=X_{E_n}$  for all $n>m$, which implies $Z^*=Z_\nu$.

3\grd.\ 
Now we prove that
\begin{equation} 
\label{CC6}
Z^*= \tilde Z_\nu\qt{$ P_x$-a.s. on $\A_m$}.
\end{equation}
 Consider $h^0=h_\nu-\tilde h_\nu$ and $Z^0=Z_\nu^*-\tilde Z_\nu$.
If  $y\in \pD\cap\pE$, then
\begin{equation}
\label{M5.8}
k_E(x,y)=k_D(x,y)+\Pi_x\{\tau_D<\tau_E,k_E(\xi_{\tau_D},y)\}.
\end{equation}
 Therefore 
\begin{equation}
\label{Mhh}
h^0(x)=\Pi_x\{\xi_{\tau_D}\in \pD\cap E, h_\nu(\xi_{\tau_D})\}.
\end{equation}
This is a harmonic function in $D$. It vanishes on  $\GG_m=\pD\cap D^*_m=\pE\cap D^*_m$.

We claim that, for every $\ee>0$ and every $m$, $h^0<\ee$ on $\GG_{m,n}=\pE_n\cap D_m^*$ for all sufficiently large $n$. [If this is not true, then there exists a  sequence $n_i\to\infty$ such that $z_{n_i}\in
\GG_{m,n_i}$ and $h^0(z_{n_i})\ge\ee$. If $z$ is a limit point of $z_{n_i}$, then $z\in
\GG_m$ and $h^0(z)\ge \ee$.]

All measures $X_{D_n}$ are concentrated, $P_x$-a.s., on $\RR_D$.
Therefore $\A_m$ implies that they are concentrated, $P_x$-a.s., on $D_m^*$. Since 
$\GG_{m,n}\subset D_m^*$, we conclude that, for all sufficiently large $n$, $\0$. It follows from \eqref{M5aa} that $Z_\nu<\infty$ $P_x$-a.s. and therefore $P_x\{\A, Z_\nu<\infty\}>0$. By Proposition \ref{Proposition B2.1},  \eqref{CC2} follows from \eqref{CC4} .
\end{proof}


\subsection{}
We also need the following  result  (see  \cite{Dy04a}, Lemma 3.2).
\begin{thm}
\label{Theorem 3.2}
Suppose that $D\subset E$ are smooth open sets.
Denote by $\tilde \F$ the $\myss$-algebra in $\OO$ generated  by the sets 
$\{s<\tau_D,\xi_s\in B\}$  where $s\ge 0, B\in\B(E)$.  
 We have
\begin{equation}
\label{5.2.1}
\tilde\Pi_x^y Y=\Pi_x^y\{\tau_D=\tau_E,Y\}
\end{equation}
for all $x\in D, y\in\pE\cap\pD$ and for all $Y\in\tilde\F$.
\end{thm}
  
\begin{corol}
\label{Corollary 3.1}
If
\begin{equation}
\label{5.13}
F_t=\exp\big[-\int_0^t a(\xi_s)\ ds\big]
\end{equation}
where $a$ is a positive continuous function on $[0,\infty)$, then, for 
$y\in \pD\cap\pE,$
\begin{equation}
\label{5.14}
\tilde\Pi_x^y F_{\tau_D}=\Pi_x^y\{\tau_D=\tau_E, F_{\tau_E}\}.
\end{equation} 
\end{corol}

Indeed, it is easy to see that $F_{\tilde\tau}\in\tilde\F$.

\section{Equations connecting $P_x$ and ${\mathbb{N}}_x$ with $\Pi_x^\nu$}
\subsection{}
\begin{thm}
\label{Theorem 4.1}
Let $Z_\nu=\SBV(h_\nu), Z_u=\SBV(u)$ where $\nu\in\N_1^E$ and $u\in \U(E)$. Then
\begin{equation}
\label{jCR}
P_x Z_\nu e^{-Z_u}=e^{-u(x)}\Pi^\nu_x e^{-\Phi(u)}
\end{equation}
and
\begin{equation}
\label{d1}
\NN_x Z_\nu e^{-Z_u}=\Pi^\nu_x e^{-\Phi(u)}
\end{equation}
where 
\begin{equation}
\label{1.10}
\Phi(u)=\int_0^{\tau_E}\psi'[u(\xi_t)]dt.
\end{equation}
\end{thm}
\begin{proof}
Formula \eqref{jCR}  follows from Theorem 3.1 in Chapter 9 of \cite{Dy02}. To prove \eqref{d1}, we observe that, for every
 $\myll
\ge 0$, $\myll Z_\nu+Z_u=\SBV(v)$ where $v=\myll  h_\nu+u\in \U^-(E)$ and therefore, by \eqref{b2.2}, 
\begin{equation}
\label{d2}
\NN_x(1-e^{-\myll Z_\nu-Z_u})=-\log P_xe^{-\myll Z_\nu-Z_u}.
\end{equation}
 By taking the derivatives with respect to $\myll$ at $\myll=0$,\footnote{The 
differentiation under the integral signs is justified by \eqref{5aa}.} 
we get 
\begin{equation}
\label{aaa}
\NN_x Z_\nu e^{-Z_u}= P_x Z_\nu  e^{-Z_u}/P_x e^{-Z_u}.
\end{equation}
By Theorem 1.1 of Chapter 9 in \cite{Dy02},
\begin{equation}
\label{bbb}
P_x e^{-Z_u}=e^{-u(x)}.
\end{equation}
Therefore  \eqref{d2}   follows from \eqref{jCR}, \eqref{aaa} and \eqref{bbb}.
\end{proof}

\begin{thm}
\label{Theorem uu1}
Suppose that   $D\subset  E$ are bounded smooth open sets  and $\LL, L, D^*$ are the sets introduced in Theorem \ref{Theorem A1.1}. Let $\nu$ be a finite measure on  $\pD\cap\pE$, $x\in E$ and $\E_x(\nu)<\infty$.  
Put
\begin{equation}
\label{5.10}
\begin{split}
w_\LL(x)&=\NN_x\{\RR_D\cap\LL\neq\emptyset\},\\
v_s(x)&=w_\LL(x)+\NN_x\{\RR_D\cap\LL=\emptyset,1-e^{-sZ_\nu}\}
\end{split}
\end{equation}
for $x\in D$ and let $w_\LL(x)=v_s(x)=0$ for $x\in E\setminus D$.
For every $x\in E$, we have
\begin{equation}
\label{5.11}
\NN_x\{\RR_E\subset D^*, Z_\nu\}
=\Pi_x^\nu\{A, e^{-\Phi(w_\LL)}\},
\end{equation}
\begin{equation}
\label{5.12}
\NN_x\{\RR_E\subset D^*, Z_\nu\neq 0\}
=\int_0^\infty\Pi_x^\nu \{A, e^{-\Phi(v_s)}\}ds
\end{equation}
where $\Phi$ is defined by \eqref{1.10} and
 \begin{equation}
\label{uu3}
A=\{\tau_E=\tau_D\}=\{\xi_t\in D\text{ for all } t<\tau_E\}.
\end{equation}
\end{thm}
 
\begin{rmk}
 Since $\E_x(\nu)<\infty$, $\nu$ belongs to $\N_x^E$ and to $\N_x^D$.
\end{rmk}

\begin{proof}
1\grd.\ If $x\in E\setminus D$, then, 
$\NN_x$-a.s., $\RR_E$ is not a subset of $D^*$ 
because $\RR_E$ contains supports of  $X_O$ for all neighborhoods $O$ of $x$ 
and we can choose $O$ such that $\bar O\cap D^*=\emptyset$. On the other hand, 
$\Pi_x^\nu(A)=0$. Therefore \eqref{5.11} and \eqref{5.12} hold independently of  values of 
$w_\LL
$ and $v_s$. 

2\grd.\ Now we assume that $x\in D$. 
 Put $\A=\{\RR_D\subset D^*\}$. We claim that
\begin{equation*}
\A=\{\RR_E\subset D^*\} \qt{$\NN_x$-a.s.}
\end{equation*}
 Indeed, $ \{\RR_E\subset D^*\}\subset \A$ because $\RR_D\subset\RR_E$. By Theorem \ref{Theorem 3.1}, $\A\subset\{\RR_D=\RR_E\}$ $\NN_x$-a.s. Hence, $\A\subset\{\RR_E\subset D^*\}$.

By Theorem \ref{Theorem 3.1},  $\RR_D=\RR_E$ and $Z_\nu=\tilde Z_\nu$ $\NN_x$-a.s. on $\A$. Therefore
\begin{equation}
\label{5.17}
\begin{split}
\NN_x\{\RR_E\subset D^*,Z_\nu\}&= \NN_x\{\A,Z_\nu\}
=\NN_x\{\A,\tilde Z_\nu\},\\
\NN_x\{\RR_E\subset D^*,Z_\nu e^{-sZ_\nu}\}&=
\NN_x\{\A,Z_\nu e^{-sZ_\nu}\}
=\NN_x\{\A,\tilde Z_\nu e^{-s\tilde Z_\nu}\}.
\end{split}
\end{equation}
Formula \eqref{5.10} defines two elements of $\U(D)$.
The stochastic boundary value $Z_\LL$ of $w_\LL$ in $D$ is equal to
$\infty1_{\A^c}$ (Remark 1.2 on p.~133 in \cite{Dy02}) and therefore
\begin{equation}
\label{5.18}
e^{-Z_\LL}=1_\A.
\end{equation}
By \eqref{B5.1} and \eqref{5.1c}, 
$v_s(x)=-\log P_x\{\RR_D\cap\LL=\emptyset,e^{-sZ_\nu}\}$ and, by Remark 2.1 
on p.~137 in \cite{Dy02}, the stochastic boundary value $Z^s$ of $v_s$ in $D$ is equal to
$Z_\LL+s\tilde Z_\nu$. Hence,
\begin{equation}
\label{5.19}
e^{-Z^s}=1_\A e^{-s\tilde Z_\nu}.
\end{equation}
By \eqref{5.17}, \eqref{5.18} and \eqref{5.19},
\begin{equation}
\label{EE1}
\NN_x\{\A,Z_\nu\}=\NN_x\{\tilde Z_\nu e^{-Z_\LL}\}
\end{equation} 
and
\begin{equation}
\label{EE2}
\NN_x\{\A,Z_\nu e^{-sZ_\nu}\}=\NN_x \{\tilde Z_\nu e^{-Z^s}\}.
\end{equation}
By applying formula \eqref{d1} to $\tilde Z_\nu$ and to the restriction  of $w
_\LL$ to $D$, we conclude from \eqref{EE1} that
\begin{equation}
\label{5.20}
\NN_x\{\A,Z_\nu\}=\tilde\Pi_x^\nu
\exp\big[-\int_0^{\tau_D}\psi'[w_\LL(\xi_s)]ds\big]
\end{equation}
and, by Corollary \ref{Corollary 3.1},
\begin{equation}
\label{5.20a}
\NN_x\{\A,Z_\nu\}=\Pi_x^\nu\{A,  e^{-\Phi(w_\LL)}\}.
\end{equation}
Analogously, \eqref{d1} applied to the restriction  of $v_s$ to $D$, in 
combination with \eqref{EE2} and \eqref{5.14}, yields 
\begin{equation}
\label{5.21}
\NN_x\{\A,Z_\nu e^{-sZ_\nu}\}
=\Pi_x^\nu\{A, e^{-\Phi(v_s)}\}.
\end{equation}

Formula \eqref{5.11} follows from \eqref{5.20a} and formula \eqref{5.12} follows from \eqref{5.21} because
\begin{equation}
\label{5.22}
\NN_x\{\A,Z_\nu\neq 0\}
=\lim\limits_{t\to\infty}\NN_x\{\A,1-e^{-tZ_\nu}\}
\end{equation}
and 
\begin{equation}
\label{5.23}
1-e^{-tZ_\nu}=\int_0^t Z_\nu e^{-sZ_\nu}ds.
\end{equation}
\end{proof}

\section{Proof of Theorem \ref{Theorem A1.1}}
We  use the following two elementary inequalities:

\begin{asma}
\item\label{ua}
For all $a,b\ge 0$ and $0<\bb<1$,
\begin{equation}
\label{u2}
(a+b)^\bb\le a^\bb+b^\bb.
\end{equation}
\begin{proof}
It is sufficient to prove \eqref{u2} for $a=1$. Put $f(t)=(1+t)^\bb-t^\bb$. Note that $f(0)=1$ and $f'(t)\le 0$ for $t>0$.  Hence $f(t)\le 1$ for $t\ge 0$.
\end{proof}

\item\label{ub}
For every finite measure $M$, every positive measurable function $Y$ and every $\bb>0$,
\[
M(Y^{-\bb})\ge M(1)^{1+\bb}(MY)^{-\bb}.
\]

 Indeed $f(y)=y^{-\bb}$ is a convex function on $\R_+$, and we get \ref{ub} by applying Jensen's inequality to the probability measure $M/M(1)$.
\end{asma}

\begin{proof}[Proof of Theorem \ref{Theorem A1.1}]
1\grd.\ 
If $x\in E\setminus D$, then, $\NN_x$-a.s., $\RR_E$ is not a subset of
$D^*$ (see the proof of Theorem \ref{Theorem uu1}). Hence, both sides
of \eqref{A1.8} vanish.

2\grd.\ Suppose $x\in D$. By \eqref{b2.2},
 $\NN_x(1-e^{-sZ_\nu})=u_{s\nu}$. Thus \eqref{5.10} implies  $v_s\le w_\LL + u_{s\nu}$. Therefore, by \ref{ua},  $v_s^{\myaa-1}\le w
_\LL^{\myaa-1}+u_{s\nu}^{\myaa-1}$ and, since $u_{s\nu}\le h_{s\nu}=sh_\nu$, 
$\Phi(v_s)\le\Phi(w_\LL)+s^{\myaa-1}\Phi(h_\nu)$. 

Put $\A=\{\RR_E\subset D^*\}$.
It follows from \eqref{5.12}  that 
\begin{equation}
\label{u3}
\NN_x\{\A, Z_\nu\neq 0\}\ge
\Pi_x^{\nu}\{A, \int_0^\infty e^{-\Phi(w_\LL)-s^{\myaa-1}\Phi(h_\nu)}ds\}.
\end{equation}
Note that $\int_0^\infty e^{-as^\bb}ds=Ca^{-1/\bb}$ where 
$C=\int_0^\infty e^{-t^\bb} dt$.
Therefore  \eqref{u3}  implies 
\begin{equation}
\label{u1}
\NN_x\{\A, Z_\nu\neq 0\}\ge
C\Pi_x^\nu\{A, e^{-\Phi(w_\LL)} \Phi(h_\nu)^{-1/(\myaa-1)}\}.
\end{equation}
The right side in \eqref{u1} is equal to $CM(Y^{-\bb})$  where $\bb=1/(\myaa-1), Y=
\Phi(h_\nu)$ and $M$ is the measure with the density 
$1_Ae^{-\Phi(w_\LL)}$ with respect to $\Pi_x^\nu$. We get from \eqref{u1} 
and \ref{ub} that
\begin{multline*}
\NN_x\{\A, Z_\nu\neq 0\}\ge CM(1)^{1+\bb}(MY)^{-\bb}\\=
C[\Pi_x^\nu \{A,e^{-\Phi(w_\LL)}\}]^{\myaa/(\myaa-1)} [\Pi_x^\nu\{A,e^{-\Phi(w_\LL)}\Phi(h_\nu)\}]^{-1/(\myaa-1)}.
\end{multline*}
By \eqref{5.11}, $\Pi_x^\nu \{A,e^{-\Phi(w_\LL)}\} =\NN_x\{\RR_E\subset D^*,Z_\nu\}$ and, since 
$\Pi_x^\nu\{A,e^{-\Phi(w_\LL)}\Phi(h_\nu)\}\le \Pi_x^\nu\Phi(h_\nu)$,
we have
\begin{equation}
\label{nu}
\NN_x\{\A, Z_\nu\neq 0\}\ge C[\NN_x\{\RR_E\subset D^*,Z_\nu\}]^{\myaa/(\myaa-1)} 
[\Pi_x^\nu\Phi(h_\nu)]^{-1/(\myaa-1)}.
\end{equation}

3\grd.\ 
By the definition of $h$-transform,
for every  $f\in\B(E)$  and every $h\in\myH(E)$,
\begin{equation*}
\Pi_x^h \int_0^{\tau_E}f(\xi_t)dt=\int_0^\infty\Pi_x^h\{t<\tau_E,f(\xi_t)\}dt=
\int_0^\infty \Pi_x \{t<\tau_E,f(\xi_t)h(\xi_t)\} dt.
\end{equation*}
 By taking $f=\myaa h_\nu^{\myaa-1}$ and $h=h_\nu$ we get
\begin{equation}
\label{u5}
\Pi_x^\nu\Phi(h_\nu)=\myaa \E_x(\nu).
\end{equation}
Formula \eqref{A1.8} follows from  \eqref{nu} and \eqref{u5}.
\end{proof}


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\end{document}