# Absolute continuity of the Super-Brownian motion with infinite mean

Rustam Mamin  Faculty of Mathematics, Technion—Israel Institute of Technology,
Haifa 32000, Israel<EMAIL_ADDRESS>and  Leonid Mytnik  Faculty of
Industrial Engineering & Management, Technion—Israel Institute of Technology,
Haifa 32000, Israel<EMAIL_ADDRESS>
(Date: December 2020)

###### Abstract.

In this work we prove that for any dimension $d\geq 1$ and any
$\gamma\in(0,1)$ super-Brownian motion corresponding to the log-Laplace
equation

$\begin{split}\frac{\partial v(t,x)}{\partial t}&=\frac{1}{2}\bigtriangleup
v(t,x)+v^{\gamma}(t,x),\>(t,x)\in\mathbb{R}_{+}\times\mathbb{R}^{d},\\\
v(0,x)&=f(x)\end{split}$

is absolutely continuous with respect to the Lebesgue measure at any fixed
time $t>0$. Our proof is based on properties of solutions of the log-Laplace
equation. We also prove that when initial datum $v(0,\cdot)$ is a finite, non-
zero measure, then the log-Laplace equation has a unique, continuous solution.
Moreover this solution continuously depends on initial data.

###### Key words and phrases:

Superprocesses, stable branching

###### 2000 Mathematics Subject Classification:

Primary 60G57, 60J68; secondary 60J80

LM is supported in part by ISF grant No. ISF 1704/18.

##  1\. Introduction and main result

This paper is devoted to studying regularity properties of the super-Brownian
motion with stable branching mechanism with infinite mean.

Let us start with a notation. For a measure $\mu$ on $\mathbb{R}^{d}$ and a
function $f$ on $\mathbb{R}^{d}$ let $\langle\mu,f\rangle$ or $\langle
f,\mu\rangle$ denote the integral of a function $f$ with respect to a measure
$\mu$ (whenever it is well defined):

$\langle
f,\mu\rangle=\langle\mu,f\rangle\equiv\int_{\mathbb{R}^{d}}f(x)\mu(\mathrm{d}x).$

Let $\gamma\in(0,2]\setminus\\{1\\}$. The super-Brownian motion with
$\gamma$-stable branching mechanism, $X=\\{X_{t},t\geq 0\\}$, is a Markov
measure-valued process on $\mathbb{R}^{d}$ which is characterized as follows:
for any finite measure $\mu$ and a nonnegative not identically zero function
$f\in L^{\infty}(\mathbb{R}^{d})$,

(1.1) $E_{\mu}\left(e^{-\langle
X_{t},\phi\rangle}\right)=E\left(\left.e^{-\langle
X_{t},\phi\rangle}\right|X_{0}=\mu\right)=e^{-\langle\mu,v(t,\cdot)\rangle},\quad\forall
t\geq 0.$

Here $v$ is a solution to the so-called log-Laplace equation:

(1.2)
$v(t,x)=\left(S_{t}f\right)(x)-\int_{0}^{t}(S_{t-s}v^{\gamma}(s,\cdot))(x)\mathrm{d}s,\quad(t,x)\in\mathbb{R}_{+}\times\mathbb{R}^{d},$

if $\gamma\in(1,2]$, and for $\gamma\in(0,1)$, the sign in front of the non-
linear term is reversed:

(1.3)
$v(t,x)=\left(S_{t}f\right)(x)+\int_{0}^{t}(S_{t-s}v^{\gamma}(s,\cdot))(x)\mathrm{d}s,\quad(t,x)\in\mathbb{R}_{+}\times\mathbb{R}^{d}.$

Here and for the rest of the paper $\\{S_{t}\\}_{t\geq 0}$ denotes a
transition semigroup of Brownian motion whose generator is Laplacian
$\frac{1}{2}\Delta$ in $\mathbb{R}^{d}$. Clearly

$S_{t}f(x)=\int_{\mathbb{R}^{d}}f(y)p_{t}(x-y)\,dy,\quad t\geq 0,$

where $\\{p_{t}(x),t\geq 0,x\in\mathbb{R}^{d}\\}$ is the transition density of
the Brownian motion. $\\{S_{t}\\}_{t\geq 0}$ describes the underlying Brownian
motion of $X$, whereas its continuous-state branching mechanism described by
$v\mapsto\pm v^{\gamma},v\geq 0$. These equations were considered by Watanabe
in [18] for a more general “motion” operator. As for the case of super-
Brownian motion with $\gamma$-stable branching mechanism (in what follows we
will call it $\gamma$-super-Brownian motion) it is well known that for
$\gamma\in(1,2]$ in dimensions $d<\frac{2}{\gamma-1}$ at any fixed time $T>0$,
the measure $X_{t}=X_{t}(dx)$ is absolutely continuous with probability one
(cf. [5]). By an abuse of notation, we sometimes denote a version of the
density function of the measure $X_{t}=X_{t}(dx)$ by the same symbol,
$X_{t}(dx)=X_{t}(x)\,dx$. It is even known that for $d=1$, $\gamma\in(1,2]$,
at fixed times $t$, there is a continuous version of the density in $x$
variable (see [12]), and for $\gamma=2$, and again $d=1$, there exists even
jointly space-time continuous version of the density (see [9], [16]). More
detailed regularity properties of the densities of superprocesses with stable
branching mechanism with possibly more general motion have been studied in
[6], [7], [14], [13].

This paper is devoted to deriving absolute continuity of $X$ for the case of
$\gamma\in(0,1)$. It is easy to check that in this case $E(\langle
X_{t},1\rangle)=\infty$, for $t>0$, which adds some technical dificulties for
the proofs.

Before we state the main result of this paper we need to introduce some
notation. Let $E$ be any Polish space. Let $C(E)$ and $B(E)$ be respectively
the spaces of continuous and _Borel_ measurable functions on space $E$. If
$F(E)$ is a space of real-valued functions on $E$ we define the following
subspaces of $F(E)$. $F_{b}(E)$ (respectively $F^{+}(E),F_{c}(E),F_{bc}(E)$)
denotes the subspace of bounded (respectively positive, with compact support,
bounded with compact support) functions. For example,
$B^{+}_{bc}(\mathbb{R}^{d})$ denotes a set of positive, bounded, Borel
measurable functions with compact support on $\mathbb{R}^{d}$.

Now let us define the explosion time of the superprocess.

######  Definition 1.1 (Time of explosion).

Let $d\geq 1$ and let $\\{X_{t}\\}_{t\geq 0}$ be a super-Brownian motion with
non-random initial state $X_{0}$. Given nonnegative continuous function $f$ on
$\mathbb{R}^{d}$, we define the time of explosion $T(X_{0},f)$ of $X$ as
follows

$T(X_{0},f)\equiv\inf\left\\{t\geq 0:\bigcap_{n=1}^{\infty}\left\\{\langle
X_{t},f\rangle\geq n\right\\}\right\\}.$

Now we are able to state the main result of the paper.

######  Theorem 1.2 (Absolute continuity).

Let $d\geq 1$ and $0<\gamma<1$. Let $\\{X_{t}\\}_{t\geq 0}$ be a
$\gamma$-super-Brownian motion with non-random initial state $X_{0}$ being a
finite measure on $\mathbb{R}^{d}$. For each $t>0$, $X_{t}(\mathrm{d}x)$ is
$P-a.s.$ absolutely continuous on the event $\\{t<T(X_{0},1)\\}$.

The proof of this theorem is based on the properties of solutions to the log-
Laplace equation corresponding to the process $\\{X_{t}\\}_{t\geq 0}$. These
properties are stated in Theorem 2.2. This theorem extends results of Aguirre
and Escobedo (see [1]) for the case of non-zero measure-valued initial
conditions. The entire Section 4 is devoted to the proof of these properties.
In Section 3.1 we will prove that for any nonnegative, non-zero continuous
function $f$ on $\mathbb{R}^{d}$,

$\displaystyle T(X_{0},f)=T(X_{0},1),\>P-\mathrm{a.s.}$

This property allows us to define the density of the superprocess
$\\{X_{t}\\}_{t\geq 0}$ for a fixed time $t>0$. In Section 3.2 we conclude the
proof of Theorem 1.2 — the main result of the paper.

##  2\. Semilinear heat equation

For the rest of the paper fix $\gamma\in(0,1)$ and arbitrary dimension $d\geq
1$. One of the main tools for investigating the $\gamma$-super-Brownian motion
is the log-Laplace equation

(2.1)
$v(t,x)=\left(S_{t}f\right)(x)+\int_{0}^{t}(S_{t-s}v^{\gamma}(s,\cdot))(x)\mathrm{d}s.$

Usually in the literature (2.1) is studied for $f$ being a non-negative
function. In the sequel we will consider (2.1) also with $f$ being a measure.

Before we discuss properties of (2.1) we need to introduce some notation. For
a topological space $S$, $\mathcal{B}(S)$ will denote the $Borel$
$\sigma\textrm{-algebra}$ on the space $S$.

We denote by $L^{p,w}(\mathbb{R}^{d})$ (for $p=1$ or $p=\infty$) a Banach
space of (equivalence classes of) measurable functions on $\mathbb{R}^{d}$
with the norms:

$\displaystyle\|f\|_{1,w}$
$\displaystyle\equiv\int_{\mathbb{R}^{d}}|w(x)f(x)|\mathrm{d}x,\>\>\textnormal{for}\>\>p=1$
$\displaystyle\left\|f\right\|_{\infty,\omega}$
$\displaystyle\equiv\inf\\{M:Leb(x:|w(x)f(x)|>M)=0\\},\>\textnormal{for}\>p=\infty,$

where

$\displaystyle w(x)\equiv
C_{w}e^{-|x|},\>\>\int_{\mathbb{R}^{d}}w(x)\mathrm{d}x=1,$

and $Leb$ denotes the Lebesgue measure on $\mathbb{R}^{d}$.
$L^{p,w}_{+}(\mathbb{R}^{d})$ (respectively $L^{p}_{+}(\mathbb{R}^{d})$) will
denote the nonnegative elements of $L^{p,w}(\mathbb{R}^{d})$ (respectively
$L^{p}(\mathbb{R}^{d})$).

Given $L^{p,w}(\mathbb{R}^{d})$ (for $p=1$ or $p=\infty$) we define the Banach
space $L^{\infty}_{loc}((0,\infty),L^{p,w}(\mathbb{R}^{d}))$ of (equivalent
classes of) measurable functions on $(0,\infty)\times\mathbb{R}^{d}$ as
follows: $f\in
L^{\infty}_{loc}\left((0,\infty),L^{p,w}(\mathbb{R}^{d})\right)$ if and only
if $f(t,\cdot)\in L^{p,w}(\mathbb{R}^{d})$ for any fixed $t$ and

$t\to\|f(t)\|_{p,w}\in L^{\infty}([a,b])$

for any compact interval $[a,b]\in(0,\infty)$. Similarly
$L^{\infty}_{loc}((0,\infty),L^{p,w}_{+}(\mathbb{R}^{d}))$ is defined.

By $M_{F}(E)$ (respectively $M_{F,S}(E)$) we denote the space of finite
measures and finite (respectively finite signed) signed measures on a Polish
space $E$ equipped with the topology of the weak convergence. We write

$\mu_{n}\overset{w}{\Longrightarrow}\mu,\quad\text{as $n\rightarrow\infty$,}$

if the sequence $\\{\mu\\}_{n=1}^{\infty}$ of finite measures or finite signed
measures weakly converges to a finite measure $\mu$.

If $F$ is a set of functions or measures then $\overset{\circ}{F}$ denotes
this set without zero element, that is $\overset{\circ}{F}=F\setminus\\{0\\}$.
If $F$ is a topological space then, topology of $\overset{\circ}{F}$ is
inherited from $F$. For example ${\overset{\circ}{M}}_{F}(E)$ is a space of
finite non-zero measures on Polish space $E$ with the topology inherited from
$M_{F}(E)$.

With all this notation at hand we can get back to (2.1). Equation (2.1) was
studied by Aguirre and Escobedo in [1]). We state some of their results in the
following theorem.

######  Theorem 2.1 (Aguirre and Escobedo [1]).

For any not identically zero $f\in L^{\infty,w}_{+}(\mathbb{R}^{d})$ there
exists the unique solution $v(t,x,f)$ of equation (2.1) such that

  * $(1)$

$v(\cdot,\cdot,f)\in C^{+}\left((0,\infty)\times\mathbb{R}^{d}\right)\cap
L^{\infty}_{loc}\left((0,\infty),L^{\infty,w}_{+}(\mathbb{R}^{d})\right)$;

  * $(2)$

$v(t,x,f)>((1-\gamma)t)^{1/(1-\gamma)},\>(t,x)\in(0,\infty)\times\mathbb{R}^{d}$,

  * $(3$

for $i=1,2$, let $v(t,x,f_{i})$ be the solution to (2.1) with initial
condition $v(0,\cdot,f_{i})=f_{i}$. If $f_{1}(x)\leq f_{2}(x),\>a.e.\>x$, then

$\displaystyle v(t,x,f_{1})\leq
v(t,x,f_{2}),\>\forall(t,x)\in(0,\infty)\times\mathbb{R}^{d}.$

  * $(4)$

$\lim_{t\to 0}v(t,\cdot,f)=f,\>\textnormal{for}\>a.e.\>x\in\mathbb{R}^{d}$;

  * $(5)$

for any fixed $(t,x)\in(0,\infty)\times\mathbb{R}^{d}$ the mapping

$v(t,x,\cdot):L^{\infty,w}_{+}(\mathbb{R}^{d})\mapsto\mathbb{R}_{++}$

is continuous. Here $\mathbb{R}_{++}\equiv(0,\infty).$

######  Remark.

In fact, Aguirre and Escobedo prove the above theorem for a more general class
of initial data.

We extend the results in Theorem 2.2 for the case of not identically zero
measure-valued initial conditions. Consider the following equation:

(2.2)
$v(t,x)=(S_{t}\mu)(x)+\int_{0}^{t}(S_{t-s}v^{\gamma}(\cdot,s))\mathrm{d}s,\>x\in\mathbb{R}^{d},\>t>0,$

where $\mu\in{{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})}$, and again $d\geq 1$
is an arbitrary dimension. We set
$S_{t}\mu(x)=\int_{\mathbb{R}}^{d}p_{t}(x-y)\mu(dy),x\in\mathbb{R}^{d}$. In
order to stress dependence of the solutions of this equation on initial data
we will sometimes write $v(t,x,\mu)$. In what follows we will also use the
following notation for solutions of (2.2):

(2.3) $V_{t}(\mu)(x)\equiv v(t,x,\mu),\quad t>0,x\in\mathbb{R}^{d},$

for $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$ or being a non-negative,
not identically zero function.

Before we state the main result of this section, let us define the constant
$\gamma^{\prime}$ in terms of $\gamma$ as follows:

$\gamma^{\prime}=\frac{1}{1-\gamma}.$

######  Theorem 2.2 (Existence, uniqueness and dependence on initial data).

For

any $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$, equation (2.2) has the
unique solution $v(t,x)$ such that

$v(\cdot,\cdot,\mu)\in
L_{loc}^{\infty}\left((0,\infty),L^{1,w}_{+}(\mathbb{R}^{d})\right)\cap
C^{+}((0,\infty)\times\mathbb{R}^{d})$

and

(2.4) $\left((1-\gamma)t\right)^{\gamma^{\prime}}<v(t,x,\mu)\leq
e^{t}\left(S_{t}\mu\right)(x)+e^{t},\>0<t<\infty.$

Moreover, this solution continuously depends on initial data: if a sequence
$\\{\mu_{n}\\}_{n=1}^{\infty}$ from ${\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$
converges weakly to $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$ then

$\lim_{n\to\infty}v(t,x,\mu_{n})\to v(t,x,\mu),$

for any $(t,x)\in(0,\infty)\times\mathbb{R}^{d}$.

######  Remark 2.3.

Since for any $t\in(0,\infty)$,

$\displaystyle e^{t}\left(S_{t}\mu\right)(x)+e^{t}\in
L^{1,w}(\mathbb{R}^{d}),$

it easily follows from inequality (2.4) that the sequence of solutions

$\\{v(\cdot,\cdot,\mu_{n})\\}_{n=1}^{\infty}$ which also converges to
$v(\cdot,\cdot,\mu)$ in
$L_{loc}^{\infty}\left(((0,\infty),L^{1,w}_{+}(\mathbb{R}^{d})\right)\cap
C^{+}((0,\infty)\times\mathbb{R}^{d})$.

The proof of the next lemma is trivial and hence it is omitted.

######  Lemma 2.4.

Let $\mu\in M_{F}(\mathbb{R}^{d})$. Then, for any $t\in(0,\infty)$,

$\left(S_{t}\mu\right)(x)\leq\frac{\mu(\mathbb{R}^{d})}{(2\pi
t)^{d/2}},\>\forall x\in\mathbb{R}^{d}.$

Now we are ready to state the corollary to Theorem 2.2.

######  Corollary 2.5.

Let
$\\{\mu_{n}\\}_{n=1}^{\infty}\subset{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$
be a sequence of measures that converges weakly to
$\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$ and let
$v(\cdot,\cdot,\mu_{n})$ be the corresponding solutions of (2.2). Then, for
any $\chi\in M_{F}(\mathbb{R}^{d})$ and $t\in(0,\infty)$,

(2.5)
$\lim_{n\to\infty}\int_{\mathbb{R}^{d}}v(t,x,\mu_{n})\chi(\mathrm{d}x)=\int_{\mathbb{R}^{d}}v(t,x,\mu)\chi(\mathrm{d}x).$

###### Proof.

By Theorem 2.2 we have

$v(t,x,\mu)\leq
e^{t}\left(S_{t}\mu\right)(x)+e^{t},n=1,2,\ldots,\>0<t<\infty.$

Since the sequence $\\{\mu_{n}\\}_{n=1}^{\infty}$ converges weakly to $\mu$,
then by Lemma 2.4

$\begin{split}\sup_{x\in\mathbb{R}^{d}}\left(e^{t}\left(S_{t}\mu\right)(x)+e^{t}\right)\leq
e^{t}\left(\frac{1}{(2\pi t)^{d/2}}\sup_{n\geq
1}\mu_{n}(\mathbb{R}^{d})+1\right)<\infty.\end{split}$

Thus, we conclude that the sequence $\\{v(t,\cdot,\mu_{n})\\}_{n=1}^{\infty}$
is bounded. Also by Theorem 2.2 $\\{v(t,\cdot,\mu_{n})\\}_{n=1}^{\infty}$
converges pointwise to $v(t,\cdot,\mu)$. Hence, by the bounded convergence
theorem, (2.5) follows. ∎

Theorem 2.2 will be proved in Section 4.

##  3\. Proof of Theorem 1.2

In this section we prove the main result of this paper — absolute continuity
of the super-Brownian motion $X$ with the branching mechanism $v\mapsto
v^{\gamma}$, for $\gamma\in(0,1)$.

In Section 3.1 we investigate explosion time for the $\gamma$-super-Brownian:
this is necessary for the proof of Theorem 1.2 that will be concluded in
Section 3.2.

###  3.1. Explosion times

As we will see for any $t>0$, the $\gamma$-super-Brownian motion
$X=\\{X_{t}\\}_{t\geq 0}$ explodes by time $t$ with non-zero probability. In
this section we investigate the distribution of the explosion times. We assume
that $X$ is defined on probability space
$(\Omega,P,\mathcal{F},\mathcal{F}_{t})$ and adapted to filtration
$\\{\mathcal{F}_{t}\\}_{t\geq 0}$. We also assume that the initial state
$X_{0}$ of $X$ is a non-random finite measure.

######  Remark 3.1.

By Corollary $4.3.2$, in [2], it is easy to show that $\\{X_{t}\\}_{t\geq 0}$
is a Feller process and therefore it has a strong Markov property.

The next lemma states the elementary properties of the explosion times. The
proofs are simple and easily follow from the definition, so they are omitted.

######  Lemma 3.2.

  1. (1)

For any function $f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$,
and any $a\in(0,\infty)$,

$\displaystyle T(X_{0},f)=T(X_{0},af),\>P-\textnormal{a.s.}$

  2. (2)

For any $f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$,

$T(X_{0},1)\leq T(X_{0},f),\>P-\textnormal{a.s.}$

In the next lemma we will show that for any $t\geq T(X_{0},f)$, one has
$X_{t}(f)=\infty$. Before we proceed, let us recall that the Laplace transform
of $\\{X_{t}\\}_{t\geq 0}$ is given by

(3.1) $E\left(e^{-\langle X_{t},f\rangle}\right)=e^{-\langle
X_{0},V_{t}(f)\rangle},\>f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d}),$

where $\\{V_{t}(f)\\}_{t\geq 0}$ solves log-Laplace equation (2.1).

######  Lemma 3.3.

For any $t>0$, $f\in\ \overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$,

$\left\\{T(X_{0},f)\leq
t\right\\}=\left\\{X_{t}(f)=\infty\right\\},\>P-\textnormal{a.s.}.$

###### Proof.

The $P$-a.s. inclusion
$\left\\{X_{t}(f)=\infty\right\\}\subset\\{T(X_{0},f)\leq t\\}$ is trivial.
Now let us show $\\{T(X_{0},f)\leq t\\}\subset\\{X_{t}(f)=\infty\\}$, $P$-a.s.
It is enough to verify that

(3.2) $E\left(e^{-\langle X_{t},f\rangle}1_{\left\\{T(X_{0},f)\leq
t\right\\}}\right)=0.$

Define the stopping time

$\displaystyle T_{n}(X_{0},f)\equiv\inf\left\\{t\geq 0,X_{t}(f)=n\right\\}.$

Clearly $T_{n}(X_{0},f)\to T(X_{0},f),\>P$-a.s., as $n\to\infty$. Then, for
any $\delta\in(0,t)$ arbitrarily small,

$\displaystyle E$ $\displaystyle\left(e^{-\langle
X_{t},f\rangle}1_{\left\\{T(X_{0},f)\leq t-\delta\right\\}}\right)=$
$\displaystyle=E\left(e^{-\langle X_{t},f\rangle}1_{\left\\{T(X_{0},f)\leq
t-\delta\right\\}}1_{\left\\{T_{n}(X_{0},f)\leq t-\delta\right\\}}\right)$
$\displaystyle\leq E\left(e^{-\langle
X_{t},f\rangle}1_{\left\\{T_{n}(X_{0},f)\leq t-\delta\right\\}}\right)$ (3.3)
$\displaystyle=E\left(E\left(e^{-\langle
X_{t},f\rangle}|\mathcal{F}_{T_{n}(X_{0},f)}\right)1_{\left\\{T_{n}(X_{0},f)\leq
t-\delta\right\\}}\right)$ (3.4) $\displaystyle=E\left(e^{-\langle
X_{T_{n}(X_{0},f)},V_{t-T_{n}(X_{0},f)}(f)\rangle}1_{\left\\{T_{n}(X_{0},f)\leq
t-\delta\right\\}}\right).$

Here, in (3.3), we used the strong Markov Property (see Remark 3.1). Fix
$c(\delta)>0$ sufficiently small such that
$c(\delta)f(x)\leq((1-\gamma)t)^{\gamma^{\prime}}$ for all $t\geq\delta$,
$x\in\mathbb{R}^{d}$. Then by Theorem 2.1 (see also Lemma $2.2$ in [1]) we
have

$\displaystyle c(\delta)f(x)\leq((1-\gamma)t)^{\gamma^{\prime}}\leq
V_{t}(f)(x),\>\forall t\geq\delta,x\in\mathbb{R}^{d}.$

Therefore the expression (3.4) can be bounded from the above by

$\displaystyle E\left(e^{-c(\delta)n}1_{\left\\{T_{n}(X_{0},f)\leq
t-\delta\right\\}}\right).$

By the dominated convergence theorem this expression tends to zero, as
$n\to\infty$ and we get

$\displaystyle E\left(e^{-\langle X_{t},f\rangle}1_{\left\\{T(X_{0},f)\leq
t-\delta\right\\}}\right)=0,\forall t>0.$

Now take $\delta\searrow 0$ and by the monotone convergence theorem we get
(3.2) and this completes the proof. ∎

The following corollary is immediate.

######  Corollary 3.4.

Let $f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$. Then

(3.5) $\displaystyle E\left(e^{-\langle
X_{t},f\rangle}\right)=E\left(e^{-\langle
X_{t},f\rangle}1_{\left\\{T(X_{0},f)>t\right\\}}\right),\>\forall t>0.$

Now, let us calculate the distribution of $T(X_{0},1)$ — the distribution of
the explosion time of the total mass of the super-Brownian motion $X$. By
Corollary 3.4 and (3.1), we get

(3.6) $\begin{split}P(t<T(1,X_{0}))&=\lim_{a\searrow
0}E\left(1_{\\{t<T(1,X_{0})\\}}\left(e^{-a\langle
X_{t},1\rangle}\right)\right)\\\ &=\lim_{a\searrow 0}e^{-\langle
V_{t}(a),X_{0}\rangle}\\\ &=\lim_{a\searrow 0}\exp\left(-\langle
X_{0},1\rangle\big{(}a^{1-\gamma}+t(1-\gamma)\big{)}^{\gamma^{\prime}}\right)\\\
&=\exp\left(-\langle
X_{0},1\rangle\big{(}t(1-\gamma)\big{)}^{\gamma^{\prime}}\right),\end{split}$

where the third equality follows from the fact that $V_{t}(a)$ is a solution
of the ordinary differential equation

$\left\\{\begin{array}[]{ll}\displaystyle{\frac{\mathrm{d}v(t)}{\mathrm{d}t}}&=v^{\gamma}(t),\quad
t\geq 0,\\\\[10.0pt] v(0)&=a,\end{array}\right.$

and hence

(3.7) $\displaystyle
V_{t}(a)(x)=(a^{1-\gamma}+t(1-\gamma))^{\gamma^{\prime}},\quad\forall
x\in\mathbb{R}^{d},t\geq 0.$

Then

(3.8) $F_{T(1,X_{0})}(t)=P(t\geq T)=1-\exp\left(-\langle
X_{0},1\rangle\big{(}t(1-\gamma)\big{)}^{\gamma^{\prime}}\right).$

But what about other test functions $f$? What is the law of $\langle
X_{t},f\rangle$ for a general
$f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$? The answer is
given in the following lemma. In what follows, in order to simplify notation,
we often write $T(f)$ instead of $T(f,X_{0})$.

######  Lemma 3.5.

For any $f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$ random
variable $T(f)$ has the same distribution as $T(1)$:

$F_{T(f)}(t)=P(t\geq T(f))=1-\exp\left(-\langle
X_{0},1\rangle\big{(}t(1-\gamma)\big{)}^{\gamma^{\prime}}\right),\quad t\geq
0.$

###### Proof.

Fix an arbitrary $f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$,
and $t>0$. Then we have

(3.9) $\begin{split}P(t<T(f))&=\lim_{a\searrow
0}E\left(1_{\\{t<T(f)\\}}e^{-\langle X_{t},af\rangle}\right)\\\
&=\lim_{a\searrow 0}E\left(e^{-\langle X_{t},af\rangle}\right)\\\
&=\lim_{a\searrow 0}e^{-\langle V_{t}(af),X_{0}\rangle},\end{split}$

where the second equality follows by Corollary 3.4. By Theorem 2.1 (see also
Lemma $2.2$ in [1]), we get

(3.10) $\left((1-\gamma)t\right)^{\gamma^{\prime}}<V_{t}(af)(x),\>\forall
a>0,t>0,x\in\mathbb{R}^{d}.$

Using (3.10), we get

(3.11) $\displaystyle\exp\left(-\left<X_{0},V_{t}(af)\right>\right)$
$\displaystyle\leq\exp\left(-\left<X_{0},1\right>\left(t(1-\gamma)\right)^{\gamma^{\prime}}\right).$

By Lemma 3.2(2) we have $T(1)\leq T(f)$, $P$-a.s. By this, (3.8), (3.9), and
(3.11) we obtain

$\displaystyle P(t<T(1))$ $\displaystyle\leq P(t<T(f))$
$\displaystyle\leq\exp\left(-\left<X_{0},1\right>\left(t(1-\gamma)\right)^{\gamma^{\prime}}\right)$
$\displaystyle=P(t<T(1)).$

Thus, we get $P(t<T(f))=P(t<T(1))$. Since $t>0$ was arbitrary, we are done. ∎

The next lemma is a consequence of the first two lemmas in this section.

######  Lemma 3.6.

For any $f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$,

$P\left(T(1))\neq T(f)\right)=0.$

###### Proof.

By Lemma 3.2(2), $T(1)\leq T(f)$, $P$-a.s. and, by Lemma 3.5, $T(1)$ and
$T(f)$ have the same distribution, hence the result follows. ∎

######  Corollary 3.7.

For any $f\in\overset{\circ}{L}\mbox{}^{\infty}_{+}(\mathbb{R}^{d})$,

$\displaystyle E\left(e^{-\langle X_{t},f\rangle}\right)=E\left(e^{-\langle
X_{t},f\rangle}1_{\left\\{t<T(1)\right\\}}\right)=e^{-\langle
X_{0},V_{t}(f)\rangle},\;\;t>0.$

###  3.2. Proof of Theorem 1.2

We begin this subsection with the following remark.

######  Remark 3.8.

By Lemma $3.4.2.1$ in [3], any random measure $Y\in M_{F}(\mathbb{R}^{d})$ can
be decomposed into its absolutely continuous $Y^{ac}$ and singular $Y^{s}$
parts with respect to the Lebesgue measure:
$Y(\omega,\mathrm{d}x)=Y^{ac}(\omega,\mathrm{d}x)+Y^{s}(\omega,\mathrm{d}x)$.
By the definition of $T(1)$, $X_{t}$ is a finite measure on
$\Omega\cap\\{t<T(1)\\}$. Hence on the set $\Omega\cap\\{t<T\\}$, $X_{t}$ can
be decomposed into absolutely continuous and singular parts

$\displaystyle
X_{t}(\omega,\mathrm{d}x)=X_{t}^{ac}(\omega,\mathrm{d}x)+X_{t}^{s}(\omega,\mathrm{d}x).$

The next lemma is used in the proof of measurability of density. Its proof is
a simple adaptation of the proof of Theorem $1.8$ from [11] and therefore it
is omitted.

######  Lemma 3.9.

For any $f\in B^{+}_{bc}(\mathbb{R}^{d})$ and any fixed $t\in(0,\infty)$, the
map $(\omega,z)\mapsto\langle X_{t}(\omega),f(z-\cdot)\rangle$ is a measurable
map from

${\left(\Omega\cap\\{t<T(1)\\},\mathcal{F}\cap\\{t<T(1)\\}\right)\times\left(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d})\right)}$
to $\mathbb{R}_{+}$.

For studying differentiability properties of $X_{t}$, let us introduce a
sequence of functions $\\{\delta^{n}(\cdot)\\}_{n=1}^{\infty}$ defined as

$\delta^{n}(x)=\left\\{\begin{array}[]{ll}1/Leb(B_{1/n}(0)),&\text{if
}\left|x\right|\leq\frac{1}{n};\\\ 0,&\text{otherwise.}\end{array}\right.$

Here $B_{1/n}(0)$ is a closed ball of radius $1/n$, centered at the origin.
Notice that the sequence $\\{\delta^{n}(z-\cdot)\\}_{n=1}^{\infty}$ converges
to Dirac $\delta$-function with support at point $z$.

######  Lemma 3.10.

On $\left(\Omega\cap\\{t<T(1)\\}\right)\times\mathbb{R}^{d}$,
$P(\mathrm{d}\omega)\mathrm{d}z$ \- a.e. there exists a limit

$\widetilde{\eta}^{ac}_{t}(\omega,z)=\lim_{n\to\infty}\langle
X_{t}(\omega),\delta^{n}(z-\cdot)\rangle.$

The random function $\widetilde{\eta}^{ac}_{t}$ is a version of the Radon-
Nikodym derivative of $X_{t}$ on $\left(\Omega\cap\\{t<T(1)\\}\right)$.
Moreover $\widetilde{\eta}^{ac}_{t}$ is a measurable map from

$\left(\Omega\cap\\{t<T(1)\\},\mathcal{F}\cap\\{t<T(1)\\}\right)\times\left(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d})\right)$
to $\mathbb{R}_{+}$.

###### Proof.

By the Lebesgue density theorem (see [17], Theorem $7.14$), for $P$-a.s.
$\omega\in\left(\Omega\cap\\{t<T(1)\\}\right)$, there exists a limit

(3.12) $\widetilde{\eta}^{ac}_{t}(\omega,z)=\lim_{n\to\infty}\langle
X_{t}(\omega),\delta^{n}(z-\cdot)\rangle$

for all $z\in R^{d}\setminus N(\omega)$ where $N(\omega)$ is a Borel subset of
the Lebesgue measure zero and $\widetilde{\eta}^{ac}_{t}$ is a Radon-Nikodym
derivative with respect to the Lebesgue measure. It is easy to see that
convergence in (3.12) takes place $P(\mathrm{d}\omega)\mathrm{d}z$-a.e. We set
$\widetilde{\eta}^{ac}_{t}(\omega,z)$ to be zero at points $(\omega,z)$ where
the limit does not exist.

By Lemma 3.9, for each $n=1,2,\ldots$, $\langle
X_{t}(\omega),\delta^{n}(z-\cdot)\rangle$ is measurable and the measurability
of $\widetilde{\eta}^{ac}_{t}(\omega,z)$ follows from
$P(\mathrm{d}\omega)\mathrm{d}z$-a.e. convergence. ∎

Function $\widetilde{\eta}^{ac}_{t}(\omega,z)$ is defined on
$\left(\Omega\cap\\{t<T(1)\\}\right)$. Function $\eta^{ac}_{t}(\omega,z)$ is
an extension of function $\widetilde{\eta}^{ac}_{t}(\omega,z)$ to entire
$\Omega$:

(3.13)
$\eta^{ac}_{t}(\omega,z)=\left\\{\begin{array}[]{ll}\widetilde{\eta}^{ac}_{t}(\omega,z)&\text{if
}\omega\in\left(\Omega\cap\\{t<T(1)\\}\right),\\\ \infty&\text{otherwise.
}\end{array}\right.$

Recall that for any $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$,
$\\{V_{t}(\mu)\\}_{t>0}$ denotes the solution to (2.2).

######  Lemma 3.11.

For every $t\in(0,\infty)$ the equality

$E\left(1_{\\{t<T(1)\\}}\exp\left(-\sum_{i=1}^{N}a_{i}\eta^{ac}(z_{i})\right)\right)=\exp\left(-\left(\left\langle
X_{0},V_{t}\bigg{(}\sum_{i=0}^{N}a_{i}\delta(z_{i}-\cdot)\right)\right\rangle\right)$

holds for almost every $\\{z_{i}\\}_{i=1}^{N}\subset\mathbb{R}^{d}$ and any
$\\{a_{i}\\}_{i=1}^{N}\subset\mathbb{R}_{++}\,$.

###### Proof.

Let $\phi(z_{1},z_{2},\ldots,z_{N})$ be any function in
$C^{+}_{b}(\mathbb{R}^{d\times N})\cap L^{1}(\mathbb{R}^{d\times N})$. By
Corollary 3.7 we have

$E\left(1_{\\{t<T(1)\\}}e^{-\left<X_{t},\sum_{i=1}^{N}a_{i}\delta^{n}(z_{i}-\cdot)\right>}\right)=e^{-\left<X_{0},V_{t}\left(\sum_{i=1}^{N}a_{i}\delta^{n}(z_{i}-\cdot)\right)\right>}.$

Let us multiply both parts of this equation by the function
$\phi(z_{1},z_{2},\ldots,z_{N})$, integrate over $\mathbb{R}^{d\times N}$ and
take the limit

(3.14) $\begin{split}\lim_{n\to\infty}&\int_{\mathbb{R}^{d\times
N}}E\left(1_{\\{t<T(1)\\}}e^{-X_{t}(\sum_{i=1}^{N}a_{i}\delta^{n}(z_{i}-\cdot))}\right)\phi(z_{1},z_{2},\ldots,z_{N})\mathrm{d}z_{1}\mathrm{d}z_{2}\ldots\mathrm{d}z_{N}\\\
&=\lim_{n\to\infty}\int_{\mathbb{R}^{d\times N}}e^{-\langle
V_{t}(\sum_{i=1}^{N}a_{i}\delta^{n}(z_{i}-\cdot)),X_{0}\rangle}\phi(z_{1},z_{2},\ldots,z_{N})\mathrm{d}z_{1}\mathrm{d}z_{2}\ldots\mathrm{d}z_{N}.\end{split}$

By Lemma 3.10, the limit

$\lim_{n\to\infty}\left<X_{t}(\omega),\sum_{i=1}^{N}a_{i}\delta^{n}(z_{i}-\cdot)\right>=\sum_{i=1}^{N}a_{i}\eta^{ac}_{t}(\omega,z_{i})$

exists almost everywhere on $\Omega\cap\\{t<T(1)\\}\times\mathbb{R}^{N\times
d}$ with respect to the measure
$P(\mathrm{d}\omega)\phi(z_{1},z_{2},\ldots,z_{N})\mathrm{d}z_{1}\mathrm{d}z_{2}\ldots\mathrm{d}z_{N}$.
Therefore, by the bounded convergence theorem, we get following limit on the
left-hand side of (3.14):

(3.15) $\begin{split}&\lim_{n\to\infty}\int_{\mathbb{R}^{d\times
N}}E\left(1_{\\{t<T(1)\\}}e^{-\langle
X_{t},\sum_{i=1}^{N}a_{i}\delta^{n}(z_{i}-\cdot)\rangle}\right)\phi(z_{1},z_{2},\ldots,z_{N})\mathrm{d}z_{1}\mathrm{d}z_{2}\ldots\mathrm{d}z_{N}\\\
&=\int_{\mathbb{R}^{d\times
N}}E\left(1_{\\{t<T(1)\\}}e^{-\sum_{i=1}^{N}a_{i}\eta^{ac}_{t}(z_{i})}\right)\phi(z_{1},z_{2},\ldots,z_{N})\mathrm{d}z_{1}\mathrm{d}z_{2}\ldots\mathrm{d}z_{N}.\end{split}$

Now let us take care of the right-hand side of (3.14). Since $X_{0}$ is a
finite, non-random measure, then by Corollary 2.5, the right hand side of
equation (3.15) also converges:

(3.16) $\begin{split}\lim_{n\to\infty}&\int_{\mathbb{R}^{d\times
N}}e^{-\langle
V_{t}(\sum_{i=1}^{N}a_{i}\delta^{n}(z_{i}-\cdot)),X_{0}\rangle}\phi(z_{1},z_{2},\ldots,z_{N})\mathrm{d}z_{1}\mathrm{d}z_{2}\ldots\mathrm{d}z_{N}\\\
&=\int_{\mathbb{R}^{d\times N}}e^{-\langle
V_{t}(\sum_{i=1}^{N}a_{i}\eta^{ac}(z_{i}-\cdot)),X_{0}\rangle}\phi(z_{1},z_{2},\ldots,z_{N})\mathrm{d}z_{1}\mathrm{d}z_{2}\ldots\mathrm{d}z_{N}.\end{split}$

Now, since $\phi$ was chosen arbitrarily, we can combine (3.14), (3.15) and
(3.16) and get

$\begin{split}E\left(1_{\\{t<T(1)\\}}e^{-\sum_{i=1}^{N}a_{i}\eta^{ac}_{t}(z_{i})}\right)&=e^{-\langle
V_{t}(\sum_{i=1}^{N}a_{i}\delta(z_{i}-\cdot)),X_{0}\rangle}\\\ &\text{for
Lebesgue almost
every}\\{z_{i}\\}_{i=1}^{N}\>\text{in}\>\mathbb{R}^{d}.\end{split}$

∎

######  Lemma 3.12.

Let $\phi\in C^{+}_{b}(\mathbb{R}^{d})\cap L^{1}(\mathbb{R}^{d})$ and let
$\\{\xi_{n}\\}_{n=1}^{\infty}$ be a sequence of i.i.d. random variables
defined on some probability space
$(\Omega^{\prime},\mathcal{F}^{\prime},P^{\prime})$ with the probability
density function

$g^{r}_{\xi}(x)=\left\\{\begin{array}[]{ll}1/Leb(B_{r}(0)),&\textnormal{if
}|x|\leq r\\\ 0,&\textnormal{elsewhere. }\end{array}\right.$

Then, for any $f\in L^{1}(\mathbb{R}^{d})$,

$\begin{split}\lim_{N\to\infty}\frac{Leb(B_{r}(0))}{N}\sum_{i=1}^{N}\phi(\xi_{i})f(\xi_{i})&=Leb(B_{r}(0))\int_{\Omega^{\prime}}\phi(\omega^{\prime})f(\xi_{1}(\omega))P^{\prime}(\mathrm{d}\omega^{\prime})\\\
&=\int_{\mathbb{R}^{d}}\phi(x)f(x)1_{B_{r}(0)}(x)\mathrm{d}x,\>P^{\prime}-\textnormal{a.s.}\end{split}$

This also implies that

$\lim_{N\to\infty}\frac{Leb(B_{r}(0))}{N}\sum_{i=1}^{N}\phi(\xi_{i})\delta(\xi_{i}-\cdot)\overset{w}{\Longrightarrow}\phi(x)1_{B_{r}(0)}(x)\mathrm{d}x,\>P^{\prime}-\textnormal{
a.s.}$

###### Proof.

It is obvious that $\phi f\in L^{1}(\mathbb{R}^{d})$ and the rest follows from
the law of large numbers. ∎

######  Lemma 3.13.

For any $f\in\overset{\circ}{C}\mbox{}^{+}_{b}(\mathbb{R}^{d}),t>0$,

$E\left(1_{\\{t<T(1)\\}}\exp\left(-\int_{\mathbb{R}^{d}}\eta^{ac}_{t}(z)f(x)\mathrm{d}x\right)\right)=E\left(1_{\\{t<T(1)\\}}\exp\left(-\int_{\mathbb{R}^{d}}X_{t}(\mathrm{d}x)f(x)\right)\right).$

###### Proof.

We augment our probability space $(\Omega,\mathcal{F},P(\mathrm{d}\omega))$ by
taking the Cartesian product with another probability space
$(\Omega^{\prime},\mathcal{F}^{\prime},P^{\prime}(\mathrm{d}\omega^{\prime}))$:

$(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{P})\equiv(\Omega\times\Omega^{\prime},\mathcal{F}\times\mathcal{F}^{\prime},P(\mathrm{d}\omega)P^{\prime}(\omega)).$

We also denote expectations on these spaces by $E$, $E^{\prime}$ and
$\tilde{E}$ respectively. Let $C^{++}_{b}(\mathbb{R}^{d})$ denote the space of
bounded continuous functions on $\mathbb{R}^{d}$ such that for any $f\in
C^{++}_{b}(\mathbb{R}^{d})$, we have $\inf_{x\in\mathbb{R}^{d}}f(x)>0$. Let us
fix an arbitrary $f\in C^{++}_{b}(\mathbb{R}^{d})$ and a positive integer $n$.

By the Borel theorem (see [8], Thm 3.19, p. 55) , for each $n\geq 1$ we can
build on the probability space
$(\Omega^{\prime},\mathcal{F}^{\prime},P^{\prime}(\mathrm{d}\omega^{\prime}))$
a sequence $\\{\xi_{i}^{n}(\omega)\\}_{i=1}^{\infty}$ of i.i.d. random
variables with the density function

$g_{\xi}^{n}(x)=\left\\{\begin{array}[]{ll}1/Leb(B_{n}(0))&\text{if }|x|\leq
n\\\ 0&\text{elsewhere. }\end{array}\right.$

By Lemma 3.11 we get, that the equality

$\displaystyle
E\left(1_{\\{t<T(1)\\}}\exp\bigg{(}-\frac{Leb(B_{n}(0))}{N}\sum_{i=1}^{N}f(z_{i})\eta^{ac}(z_{i})\bigg{)}\right)$
$\displaystyle\;\;\;=\exp\bigg{(}-\bigg{<}X_{0},V_{t}\bigg{(}\frac{Leb(B_{n}(0))}{N}\sum_{i=1}^{N}f(z_{i})\delta(z_{i}-\cdot)\bigg{)}\bigg{>}\bigg{)}$

holds for Lebesgue almost every $\\{z_{i}\\}_{i=1}^{N}$ in $\mathbb{R}^{d}$.
By changing $\\{z_{i}\\}_{i=1}^{N}$ to $\\{\xi_{i}^{n}\\}_{i=1}^{N}$, we
obtain

(3.17)
$\begin{split}&E\left(1_{\\{t<T(1)\\}}\exp\bigg{(}-\frac{Leb(B_{n}(0))}{N}\sum_{i=1}^{N}f(\xi_{i})1_{B_{n}(0)}(\xi_{i}^{n})\eta^{ac}(\xi_{i}^{n})\bigg{)}\right)\\\
&\;\;\;=\exp\bigg{(}-\bigg{<}X_{0},V_{t}\bigg{(}\frac{Leb(B_{n}(0))}{N}\sum_{i=1}^{N}f(\xi_{i})1_{B_{n}(0)}(\xi_{i}^{n})\delta(\xi_{i}^{n}-\cdot)\bigg{)}\bigg{>}\bigg{)},\>P^{\prime}-\mathrm{a.s.}\end{split}$

By taking limits $N\to\infty$ on both sides of (3.17), as well as using
Corollary 2.5 and Lemma 3.12 we get the equality

$\begin{split}&E\left(1_{\\{t<T(1)\\}}\exp\bigg{(}-\int_{\mathbb{R}^{d}}\eta_{t}^{ac}(x)f(x)1_{B_{n}(0)}(x)\mathrm{d}x\bigg{)}\right)\\\
&\;\;\;=\exp\bigg{(}-\bigg{<}X_{0},V_{t}(f1_{B_{n}(0)})\bigg{>}\bigg{)},\>P^{\prime}-\text{a.s.}.\end{split}$

Since both sides of the above equation are constants, we can drop
$P^{\prime}$-a.s., and get

(3.18)
$E\left(1_{\\{t<T(1)\\}}\exp\bigg{(}-\int_{\mathbb{R}^{d}}\eta^{ac}(x)f)1_{B_{n}(0)}(x)\mathrm{d}x\bigg{)}\right)=\exp\bigg{(}-\bigg{<}X_{0},V_{t}(f1_{B_{n}(0)})\bigg{>}\bigg{)}.$

By Theorem $2.8$ in [1],

$V_{t}(f1_{B_{n}(0)})\leq V_{t}(f1_{B_{n+1}(0)})$

and

(3.19) $\lim_{n\to\infty}V_{t}(f1_{B_{n}(0)})=V_{t}(f).$

Now we take limits, as $n\to\infty$ on both sides of (3.18), use the monotone
convergence theorem and (3.19) to get

(3.20)
$E\left(1_{\\{t<T(1)\\}}\exp\bigg{(}-\int_{\mathbb{R}^{d}}\eta^{ac}_{t}(z)f(x)\mathrm{d}x\bigg{)}\right)=\exp\left(-\left\langle
X_{0},V_{t}(f)\right\rangle\right).$

Since any function in $\overset{\circ}{C}\mbox{}^{+}_{b}(\mathbb{R}^{d})$ can
be approximated boundedly pointwise by functions from
$C^{++}_{b}(\mathbb{R}^{d})$, we can again apply the dominated convergence
theorem and obtain that the equality (3.20) holds for any
$f\in\overset{\circ}{C}\mbox{}^{+}_{b}(\mathbb{R}^{d})$. Recall, that

$E\left(1_{\\{t<T(1)\\}}\exp\left(-\left\langle
X_{t},f\right\rangle\right)\right)=\exp\left(-\left\langle
V_{t}(f),X_{0}\right\rangle\right),$

and we are done. ∎

Now we are ready to conclude the proof of the main result.

###### Proof of Theorem 1.2.

Fix arbitrary $t>0$. By Corollary 3.7, Lemma 3.13, for every
$f\in\overset{\circ}{C}\mbox{}^{+}_{b}(\mathbb{R}^{d})$,

(3.21) $\begin{split}&E\left(1_{\\{t<T(1)\\}}\exp\left(-\left\langle
X_{t},f\right\rangle\right)\right)\\\
&\quad\quad=E\left(1_{\\{t<T(1)\\}}\exp\left(-\int_{\mathbb{R}^{d}}\eta^{ac}_{t}(x)f(x)\mathrm{d}x\right)\right).\end{split}$

This equation implies, that, on the event $\left\\{t<T(1)\right\\}$,

$\int_{\mathbb{R}^{d}}X_{t}(\mathrm{d}x)f(x)\overset{d}{=}\int_{\mathbb{R}^{d}}\eta^{ac}_{t}(x)f(x)\mathrm{d}x,$

where $\overset{d}{=}$ means equality in distribution. By Lemma 3.10 and the
definition of $\eta_{t}^{ac}$, $\eta_{t}^{ac}$ is a version of the Radon-
Nikodym derivative of $X_{t}(\mathrm{d}x)$ on $\left\\{t<T(1)\right\\}$.
Therefore, on $\left\\{t<T(1)\right\\}$,

(3.22)
$\int_{\mathbb{R}^{d}}X_{t}(\mathrm{d}x)f(x)\geq\int_{\mathbb{R}^{d}}\eta^{ac}_{t}(x)f(x)\mathrm{d}x,\>P-\mathrm{a.s.}$

Equations (3.22) and (3.2) imply that

$\int_{\mathbb{R}^{d}}X_{t}(\mathrm{d}x)f(x)=\int_{\mathbb{R}^{d}}\eta^{ac}_{t}(x)f(x)\mathrm{d}x,\>P-\mathrm{a.s.}\>\textnormal{on}\>\left\\{t<T(1)\right\\}.$

Since $f\in\overset{\circ}{C}\mbox{}^{+}_{b}(\mathbb{R}^{d})$ was arbitrary,
this completes the proof of the theorem. ∎

##  4\. Proof of Theorem 2.2

Many steps in the proof follow the lines from [1]. However since the initial
conditions are measures, modifications are required.

###  4.1. Existence of solutions

We now prove the existence of a solution to equation (2.2) by the Picard
iterations. Let $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$, and

(4.1)
$w(x,t,\mu)=S_{t}\mu+\int_{0}^{t}(S_{t-s}\Psi(w(s,\cdot,\mu)))(x)\mathrm{d}s,\>x\in\mathbb{R}^{d},\>t>0,$

be an integral evolution equation such that $\Psi$ is some non-negative
function defined on $\mathbb{R}_{+}$. Recall that the Picard iterations for
this equation are defined by induction as follows:

(4.2) $\begin{split}w_{1}(x,t,\mu)&=(S_{t}\mu)(x),\\\
w_{n+1}(x,t,\mu)&=(S_{t}\mu)(x)+\int_{0}^{t}(S_{t-s}\Psi(w_{n}(s,\cdot,\mu)))(x)\mathrm{d}s,\>x\in\mathbb{R}^{d},\>t>0\\\
&\hskip 71.13188ptn=1,2,\ldots\end{split}$

Notice that (2.2) is a particular case of (4.1) with
$\Psi(\lambda)=\lambda^{\gamma}$. It is obvious that for $0<t<\infty$, the
Picard iterations (4.2) form the non-decreasing sequence: $w_{n}(x,t,\mu)\leq
w_{n+1}(x,t,\mu)$. In the next lemma we derive some properties of the Picard
iterations.

######  Lemma 4.1.

Let $\\{v_{n}(x,t,\mu)\\}_{n=1}^{\infty}$ be a sequence of Piccard iterations
corresponding to (2.2). Then for every $n=1,2,\ldots$ and any
$0<t<\infty,x\in\mathbb{R}^{d}$, the following inequalities hold:

  * $(1)$

$0\leq v_{n}(t,x,\mu)\leq e^{t}(S_{t}\mu)(x)+e^{t}$,

  * $(2)$

$v^{\gamma}_{n}(t,x,\mu)\leq e^{t}(S_{t}\mu)(x)+e^{t}$.

###### Proof.

First note that $v_{n},n\geq 1,$ are non-negative by construction. Let $\mu\in
M_{F}(\mathbb{R}^{d})$ and let us consider a linear integral equation

(4.3)
$u(t,x,\mu)=(S_{t}\mu)(x)+\int_{0}^{t}\big{(}S_{t-s}(u(s,\cdot,\mu)+1)\big{)}(x)\mathrm{d}s,\>x\in\mathbb{R}^{d},t>0.$

Let $\\{u_{n}(x,t,\mu)\\}_{n=1}^{\infty}$ be corresponding Picard iterations.
Note that (4.3) is a particular case of (4.1) with $\Psi(\lambda)=\lambda+1$.
Since $\lambda^{\gamma}\leq\lambda+1$ for $\gamma\in(0,1)$, one can easily see
that $v_{n}(t,x,\mu)\leq u_{n}(t,x,\mu)$, for all $n\geq 1$.

On the other hand by direct calculations, one gets that for any
$t\in(0,\infty)$

$\lim_{n\to\infty}u_{n}(x,t,\mu)\nearrow
e^{t}(S_{t}\mu)(x)+e^{t}-1,\quad\text{as $n\to\infty$,}$

and the first inequality of the lemma follows. The second inequality is a
consequence of the first one and the inequality
$\lambda^{\gamma}\leq\lambda+1$. ∎

######  Proposition 4.2 (Existence).

Let $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$. Then the integral
equation (2.2) has a solution $v(\cdot,\cdot)$ which is a limit of Picard
iterations and, for any $\phi\in L^{1}(\mathbb{R}^{d})\cap
C_{b}(\mathbb{R}^{d})$

(4.4) $\lim_{t\to
0}\int_{\mathbb{R}^{d}}v(t,x)\phi(x)\mathrm{d}x=\int_{\mathbb{R}^{d}}\phi(x)\mu(\mathrm{d}x).$

Moreover $v(\cdot,\cdot)$ satisfies the following inequalities for any
$(t,x)\in(0,\infty)\times\mathbb{R}^{d}$:

(4.5) $\displaystyle v(t,x)$ $\displaystyle\leq e^{t}(S_{t}\mu)(x)+e^{t},$
(4.6) $\displaystyle v^{\gamma}(t,x)$ $\displaystyle\leq
e^{t}(S_{t}\mu)(x)+e^{t}.$

###### Proof.

Let $\\{v_{n}(t,x)\\}_{n=1}^{\infty}$ be a sequence of Picard iterations
corresponding to equation (2.2). By the previous discussion for any
$t\in(0,\infty)$, $\\{v_{n}(t,\cdot)\\}_{n=1}^{\infty}$ form a non-decreasing
sequence and by Lemma 4.1 we have

(4.7) $v_{n}(t,x)\leq e^{t}(S_{t}\mu)(x)+e^{t}.$

Lemma 2.4 tells us that for every $t>0$, $(S_{t}\mu)(\cdot)$ is bounded. Thus,
for any $(t,x)$ in $(0,\infty)\times\mathbb{R}^{d}$, the sequence
$\\{v_{n}(t,x)\\}_{n=1}^{\infty}$ is non-decreasing and bounded. Consequently
there exists a bounded limit $v(t,x)=\lim_{n\to\infty}v_{n}(t,x)$.
Inequalities (4.5) and (4.6) follow from existence of the limit and (4.7).

Now consider the sequence of equations which defines the Picard iterations:

(4.8)
$\begin{split}&v_{l+1}(x,t)=\int_{\mathbb{R}^{d}}p_{t-s}(x-y)\mu(\mathrm{d}y)+\int_{0}^{t}\int_{\mathbb{R}^{d}}p_{t}(x-y)v^{\gamma}_{l}(s,y)\mathrm{d}y\mathrm{d}s,\\\
&\hskip 170.71652ptl=1,2,\ldots.\end{split}$

We have already proved that the left side of (4.8) converges boundedly
pointwise to $v(t,x)$. From the monotone convergence theorem it follows that
the right side converges to

$\int_{\mathbb{R}^{d}}p_{t-s}(x-y)\mu(\mathrm{d}y)+\int_{0}^{t}\int_{\mathbb{R}^{d}}p_{t}(x-y)v^{\gamma}(s,y))\mathrm{d}y\mathrm{d}s.$

Thus $v(t,x)$ satisfies equation (2.2) for all $x\in\mathbb{R}^{d},\>t>0$.

Now let us verify (4.4). In the following discussion we can assume without
loss of generality that $\phi\in C^{+}_{b}(\mathbb{R}^{d})\cap
L^{1}(\mathbb{R}^{d})$. Since the family of functions
$\\{p_{t}(\cdot)\\}_{t>0}$ builds up the _Dirac family_ (see [10, pp. 284-287,
348]), we can easily conclude that

(4.9) $\begin{split}\lim_{t\searrow 0}\langle
S_{t}\mu,\phi\rangle&=\langle\mu,\phi\rangle.\end{split}$

Now let us prove

(4.10) $\lim_{t\to
0}\int_{\mathbb{R}^{d}}\left(\int_{0}^{t}\left(S_{t-s}v^{\gamma}(s,\cdot))(x)\right)\mathrm{d}s\right)\phi(x)\mathrm{d}x=0.$

First, using the inequality (4.5) we obtain

$\begin{split}\int_{0}^{t}(S_{t-s}v^{\gamma}(s))(x)\mathrm{d}s&\leq\int_{0}^{t}\big{(}S_{t-s}\left(e^{s}S_{s}\mu+e^{s}\right)\mathrm{d}s\\\
&=(e^{t}-1)(S_{t}\mu)(x)+(e^{t}-1).\end{split}$

Using the above bound we get:

$\begin{split}&\lim_{t\searrow
0}\int_{\mathbb{R}^{d}}\left(\int_{0}^{t}(S_{t-s}v^{\gamma}(s))(x)\mathrm{d}s\right)\phi(x)\mathrm{d}x\\\
&\leq\lim_{t\searrow
0}\int_{\mathbb{R}^{d}}\left((e^{t}-1)(S_{t}\mu)(x)+(e^{t}-1)\right)\phi(x)\mathrm{d}x\\\
&=\lim_{t\searrow
0}(e^{t}-1)(\mu(\mathbb{R}^{d})\left\|\phi\right\|_{\infty}+\|\phi\|_{1})=0,\end{split}$

and (4.10) follows. Equations (4.10) and (4.9) imply (4.4) and this completes
the proof of the proposition. ∎

######  Corollary 4.3.

Let $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$ and let
$v(\cdot,\cdot,\mu)$ be a solution of (2.2) obtained as a limit of the Picard
iterations in Proposition 4.2. Then $v(\cdot,\cdot,\mu)\in
L_{loc}^{\infty}\left(((0,\infty),L^{1,w}_{+}(\mathbb{R}^{d})\right)$.

###### Proof.

Let $v(\cdot,\cdot,\mu)$ be a solution constructed in Proposition 4.2. Using
the bound (4.5), it is easy to derive the result by standard Gaussian bounds.
∎

###  4.2. Continuity of solutions

In this section we will prove the continuity of the solution obtained in
Proposition 4.2.

We start with the technical lemma, whose proof is pretty standard, and
therefore it is omitted.

######  Lemma 4.4.

Fix $0<T_{1}<T_{2}$ and $r>0$. Let
$\\{p_{s}(\cdot+z),\>s\in\left[T_{1},T_{2}\right],\>|z|\leq r\\}$ be a family
of functions, where $p_{s}(\cdot)$ is a standard Gaussian kernel on
$\mathbb{R}^{d}$. Then, there exists a constant $K$, such that

$p_{s}(x+z)\leq Kp_{2T_{2}}(x),\>\forall s\in[T_{1},T_{2}],\>|z|\leq
r,x\in\mathbb{R}^{d}.$

Now we are ready to state and prove the main proposition of Section 4.2.

######  Proposition 4.5 (Continuity).

Let $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$ and let $v(\cdot,\cdot)$
be a solution of (2.2) obtained as a limit of Picard iterations in Proposition
4.2. Then $v(\cdot,\cdot)\in C^{+}((0,\infty)\times\mathbb{R}^{d})$.

###### Proof.

By construction the solution is clearly non-negative. Now, let us fix a point
$(t,x)\in(0,\infty)\times\mathbb{R}^{d}$ and an arbitrary $\epsilon>0$. Let
$\delta_{i}>0,i=1,2,3,\>\,\delta_{1}<\delta_{3}<t/10$. In what follows we will
show that $\delta_{1}$, $\delta_{2}$ and $\delta_{3}$ can be chosen
sufficiently small so that if $|\Delta t|<\delta_{1}$ and $|\Delta
x|<\delta_{2}$ then

(4.11) $\displaystyle\left|v(t+\Delta t,x+\Delta
x)-v(t,x)\right|\leq\epsilon.$

We will bound absolute value of difference $v(t+\Delta t,x+\Delta x)-v(t,x)$
only for the case of $\Delta t\geq 0$, since the case of $\Delta t<0$ can be
treated similarly. We split the difference $v(t+\Delta t,x+\Delta x)-v(t,x)$
as follows:

$\displaystyle v(t+\Delta t,x+\Delta x)-v(x,t)$ $\displaystyle=I(\Delta
t,\Delta x)+J_{1}(\Delta t,\Delta x)-J_{2}(\Delta t,\Delta x)+J_{3}(\Delta
t,\Delta x)$ $\displaystyle+J_{4}(\Delta t,\Delta x)-J_{5}(\Delta t,\Delta
x).$

Here

$\displaystyle I(\Delta t,\Delta x)$ $\displaystyle=(S_{t+\Delta
t}\mu)(x+\Delta x)-(S_{t}\mu)(x),$ $\displaystyle J_{1}(\Delta t,\Delta x)$
$\displaystyle=\int_{t-\delta_{3}}^{t+\Delta t}\big{(}S_{t+\Delta
t-s}v^{\gamma}(s)\big{)}(x+\Delta x)\mathrm{d}s,$ $\displaystyle J_{2}(\Delta
t,\Delta x)$
$\displaystyle=\int_{t-\delta_{3}}^{t}\big{(}S_{t-s}v^{\gamma}(s)\big{)}(x)\mathrm{d}s,$
$\displaystyle J_{3}(\Delta t,\Delta x)$
$\displaystyle=\int_{\delta_{3}}^{t-\delta_{3}}\big{(}S_{t+\Delta
t-s}v^{\gamma}(s)\big{)}(x+\Delta
x)\mathrm{d}s-\int_{\delta_{3}}^{t-\delta_{3}}\big{(}S_{t-s}v^{\gamma}(s)\big{)}(x)\mathrm{d}s,$
$\displaystyle J_{4}(\Delta t,\Delta x)$
$\displaystyle=\int_{0}^{\delta_{3}}\big{(}S_{t+\Delta
t-s}v^{\gamma}(s)\big{)}(x+\Delta x)\mathrm{d}s,$ $\displaystyle J_{5}(\Delta
t,\Delta x)$
$\displaystyle=\int_{0}^{\delta_{3}}\big{(}S_{t-s}v^{\gamma}(s)\big{)}(x)\mathrm{d}s.$

Note that the integrals $J_{1},J_{2},J_{4}$ and $J_{5}$ have the same form:

(4.12)
$J_{*}=\int_{t_{1}}^{t_{2}}\big{(}S_{t_{3}-s}v^{\gamma}(s)\big{)}(z)\mathrm{d}s,$

for appropriate $t_{1},t_{2},t_{3}\geq 0$ and $z\in\mathbb{R}^{d}$. From the
definitions of $\delta_{1}$, $\delta_{3}$ and $\Delta t$, it follows that
$t_{1},t_{2}$ and $t_{3}$ in (4.12) can vary but satisfy the inequalities
$t_{1}<t_{2}$, $t\leq t_{3}$ and $t_{2}-t_{1}\leq 2\delta_{3}$ hold. Let us
bound $J_{*}$ from above. By Lemma 2.4 and Proposion 4.2, we easily get

(4.13)
$\begin{split}J_{*}&=\int_{t_{1}}^{t_{2}}\big{(}S_{t_{3}-s}v^{\gamma}(s)\big{)}(z)\mathrm{d}s\\\
&\leq\int_{t_{1}}^{t_{2}}\left(S_{t_{3}-s}(e^{s}(S_{s}\mu+1))\right)(z)\mathrm{d}s\\\
&=\int_{t_{1}}^{t_{2}}e^{s}\left(\left(S_{t_{3}}\mu\right)(z)+1\right)\mathrm{d}s\\\
&=(e^{t_{2}}-e^{t_{1}})\left(\left(S_{t_{3}}\mu\right)(z)+1\right)\\\
&\leq\frac{(e^{t_{2}}-e^{t_{1}})\left(\mu(\mathbb{R}^{d})+1\right)}{(2\pi
t_{3})^{d/2}}\\\
&\leq\frac{(e^{t_{2}}-e^{t_{1}})\left(\mu(\mathbb{R}^{d})+1\right)}{(2\pi
t)^{d/2}}\end{split}$

where the last inequality follows from $t\leq t_{3}$. Recall that
$t_{2}-t_{1}\leq 2\delta_{3}$, and so by (4.13) we can choose $\delta_{3}$
sufficiently small so that, for $i=1,2,4,5$

(4.14) $\displaystyle J_{i}(\Delta t,\Delta x)\leq\epsilon/10,\>\Delta
t<\delta_{1}<\delta_{3}.$

Let us fix such $\delta_{3}$. Let us recall that $\Delta
t<\delta_{1}<\delta_{3}$. Now we will handle $J_{3}(\Delta t,\Delta x)$. Write
$J_{3}$ as $J_{3}(\Delta t,\Delta x)=J_{31}(\Delta t,\Delta x)-J_{32}$, where

(4.15) $\displaystyle J_{31}(\Delta t,\Delta x)$
$\displaystyle=\int_{\delta_{3}}^{t-\delta_{3}}\int_{\mathbb{R}^{d}}p_{t+\Delta
t-s}(x+\Delta x-y)v^{\gamma}(s,y)\mathrm{d}y\mathrm{d}s,$ (4.16)
$\displaystyle J_{32}$
$\displaystyle=\int_{\delta_{3}}^{t-\delta_{3}}\int_{\mathbb{R}^{d}}p_{t-s}(x-y)v^{\gamma}(s,y)\mathrm{d}y\mathrm{d}s.$

By Lemma 4.4 and Proposition 4.2, we immediately get that there exists
$K=K(\delta_{1},\delta_{1},\delta_{3},t)$ such that

$\displaystyle p_{t+\Delta t-s}(x+\Delta x-y)v^{\gamma}(s,y)$
$\displaystyle\leq Kp_{2t}(x-y)e^{s}((S_{s}\mu)(y)+1)$
$\displaystyle\forall\Delta
t\in(0,\delta_{1}),s\in(\delta_{3},t-\delta_{3}),|\Delta x|<\delta_{2}.$

It is easy to verify that

$\displaystyle\int_{\delta_{3}}^{t-\delta_{3}}\int_{\mathbb{R}^{d}}Kp_{2t}(x-y)e^{s}\left(\left(S_{s}\mu\right)(y)+1\right)\mathrm{d}y\mathrm{d}s<\infty.$

Therefore we can use the dominated convergence theorem and obtain:

(4.17) $\begin{split}\lim_{\begin{subarray}{c}\Delta t\rightarrow 0\\\ \Delta
x\rightarrow 0\end{subarray}}J_{31}(\Delta t,\Delta
x)&=\lim_{\begin{subarray}{c}\Delta t\longrightarrow 0\\\ \Delta
x\longrightarrow
0\end{subarray}}\int_{\delta_{3}}^{t-\delta_{3}}\int_{\mathbb{R}^{d}}p_{t+\Delta
t-s}(x+\Delta x-y)v^{\gamma}(s,y)\mathrm{d}y\mathrm{d}s\\\
&=\int_{\delta_{3}}^{t-\delta_{3}}\int_{\mathbb{R}^{d}}p_{t-s}(x-y)v^{\gamma}(s,y)\mathrm{d}y\mathrm{d}s\\\
&=J_{32}.\end{split}$

Similarly we show

(4.18) $\lim_{\begin{subarray}{c}\Delta t\longrightarrow 0\\\ \Delta
x\longrightarrow 0\end{subarray}}(S_{t+\Delta t}\mu)(x+\Delta
x)=(S_{t}\mu)(x),$

and thus from (4.17) , (4.18) and the definition of $I(\Delta t,\Delta
x),J_{3}(\Delta t,\Delta x)$ we get

(4.19) $\lim_{\begin{subarray}{c}\Delta t\longrightarrow 0\\\ \Delta
x\longrightarrow 0\end{subarray}}I(\Delta t,\Delta x)+J_{3}(\Delta t,\Delta
x)=0.$

This implies that there exist $\delta_{1},\delta_{2}\in(0,\delta_{3})$
sufficiently small such that for $|\Delta t|<\delta_{1}$ and $|\Delta
x|<\delta_{2}$

(4.20) $\displaystyle\left|J_{3}(\Delta t,\Delta x)\right|+\left|J_{3}(\Delta
t,\Delta x)\right|\leq\epsilon/2.$

Thus we get from (4.14) and (4.20) that

$\displaystyle\left|v(t+\Delta t,x+\Delta x)-v(t,x)\right|$
$\displaystyle=|I(t+\Delta t,x+\Delta x)+J_{1}(t+\Delta t,x+\Delta
x)-J_{2}(t+\Delta t,x+\Delta x)$ $\displaystyle\quad\>+J_{3}(t+\Delta
t,x+\Delta x)+J_{4}(t+\Delta t,x+\Delta x)-J_{5}(t+\Delta t,x+\Delta x)|$
$\displaystyle\leq\left|I(t+\Delta t,x+\Delta x)\right|+\left|J_{1}(t+\Delta
t,x+\Delta x)\right|+\left|J_{2}(t+\Delta t,x+\Delta x)\right|$
$\displaystyle\quad\>+\left|J_{3}(t+\Delta t,x+\Delta
x)\right|+\left|J_{4}(t+\Delta t,x+\Delta x)\right|+\left|J_{5}(t+\Delta
t,x+\Delta x)\right|$ $\displaystyle\leq\epsilon,\>\forall\Delta t,\Delta
x:\>\Delta t\in(0,\delta_{1}),|\Delta x|<\delta_{2}.$

Since $\epsilon>0$ was arbitrary we are done. ∎

###  4.3. Uniqueness of solutions

The proof of uniqueness is again based on proofs in [1] which are adjusted to
our case. Let us recall that $\gamma^{\prime}=1/(1-\gamma).$

We start with the technical lemma, whose proof is pretty standard, and
therefore it is omitted.

######  Lemma 4.6.

Let $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$. Then, for every $t>0$,
there exist two positive constants $c(t)$, $a(t)$ such that following
inequality holds:

$(S_{t}\mu)(x)\geq c(t)\cdot e^{-a(t)|x|^{2}},\>\forall x\in\mathbb{R}^{d}.$

In the next lemma we proof an important lower bound.

######  Lemma 4.7.

Let $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$ and $u(t,x)$ be a non-
negative function on $(0,\infty)\times\mathbb{R}^{d}$ such that, for any
$t\in(0,\infty)$ and any $x\in\mathbb{R}^{d}:$

$v(t,x)\geq
S_{t}\mu+\int_{0}^{t}\left(S_{t-s}v^{\gamma}(s)\right)(x)\mathrm{d}s.$

Then

(4.21) $\displaystyle
v(t,x)>\left(\left(1-\gamma\right)t\right)^{\gamma^{\prime}},\>\forall(t,x)\in(0,\infty)\times\mathbb{R}^{d}.$

###### Proof.

Let us fix an arbitrary $t_{0}>0$ and define

$\displaystyle\tilde{v}(t)\equiv v(t+t_{0}),\>\forall t\geq 0.$

Using this definition one can easily check that

$\displaystyle\tilde{v}(t)$ $\displaystyle\geq
S_{t}\tilde{v}_{0}+\int_{0}^{t}S_{t-s}\tilde{v}^{\gamma}(s)\mathrm{d}s,\quad
t\geq 0,$

where $\tilde{v}_{0}=\tilde{v}(0)\geq S_{t_{0}}\mu$, and the last inequality
follows by definition of $\tilde{v}$ and assumptions on $v$. Thus, by Lemma
4.6 there exists $a(t_{0})>0$ and $c(t_{0})>0$ such that

$\displaystyle\tilde{v}_{0}(x)\geq\left(S_{t_{0}}\mu\right)(x)\geq
c(t_{0})e^{-a(t_{0})|x|^{2}},\>\forall x\in\mathbb{R}^{d}.$

By Lemma $2.2$ in [1] we have

$\tilde{v}(t,x)\geq((1-\gamma)t)^{\gamma^{\prime}},\forall(t,x)\in(0,\infty)\times\mathbb{R}^{d}.$

Since $t_{0}\geq 0$ was arbitrary, we get

$\displaystyle
v(t,x)\geq((1-\gamma)t)^{\gamma^{\prime}},\forall(t,x)\in(0,\infty)\times\mathbb{R}^{d},$

and we are done. ∎

######  Lemma 4.8 (Comparison lemma).

Let

$v,u\in L_{loc}^{\infty}\left((0,\infty),L^{1,w}(\mathbb{R}^{d})\right)\cap
C((0,\infty)\times\mathbb{R}^{d})$

be non-negative functions such that, for all $t>0$,

$\begin{split}u(t)\geq
S_{t}\nu+\int_{0}^{t}S_{t-s}u^{\gamma}(s)\mathrm{d}s,\\\ v(t)\leq
S_{t}\mu+\int_{0}^{t}S_{t-s}v^{\gamma}(s)\mathrm{d}s.\end{split}$

Here $\mu$, $\nu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$ are such that

$\nu(f)\geq\mu(f),\forall f\in C^{+}_{b}(\mathbb{R}^{d}).$

Then

$u(t,x)\geq v(t,x),\>\>\textnormal{for
all}\>(t,x)\in(0,\infty)\times\mathbb{R}^{d}.$

###### Proof.

Define

$g(t)\equiv v(t)-u(t).$

We will now prove that $g_{+}(t)\equiv\max(g(t),0)=0$.

Fix arbitrary $T>0$. We use the condition $\nu\geq\mu$ and an elementary
inequality $(a^{\gamma}-b^{\gamma})\leq((a-b)_{+})^{\gamma}$ to get

$\displaystyle g(t)$ $\displaystyle\leq
S_{t}(\mu-\nu)+\int_{0}^{t}S_{t-s}(v^{\gamma}(s)-u^{\gamma}(s))\mathrm{d}s$
$\displaystyle\leq$
$\displaystyle\int_{0}^{t}S_{t-s}(v^{\gamma}(s)-u^{\gamma}(s))_{+}\mathrm{d}s$
(4.22) $\displaystyle\leq$
$\displaystyle\int_{0}^{t}S_{t-s}\left(\left(g_{+}(s)\right)^{\gamma}\right)\mathrm{d}s.$

From this point the proof follows the proof of Theorem 2.8 in [1] while using
Lemma 4.7 whenever necessary. We left the details to the reader. ∎

The uniqueness for (2.2) follows easily from the above comparison Lemma 4.8.

######  Proposition 4.9 (Uniqueness).

Let $\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$. There is at most one
solution to (2.2) which belongs to
$L_{loc}^{\infty}\left([0,\infty),L^{1,w}_{+}(\mathbb{R}^{d})\right)\cap
C^{+}((0,\infty)\times\mathbb{R}^{d})$.

###### Proof.

Suppose there exist two functions $v,u\in
L_{loc}^{\infty}\left([0,\infty),L^{1,w}_{+}(\mathbb{R}^{d})\right)\cap
C^{+}((0,\infty)\times\mathbb{R}^{d})$ that solve equation (2.2) for the same
initial measure $\mu$. Then by Lemma 4.8, $v(t,x)\geq u(t,x)$ and $u(t,x)\geq
v(t,x)$ for any $(t,x)\in(0,\infty)\times\mathbb{R}^{d}$, and thus $u=v$ and
we are done. ∎

###  4.4. Continuous dependence of solutions on initial data

In the previous sections we proved the existence and uniqueness of solutions
to equation (2.2), or looking from different perspective we proved for every
$(t,x)\in(0,\infty)\times\mathbb{R}^{d}$ the existence of the mapping

$v(t,x,\cdot):{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})\to\mathbb{R}_{++}.$

Here $v(t,x,\mu)$ is a solution to equation (2.2) with initial datum $\mu$.
In this section we will prove the continuity of this mapping.

######  Lemma 4.10.

For any $(t,x)\in(0,\infty)\times\mathbb{R}^{d}$, the mapping

$v(t,x,\cdot):{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})\mapsto\mathbb{R}_{++}$

is concave, i.e.

$\begin{split}v(t,x,\lambda\mu+(1-\lambda)\nu)\geq&\lambda
v(t,x,\mu)+(1-\lambda)v(t,x,\nu),\\\
&\forall\lambda\in(0,1),\>\>\forall(t,x)\in(0,\infty)\times\mathbb{R}^{d}.\end{split}$

###### Proof.

Since the function $x\to x^{\gamma}$ is concave, for any positive $a,b$ and
$\lambda\in(0,1)$, we have

(4.23) $\lambda a^{\gamma}+(1-\lambda)b^{\gamma}\leq(\lambda
a+(1-\lambda)b)^{\gamma}.$

Let us fix an arbitrary $\lambda\in(0,1)$ and define $u(t,z,\mu,\nu,\lambda)$
as follows

$u(t,x,\mu,\nu,\lambda)\triangleq\lambda v(t,\mu)+(1-\lambda)v(t,\nu).$

Then we have

$\begin{split}u(t,x,\mu,\nu,\lambda)&=\lambda
v(t,x,\mu)+(1-\lambda)v(t,x,\nu)\\\
&=\big{(}S_{t}\sigma\big{)}(x)+\int_{0}^{t}\Big{(}S_{t-s}\Big{(}\lambda
v^{\gamma}(s,\mu)+(1-\lambda)v^{\gamma}(s,\nu)\Big{)}\Big{)}(x)\mathrm{d}s\\\
&\leq\big{(}S_{t}\sigma\big{)}(x)+\int_{0}^{t}\Big{(}S_{t-s}\Big{(}\lambda
v(s,\mu)+(1-\lambda)v(s,\nu)\Big{)}^{\gamma}\Big{)}(x)\mathrm{d}s\\\
&=\big{(}S_{t}\sigma\big{)}(x)+\int_{0}^{t}\Big{(}S_{t-s}(u^{\gamma}(s,\mu,nu,\lambda)\Big{)}(x)\mathrm{d}s,\end{split}$

where the above inequality follows from (4.23), and we set
$\sigma=\lambda\mu+(1-\lambda)\nu$. Hence we obtained

(4.24)
$u(t,x)\leq\left(S_{t}\sigma\right)(x)+\int_{0}^{t}\left(S_{t-s}u^{\gamma}(s)\right)(x)\mathrm{d}s.$

We now recall that by definition

(4.25)
$v(t,x,\sigma)=\left(S_{t}\sigma\right)(x)+\int_{0}^{t}\left(S_{t-s}v(s,\sigma)^{\gamma}\right)(x)\mathrm{d}s.$

and it is left to use comparison Lemma 4.8. ∎

Before we prove continuity let us recall that the space of signed finite
measures on $\mathbb{R}^{d}$, $M_{F,S}(\mathbb{R}^{d})$, is a topological
vector space with topology of weak convergence.

######  Proposition 4.11.

For any fixed $(t,x)\in(0,\infty)\times\mathbb{R}^{d}$, the mapping

$v(t,x,\cdot):{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})\mapsto\mathbb{R}_{++}$

is continuous.

######  Remark 4.12.

It follows from the above proposition that the weak convergence of initial
measures implies pointwise convergence of solutions to equation (2.2).

###### Proof.

By Lemma 4.10 for any $(t,x)\in(0,\infty)\times\mathbb{R}^{d}$, the mapping
$\mu\mapsto v(t,x,\cdot)$ is concave, and $v(t,x,\mu)\geq 0$ for any
$\mu\in{\overset{\circ}{M}}_{F}(\mathbb{R}^{d})$. Hence, by Lemma $2.1$ in
[4], mapping $\mu\mapsto v(t,x,\mu)$ is continuous. ∎

Now we are able to prove Theorem 2.2.

###### Proof of Theorem 2.2.

The statement of the theorem follows from Propositions 4.2, 4.5, 4.9,
Corollary 4.3 and Proposition 4.11. ∎

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