# Parabolic Littlewood-Paley inequality for $\phi(-\Delta)$-type operators and
applications to Stochastic integro-differential equations

Ildoo Kim  Department of Mathematics, Korea University, 1 Anam-dong, Sungbuk-
gu, Seoul, 136-701, Republic of Korea<EMAIL_ADDRESS>,  Kyeong-Hun Kim
Department of Mathematics, Korea University, 1 Anam-dong, Sungbuk-gu, Seoul,
136-701, Republic of Korea<EMAIL_ADDRESS>and  Panki Kim  Department
of Mathematical Sciences and Research Institute of Mathematics, Seoul National
University, Building 27, 1 Gwanak-ro, Gwanak-gu Seoul 151-747, Republic of
Korea<EMAIL_ADDRESS>
###### Abstract.

In this paper we prove a parabolic version of the Littlewood-Paley inequality
(1.3) for the operators of the type $\phi(-\Delta)$, where $\phi$ is a
Bernstein function. As an application, we construct an $L_{p}$-theory for the
stochastic integro-differential equations of the type
$du=(-\phi(-\Delta)u+f)\,dt+g\,dW_{t}$.

###### Key words and phrases:

Parabolic Littlewood-Paley inequality, Stochastic partial differential
equations, Integro-differential operators, Lévy processes, Estimates of
transition functions

###### 2010 Mathematics Subject Classification:

42B25, 26D10, 60H15, 60G51, 60J35

The research of the second author was supported by Basic Science Research
Program through the National Research Foundation of Korea(NRF) funded by the
Ministry of Education, Science and Technology (20120005158)

The research of the third author was supported by Basic Science Research
Program through the National Research Foundation of Korea(NRF) grant funded by
the Korea government(MEST)(2011-000251)

##  1\. Introduction

The operators we are considering in this article are certain functions of the
Laplacian. To be more precise, recall that a function
$\phi:(0,\infty)\to(0,\infty)$ such that $\phi(0+)=0$ is called a Bernstein
function if it is of the form

$\phi(\lambda)=b\lambda+\int_{(0,\infty)}(1-e^{-\lambda
t})\,\mu(dt)\,,\quad\lambda>0\,,$

where $b\geq 0$ and $\mu$ is a measure on $(0,\infty)$ satisfying
$\int_{(0,\infty)}(1\wedge t)\,\mu(dt)<\infty$, called the Lévy measure. By
Bochner’s functional calculus, one can define the operator
$\phi(\Delta):=-\phi(-\Delta)$ on $C_{b}^{2}({\mathbb{R}}^{d})$, which turns
out to be an integro-differential operator

$b\Delta f(x)+\int_{{\mathbb{R}}^{d}}\left(f(x+y)-f(x)-\nabla f(x)\cdot
y{\mathbf{1}}_{\\{|y|\leq 1\\}}\right)\,J(y)\,dy\,,$ (1.1)

where $J(x)=j(|x|)$ with $j:(0,\infty)\to(0,\infty)$ given by

$j(r)=\int_{0}^{\infty}(4\pi t)^{-d/2}e^{-r^{2}/(4t)}\,\mu(dt)\,.$

It is also known that the operator $\phi(\Delta)$ is the infinitesimal
generator of the $d$-dimensional subordinate Brownian motion. Let
$S=(S_{t})_{t\geq 0}$ be a subordinator (i.e. an increasing Lévy process
satisfying $S_{0}=0$) with Laplace exponent $\phi$, and let $W=(W_{t})_{t\geq
0}$ be a Brownian motion in ${\mathbb{R}}^{d}$, $d\geq 1$, independent of $S$
with
${\mathbb{E}}_{x}\left[e^{i\xi(W_{t}-W_{0})}\right]=e^{-t{|\xi|^{2}}},\xi\in{\mathbb{R}}^{d},t>0$.
Then $X_{t}:=W_{S_{t}}$, called the subordinate Brownian motion, is a
rotationally invariant Lévy process in ${\mathbb{R}}^{d}$ with characteristic
exponent $\phi(|\xi|^{2})$, and for any $f\in C^{2}_{b}(\mathbb{R}^{d})$

$\phi(\Delta)f(x)=\lim_{t\to 0}\frac{1}{t}[\mathbb{E}_{x}f(X_{t})-f(x)].$
(1.2)

For instance, by taking $\phi(\lambda)=\lambda^{\alpha/2}$ with
$\alpha\in(0,2)$, we get the fractional laplacian
$\Delta^{\alpha/2}:=-(-\Delta)^{\alpha/2}$ which is the infinitesimal
generator of a rotationally symmetric $\alpha$-stable process in
${\mathbb{R}}^{d}$.

In this article we prove a parabolic Littlewood-Paley inequality for
$\phi(\Delta)$:

######  Theorem 1.1.

Let $\phi$ be a Bernstein function, $T_{t}$ be the semigroup corresponding to
$\phi(\Delta)$ and $H$ be a Hilbert space. Suppose that $\phi$ satisfies

(H1)$:$ $\exists$ constants $0<\delta_{1}\leq\delta_{2}<1$ and $a_{1},a_{2}>0$
such that

$a_{1}\lambda^{\delta_{1}}\phi(t)\leq\phi(\lambda t)\leq
a_{2}\lambda^{\delta_{2}}\phi(t),\quad\lambda\geq 1,t\geq 1\,;$

(H2)$:$ $\exists$ constants $0<\delta_{3}\leq 1$ and $a_{3}>0$ such that

$\phi(\lambda t)\leq a_{3}\lambda^{\delta_{3}}\phi(t),\quad\lambda\leq 1,t\leq
1\,.$

Then for any $p\in[2,\infty),\,T\in(0,\infty)$ and $f\in
C_{0}^{\infty}(\mathbb{R}^{d+1},H)$,

$\displaystyle\int_{\mathbb{R}^{d}}\int_{0}^{T}[\int_{0}^{t}|\phi(\Delta)^{1/2}T_{t-s}f(s,\cdot)(x)|_{H}^{2}ds]^{p/2}dtdx\leq
N\int_{\mathbb{R}^{d}}\int_{0}^{T}|f(t,x)|^{p}_{H}~{}dtdx,$ (1.3)

where the constant $N$ depends only on $d,p,T,a_{i}$ and $\delta_{i}$
$(i=1,2,3)$.

(H1) is a condition on the asymptotic behavior of $\phi$ at infinity and it
governs the behavior of the corresponding subordinate Brownian motion $X$ for
small time and small space. (H2) is a condition about the asymptotic behavior
of $\phi$ at zero and it governs the behavior of the corresponding subordinate
Brownian motion $X$ for large time and large space. Note that it follows from
the second inequality in (H1) that $\phi$ has no drift, i.e., $b=0$ in (1.1).
It also follows from (H2) that $\phi(0+)=0$.

Using the tables at the end of [20], one can construct a lot of explicit
examples of Bernstein functions satisfying (H1)–(H2). Here are a few of them:

  * (1)

$\phi(\lambda)=\lambda^{\alpha}+\lambda^{\beta}$, $0<\alpha<\beta<1$;

  * (2)

$\phi(\lambda)=(\lambda+\lambda^{\alpha})^{\beta}$, $\alpha,\beta\in(0,1)$;

  * (3)

$\phi(\lambda)=\lambda^{\alpha}(\log(1+\lambda))^{\beta}$, $\alpha\in(0,1)$,
$\beta\in(0,1-\alpha)$;

  * (4)

$\phi(\lambda)=\lambda^{\alpha}(\log(1+\lambda))^{-\beta}$, $\alpha\in(0,1)$,
$\beta\in(0,\alpha)$;

  * (5)

$\phi(\lambda)=(\log(\cosh(\sqrt{\lambda})))^{\alpha}$, $\alpha\in(0,1)$;

  * (6)

$\phi(\lambda)=(\log(\sinh(\sqrt{\lambda}))-\log\sqrt{\lambda})^{\alpha}$,
$\alpha\in(0,1)$.

For example, the subordinate Brownian motion corresponding to the example (1)
$\phi(\lambda)=\lambda^{\alpha}+\lambda^{\beta}$ is the sum of two independent
symmetric $\alpha$ and $\beta$ stable processes. In this case the
characteristic exponent is

$\Phi(\theta)=|\theta|^{\alpha}+|\theta|^{\beta}\,,\
\theta\in{\mathbb{R}}^{d}\,,\quad 0<\beta<\alpha<2\,,$

and its infinitesimal generator is
$-(-\Delta)^{\beta/2}-(-\Delta)^{\alpha/2}$.

We remark here that relativistic stable processes satisfy (H1)–(H2) with
$\delta_{3}=1$; Suppose that $\alpha\in(0,2)$, $m>0$ and define

$\phi_{m}(\lambda)=(\lambda+m^{2/\alpha})^{\alpha/2}-m.$

The subordinate Brownian motion corresponding to $\phi_{m}$ is a relativistic
$\alpha$-stable process on $\mathbb{R}^{d}$ with mass $m$ whose characteristic
function is given by

$\exp(-t((|\xi|^{2}+m^{2/\alpha})^{\alpha/2}-m)),\qquad\xi\in\mathbb{R}^{d}.$

The infinitesimal generator is $m-(m^{2/\alpha}-\Delta)^{\alpha/2}$.

Note that when $m=1$, this infinitesimal generator reduces to
$1-(1-\Delta)^{\alpha/2}$. Thus the $1$-resolvent kernel of the relativistic
$\alpha$-stable process with mass $1$ on $\mathbb{R}^{d}$ is just the Bessel
potential kernel. When $\alpha=1$, the infinitesimal generator reduces to the
so-called free relativistic Hamiltonian $m-\sqrt{-\Delta+m^{2}}$. The operator
$m-\sqrt{-\Delta+m^{2}}$ is very important in mathematical physics due to its
application to relativistic quantum mechanics. We emphasize that the present
article covers this case.

The parabolic Littlewood-Paley inequality (1.3) was first proved by Krylov
([14, 15]) for the case $\phi(\Delta)=\Delta$ with $N=N(p)$ depending only on
$p$. In this case, if $f$ depends only on $x$ and $H=\mathbb{R}$ then (1.3)
leads to the the classical (elliptic) Littlewood-Paley inequality (cf. [23]):

$\int_{\mathbb{R}^{d}}\left(\int^{\infty}_{0}|\nabla
T_{t}f|^{2}dt\right)^{p/2}dx\leq N(p)\|f\|^{p}_{p},\quad\quad\forall\,\,f\in
L_{p}(\mathbb{R}^{d}).$

Recently, (1.3) was proved for the fractional Laplacian $\Delta^{\alpha/2}$,
$\alpha\in(0,2)$, in [2, 8]. Also, in [17] similar result was proved for the
case $J=J(t,y)=m(t,y)|y|^{-d-\alpha}$ in (1.1), where $\alpha\in(0,2)$ and
$m(t,y)$ is a bounded smooth function satisfying $m(t,y)=m(t,y|y|^{-1})$ (i.e.
homogeneous of degree zero) and $m(t,y)>c>0$ on a set $\Gamma\subset S^{d-1}$
of a positive Lebesgue measure. We note that even the case
$\phi(\lambda)=\lambda^{\alpha}+\lambda^{\beta}$ ($\alpha\neq\beta$) is not
covered in [17].

Our motivation of studying (1.3) is that (1.3) is the key estimate for the
$L_{p}$-theory of the corresponding stochastic partial differential equations.
For example, Krylov’s result ([14, 15]) for $\Delta$ is related to the
$L_{p}$-theory of the second-order stochastic partial differential equations.
Below we briefly explain the reason for this. See [9, 16] or Section 6 of this
article for more details. Consider the stochastic integro-differential
equation

$du=(\phi(\Delta)u+h)\,dt+\sum_{k=1}^{\infty}f^{k}dw^{k}_{t},\quad u(0,x)=0.$
(1.4)

Here $f=(f^{1},f^{2},\cdots)$ is an $\ell_{2}$-valued random function of
$(t,x)$, and $w^{k}_{t}$ are independent one-dimensional Wiener processes
defined on a probability space $(\Omega,P)$. Considering $u-w$, where
$w(t):=\int^{t}_{0}T_{t-s}h(s)ds$, we may assume $h=0$ (see Section 6). It
turns out that if $f=(f^{1},f^{2},\cdots)$ satisfies certain measurability
condition, the solution of this problem is given by

$u(t,x)=\sum_{k=1}^{\infty}\int^{t}_{0}T_{t-s}f^{k}(s,\cdot)(x)dw^{k}_{s}.$

By Burkholder-Davis-Gundy inequality,

$\displaystyle\mathbb{E}\int^{T}_{0}\|\phi(\Delta)^{1/2}u(t,\cdot)\|^{p}_{L_{p}}dt$
(1.5) $\displaystyle\leq$ $\displaystyle
N(p)\,\mathbb{E}\int^{T}_{0}\int_{\mathbb{R}^{d}}\left[\int^{t}_{0}|\phi(\Delta)^{1/2}T_{t-s}f(s,\cdot)(x)|^{2}_{\ell_{2}}ds\right]^{p/2}dxdt.$

Actually if $f$ is not random, then $u$ becomes a Gaussian process and the
reverse inequality of (1.5) also holds. Thus to prove $\phi(\Delta)^{1/2}u\in
L_{p}$ and to get a legitimate start of the $L_{p}$-theory of equation (1.4),
one has to estimate the right-hand side of (1.5) (or the left-hand side of
(1.3)). We will also see that (1.3) yields the uniqueness and existence of
equation (1.4) in certain Banach spaces.

The key of our approach is estimating the sharp function $(v)^{\sharp}(t,x)$
of
$v(t,x):=[\int_{0}^{t}|\phi(\Delta)^{1/2}T_{t-s}f(s,\cdot)(x)|_{H}^{2}ds]^{1/2}$:

$(v)^{\sharp}(t,x):=\sup_{(t,x)\in\mathcal{Q}}-\int_{\mathcal{Q}}|v-v_{\mathcal{Q}}|dtdx,$
(1.6)

where $v_{\mathcal{Q}}:=-\int_{\mathcal{Q}}v\;dxdt$ is the average of $v$ over
$Q$ and the supremum is taken for all cubes $Q$ containing $(t,x)$ of the type
$\mathcal{Q}_{c}(r,y):=(r-\phi(c^{-2})^{-1},r+\phi(c^{-2})^{-1})\times
B_{c}(y)$. We control $(v)^{\sharp}(t,x)$ in terms of the maximal functions of
$|f|_{H}$, and then apply Fefferman-Stein and Hardy-Littlewood theorems to
prove (1.3). The operators considered in [14, 8, 17] have simple scaling
properties, and so to estimate the mean oscillation
$-\int_{\mathcal{Q}}|v-v_{\mathcal{Q}}|dtdx$ in (1.6), it was enough to
consider the only case $\mathcal{Q}=\mathcal{Q}_{1}(0,0)$, that is the case
$c=1$ and $(r,y)=(0,0)$. However, in our case, due to the lack of the scaling
property, it is needed to consider the mean oscillation
$-\int_{\mathcal{Q}}|v-v_{\mathcal{Q}}|dtdx$ on every $Q_{c}(r,y)$ containing
$(t,x)$. This causes serious difficulties as can be seen in the proofs of
Lemmas 5.2–5.5. Our estimation of $-\int_{\mathcal{Q}}|v-v_{\mathcal{Q}}|dtdx$
relies on the upper bounds of $\phi(\Delta)^{n/2}D^{\beta}p(t,x)$, which are
obtained in this article. Here $\beta$ is an arbitrary multi-index,
$n=0,1,2,\cdots$ and $p(t,x)$ is the density of the semigroup $T_{t}$
corresponding to $\phi(\Delta)$.

The article is organized as follows. In Section 2 we give upper bounds of the
density $p(t,x)$. Section 3 contains various properties of Bernstein functions
and subordinate Brownian motions. In Section 4 we establish upper bounds of
the fractional derivatives of $p(t,x)$ in terms of $\phi$. Using these
estimates we give the proof of of Theorem 1.1 in Section 5. In Section 6 we
apply Theorem 1.1 and construct an $L_{p}$-theory for equation (1.4).

We finish the introduction with some notation. As usual $\mathbb{R}^{d}$
stands for the Euclidean space of points $x=(x^{1},...,x^{d})$,
$B_{r}(x):=\\{y\in\mathbb{R}^{d}:|x-y|<r\\}$ and $B_{r}:=B_{r}(0)$. For
$i=1,...,d$, multi-indices $\beta=(\beta_{1},...,\beta_{d})$,
$\beta_{i}\in\\{0,1,2,...\\}$, and functions $u(x)$ we set

$u_{x^{i}}=\frac{\partial u}{\partial x^{i}}=D_{i}u,\quad
D^{\beta}u=D_{1}^{\beta_{1}}\cdot...\cdot
D^{\beta_{d}}_{d}u,\quad|\beta|=\beta_{1}+...+\beta_{d}.$

We write $u\in C^{\infty}_{0}(X,Y)$ if $u$ is a $Y$-valued infinitely
differentiable function defined on $X$ with compact support. By
$C^{2}_{b}(\mathbb{R}^{d})$ we denote the space of twice continuously
differentiable functions on $\mathbb{R}^{d}$ with bounded derivatives up to
order $2$. We use “$:=$” to denote a definition, which is read as “is defined
to be”. We denote $a\wedge b:=\min\\{a,b\\}$, $a\vee b:=\max\\{a,b\\}$. If we
write $N=N(a,\ldots,z)$, this means that the constant $N$ depends only on
$a,\ldots,z$. The constant $N$ may change from location to location, even
within a line. By $\mathcal{F}$ and $\mathcal{F}^{-1}$ we denote the Fourier
transform and the inverse Fourier transform, respectively. That is, for a
suitable function $f$,
$\mathcal{F}(f)(\xi):=\int_{\mathbb{R}^{d}}e^{-ix\cdot\xi}f(x)dx$ and
$\mathcal{F}^{-1}(f)(x):=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}e^{i\xi\cdot
x}f(\xi)d\xi$. Finally, for a Borel set $A\subset\mathbb{R}^{d}$, we use $|A|$
to denote its Lebesgue measure.

##  2\. Upper bounds of $p(t,x)$

In this section we give upper bounds of the density $p(t,x)$ of the semigroup
$T_{t}$ corresponding to $\phi(\Delta)$. We give the result under slightly
more general setting. We will assume that $Y$ is a rotationally symmetric Lévy
process with Lévy exponent $\Psi_{Y}(\xi)$. Because of rotational symmetry,
the function $\Psi_{Y}$ is positive and depends on $|\xi|$ only. Accordingly,
by a slight abuse of notation we write $\Psi_{Y}(\xi)=\Psi_{Y}(|\xi|)$ and get

${\mathbb{E}}_{x}\left[e^{i\xi\cdot(Y_{t}-Y_{0})}\right]=e^{-t\Psi_{Y}(|\xi|)},\quad\quad\mbox{
for every }x\in{\mathbb{R}}^{d}\mbox{ and }\xi\in{\mathbb{R}}^{d}.$ (2.1)

We assume that the transition probability $\mathbb{P}(Y_{t}\in dy)$ is
absolutely continuous with respect to Lebesgue measure in ${\mathbb{R}}^{d}$.
Thus there is a function $p_{Y}(t,r)$, $t>0,r\geq 0$ such that

$\mathbb{P}(Y_{t}\in dy)=p_{Y}(t,|y|)dy.$

Note that $r\to\Psi_{Y}(r)$ and $r\to p_{Y}(t,r)$ may not be monotone in
general. We first consider the following mild condition on $\Psi_{Y}$.

(A1): There exists a positive function $h$ on $[0,\infty)$ such that for every
$t,\lambda>0$

${\Psi_{Y}(\lambda t)}/{\Psi_{Y}(t)}\leq
h(\lambda)\quad\text{and}\quad\int_{0}^{\infty}e^{-r^{2}/2}r^{d-1}h(r)dr<\infty.$

Note that by Lemma 3.1 below, (A1) always holds with
$h(\lambda)=1\vee\lambda^{2}$ for every subordinate Brownian motion. Moreover,
by [7, Lemma 3 and Proposition 11], (A1) always holds with
$h(\lambda)=24(1+\lambda^{2})$ for rotationally symmetric unimodal Lévy
process (i.e., $r\to p_{Y}(t,r)$ is decreasing for all $t>0$).

Recall that

$e^{-|z|^{2}}=(4\pi)^{-d/2}\int_{{\mathbb{R}}^{d}}e^{i\xi\cdot
z}e^{-|\xi|^{2}/4}d\xi.$

Using this and (2.1) we have for $\lambda>0$

$\displaystyle{\mathbb{E}}_{0}[e^{-\lambda|Y_{t}|^{2}}]=(4\pi)^{-d/2}\int_{{\mathbb{R}}^{d}}{\mathbb{E}}_{0}[e^{i\sqrt{\lambda}\xi\cdot
Y_{t}}]e^{-|\xi|^{2}/4}d\xi$
$\displaystyle=(4\pi)^{-d/2}\int_{{\mathbb{R}}^{d}}e^{-t\Psi_{Y}(\sqrt{\lambda}|\xi|)}e^{-|\xi|^{2}/4}d\xi.$
(2.2)

Thus

$\displaystyle{\mathbb{E}}_{0}[e^{-\lambda|Y_{t}|^{2}}-e^{-2\lambda|Y_{t}|^{2}}]=(4\pi)^{-d/2}\int_{{\mathbb{R}}^{d}}(e^{-t\Psi_{Y}(\sqrt{\lambda}|\xi|)}-e^{-t\Psi_{Y}(\sqrt{2\lambda}|\xi|)})e^{-|\xi|^{2}/4}d\xi.$
(2.3)

For $t,\lambda>0$, let

$g_{t}(\lambda):=\int_{0}^{\infty}(e^{-t\Psi_{Y}(\sqrt{\lambda}r)}-e^{-t\Psi_{Y}(\sqrt{2\lambda}r)})e^{-r^{2}/4}r^{d-1}dr,$

which is positive by (2.3).

######  Lemma 2.1.

Suppose that (A1) holds. Then there exists a constant $N=N(h,d)$ such that for
every $t,v>0$

$g_{t}(v^{-1})\leq Nt\Psi_{Y}(v^{-1/2}).$

###### Proof.

By (A1) we have

$\displaystyle\frac{1}{t\Psi_{Y}(v^{-1/2})}\leq\frac{\Psi_{Y}(\sqrt{2}v^{-1/2}r)+\Psi_{Y}(v^{-1/2}r)}{\Psi_{Y}(v^{-1/2})}\frac{1}{t|\Psi_{Y}(\sqrt{2}v^{-1/2}r)-\Psi_{Y}(v^{-1/2}r)|}$
$\displaystyle\leq$
$\displaystyle(h(\sqrt{2}r)+h(r))\frac{1}{t|\Psi_{Y}(\sqrt{2}v^{-1/2}r)-\Psi_{Y}(v^{-1/2}r)|}.$

Thus using the inequality $|e^{-a}-e^{-b}|\leq|a-b|$, $a,b>0$

$\displaystyle\frac{g_{t}(v^{-1})}{t\Psi_{Y}(v^{-1/2})}$ $\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\frac{|e^{-t\Psi_{Y}(v^{-1/2}r)}-e^{-t\Psi_{Y}(\sqrt{2}v^{-1/2}r)}|}{t|\Psi_{Y}(\sqrt{2}v^{-1/2}r)-\Psi_{Y}(v^{-1/2}r)|}e^{-r^{2}/4}r^{d-1}(h(\sqrt{2}r)+h(r))dr$
$\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}e^{-r^{2}/4}r^{d-1}(h(\sqrt{2}r)+h(r))dr<\infty.$

Therefore the lemma is proved. $\Box$

Recall that $\mathbb{P}_{0}(Y_{t}\in dy)=p_{Y}(t,|y|)dy$. We now consider the
following mild condition on $p_{Y}(t,r)$.

(A2): For every $T\in(0,\infty]$, there exists a constant $c=c(T)>0$ such that
for every $t\in(0,T)$

$\displaystyle p_{Y}(t,r)\leq cp_{Y}(t,s)\qquad\forall r\geq s\geq 0.$ (2.4)

Obviously (2.4) always holds on all $t>0$ for rotationally symmetric unimodal
Lévy process.

######  Theorem 2.2.

Suppose that $Y$ is a rotationally symmetric Lévy process with Lévy exponent
$\Psi_{Y}(\xi)$ satisfying (A1). Assume that $\mathbb{P}(Y_{t}\in
dy)=p_{Y}(t,|y|)dy$ and (A2) holds. Then for every $T>0$, there exists a
constant $N=N(T,c,d,h)>0$ such that

$p_{Y}(t,r)\,\leq\,N\,t\,r^{-d}\Psi_{Y}(r^{-1}),\quad(t,r)\in(0,T]\times[0,\infty).$

###### Proof.

Fix $t\in(0,T]$. For $r\geq 0$ define $f_{t}(r)=r^{d/2}p_{Y}(t,r^{1/2}).$ By
(A2), for $r\geq 0$,

$\displaystyle\mathbb{P}_{0}(\sqrt{r/2}<|Y_{t}|<\sqrt{r})=\int_{\sqrt{r/2}<|y|<\sqrt{r}}p_{Y}(t,|y|)dy$
$\displaystyle\geq|B(0,1)|(1-2^{-d/2})c^{-1}r^{d/2}p_{Y}(t,r^{1/2})=|B(0,1)|(1-2^{-d/2})c^{-1}f_{t}(r).$
(2.5)

Denoting $\mathcal{L}f_{t}(\lambda)$ the Laplace transform of $f_{t}$, we have

$\displaystyle\mathcal{L}f_{t}(\lambda)\leq
N\int_{0}^{\infty}\mathbb{P}_{0}(\sqrt{r/2}<|Y_{t}|<\sqrt{r})e^{-\lambda
r}dr=N{\mathbb{E}}_{0}\int_{|Y_{t}|^{2}}^{2|Y_{t}|^{2}}e^{-\lambda r}dr$
$\displaystyle=N\lambda^{-1}{\mathbb{E}}_{0}[e^{-\lambda|Y_{t}|^{2}}-e^{-2\lambda|Y_{t}|^{2}}]=N\lambda^{-1}g_{t}(\lambda),\quad\lambda>0$
(2.6)

from (2.3).

Using (A2) again, we get that, for any $v>0$

$\displaystyle\mathcal{L}f_{t}(v^{-1})=$
$\displaystyle\int^{\infty}_{0}e^{-av^{-1}}f_{t}(a)\,da=v\int^{\infty}_{0}e^{-s}f_{t}\left(sv\right)ds$
$\displaystyle\geq$ $\displaystyle
v\int^{1}_{1/2}e^{-s}f_{t}\left(sv\right)ds=v\int^{1}_{1/2}e^{-s}s^{d/2}v^{d/2}p_{Y}\left(t,s^{1/2}v^{1/2}\right)ds$
$\displaystyle\geq$ $\displaystyle
c^{-1}v2^{-d/2}v^{d/2}p_{Y}\left(t,v^{1/2}\right)\int^{1}_{1/2}e^{-s}ds=c^{-1}2^{-d/2}vf_{t}\left(v\right)\left(\int^{1}_{1/2}e^{-s}ds\right).$

Thus

$\displaystyle f_{t}\left(v\right)\leq
c2^{d/2}\frac{v^{-1}\mathcal{L}f_{t}(v^{-1})}{e^{-1/2}-e^{-1}}.$ (2.7)

Now combining (2.6) and (2.7) with Lemma 2.1 we conclude

$p_{Y}(t,r)=r^{-d}f_{t}(r^{2})\leq Nr^{-d-2}\mathcal{L}f_{t}(r^{-2})\leq
Nr^{-d}g_{t}(r^{-2})\leq Ntr^{-d}\Psi_{Y}(r^{-1}).$

$\Box$

##  3\. Bernstein functions and subordinate Brownian motion

Let $S=(S_{t}:t\geq 0)$ be a subordinator, that is, an increasing Lévy process
taking values in $[0,\infty)$ with $S_{0}=0$. A subordinator $S$ is completely
characterized by its Laplace exponent $\phi$ via

${\mathbb{E}}[\exp(-\lambda S_{t})]=\exp(-t\phi(\lambda))\,,\quad\lambda>0.$

The Laplace exponent $\phi$ can be written in the form (cf. [1, p. 72])

$\displaystyle\phi(\lambda)=b\lambda+\int_{0}^{\infty}(1-e^{-\lambda
t})\,\mu(dt)\,.$ (3.1)

Here $b\geq 0$, and $\mu$ is a $\sigma$-finite measure on $(0,\infty)$
satisfying

$\int_{0}^{\infty}(t\wedge 1)\,\mu(dt)<\infty\,.$

We call the constant $b$ the drift and $\mu$ the Lévy measure of the
subordinator $S$.

A smooth function $g:(0,\infty)\to[0,\infty)$ is called a Bernstein function
if

$(-1)^{n}D^{n}g\leq 0,\quad\forall n\in\mathbb{N}.$

It is well known that a nonnegative function $\phi$ on $(0,\infty)$ is the
Laplace exponent of a subordinator if and only if it is a Bernstein function
with $\phi(0+)=0$ (see, for instance, Chapter 3 of [20]). By the concavity,
for any Bernstein function $\phi$,

$\phi(\lambda t)\leq\lambda\phi(t)\qquad\text{ for all }\lambda\geq 1,t>0\,,$
(3.2)

implying

$\frac{\phi(v)}{v}\leq\frac{\phi(u)}{u}\,,\quad 0<u\leq v\,.$ (3.3)

Clearly (3.2) implies the following.

######  Lemma 3.1.

Let $\phi$ be a Bernstein function. Then for all $\lambda,t>0$,
$1\wedge\lambda\leq{\phi(\lambda t)}/{\phi(t)}\leq 1\vee\lambda$.

The following will be used in section 6 to control the deterministic part of
equation (1.4).

######  Lemma 3.2.

For each nonnegative integer $n$, there is a constant $N(n)$ such that for
every Bernstein function with the drift $b=0$,

$\displaystyle\frac{\lambda^{n}|D^{n}\phi(\lambda)|}{\phi(\lambda)}\leq
N(n),\quad\forall\lambda.$ (3.4)

###### Proof.

The statement is trivial if $n=0$. So let $n\geq 1$. Due to (3.1),

$|D^{n}\phi(\lambda)|=\int_{0}^{\infty}t^{n}e^{-\lambda t}\mu(dt).$

Use $t^{n}e^{-t}\leq N(1-e^{-t})$ to conclude

$\displaystyle\lambda^{n}|D^{n}\phi(\lambda)|\leq\int_{0}^{\infty}(\lambda
t)^{n}e^{-\lambda t}\mu(dt)\leq N\int_{0}^{\infty}(1-e^{-\lambda t})\mu(dt).$

This obviously leads to (3.4). $\Box$

Throughout this article, we assume that $\phi$ is a Bernstein functions with
the drift $b=0$ and $\phi(1)=1$. Thus

$\displaystyle\phi(\lambda)=\int_{0}^{\infty}(1-e^{-\lambda t})\,\mu(dt).$

Let $d\geq 1$ and $W:=(W_{t}:t\geq 0)$ be a $d$-dimensional Brownian motion
with $W_{0}=0$. Then

${\mathbb{E}}\left[e^{i\xi\cdot
W_{t}}\right]=e^{-t|\xi|^{2}},\qquad\forall\xi\in{\mathbb{R}}^{d},\,\,t>0$

and $W$ has the transition density

$q(t,x,y)=q_{d}(t,x,y)=(4\pi t)^{-d/2}e^{-\frac{|x-y|^{2}}{4t}}\,,\quad
x,y\in{\mathbb{R}}^{d},\ t>0\,.$

Let $X=(X_{t}:t\geq 0)$ denote the subordinate Brownian motion defined by
$X_{t}:=W_{S_{t}}$. Then $X_{t}$ has the characteristic exponent
$\Psi(x)=\phi(|x|^{2})$ and has the transition density

$p(t,x)=p_{d}(t,x):=\int_{\mathbb{R}^{d}}e^{i\xi\cdot
x}e^{-t\phi(|\xi|^{2})}d\xi.$ (3.5)

For $t\geq 0$, let $\eta_{t}$ be the distribution of $S_{t}$. That is, for any
Borel set $A\subset[0,\infty)$, $\eta_{t}(A)=\mathbb{P}(S_{t}\in A)$. Then we
have

$p(t,x)=p_{d}(t,x)=\int_{(0,\infty)}(4\pi
s)^{-d/2}\exp\left(-\frac{|x|^{2}}{4s}\right)\,\eta_{t}(ds)$ (3.6)

(see [11, Section 13.3.1]). Thus $p(t,x)$ is smooth in $x$.

The Lévy measure $\Pi$ of $X$ is given by (see e.g. [19, pp. 197–198])

$\Pi(A)=\int_{A}\int_{0}^{\infty}p(t,x)\,\mu(dt)\,dx=\int_{A}J(x)\,dx\,,\quad
A\subset{\mathbb{R}}^{d}\,,$

where

$J(x):=\int_{0}^{\infty}p(t,x)\,\mu(dt)$ (3.7)

is the Lévy density of $X$. Define the function $j:(0,\infty)\to(0,\infty)$ as

$j(r)=j_{d}(r):=\int_{0}^{\infty}(4\pi)^{-d/2}t^{-d/2}\exp\left(-\frac{r^{2}}{4t}\right)\,\mu(dt)\,,\quad
r>0.$ (3.8)

Then $J(x)=j(|x|)$ and

$\Psi(\xi)=\int_{{\mathbb{R}}^{d}}(1-\cos(\xi\cdot y))j(|y|)dy.$ (3.9)

Note that the function $r\mapsto j(r)$ is strictly positive, continuous and
decreasing on $(0,\infty)$.

The next lemma is an extension of [13, Lemma 3.1].

######  Lemma 3.3.

There exists a constant $N>0$ depending only on $d$ such that

$j(r)\leq N\,r^{-d}\phi(r^{-2})\,,\qquad\forall r>0.$

###### Proof.

By Lemma 3.1, (A1) holds with $h(\lambda)=1\vee\lambda^{2}$, and (A2) holds
with $c=1$ since $r\to p(t,r)$ is decreasing. Thus by Theorem 2.2, we have

$p(t,r)\leq Ntr^{-d}\phi(r^{-2})\quad\forall t,r>0$ (3.10)

where $N>0$ depends only on $d$. The lemma now follows from (3.10) and (1.2).
Indeed, by (1.2) and Section 4.1 in [21] that for $f\in
C^{2}_{0}({\mathbb{R}}^{d}\setminus\\{0\\})$ (the set of $C^{2}$-functions on
$\mathbb{R}^{d}\setminus\\{0\\}$ with compact support), we have

$\lim_{t\to
0}\frac{1}{t}\int_{{\mathbb{R}}^{d}}p(t,|y|)f(y)dy=\phi(\Delta)f(0)=\int_{{\mathbb{R}}^{d}}j(|y|)f(y)dy.$
(3.11)

We fix $r>0$ and choose a $f\in C^{2}_{0}(\mathbb{R}^{d}\setminus\\{0\\})$
such that $f=1$ on $B(0,r)\setminus B(0,r/2)$ and $f=0$ on $B(0,2r)^{c}\cup
B(0,r/4)$. Note that since $s\to j(s)$ is decreasing, we have

$\displaystyle
r^{d}j(r)\leq\frac{d2^{d}}{2^{d}-1}\int_{r/2}^{r}j(s)s^{d-1}ds\leq
N\int_{B(0,r)\setminus B(0,r/2)}j(|y|)dy$ $\displaystyle\leq
N\int_{{\mathbb{R}}^{d}}j(|y|)f(y)dy$

where $N>0$ depends only on $d$. Thus by (3.10) and (3.11), we see that

$\displaystyle r^{d}j(r)\leq N\lim_{t\to
0}\frac{1}{t}\int_{{\mathbb{R}}^{d}}p(t,|y|)f(y)dy\leq
N\int_{{\mathbb{R}}^{d}}|y|^{-d}\phi(|y|^{-2})f(y)dy$ $\displaystyle\leq
N\int_{B(0,2r)\setminus B(0,r/4)}|y|^{-d}\phi(|y|^{-2})dy\leq
N\int_{B(0,2r)\setminus B(0,r/4)}|y|^{-d}dy\phi(4r^{-2})$ $\displaystyle\leq
N\int^{2r}_{r/4}r^{-1}dr\phi(4r^{-2})\leq N\phi(4r^{-2})$

where $N>0$ depends only on $d$. Now the lemma follows immediately by (3.2).
$\Box$

For $a>0$, we define $\phi^{a}(\lambda)=\phi(\lambda a^{-2})/\phi(a^{-2})$.
Then $\phi^{a}$ is again a Bernstein function satisfying $\phi^{a}(1)=1$. We
will use $\mu^{a}(dt)$ to denote the Lévy measure of $\phi^{a}$ and
$S^{a}=(S^{a}_{t})_{t\geq 0}$ to denote a subordinator with Laplace exponent
$\phi^{a}$.

Assume that $S^{a}=(S^{a})_{t\geq 0}$ is independent of the Brownian motion
$W$. Let $X^{a}=(X^{a}_{t})_{t\geq 0}$ be defined by
$X^{a}_{t}:=W_{S^{a}_{t}}$. Then $X^{a}$ is a rotationally invariant Lévy
process with characteristic exponent

$\Psi^{a}(\xi)=\phi^{a}(|\xi|^{2})=\frac{\phi(a^{-2}|\xi|^{2})}{\phi(a^{-2})}=\frac{\Psi(a^{-1}\xi)}{\phi(a^{-2})}\,,\quad\xi\in{\mathbb{R}}^{d}\,.$
(3.12)

This shows that $\\{X_{t}^{a}-X^{a}_{0}\\}_{t\geq 0}$ is identical in law to
the process $\\{a^{-1}(X_{t/\phi(a^{-2})}-X_{0})\\}_{t\geq 0}$. $X^{1}$ is
simply the process $X$.

Since, by (3.9) and (3.12),

$\Psi^{a}(\xi)=\frac{1}{\phi(a^{-2})}\int_{{\mathbb{R}}^{d}}(1-\cos(a^{-1}\xi\cdot
y))j(|y|)dy=\frac{a^{d}}{\phi(a^{-2})}\int_{{\mathbb{R}}^{d}}(1-\cos(\xi\cdot
z))j(a|z|)dz,$ (3.13)

the Lévy measure of $X^{a}$ has the density $J^{a}(x)=j^{a}(|x|)$, where
$j^{a}$ is given by

$\displaystyle j^{a}(r):=a^{d}\phi(a^{-2})^{-1}j(ar)\,.$ (3.14)

We use $p^{a}(t,x,y)=p^{a}(t,x-y)$ to denote the transition density of
$X^{a}$. Recall that the process $\\{a^{-1}(X_{t/\phi(a^{-2})}-X_{0}):t\geq
0\\}$ has the same law as $\\{X^{a}_{t}-X^{a}_{0}:t\geq 0\\}$. In terms of
transition densities, this can be written as

$p^{a}(t,x,y)=a^{d}p(\frac{t}{\phi(a^{-2})},ax,ay),\qquad(t,x,y)\in(0,\infty)\times{\mathbb{R}}^{d}\times{\mathbb{R}}^{d}.$

Thus

$p(t,x)=p^{1}(t,x)=a^{-d}p^{a}(t\phi(a^{-2}),a^{-1}x),\qquad(t,x)\in(0,\infty)\times{\mathbb{R}}^{d}.$
(3.15)

Denote

$a_{t}:=\frac{1}{\sqrt{\phi^{-1}(t^{-1})}}.$

From (3.15), we see that

$p(t,x)=(a_{t})^{-d}p^{a_{t}}(1,(a_{t})^{-1}x),\qquad(t,x)\in(0,\infty)\times{\mathbb{R}}^{d}.$

Let $\beta>0$. For appropriate functions $f=f(x)$, define

$T_{t}f(x):=(p(t,\cdot)\ast
f(\cdot))(x)=\int_{\mathbb{R}^{d}}p(t,x-y)f(y)dy,\quad t>0,$
$\phi(\Delta)^{\beta}f:=-\phi(-\Delta)^{\beta}f:=\mathcal{F}^{-1}(\phi(|\xi|^{2})^{\beta}\mathcal{F}(f))(x),\quad
t>0.$

In particular, if $\beta=1$ and $f\in C_{b}^{2}({\mathbb{R}}^{d})$ then we
have

$\displaystyle\phi(\Delta)f(x)=-\phi(-\Delta)f(x)=\int_{{\mathbb{R}}^{d}}\left(f(x+y)-f(x)-\nabla
f(x)\cdot y{\mathbf{1}}_{\\{|y|\leq 1\\}}\right)\,j(|y|)\,dy$
$\displaystyle=\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|y|>\varepsilon\\}}(f(x+y)-f(x))j(|y|)\,dy$
(3.16)

(see Section 4.1 in [21]).

Recall that
$\phi^{a_{t}}(\lambda):=\phi(\lambda(a_{t})^{-2})/\phi((a_{t})^{-2})$. Since
$t\phi(a_{t}^{-2})=1$, by (3.5)

$\displaystyle\phi(\Delta)^{1/2}p(t,\cdot)(x)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}\phi(|\xi|^{2})^{1/2}e^{ix\xi}e^{-t\phi(|\xi|^{2})}d\xi$
(3.17) $\displaystyle=$ $\displaystyle
t^{-1/2}\int_{\mathbb{R}^{d}}(\phi(|\xi|^{2})/\phi(a_{t}^{-2}))^{1/2}e^{ix\xi}e^{-\phi(|\xi|^{2})/\phi(a_{t}^{-2})}d\xi$
$\displaystyle=$ $\displaystyle
t^{-1/2}(a_{t})^{-d}\int_{\mathbb{R}^{d}}\phi^{a_{t}}(|\xi|^{2})^{1/2}e^{i(a_{t})^{-1}x\xi}e^{-\phi^{a_{t}}(|\xi|^{2})}d\xi$
$\displaystyle=$
$\displaystyle(a_{t})^{-d}t^{-1/2}\phi^{a_{t}}(\Delta)^{1/2}p^{a_{t}}(1,\cdot)((a_{t})^{-1}x).$

By [20, Corollary 3.7 (iii)], $\phi^{a}(\lambda)^{1/2}$ is also a Bernstein
function. Thus $\phi^{a}(\lambda)^{1/2}=\int_{0}^{\infty}(1-e^{-\lambda
t})\,\widehat{\mu}^{a}(dt)$ where $\widehat{\mu}^{a}$ is the Lévy measure of
$\phi^{a}(\lambda)^{1/2}$. Let

$\widehat{j}^{a}(r):=\int^{\infty}_{0}(4\pi
t)^{-d/2}e^{-r^{2}/(4t)}\widehat{\mu}^{a}(dt),\qquad r,a>0$

and $\widehat{j}(r):=\widehat{j}^{1}(r)$. Then, by (3.14)

$\displaystyle\widehat{j}^{a}(r)=a^{d}\phi(a^{-2})^{-1/2}\widehat{j}(ar)\,,\qquad
r,a>0.$ (3.18)

As (3.16), for every $f\in C_{b}^{2}({\mathbb{R}}^{d})$,

$\displaystyle\phi^{a}(\Delta)^{1/2}f(x)$ $\displaystyle:=$
$\displaystyle-\phi^{a}(-\Delta)^{1/2}f(x)$ (3.19) $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{d}}\left(f(x+y)-f(x)-\nabla f(x)\cdot
y{\mathbf{1}}_{\\{|y|\leq r\\}}\right)\,\widehat{j}^{a}(|y|)\,dy$
$\displaystyle=$ $\displaystyle\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|y|>\varepsilon\\}}(f(x+y)-f(x))\widehat{j}^{a}(|y|)\,dy$

Clearly by Lemma 3.3 we have the following. We emphasize that the constant
does not depend on neither $\phi$ nor $a$.

######  Lemma 3.4.

There exists a constant $N>1$ depending only on $d$ such that for any $a>0$
and $x\neq 0$

$\widehat{j}^{a}(r)\leq Nr^{-d}(\phi^{a}(r^{-2}))^{1/2}\,,\qquad\forall r>0.$

Recall conditions (H1) and (H2):

(H1): There exist constants $0<\delta_{1}\leq\delta_{2}<1$ and $a_{1},a_{2}>0$
such that

$a_{1}\lambda^{\delta_{1}}\phi(t)\leq\phi(\lambda t)\leq
a_{2}\lambda^{\delta_{2}}\phi(t)\quad\lambda\geq 1,t\geq 1\,.$

(H2): There exist constants $0<\delta_{3}\leq 1$ and $a_{3}>0$ such that

$\phi(\lambda t)\leq a_{3}\lambda^{\delta_{3}}\phi(t),\quad\lambda\leq 1,t\leq
1\,.$

By taking $t=1$ in (H1) and (H2) and using Lemma 3.1, we get that if (H1)
holds then

$a_{1}\lambda^{\delta_{1}}\leq\phi(\lambda)\leq
a_{2}\lambda^{\delta_{2}}\,,\quad\lambda\geq 1\,,$ (3.20)

and, if (H2) holds then

$\lambda\leq\phi(\lambda)\leq a_{3}\lambda^{\delta_{3}}\,,\quad\lambda\leq
1\,.$ (3.21)

Also, if (H1) holds we have

$a_{1}\lambda^{\delta_{1}}\phi^{a}(t)\leq\phi^{a}(\lambda t)\leq
a_{2}\lambda^{\delta_{2}}\phi^{a}(t)\,,\quad\lambda\geq 1,t\geq a^{2}\,.$
(3.22)

Thus, by taking $t=1$ in (3.22), if (H1) holds and $a\leq 1$ we get

$a_{1}\lambda^{\delta_{1}}\leq\phi^{a}(\lambda)\leq
a_{2}\lambda^{\delta_{2}}\,,\quad\lambda\geq 1.$

Thus, if (H1) holds

$a_{1}(T^{-2}\wedge
1)\lambda^{\delta_{1}}\leq\phi^{a}(\lambda)\leq\frac{a_{2}}{\phi(T^{-2})\wedge
1}\lambda^{\delta_{2}}\,,\quad a\in(0,T],\,\lambda\geq 1.$ (3.23)

In fact, if $T>1$ and $1\leq a\leq T$ then for $\lambda\geq 1$

$\phi^{a}(\lambda)=\frac{\phi(\lambda
a^{-2})}{\phi(a^{-2})}\leq\frac{\phi(\lambda)}{\phi(a^{-2})}\leq\frac{a_{2}\lambda^{\delta_{2}}}{\phi(T^{-2})}$

and using Lemma 3.1

$\phi^{a}(\lambda)=\frac{\phi(\lambda
a^{-2})}{\phi(a^{-2})}\geq\frac{\phi(\lambda
a^{-2})}{\phi(1)}=\frac{\phi(\lambda a^{-2})}{\phi(\lambda)}\phi(\lambda)\geq
a^{-2}\phi(\lambda)\geq T^{-2}\phi(\lambda)\geq
T^{-2}a_{1}\lambda^{\delta_{1}}.$

Recall that $p(t,x)$ is the transition density of $X_{t}$.

######  Corollary 3.5.

Suppose (H1) holds. Then for each $T>0$ there exists a constant
$N=N(T,d,\phi)>0$ such that for $(t,x)\in(0,T]\times{\mathbb{R}}^{d}$,

$p(t,x)\leq N\left(\left(\phi^{-1}(t^{-1})\right)^{d/2}\wedge
t\frac{\phi(|x|^{-2})}{|x|^{d}}\right).$ (3.24)

Proof. The corollary follows from Theorem 2.2 and the first display on page
1073 of [3]. Also one can see from (3.5) and (3.15) that

$\displaystyle
p(t,x)=(a_{t})^{-d}p^{a_{t}}(1,a_{t}^{-1}x)\leq(a_{t})^{-d}\int_{\mathbb{R}^{d}}e^{-\phi^{a_{t}}(|\xi|^{2})}d\xi$
$\displaystyle\leq$
$\displaystyle(a_{t})^{-d}\int_{|\xi|<1}d\xi+(a_{t})^{-d}\int_{|\xi|\geq
1}e^{-\phi^{a_{t}}(|\xi|^{2})}d\xi$ $\displaystyle\leq$ $\displaystyle
N(a_{t})^{-d}\left(1+\int_{|\xi|\geq 1}e^{-a_{1}(a_{T}^{-2}\wedge
1)|\xi|^{2\delta_{1}}}d\xi\right).$

where (3.23) is used in the last inequality. $\Box$

######  Remark 3.6.

If there exist constants $0<\delta_{3}\leq\delta_{4}<1$ and $a_{3},a_{4}>0$
such that

$a_{4}\lambda^{\delta_{4}}\phi(t)\leq\phi(\lambda t)\leq
a_{3}\lambda^{\delta_{3}}\phi(t),\quad\lambda\leq 1,t\leq 1\,,$ (3.25)

then, as in [13] the subordinate Brownian motion $X$ satisfies conditions
(1.4), (1.13) and (1.14) from [5]. Thus, in fact, by [5], if (H1) and (3.25)
hold we have the sharp two-sided estimates for all $t>0$

$N^{-1}\left(\left(\phi^{-1}(t^{-1})\right)^{d/2}\wedge
t\frac{\phi(|x|^{-2})}{|x|^{d}}\right)\leq p(t,x)\leq
N\left(\left(\phi^{-1}(t^{-1})\right)^{d/2}\wedge
t\frac{\phi(|x|^{-2})}{|x|^{d}}\right),$ (3.26)

where $N=N(\phi)>1$.

On the other hand, when $\delta_{3}=\delta_{4}=1$ in (3.25), (3.26) does not
hold and (3.24) is not sharp. For example, see [4, (2.2), (2.4) and Theorem
4.1].

For the rest of this article we assume that (H1) and (H2) hold. Now we further
discuss the scaling. If $0<r<1<R$, using (3.3), (3.20) and (3.21), we have

$\frac{\phi(R)}{\phi(r)}\leq\frac{R}{r}\quad\text{ and
}\quad\frac{\phi(R)}{\phi(r)}\geq\frac{a_{1}}{a_{3}}\frac{R^{\delta_{1}}}{r^{\delta_{3}}}\geq\frac{a_{1}}{a_{3}}\left(\frac{R}{r}\right)^{\delta_{1}\wedge\delta_{3}}.$

Combining these with (H1) and (H2) we get

$\frac{a_{1}}{a_{3}}\left(\frac{R}{r}\right)^{\delta_{1}\wedge\delta_{3}}\leq\frac{\phi(R)}{\phi(r)}\leq\frac{R}{r},\quad
0<r<R<\infty\,.$ (3.27)

Now applying this to $\phi^{a}$, we get

$\frac{a_{1}}{a_{3}}\left(\frac{R}{r}\right)^{\delta_{1}\wedge\delta_{3}}\leq\frac{\phi^{a}(R)}{\phi^{a}(r)}\leq\frac{R}{r},\quad
a>0,\ 0<r<R<\infty\,.$ (3.28)

Next two lemmas will be used several times in this article.

######  Lemma 3.7.

Assume (H1) and (H2). Then there exists a constant
$N=N(\delta_{1},\delta_{3})$ such that for all $\lambda>0$

$\int_{\lambda^{-1}}^{\infty}r^{-1}\phi(r^{-2})\,dr\leq N\phi(\lambda^{2}).$

###### Proof.

Changing variable $r\to\lambda^{-1}r$, from (3.27) we have

$\displaystyle\int_{\lambda^{-1}}^{\infty}r^{-1}\phi(r^{-2})\,dr~{}=~{}\int_{1}^{\infty}r^{-1}\phi(\lambda^{2}r^{-2})\,dr$
$\displaystyle=$
$\displaystyle\int_{1}^{\infty}r^{-1}\phi(\lambda^{2}r^{-2})\frac{\phi(\lambda^{2})}{\phi(\lambda^{2})}\,dr$
$\displaystyle\leq$
$\displaystyle\int_{1}^{\infty}r^{-1-2\delta_{1}\wedge\delta_{3}}\,dr\phi(\lambda^{2}).$

$\Box$

######  Corollary 3.8.

Assume (H1) and (H2). Then there exists a constant
$N=N(\delta_{1},\delta_{3})$ such that for all $\lambda>0$

$\int_{\lambda^{-1}}^{\infty}r^{-1}(\phi(r^{-2}))^{1/2}\,dr\leq
N(\phi(\lambda^{2}))^{1/2},$

###### Proof.

Since $(\phi(\lambda))^{1/2}$ also satisfies conditions (H1) and (H2) with
different $\delta_{1},\delta_{3}>0$, we get this corollary directly from the
previous lemma. $\Box$

##  4\. Upper bounds of $|\phi(\Delta)^{n/2}D^{\beta}p(t,x)|$

In this section we give upper bounds of $|\phi(\Delta)^{n/2}D^{\beta}p(t,x)|$
for any $n=0,1,2,\cdots$ and multi-index $\beta$.

Recall $a_{t}:=(\phi^{-1}(t^{-1}))^{-1/2}$ and so $t\phi(a_{t}^{-2})=1$. Thus
From (3.15) and (3.24), we have for every $t\in(0,T]$,

$\displaystyle p^{a_{t}}(1,x)$ $\displaystyle=$
$\displaystyle(a_{t})^{d}p(t,a_{t}x)$ $\displaystyle\leq$ $\displaystyle
N(T,\phi,d)\left(1\wedge
t\frac{\phi(|a_{t}x|^{-2})}{|x|^{d}}\right)=N\left(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})}{|x|^{d}}\right).$

######  Lemma 4.1.

For any constant $T>0$ there exists a constant $N=N(T,d,\phi)$ so that for
every $t\in(0,T]$

$\displaystyle|\nabla p^{a_{t}}(1,x)|\leq
N|x|\left(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})}{|x|^{d+2}}\right),$
$\displaystyle\sum_{i,j}|\partial_{x_{i},x_{j}}p^{a_{t}}(1,x)|$
$\displaystyle\leq
N|x|^{2}\left(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})}{|x|^{d+4}}\right)+N\left(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})}{|x|^{d+2}}\right),$

and

$\displaystyle\sum_{|\beta|\leq n}|D^{\beta}p^{a_{t}}(1,x)|$
$\displaystyle\leq N\sum_{n-2m\geq
0,m\in\mathbb{N}\cup\\{0\\}}|x|^{n-2m}\left(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})}{|x|^{d+2(n-m)}}\right).$

###### Proof.

To distinguish the dimension, we denote

$p^{a_{t}}_{d}(1,x):=\int_{(0,\infty)}(4\pi
s)^{-d/2}\exp\left(-\frac{|x|^{2}}{4s}\right)\,\eta^{a_{t}}_{1}(ds).$

By (3.6),

$\displaystyle\partial_{x_{i}}p^{a_{t}}_{d}(1,x)$
$\displaystyle=\int_{(0,\infty)}(4\pi
s)^{-d/2}\partial_{x_{i}}\exp\left(-\frac{|x|^{2}}{4s}\right)\,\eta^{a_{t}}_{1}(ds)$
$\displaystyle=-\frac{x_{i}}{2}\int_{(0,\infty)}s^{-1}(4\pi
s)^{-d/2}\exp\left(-\frac{|x|^{2}}{4s}\right)\,\eta^{a_{t}}_{1}(ds)$
$\displaystyle=-2\pi x_{i}p^{a_{t}}_{d+2}(1,(x,0,0)),$
$\displaystyle\partial_{x_{i},x_{i}}p^{a_{t}}_{d}(1,x)$ $\displaystyle=4\pi
x_{i}^{2}p^{a_{t}}_{d+4}(1,(x,0,0,0,0))-2\pi p^{a_{t}}_{d+2}(1,(x,0,0)),$

and, for $i\not=j$,

$\displaystyle\partial_{x_{i},x_{j}}p^{a_{t}}_{d}(1,x)$ $\displaystyle=4\pi
x_{i}x_{j}p^{a_{t}}_{d+4}(1,(x,0,0,0,0)).$

Thus

$\displaystyle|\nabla p^{a_{t}}_{d}(1,x)|\leq
2\pi|x|p^{a_{t}}_{d+2}(1,(x,0,0))$

and

$\displaystyle\sum_{i,j}|\partial_{x_{i},x_{j}}p^{a_{t}}_{d}(1,x)|$
$\displaystyle\leq 4d^{2}\pi|x|^{2}p^{a+t}_{d+4}(1,(x,0,0,0,0))+2d\pi
p^{a_{t}}_{d+2}(1,(x,0,0)).$

Similarly.

$\displaystyle\sum_{i,j,k}|\partial_{x_{i},x_{j}x_{k}}p^{a_{t}}_{d}(1,x)|$
$\displaystyle\leq
N(d)[|x|^{3}p^{a_{t}}_{d+6}(1,(x,0,0,0,0,0,0))+|x|p^{a_{t}}_{d+4}(1,(x,0,0,0,0))].$

Repeating the product rule of differentiation and applying (4), we prove the
lemma. $\Box$

######  Lemma 4.2.

For any $(t,x)\in(0,T]\times\mathbb{R}^{d}$ and multi-index $\beta$,

$\displaystyle|\phi(\Delta)^{1/2}D^{\beta}p(t,\cdot)(x)|\leq\,N(d,T,|\beta|,\phi)\
\left(t^{-1/2}\left(\phi^{-1}(t^{-1})\right)^{(d+|\beta|)/2}\wedge\frac{\phi(|x|^{-2})^{1/2}}{|x|^{d+|\beta|}}\right).$

###### Proof.

First we prove the lemma when $\beta=0$. Recall

$a_{t}:=\frac{1}{\sqrt{\phi^{-1}(t^{-1})}}\leq\frac{1}{\sqrt{\phi^{-1}(T^{-1})}}.$

Note that by (3.5)

$\displaystyle|\phi^{a_{t}}(\Delta)^{1/2}p^{a_{t}}(1,\cdot)(x)|$
$\displaystyle=$
$\displaystyle|\int_{\mathbb{R}^{d}}\phi^{a_{t}}(|\xi|^{2})^{1/2}e^{ix\xi}e^{-\phi^{a_{t}}(|\xi|^{2})}d\xi|$
$\displaystyle\leq$
$\displaystyle\int_{\mathbb{R}^{d}}\phi^{a_{t}}(|\xi|^{2})^{1/2}e^{-\phi^{a_{t}}(|\xi|^{2})}d\xi$
$\displaystyle\leq$
$\displaystyle\int_{|\xi|<1}\phi^{a_{t}}(|\xi|^{2})^{1/2}d\xi+\left(\sup_{b>0}b^{1/2}e^{-b/2}\right)\int_{|\xi|\geq
1}e^{-2^{-1}\phi^{a_{t}}(|\xi|^{2})}d\xi$ $\displaystyle\leq$ $\displaystyle
N\phi^{a_{t}}(1)^{1/2}+N\int_{|\xi|\geq
1}e^{-2^{-1}\phi^{a_{t}}(|\xi|^{2})}d\xi\leq N+N\int_{|\xi|\geq
1}e^{-2^{-1}\phi^{a_{t}}(|\xi|^{2})}d\xi.$

Thus it is uniformly bounded by (3.28). By (3.19),

$\displaystyle|\phi^{a_{t}}(\Delta)^{1/2}p^{a_{t}}(1,\cdot)(x)|$
$\displaystyle=|\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|y|>\varepsilon\\}}(p^{a_{t}}(1,x+y)-p^{a_{t}}(1,x))\widehat{j}^{a_{t}}(|y|)\,dy|$
$\displaystyle\leq|p^{a_{t}}(1,x)|\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}\widehat{j}^{a_{t}}(|y|)\,dy+\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}|p^{a_{t}}(1,x+y)|\widehat{j}^{a_{t}}(|y|)\,dy$
$\displaystyle\quad+|\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|x|/2>|y|>\varepsilon\\}}\int_{0}^{1}|\nabla
p^{a_{t}}(1,x+sy)|ds|y|\widehat{j}^{a_{t}}(|y|)\,dy$
$\displaystyle:=p^{a_{t}}(1,x)\times I+II+III.$

Since from (3.18)

$\displaystyle\int_{s}^{\infty}\widehat{j}^{a}(r)r^{d-1}dr=a^{d}\phi(a^{-2})^{-1/2}\int_{s}^{\infty}\widehat{j}(ar)r^{d-1}dr=\phi(a^{-2})^{-1/2}\int_{as}^{\infty}\widehat{j}(t)t^{d-1}dt,$

we get

$\displaystyle\int_{|x|}^{\infty}\widehat{j}^{a_{t}}(r)r^{d-1}dr$
$\displaystyle=$
$\displaystyle(a_{t})^{d}\phi((a_{t})^{-2})^{-1/2}\int_{|x|}^{\infty}\widehat{j}(a_{t}r)r^{d-1}dr$
$\displaystyle=$
$\displaystyle\phi((a_{t})^{-2})^{-1/2}\int_{|x|a_{t}}^{\infty}\widehat{j}(s)s^{d-1}ds$
$\displaystyle=$
$\displaystyle\phi^{a_{t}}(|x|^{-2})^{1/2}\phi((a_{t})^{-2}|x|^{-2})^{-1/2}\int_{|x|a_{t}}^{\infty}\widehat{j}(s)s^{d-1}ds.$

In addition to this, applying Lemma 3.4 and Corollary 3.8, we see

$\phi((a_{t})^{-2}|x|^{-2})^{-1/2}\int_{|x|a_{t}}^{\infty}\widehat{j}(s)s^{d-1}ds\leq
N.$

Combining these with

$p^{a_{t}}(1,x)\leq
N(T)\left(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})}{|x|^{d}}\right),$

we get

$\displaystyle p^{a_{t}}(1,x)\times I$ $\displaystyle\leq
N(T)\phi^{a_{t}}(|x|^{-2})^{1/2}\left(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})}{|x|^{d}}\right)\leq
N(T)\frac{\phi^{a_{t}}(|x|^{-2})^{3/2}}{|x|^{d}}.$

Using the fact that $r\to j^{a_{t}}(r)$ is decreasing,

$\displaystyle II$
$\displaystyle\leq\widehat{j}^{a_{t}}(|x|/2)\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}p^{a_{t}}(1,x+y)\,dy$
$\displaystyle\leq\widehat{j}^{a_{t}}(|x|/2)\int_{\mathbb{R}^{d}}p^{a_{t}}(1,x+y)\,dy\leq\widehat{j}^{a_{t}}(|x|/2).$

Finally, from Lemma 4.1 we get

$\displaystyle III=|\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|x|/2>|y|>\varepsilon\\}}\int_{0}^{1}|\nabla
p^{a_{t}}(1,x+sy)|ds|y|\widehat{j}^{a_{t}}(|y|)\,dy$ $\displaystyle\leq
N\int_{|y|<|x|/2}\int_{0}^{1}\frac{\phi^{a_{t}}(|x+sy|^{-2})}{|x+sy|^{d+1}}ds|y|\widehat{j}^{a_{t}}(|y|)dy$
$\displaystyle\leq
N\int_{|y|<|x|/2}\int_{0}^{1}\frac{\phi^{a_{t}}((|x|-s|y|)^{-2})}{(|x|-s|y|)^{d+1}}ds|y|\widehat{j}^{a_{t}}(|y|)dy$
$\displaystyle\leq
N\int_{|y|<|x|/2}\int_{0}^{1}\frac{\phi^{a_{t}}(4|x|^{-2})}{|x|^{d+1}}ds|y|\widehat{j}^{a_{t}}(|y|)dy$
$\displaystyle\leq
N\frac{\phi^{a_{t}}(4|x|^{-2})}{|x|^{d+1}}\int_{|y|<|x|/2}|y|\widehat{j}^{a_{t}}(|y|)dy.$

By Lemma 3.4,

$\displaystyle\int_{|y|<|x|/2}|y|\widehat{j}^{a_{t}}(|y|)dy\leq
N\int_{|y|<|x|/2}|y|^{-d+1}(\phi^{a_{t}}(|y|^{-2}))^{1/2}dy$
$\displaystyle\leq N\int_{0}^{|x|}(\phi^{a_{t}}(r^{-2}))^{1/2}dr.$

Since $|\phi^{a_{t}}(\Delta)^{1/2}p^{a_{t}}(1,\cdot)(x)|$ is bounded for $x$,
we may assume that $|x|\geq 1$. So from the monotone property of
$\phi^{a_{t}}(r^{-2})$ and (3.23), we get

$\displaystyle\int_{0}^{|x|}(\phi^{a_{t}}(r^{-2}))^{1/2}dr$ $\displaystyle=$
$\displaystyle\int_{0}^{1}(\phi^{a_{t}}(r^{-2}))^{1/2}dr+\int_{1}^{|x|}(\phi^{a_{t}}(r^{-2}))^{1/2}dr$
$\displaystyle\leq$ $\displaystyle
N\left(\int_{0}^{1}r^{-\delta_{2}}dr+\int_{1}^{|x|}(\phi^{a_{t}}(r^{-2}))^{1/2}dr\right)$
$\displaystyle\leq$ $\displaystyle
N\left(|x|^{1-\delta_{2}}+|x|\phi^{a_{t}}(1)^{1/2}\right).$

Thus

$|\phi^{a_{t}}(\Delta)^{1/2}p^{a_{t}}(1,\cdot)(x)|\leq
N(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})^{1/2}}{|x|^{d}}).$

Now applying (3.17) and using the fact that $t\phi(a_{t}^{-2})=1$ and
$\phi^{a}(\lambda)=\phi(\lambda a^{-2})/\phi(a^{-2})$, we get

$\displaystyle|\phi(\Delta)^{1/2}p(t,\cdot)(x)|=(a_{t})^{-d}t^{-1/2}|\phi^{a_{t}}(\Delta)^{1/2}p^{a_{t}}(1,\cdot)((a_{t})^{-1}x)|$
$\displaystyle\leq\,N\,t^{-1/2}\,((a_{t})^{-d}\wedge\frac{\phi^{a_{t}}(|(a_{t})^{-1}x|^{-2})^{1/2}}{|x|^{d}})$
$\displaystyle=\,N\,t^{-1/2}\,((a_{t})^{-d}\wedge\frac{\phi(|x|^{-2})^{1/2}}{\phi((a_{t})^{-2})^{1/2}|x|^{d}})$
$\displaystyle=\,N\,\,(t^{-1/2}(a_{t})^{-d}\wedge\frac{\phi(|x|^{-2})^{1/2}}{(t\phi((a_{t})^{-2}))^{1/2}|x|^{d}})\,=\,N\,\,(t^{-1/2}(a_{t})^{-d}\wedge\frac{\phi(|x|^{-2})^{1/2}}{|x|^{d}}).$

The case $|\beta|=1$ is proved similarly. First, one can check that
$|\phi^{a_{t}}(\Delta)^{1/2}D^{\beta}p^{a_{t}}(1,\cdot)(x)|$ is uniformly
bounded. Note

$\displaystyle|\phi^{a_{t}}(\Delta)^{1/2}D^{\beta}p^{a_{t}}(1,\cdot)(x)|$
(4.2) $\displaystyle=$ $\displaystyle|\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|y|>\varepsilon\\}}(D^{\beta}p^{a_{t}}(1,x+y)-D^{\beta}p^{a_{t}}(1,x))\widehat{j}^{a_{t}}(|y|)\,dy|$
$\displaystyle\leq$
$\displaystyle|D^{\beta}p^{a_{t}}(1,x)|\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}\widehat{j}^{a_{t}}(|y|)\,dy$
$\displaystyle+|\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}D^{\beta}p^{a_{t}}(1,x+y)\widehat{j}^{a_{t}}(|y|)\,dy|$
$\displaystyle+|\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|x|/2>|y|>\varepsilon\\}}\int_{0}^{1}|\nabla
D^{\beta}p^{a_{t}}(1,x+sy)|ds|y|\widehat{j}^{a_{t}}(|y|)\,dy$
$\displaystyle:=$ $\displaystyle|D^{\beta}p^{a_{t}}(1,x)|\times I+II+III.$

Since $I$ and $III$ can be estimated similarly as in the case $|\beta|=0$, we
only pay attention to the estimation of $II$. We use integration by parts and
get

$\displaystyle II$ $\displaystyle\leq$
$\displaystyle\int_{|y|=|x|/2}|\widehat{j}^{a_{t}}(|x|/2)p^{a_{t}}(1,x+y)|dS$
$\displaystyle+\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}\frac{d}{dr}\widehat{j}^{a_{t}}(|y|)p^{a_{t}}(1,x+y)\,dy.$

We use notation $\widehat{j}^{a_{t}}_{d}(r)$ in place of
$\widehat{j}^{a_{t}}(r)$ to express its dimension. That is,

$\widehat{j}^{a_{t}}_{d}(r):=\int^{\infty}_{0}(4\pi
t)^{-d/2}e^{-r^{2}/(4t)}\widehat{\mu}^{a_{t}}(dt),\qquad r>0.$

By its definition, we can easily see that
$\frac{d}{dr}\widehat{j}^{a_{t}}_{d}(r)=-2\pi r\widehat{j}^{a_{t}}_{d+2}(r)$.
So from Lemmas 3.4 and 4.1 we get

$\displaystyle II$ $\displaystyle\leq
N[\frac{\phi^{a_{t}}(|x|^{-2})^{3/2}}{|x|^{d+1}}+\int_{|y|>|x|/2}|y|\widehat{j}^{a_{t}}_{d+2}(|y|)|p^{a_{t}}(1,x+y)|\,dy]$
$\displaystyle\leq
N[\frac{\phi^{a_{t}}(|x|^{-2})^{3/2}}{|x|^{d+1}}+\int_{|y|>|x|/2}\frac{\phi^{a_{t}}(|y|^{-2})^{1/2}}{|y|^{d+1}}|p^{a_{t}}(1,x+y)|\,dy]$
$\displaystyle\leq
N[\frac{\phi^{a_{t}}(|x|^{-2})^{3/2}}{|x|^{d+1}}+\frac{\phi^{a}(|x|^{-2})^{1/2}}{|x|^{d+1}}].$

Therefore we get

$|\phi^{a_{t}}(\Delta)^{1/2}p_{x^{i}}^{a_{t}}(1,\cdot)(x)|\leq
N(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})^{1/2}}{|x|^{d+1}}).$

By fact that $t\phi(a_{t}^{-2})=1$ and $\phi^{a}(\lambda)=\phi(\lambda
a^{-2})/\phi(a^{-2})$,

$\displaystyle|\phi(\Delta)^{1/2}p_{x^{i}}(t,\cdot)(x)|$ (4.3)
$\displaystyle=$
$\displaystyle\left|\int_{\mathbb{R}^{d}}(-i\xi^{i})\phi(|\xi|^{2})^{1/2}e^{ix\xi}e^{-t\phi(|\xi|^{2})}d\xi\right|$
$\displaystyle=$ $\displaystyle
t^{-1/2}\left|\int_{\mathbb{R}^{d}}(-i\xi^{i})(\phi(|\xi|^{2})/\phi(a_{t}^{-2}))^{1/2}e^{ix\xi}e^{-\phi(|\xi|^{2})/\phi(a_{t}^{-2})}d\xi\right|$
$\displaystyle=$ $\displaystyle
t^{-1/2}(a_{t})^{-d-1}\left|\int_{\mathbb{R}^{d}}(-i\xi^{i})\phi^{a_{t}}(|\xi|^{2})^{1/2}e^{i(a_{t})^{-1}x\xi}e^{-\phi^{a_{t}}(|\xi|^{2})}d\xi\right|$
$\displaystyle=$
$\displaystyle(a_{t})^{-d-1}t^{-1/2}\left|\phi^{a_{t}}(\Delta)^{1/2}p_{x^{i}}^{a_{t}}(1,\cdot)((a_{t})^{-1}x)\right|$
$\displaystyle\leq$
$\displaystyle\,N\,t^{-1/2}(a_{t})^{-d-1}\,\left(1\wedge\frac{\phi^{a_{t}}(|a_{t}^{-1}x|^{-2})^{1/2}}{|a_{t}^{-1}x|^{d+1}}\right)$
$\displaystyle=$
$\displaystyle\,N\,t^{-1/2}(a_{t})^{-d-1}\,\left(1\wedge\frac{\phi(|x|^{-2})^{1/2}}{\phi((a_{t})^{-2})^{1/2}|a_{t}^{-1}x|^{d+1}}\right)$
$\displaystyle=$ $\displaystyle\,N\,t^{-1/2}(a_{t})^{-d-1}\,\left(1\wedge
t^{1/2}a_{t}^{d+1}\frac{\phi(|x|^{-2})^{1/2}}{(t\phi((a_{t})^{-2}))^{1/2}|x|^{d+1}}\right)$
$\displaystyle=$
$\displaystyle\,N\,\,\left(t^{-1/2}(a_{t})^{-d-1}\wedge\frac{\phi(|x|^{-2})^{1/2}}{|x|^{d+1}}\right).$

Finally, we consider the case $|\beta|\geq 2$. Introduce $I,II$ and $III$ as
in (4.2). $I$ and $III$ can be estimated as in the case $|\beta|=0$. Also,
$II$ can be estimated by doing the integration by parts $|\beta|$-times. For
instance, if $|\beta|=2$,

$\displaystyle II$ $\displaystyle\leq$
$\displaystyle\int_{|y|=|x|/2}|\widehat{j}^{a_{t}}(|x|/2)p_{x^{i}}^{a_{t}}(1,x+y)|dS+\int_{|y|=|x|/2}|\frac{d}{dr}\widehat{j}^{a_{t}}(|x|/2)p^{a_{t}}(1,x+y)|dS$
$\displaystyle+\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}\frac{d^{2}}{dr^{2}}\widehat{j}^{a_{t}}(|y|)p^{a_{t}}(1,x+y)\,dy,$

where $dS$ is the surface measure on $\\{y\in\mathbb{R}^{d}:|y|=|x|/2\\}$. By
its definition, we easily see that

$\frac{d}{dr}\widehat{j}^{a_{t}}_{d}(r)=-2\pi
r\widehat{j}^{a_{t}}_{d+2}(r),\quad\frac{d^{2}}{dr^{2}}\widehat{j}^{a_{t}}_{d}(r)=-2\pi\widehat{j}^{a_{t}}_{d+2}(r)+(2\pi)^{2}r^{2}\widehat{j}^{a_{t}}_{d+4}(r).$

So from Lemma 3.4 and Lemma 4.1 and we get

$\displaystyle II$ $\displaystyle\leq
N\left(\frac{\phi^{a_{t}}(|x|^{-2})^{3/2}}{|x|^{d+2}}+\int_{|y|>|x|/2}|y|^{2}\widehat{j}_{d+4}^{a_{t}}(|y|)|p^{a_{t}}(1,x+y)|\,dy\right)$
$\displaystyle\leq
N\left(\frac{\phi^{a_{t}}(|x|^{-2})^{3/2}}{|x|^{d+2}}+\int_{|y|>|x|/2}|y|^{-(d+2)}\phi^{a_{t}}(|y|^{-2})^{1/2}|p^{a_{t}}(1,x+y)|\,dy\right)$
$\displaystyle\leq
N\left(\frac{\phi^{a_{t}}(|x|^{-2})^{3/2}}{|x|^{d+2}}+\frac{\phi^{a}(|x|^{-2})^{1/2}}{|x|^{d+2}}\right).$

Therefore we have

$|\phi^{a_{t}}(\Delta)^{1/2}p_{x^{i}x^{j}}^{a_{t}}(1,\cdot)(x)|\leq
N\left(1\wedge\frac{\phi^{a_{t}}(|x|^{-2})^{1/2}}{|x|^{d+2}}\right)$

and we are done by the scaling as in (4.3). $\Box$

We generalize Lemma 4.2 as follows.

######  Lemma 4.3.

For any $n\in\mathbb{N}$ and multi-index $\beta$, there exists a constant
$N=N(d,\phi,T,|\beta|,n)>0$ such that

$\displaystyle|\phi(\Delta)^{n/2}D^{\beta}p(t,\cdot)(x)|\leq
N\left(t^{-n/2}(\phi^{-1}(t^{-1}))^{(d+|\beta|)/2}\wedge
t^{-(n-1)/2}\frac{\phi(|x|^{-2})^{1/2}}{|x|^{d+|\beta|}}\right).$ (4.4)

###### Proof.

We use the induction. Due to the previous lemma, the statement is true if
$n=1$. Assume that the statement is true for $n-1$. We put

$\displaystyle|\phi^{a_{t}}(\Delta)^{n/2}D^{\beta}p^{a_{t}}(1,\cdot)(x)|$
$\displaystyle=$ $\displaystyle\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|y|>\varepsilon\\}}(\phi(\Delta)^{(n-1)/2}D^{\beta}p^{a_{t}}(1,x+y)-\phi(\Delta)^{(n-1)/2}D^{\beta}p^{a_{t}}(1,x))\widehat{j}^{a_{t}}(|y|)\,dy|$
$\displaystyle\leq$
$\displaystyle|\phi(\Delta)^{(n-1)/2}p^{a_{t}}(1,x)|\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}\widehat{j}^{a_{t}}(|y|)\,dy$
$\displaystyle+\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}|\phi(\Delta)^{(n-1)/2}p^{a_{t}}(1,x+y)|\widehat{j}^{a_{t}}(|y|)\,dy$
$\displaystyle+|\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|x|/2>|y|>\varepsilon\\}}\int_{0}^{1}|\phi(\Delta)^{1/2}\nabla
p^{a_{t}}(1,x+sy)|ds|y|\widehat{j}^{a_{t}}(|y|)\,dy$ $\displaystyle:=$
$\displaystyle|\phi(\Delta)^{1/2}p^{a_{t}}(1,x)|\times I+II+III$

and follow the proof of Lemma 4.2 with the result (4.4) for $n-1$. Below, we
provide details only for $II$. Let $|\beta|=0$. Then since $r\to j^{a_{t}}(r)$
is decreasing,

$\displaystyle II$
$\displaystyle\leq\widehat{j}^{a_{t}}(|x|/2)\int_{\\{y\in\mathbb{R}^{d}:\,|y|>|x|/2\\}}\phi(\Delta)^{(n-1)/2}p^{a_{t}}(1,x+y)\,dy$
$\displaystyle\leq\widehat{j}^{a_{t}}(|x|/2)\int_{\mathbb{R}^{d}}\phi(\Delta)^{(n-1)/2}p^{a_{t}}(1,x+y)\,dy\leq
N\widehat{j}^{a_{t}}(|x|/2)\leq
N\frac{\phi^{a_{t}}(|x|^{-2})^{1/2}}{|x|^{d}}.$

If $|\beta|>0$ then as in Lemma 4.2 we use integration by parts
$|\beta|$-times and get

$II\leq N\frac{\phi^{a_{t}}(|x|^{-2})^{1/2}}{|x|^{d+|\beta|}}.$

Therefore, since $|\phi^{a_{t}}(\Delta)^{n/2}D^{\beta}p^{a_{t}}(1,\cdot)(x)|$
is uniformly bounded, we have

$|\phi^{a_{t}}(\Delta)^{n/2}D^{\beta}p^{a_{t}}(1,\cdot)(x)|\leq
N\left(\frac{\phi^{a_{t}}(|x|^{-2})^{1/2}}{|x|^{d+|\beta|}}\wedge 1\right)$

and the lemma is proved by the scaling as in (4.3). $\Box$

##  5\. Proof of Theorem 1.1

Let $f\in C^{\infty}_{0}(\mathbb{R}^{d+1},H)$. For each $a\in\mathbb{R}$
denote

$u_{a}(t,x):=\mathcal{G}_{a}(t,x):=[\int_{a}^{t}|\phi(\Delta)^{1/2}T_{t-s}f(s,\cdot)(x)|_{H}^{2}ds]^{1/2},$

$u(t,x):=u_{0}(t,x)$ and $\mathcal{G}(t,x):=\mathcal{G}_{0}(t,x)$.

Here is a version of Theorem 1.1 for $p=2$.

######  Lemma 5.1.

For any $\infty\geq\beta\geq\alpha\geq-\infty$ and $\beta\geq a$,

$\displaystyle\|u_{a}\|^{2}_{L_{2}([\alpha,\beta]\times\mathbb{R}^{d})}\leq
N\||f|_{H}\|^{2}_{L_{2}([a,\beta]\times\mathbb{R}^{d})},$ (5.1)

where $N=N(d)$.

###### Proof.

By the continuity of $f$, the range of $f$ belongs to a separable subspace of
$H$. Thus by using a countable orthonormal basis of this subspace and the
Fourier transform one easily finds

$\displaystyle\|u_{a}\|^{2}_{L_{2}([\alpha,\beta]\times\mathbb{R}^{d})}$
$\displaystyle=$
$\displaystyle(2\pi)^{d}\int_{\mathbb{R}^{d}}\int_{\alpha}^{\beta}\int_{a}^{t}|\mathcal{F}\\{\phi(\Delta)^{1/2}p(t-s,\cdot)\\}(\xi)|^{2}\,|\mathcal{F}(f)(s,\xi)|^{2}_{H}dsdtd\xi$
$\displaystyle\leq$
$\displaystyle(2\pi)^{d}\int_{\mathbb{R}^{d}}\int_{a}^{\beta}\int_{\alpha}^{\beta}I_{0\leq
t-s}\phi(|\xi|^{2})e^{-2(t-s)(\phi(|\xi|^{2}))}dt|\mathcal{F}(f)(s,\xi)|^{2}_{H}dsd\xi.$

Changing $t-s\to t$, we find that the last term above is equal to

$\displaystyle(2\pi)^{d}\int_{\mathbb{R}^{d}}\int_{a}^{\beta}\int_{\alpha-s}^{\beta-s}I_{0\leq
t}\phi(|\xi|^{2})e^{-2t(\phi(|\xi|^{2}))}dt|\mathcal{F}(f)(s,\xi)|^{2}_{H}dsd\xi$
$\displaystyle\leq$
$\displaystyle(2\pi)^{d}\int_{\mathbb{R}^{d}}\int_{a}^{\beta}\int_{0}^{\infty}\phi(|\xi|^{2})e^{-2t(\phi(|\xi|^{2}))}dt|\mathcal{F}(f)(s,\xi)|^{2}_{H}dsd\xi.$

Since $\int_{0}^{\infty}\phi(|\xi|^{2})e^{-2t(\phi(|\xi|^{2}))}dt=1/2$, we
have

$\displaystyle\|u_{a}\|^{2}_{L_{2}([\alpha,\beta]\times\mathbb{R}^{d})}\leq
N\int_{a}^{\beta}\int_{\mathbb{R}^{d}}|\hat{f}(s,\xi)|^{2}_{H}~{}d\xi ds.$

The last expression is equal to the right-hand side of (5.1), and therefore
the lemma is proved. $\Box$

For $c>0$ and $(r,z)\in\mathbb{R}^{d+1}$, we denote

$B_{c}(z)=\\{y\in\mathbb{R}^{d}:|z-y|<c\\},\quad\hat{B}_{c}(z)=\prod_{i=1}^{d}(z^{i}-c/2,z^{i}+c/2),$
$I_{c}(r)=({r-\phi(c^{-2})^{-1}},\,r+\phi(c^{-2})^{-1}),\quad
Q_{c}(r,z)=I_{r}(c)\times\hat{B}_{c}(z).$

Also we denote

$Q_{c}(r)=Q_{c}(r,0),\quad\hat{B}_{c}=\hat{B}_{c}(0),\quad B_{c}=B_{c}(0).$

For a measurable function $h$ on $\mathbb{R}^{d}$, define the maximal
functions

$\mathbb{M}_{x}h(x):=\sup_{r>0}\frac{1}{|B_{r}(x)|}\int_{B_{r}(x)}|h(y)|dy,$
$\mathcal{M}_{x}h(x):=\sup_{\hat{B}_{r}(z)\ni
x}\frac{1}{|\hat{B}_{r}(z)|}\int_{\hat{B}_{r}(z)}|h(y)|dy.$

Similarly, for a measurable function $h=h(t)$ on $\mathbb{R}$,

$\mathbb{M}_{t}h(t):=\sup_{r>0}\frac{1}{2r}\int_{-r}^{r}|h(t+s)|\,ds.$

Also for a function $h=h(t,x)$, we set

$\mathbb{M}_{x}h(t,x):=\mathbb{M}_{x}(h(t,\cdot))(x),\quad\mathcal{M}_{x}h(t,x):=\mathcal{M}_{x}(h(t,\cdot))(x),$
$\mathbb{M}_{t}\mathbb{M}_{x}h(t,x)=\mathbb{M}_{t}(\mathbb{M}_{x}h(\cdot,x))(t),\quad\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}h(t,x)=\mathcal{M}_{x}(\mathbb{M}_{t}\mathbb{M}_{x}h(t,\cdot))(x).$

Since we have estimates of $p(t,x)$ only for $t\leq T$, we introduce the
following functions. Denote $\hat{p}(t,x)=p(t,x)$ if $t\in[0,T]$, and
$\hat{p}(t,x)=0$ otherwise. Define
$\hat{T}_{t}f(x)=\int_{\mathbb{R}^{d}}\hat{p}(t,y)f(x-y)dy$, and for every
$a\in\mathbb{R}$

$\hat{u}_{a}(t,x):=\hat{\mathcal{G}}_{a}(t,x):=\begin{cases}\left[\int_{a}^{t}|\phi(\Delta)^{1/2}\hat{T}_{t-s}f(s,\cdot)(x)|_{H}^{2}ds\right]^{1/2}&:\,t\geq
a\\\
\left[\int_{a}^{2a-t}|\phi(\Delta)^{1/2}\hat{T}_{2a-t-s}f(s,\cdot)(x)|_{H}^{2}ds\right]^{1/2}&:\,t<a.\end{cases}$

We use $\hat{u}(t,x)$ and $\hat{\mathcal{G}}(t,x)$ in place of
$\hat{u}_{0}(t,x)$ and $\hat{\mathcal{G}}_{0}(t,x)$ respectively. Obviously,
Lemmas 4.2 and 4.3 hold with $\hat{p}(t,x)$ instead of $p(t,x)$ (for all $t$).
Moreover,

$\hat{u}_{a}(t,x)=\hat{u}_{a}(2a-t,x)\quad\forall\,t\in\mathbb{R},\quad\quad\hat{u}_{a}(t,x)\leq
u_{a}(t,x)\quad\text{if}\,\,\,t\geq a,$ (5.2)
$\hat{u}_{a}(t,x)=u_{a}(t,x)\quad\text{if}\,\,\,t\in[a,T+a].$ (5.3)

######  Lemma 5.2.

Assume that the support of $f$ belongs to $\mathbb{R}\times B_{3dc}$. Then for
any $c>0$ and $(t,x)\in Q_{c}(r)$

$\displaystyle\int_{Q_{c}(r)}|\hat{u}_{a}(s,y)|^{2}~{}dsdy\leq
N[|r-a|+\phi(c^{-2})^{-1}]c^{d}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x),$

where $N$ depends only on $d$.

###### Proof.

Fix $(t,x)\in Q_{c}(r)$. Using (5.2) and Lemma 5.1, we get

$\displaystyle\int_{Q_{c}(r)}|\hat{u}_{a}(s,y)|^{2}~{}dsdy$
$\displaystyle\leq$ $\displaystyle\int_{(r-\phi(c^{-2})^{-1})\wedge
a}^{(r+\phi(c^{-2})^{-1})\vee
a}\int_{\mathbb{R}^{d}}|\hat{u}_{a}(s,y)|^{2}dyds$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}\left[\int_{(r-\phi(c^{-2})^{-1})\wedge
a}^{a}|\hat{u}_{a}(2a-s,y)|^{2}ds+\int_{a}^{(r+\phi(c^{-2})^{-1})\vee
a}|\hat{u}_{a}(s,y)|^{2}ds\right]dy$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{d}}\left[\int_{a}^{2a-[(r-\phi(c^{-2})^{-1})\wedge
a]}|\hat{u}_{a}(s,y)|^{2}ds+\int_{a}^{(r+\phi(c^{-2})^{-1})\vee
a}|\hat{u}_{a}(s,y)|^{2}ds\right]dy$ $\displaystyle\leq$ $\displaystyle
N\int_{B_{3dc}}\left[\int_{a}^{2a-[(r-\phi(c^{-2})^{-1})\wedge
a]}|f(s,y)|_{H}^{2}ds+\int_{a}^{(r+\phi(c^{-2})^{-1})\vee
a}|f(s,y)|_{H}^{2}ds\right]dy.$

Since $|x-y|\leq|x|+|y|\leq 4dc$ for any $(t,x)\in Q_{c}(r)$ and $y\in
B_{3dc}$, the last term above is less than or equal to constant times of

$\int_{|x-y|\leq 4dc}\left[\int_{a}^{2a-[(r-\phi(c^{-2})^{-1})\wedge
a]}|f(s,y)|_{H}^{2}ds+\int_{a}^{(r+\phi(c^{-2})^{-1})\vee
a}|f(s,y)|_{H}^{2}ds\right]dy$ $\displaystyle\leq$ $\displaystyle
Nc^{d}\left[\int_{a}^{2a-[(r-\phi(c^{-2})^{-1})\wedge
a]}\mathbb{M}_{x}|f(s,x)|_{H}^{2}ds+\int_{a}^{(r+\phi(c^{-2})^{-1})\vee
a}\mathbb{M}_{x}|f(s,x)|_{H}^{2}ds\right]$ $\displaystyle\leq$ $\displaystyle
N[|r-a|+\phi(c^{-2})^{-1}]c^{d}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

In order to explain the last inequality above, we denote

$\int_{a}^{2a-[(r-\phi(c^{-2})^{-1})\wedge
a]}\mathbb{M}_{x}|f(s,x)|_{H}^{2}ds+\int_{a}^{(r+\phi(c^{-2})^{-1})\vee
a}\mathbb{M}_{x}|f(s,x)|_{H}^{2}ds:=I+J.$

First we estimate $I$. $I=0$ if $r-\phi(c^{-2})^{-1}\geq a$. So assume
$r-\phi(c^{-2})^{-1}<a$.

If $a\leq t\leq 2a-(r-\phi(c^{-2})^{-1})$, then we can easily get

$I\leq[|r-a|+\phi(c^{-2})^{-1}]\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

If $t>2a-(r-\phi(c^{-2})^{-1})$ and $t\geq a$, then

$\displaystyle I\leq\int_{a}^{t+(t-a)}\mathbb{M}_{x}|f(s,x)|_{H}^{2}ds$
$\displaystyle\leq$ $\displaystyle
2(t-a)\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x)$ $\displaystyle\leq$
$\displaystyle
2(r+\phi(c^{-2})^{-1}-a)\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

Finally, if $t<a$, then

$\displaystyle I$ $\displaystyle\leq$
$\displaystyle\int_{t}^{2a-(r-\phi(c^{-2})^{-1})}\mathbb{M}_{x}|f(s,x)|_{H}^{2}ds\leq\int_{t+t-[2a-(r-\phi(c^{-2})^{-1})]}^{2a-(r-\phi(c^{-2})^{-1})}\mathbb{M}_{x}|f(s,x)|_{H}^{2}ds$
$\displaystyle\leq$ $\displaystyle
2([2a-(r-\phi(c^{-2})^{-1})]-t)\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x)$
$\displaystyle\leq$ $\displaystyle
4[a-(r-\phi(c^{-2})^{-1})]\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

The estimation of $J$ is similar. Therefore, the lemma is proved. $\Box$

We will use the following version of integration by parts: if
$0\leq\varepsilon\leq R\leq\infty$, and $F$ and $G$ are smooth enough then
(see [14])

$\displaystyle\int_{R\geq|z|\geq\varepsilon}F(z)G(|z|)~{}dz=-\int_{\varepsilon}^{R}G^{\prime}(\rho)(\int_{|z|\leq\rho}F(z)dz)d\rho$
$\displaystyle+G(R)\int_{|z|\leq
R}F(z)dz-G(\varepsilon)\int_{|z|\leq\varepsilon}F(z)dz.$ (5.4)

We generalize Lemma 5.2 as follows.

######  Lemma 5.3.

For any $(t,x)\in Q_{c}(r)$

$\displaystyle\int_{Q_{c}(r)}|\hat{u}_{a}(s,y)|^{2}~{}dsdy\leq
N[|r-a|+\phi(c^{-2})^{-1}]c^{d}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x),$

where $N=N(d,T,\phi)$.

###### Proof.

Take $\zeta\in C_{0}^{\infty}(\mathbb{R}^{d})$ such that $\zeta=1$ in
$B_{2dc}$, $\zeta=0$ outside of $B_{3dc}$, and $0\leq\zeta\leq 1$. Set
$\mathcal{A}=\zeta f$ and $\mathcal{B}=(1-\zeta)f$. By Minkowski’s inequality,
$\hat{\mathcal{G}}_{a}f\leq\hat{\mathcal{G}}_{a}\mathcal{A}+\hat{\mathcal{G}}_{a}\mathcal{B}$.
Since $\hat{\mathcal{G}}_{a}\mathcal{A}$ can be estimated by Lemma 5.2, assume
that $f(t,x)=0$ for $x\in B_{2dc}$. Let $e_{1}:=(1,0,\ldots,0)$ and
$s\geq\mu\geq a$. Then since $\phi(\Delta)^{1/2}\hat{p}(t,x)$ is rotationally
invariant with respect to $x$, we have

$\displaystyle|\phi(\Delta)^{1/2}\hat{T}_{s-\mu}f(\mu,\cdot)(y)|_{H}$ (5.5)
$\displaystyle=$
$\displaystyle|\int_{\mathbb{R}^{d}}\phi(\Delta)^{1/2}\hat{p}(s-\mu,|z|e_{1})f(\mu,y-z)~{}dz|_{H}$
$\displaystyle=$
$\displaystyle|\int_{0}^{\infty}\phi(\Delta)^{1/2}\hat{p}_{x^{1}}(s-\mu,\rho
e_{1})\int_{|z|\leq\rho}f(\mu,y-z)~{}d\rho|_{H}.$

For the second equality above, (5.4) is used with
$G(|z|)=\phi(\Delta)^{1/2}\hat{p}(s-\mu,|z|e_{1})$ and $F(z)=f(\mu,y-z)$.
Observe that if $y\in\hat{B}_{c}$ then for $x\in(-c/2,c/2)^{d}$

$\displaystyle|x-y|\leq 2dc,\quad B_{\rho}(y)\subset B_{2dc+\rho}(x).$

Moreover, if $|z|\leq c$, then $|y-z|\leq 2dc$ and $f(\mu,y-z)=0$. Thus by
Corollary 3.8 and Lemma 4.2,

$\displaystyle|\phi(\Delta)^{1/2}\hat{T}_{s-\mu}f(\mu,\cdot)(y)|_{H}$
$\displaystyle\leq$ $\displaystyle
N\int_{c}^{\infty}\frac{\phi(\rho^{-2})^{1/2}}{\rho^{d+1}}\int_{|z|\leq\rho}|f|_{H}(\mu,y-z)~{}dz\,d\rho$
$\displaystyle=$ $\displaystyle
N\int_{c}^{\infty}\frac{\phi(\rho^{-2})^{1/2}}{\rho^{d+1}}\int_{B_{\rho}(y)}|f|_{H}(\mu,z)~{}dz\,d\rho$
$\displaystyle\leq$ $\displaystyle
N\int_{c}^{\infty}\frac{\phi(\rho^{-2})^{1/2}}{\rho^{d+1}}\int_{B_{2dc+\rho}(x)}|f|_{H}(\mu,z)~{}dz\,d\rho$
$\displaystyle\leq$ $\displaystyle
N\int_{c}^{\infty}\frac{\phi(\rho^{-2})^{1/2}}{\rho^{d+1}}\int_{B_{2d\rho+\rho}(x)}|f|_{H}(\mu,z)~{}dz\,d\rho$
$\displaystyle\leq$ $\displaystyle
N\mathbb{M}_{x}|f|_{H}(\mu,x)\int_{c}^{\infty}\frac{\phi(\rho^{-2})^{1/2}}{\rho}~{}d\rho$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{1/2}\mathbb{M}_{x}|f|_{H}(\mu,x).$

By Jensen’s inequality
$(\mathbb{M}_{x}|f|_{H})^{2}\leq\mathbb{M}_{x}|f|_{H}^{2}$, and therefore, we
get for any $s\geq a$ and $y\in\hat{B}(c)$

$\displaystyle|\hat{u}_{a}(s,y)|^{2}$ $\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})\int_{a}^{s}\mathbb{M}_{x}|f|_{H}^{2}(\mu,x)d\mu.$

So if $r+\phi(c^{-2})^{-1}\geq s\geq a$, then we have

$\displaystyle|\hat{u}_{a}(s,y)|^{2}$ $\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})[r+\phi(c^{-2})^{-1}-(a\wedge(r-\phi(c^{-2})^{-1}))]\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x)$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})[|r-a|+\phi(c^{-2})^{-1}]\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

If $r-\phi(c^{-2})^{-1}\leq s<a$, then we get

$\displaystyle|\hat{u}_{a}(s,y)|^{2}=|\hat{u}_{a}(2a-s,y)|^{2}$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})\int_{a}^{2a-s}\mathbb{M}_{x}|f|_{H}^{2}(\mu,x)d\mu$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})[|r-a|+\phi(c^{-2})^{-1}]\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

Therefore, we get for any $(t,x)\in Q_{c}(r)$

$\displaystyle\int_{Q_{c}(r)}|\hat{u}_{a}(s,y)|^{2}~{}dsdy\leq
N[|r-a|+\phi(c^{-2})^{-1}]c^{d}\cdot\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

The lemma is proved. $\Box$

######  Lemma 5.4.

Assume $2\phi(c^{-2})^{-1}<r$. Then for any $(t,x)\in Q_{c}(r)$,

$\displaystyle\int_{Q_{c}(r)}\int_{r-\phi(c^{-2})^{-1}}^{r+\phi(c^{-2})^{-1}}|\hat{u}(s_{1},y)-\hat{u}(s_{2},y)|^{2}~{}ds_{1}ds_{2}dy$
$\displaystyle\leq
N\phi(c^{-2})^{-2}c^{d}\left[\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x)+\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x)+\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t-T,x)\right],$
(5.6)

where $N=N(d,T,\phi)$.

###### Proof.

Due to the symmetry, to estimate the left term of (5.6), we only consider the
case $s_{1}>s_{2}$. Since $r-\phi(c^{-2})^{-1}>\phi(c^{-2})^{-1}>0$, we have
$s_{2}>0$ for any $(s_{2},y)\in Q_{c}(r)$. Observe that by Minkowski’s
inequality

$\displaystyle|\hat{u}(s_{1},y)-\hat{u}(s_{2},y)|^{2}$ $\displaystyle=$
$\displaystyle|(\int_{0}^{s_{1}}|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa)^{1/2}-(\int_{0}^{s_{2}}|\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa)^{1/2}|^{2}$
$\displaystyle\leq$
$\displaystyle\int_{s_{2}}^{s_{1}}|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa$
$\displaystyle+\int_{0}^{s_{2}}|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}f(\kappa,\cdot)(y)-\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa$
$\displaystyle:=$ $\displaystyle I^{0}(s_{1},s_{2},y)+J^{0}(s_{1},s_{2},y).$

One can easily estimate $I^{0}$ using Lemma 5.3 with $a=s_{2}$ and $|r-a|\leq
2\phi(c^{-2})^{-1}$, and thus we only need to show that

$\int_{Q_{c}(r)}\int_{r-\phi(c^{-2})^{-1}}^{r+\phi(c^{-2})^{-1}}J^{0}(s_{1},s_{2},y)\,ds_{1}ds_{2}dy$

is less than or equal to the right hand side of (5.6). We divide $J^{0}$ into
two parts :

$\displaystyle
I:=\int_{0}^{r-2\phi(c^{-2})^{-1}}|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}f(\kappa,\cdot)(y)-\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa$
$\displaystyle
J:=\int_{r-2\phi(c^{-2})^{-1}}^{s_{2}}|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}f(\kappa,\cdot)(y)-\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa.$

Note that since $s_{1}\geq s_{2}\geq r-\phi(c^{-2})^{-1}$, we have $\eta
s_{1}+(1-\eta)s_{2}-\kappa\geq\phi(c^{-2})^{-1}$ for any $\eta\in[0,1]$ and
$\kappa\in[0,r-2\phi(c^{-2})^{-1}]$.

If $s_{1}-\kappa>T$ and $s_{2}-\kappa\leq T$ then

$|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}-\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}=|\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}.$

Otherwise, using $\frac{\partial}{\partial
t}\phi(\Delta)^{1/2}T_{t}f(x)=\phi(\Delta)^{3/2}T_{t}f(x)$, we get

$\displaystyle|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}f(\kappa,\cdot)(y)-\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}$
$\displaystyle\leq$
$\displaystyle(s_{1}-s_{2})^{2}\int_{0}^{1}|\phi(\Delta)^{3/2}\hat{T}_{\eta
s_{1}+(1-\eta)s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\eta.$

Therefore,

$\displaystyle I$ $\displaystyle\leq$
$\displaystyle\int_{s_{2}-T}^{s_{1}-T}|\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa$
$\displaystyle+$
$\displaystyle(s_{1}-s_{2})^{2}\int_{0}^{r-2\phi(c^{-2})^{-1}}\int_{0}^{1}|\phi(\Delta)^{3/2}\hat{T}_{\eta
s_{1}+(1-\eta)s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\eta d\kappa.$

Denote $\bar{s}=\bar{s}(\eta)=\eta s_{1}+(1-\eta)s_{2}$. As in (5.5),

$\displaystyle|\phi(\Delta)^{3/2}\hat{T}_{\bar{s}-\kappa}f(\kappa,\cdot)(y)|_{H}$
$\displaystyle=$
$\displaystyle|\int_{0}^{\infty}\phi(\Delta)^{3/2}\hat{p}_{x^{1}}(\bar{s}-\kappa,\rho
e_{1})\int_{|z|\leq\rho}f(\kappa,y-z)dzd\rho|_{H}$ $\displaystyle\leq$
$\displaystyle\mathbb{M}_{x}|f|_{H}(\kappa,y)\int_{0}^{\infty}|\phi(\Delta)^{3/2}\hat{p}_{x^{1}}(\bar{s}-\kappa,\rho
e_{1})|\rho^{d}\,d\rho.$

’Note that $\bar{s}-\kappa\geq\phi(c^{-2})^{-1}$ and so
$\phi^{-1}((\bar{s}-\kappa)^{-1})\leq c^{-2}$. By Lemma 4.3 and Corollary 3.8,

$\displaystyle\int_{0}^{\infty}|\phi(\Delta)^{3/2}\hat{p}_{x^{1}}(\bar{s}-\kappa,\rho
e_{1})|\rho^{d}\,d\rho$ $\displaystyle=$
$\displaystyle\int^{c}_{0}|\phi(\Delta)^{3/2}\hat{p}_{x^{1}}(\bar{s}-\kappa,\rho
e_{1})|\rho^{d}\,d\rho+\int^{\infty}_{c}|\phi(\Delta)^{3/2}\hat{p}_{x^{1}}(\bar{s}-\kappa,\rho
e_{1})|\rho^{d}\,d\rho$ $\displaystyle\leq$ $\displaystyle
N\int_{0}^{c}(\bar{s}-\kappa)^{-3/2}\phi^{-1}((\bar{s}-\kappa)^{-1})^{(d+1)/2}\rho^{d}d\rho+N(\bar{s}-\kappa)^{-1}\int^{\infty}_{c}\frac{\phi(\rho^{-2})^{1/2}}{\rho^{d+1}}\rho^{d}\,d\rho$
$\displaystyle\leq$ $\displaystyle
N(\bar{s}-\kappa)^{-1}\left[\phi(c^{-2})^{1/2}c^{-(d+1)}\int_{0}^{c}\rho^{d}d\rho+\phi(c^{-2})^{1/2}\right]\leq
N(\bar{s}-\kappa)^{-1}\phi(c^{-2})^{1/2}.$

Similarly, one can check

$|\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}\leq
N\phi(c^{-2})^{1/2}\mathbb{M}_{x}|f|_{H}(\kappa,y).$

Therefore, remembering $|s_{1}-s_{2}|\leq 2\phi(c^{-2})^{-1}$, we get

$\displaystyle I$ $\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-1}\int_{0}^{r-2\phi(c^{-2})^{-1}}(r-\phi(c^{-2})^{-1}-\kappa)^{-2}\mathbb{M}_{x}|f|^{2}_{H}(\kappa,y)d\kappa$
$\displaystyle+N\phi(c^{-2})\int_{s_{2}-T}^{s_{1}-T}\mathbb{M}_{x}|f|^{2}_{H}(\kappa,y)d\kappa.$

Note that $\mathbb{M}_{x}|f|^{2}_{H}(\kappa,y)$ in (5) can be replaced by
$I_{0<\kappa<r-2\phi(c^{-2})}$ times of it. Thus by integration by parts,

$\displaystyle\phi(c^{-2})^{-1}\int_{0}^{r-2\phi(c^{-2})^{-1}}(r-\phi(c^{-2})^{-1}-\kappa)^{-2}\mathbb{M}_{x}|f|^{2}_{H}(\kappa,y)d\kappa$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-1}\int_{-\infty}^{r-2\phi(c^{-2})^{-1}}(r-\phi(c^{-2})^{-1}-\kappa)^{-3}\int_{\kappa}^{r+\phi(c^{-2})^{-1}}\mathbb{M}_{x}|f|^{2}_{H}(\nu,y)d\nu
d\kappa$ $\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-1}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,y)\int_{-\infty}^{r-2\phi(c^{-2})^{-1}}\frac{r+\phi(c^{-2})^{-1}-\kappa}{({r-\phi(c^{-2})^{-1}}-\kappa)^{3}}d\kappa$
$\displaystyle=$ $\displaystyle
N\phi(c^{-2})^{-1}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,y)\int_{\phi(c^{-2})^{-1}}^{\infty}\frac{\kappa+2\phi(c^{-2})^{-1}}{\kappa^{3}}d\kappa$
$\displaystyle\leq$ $\displaystyle
N\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,y).$

Also,

$\displaystyle\phi(c^{-2})\int_{s_{2}-T}^{s_{1}-T}\mathbb{M}_{x}|f|^{2}_{H}(\kappa,y)d\kappa$
$\displaystyle\leq$
$\displaystyle\phi(c^{-2})\int_{r-\phi(c^{-2})^{-1}-T}^{r+\phi(c^{-2})^{-1}-T}\mathbb{M}_{x}|f|^{2}_{H}(\kappa,y)d\kappa$
$\displaystyle\leq$ $\displaystyle
2\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t-T,y).$

Therefore,

$I\leq
N[\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,y)+\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t-T,y)],$

where $N$ depends only on $d,T,a_{i},\delta_{i}$ $(i=1,2,3)$, and this
certainly implies

$\displaystyle\int_{\hat{B}_{c}}\int_{r-\phi(c^{-2})^{-1}}^{r+\phi(c^{-2})^{-1}}\int_{r-\phi(c^{-2})^{-1}}^{r+\phi(c^{-2})^{-1}}I~{}ds_{1}ds_{2}dy$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-2}c^{d}[\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x)+\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t-T,x)].$

It only remains to estimate $J$. Since $s_{1}\geq s_{2}$,

$\displaystyle\int_{r-2\phi(c^{-2})^{-1}}^{s_{2}}|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}f(\kappa,\cdot)(x)-\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa$
$\displaystyle\leq$ $\displaystyle
2\int_{r-2\phi(c^{-2})^{-1}}^{s_{1}}|\phi(\Delta)^{1/2}\hat{T}_{s_{1}-\kappa}f(\kappa,\cdot)(x)|_{H}^{2}d\kappa$
$\displaystyle+2\int_{r-2\phi(c^{-2})^{-1}}^{s_{2}}|\phi(\Delta)^{1/2}\hat{T}_{s_{2}-\kappa}f(\kappa,\cdot)(y)|_{H}^{2}d\kappa.$

Therefore, we are done by Lemma 5.3 with $a=r-2\phi(c^{-2})^{-1}$. $\Box$

######  Lemma 5.5.

Assume $2\phi(c^{-2})^{-1}<r$. Then for any $(t,x)\in Q_{c}(r)$,

$\displaystyle\int_{r-\phi(c^{-2})^{-1}}^{r+\phi(c^{-2})^{-1}}\int_{\hat{B}_{c}}\int_{\hat{B}_{c}}|\hat{u}(s,y_{1})-\hat{u}(s,y_{2})|^{2}~{}dy_{1}dy_{2}ds$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-1}c^{2d}[\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x)+\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x)].$

where $N=N(d,T,\phi)$.

###### Proof.

By Minkowski’s inequality,

$\displaystyle|u(s,y_{1})-u(s,y_{2})|^{2}$ $\displaystyle=$
$\displaystyle|(\int_{0}^{s}|\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,y_{1})|_{H}^{2}d\kappa)^{1/2}-(\int_{0}^{s}|\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,y_{2})|_{H}^{2}d\kappa)^{1/2}|^{2}$
$\displaystyle\leq$
$\displaystyle\int_{0}^{s}|\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,y_{1})-\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,y_{2})|_{H}^{2}d\kappa$
$\displaystyle\leq$
$\displaystyle\int_{0}^{r-2\phi(c^{-2})^{-1}}|\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,y_{1})-\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,y_{2})|_{H}^{2}d\kappa$
$\displaystyle+\int_{r-2\phi(c^{-2})^{-1}}^{s}|\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,y_{1})-\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,y_{2})|_{H}^{2}d\kappa$
$\displaystyle:=$ $\displaystyle I(s,y_{1},y_{2})+J(s,y_{1},y_{2}).$

By Lemma 5.3 with $a=r-2\phi(c^{-2})^{-1}$,

$\int_{r-\phi(c^{-2})^{-1}}^{r+\phi(c^{-2})^{-1}}\int_{\hat{B}_{c}}\int_{\hat{B}_{c}}J(s)~{}dy_{1}dy_{2}ds\leq
N\phi(c^{-2})^{-1}c^{2d}[\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

Therefore, we only need to estimate $I$. Let
$r-\phi(c^{-2})^{-1}<s<r+\phi(c^{-2})^{-1}$. Observe that for
$(s,y_{1}),(s,y_{2})\in Q_{c}(r)$

$I\leq
Nc^{2d}\int_{0}^{1}\int_{0}^{r-2\phi(c^{-2})^{-1}}|\nabla\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,\cdot)(\eta
y_{1}+(1-\eta)y_{2})|_{H}^{2}d\kappa d\eta.$ (5.8)

Recall that $\hat{p}(s-\kappa,y)=0$ if $s-\kappa>T$. Therefore if
$T+r-2\phi(c^{-2})^{-1}<s$, then

$c^{2d}\int_{0}^{1}\int_{0}^{r-2\phi(c^{-2})^{-1}}|\nabla\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,\cdot)(\eta
y_{1}+(1-\eta)y_{2})|_{H}^{2}d\kappa d\eta=0.$

So we assume $T+r-2\phi(c^{-2})^{-1}\geq s$, which certainly implies

$\displaystyle T\geq s-(r-2\phi(c^{-2})^{-1})\geq\phi(c^{-2})^{-1},\quad
c\leq\phi^{-1}(\frac{1}{T})^{-1/2}.$

Moreover from (3.20) and (3.21) we see that

$\displaystyle
c\leq\phi^{-1}(\frac{1}{T})^{-1/2}\leq\left((\frac{1}{a_{2}T})^{-1/(2\delta_{2})}\vee(\frac{1}{a_{3}T})^{-1/(2\delta_{3})}\right).$
(5.9)

Recall $a_{t}:=\left(\phi^{-1}(t^{-1})\right)^{-1/2}$ and
$t\phi(a_{t}^{-2})=1$. For simplicity, denote

$\bar{y}=\bar{y}(\eta)=\eta y_{1}+(1-\eta)y_{2}.$

As before, using (5.4), we get

$\displaystyle|\nabla\phi(\Delta)^{1/2}\hat{T}_{s-\kappa}f(\kappa,\cdot)(\bar{y})|_{H}$
$\displaystyle=$
$\displaystyle|\int_{0}^{\infty}\nabla\phi(\Delta)^{1/2}\hat{p}_{x^{1}}(s-\kappa,\rho
e_{1})\int_{|z|\leq\rho}f(\kappa,\bar{y}-z)dzd\rho|_{H}$ $\displaystyle\leq$
$\displaystyle|\int_{0}^{\infty}\nabla\phi(\Delta)^{1/2}\hat{p}_{x^{1}}(s-\kappa,\rho
e_{1})\int_{|\bar{y}-z|\leq\rho}f(\kappa,z)dzd\rho|_{H}$ $\displaystyle\leq$
$\displaystyle
N\mathbb{M}_{x}|f|_{H}(s,\bar{y})\sum_{i=1}^{d}\int_{0}^{\infty}|\phi(\Delta)^{1/2}\hat{p}_{x^{i}x^{1}}(s-\kappa,\rho
e_{1})|\rho^{d}d\rho$ $\displaystyle\leq$ $\displaystyle
N\mathbb{M}_{x}|f|_{H}(s,\bar{y})\sum_{i=1}^{d}[\int_{0}^{a_{s-\kappa}}|\phi(\Delta)^{1/2}\hat{p}_{x^{i}x^{1}}(s-\kappa,\rho
e_{1})|\rho^{d}d\rho$
$\displaystyle+\int_{a_{s-\kappa}}^{\infty}|\phi(\Delta)^{1/2}\hat{p}_{x^{i}x^{1}}(s-\kappa,\rho
e_{1})|\rho^{d}d\rho]$ $\displaystyle:=$ $\displaystyle I_{1}+I_{2}.$

By Lemma 4.2,

$\displaystyle I_{1}$ $\displaystyle\leq$ $\displaystyle
N\mathbb{M}_{x}|f|_{H}(s,\bar{y})(s-\kappa)^{-1/2}\phi^{-1}((s-\kappa)^{-1})^{(d+2)/2}\int_{0}^{a_{s-\kappa}}\rho^{d}d\rho$
(5.10) $\displaystyle\leq$ $\displaystyle
N\mathbb{M}_{x}|f|_{H}(s,\bar{y})(s-\kappa)^{-1/2}\phi^{-1}((s-\kappa)^{-1})^{1/2},$

and by Lemma 4.2 and Corollary 3.8,

$\displaystyle I_{2}$ $\displaystyle\leq$ $\displaystyle
N\mathbb{M}_{x}|f|_{H}(s,\bar{y})\int_{a_{s-\kappa}}^{\infty}\frac{\phi(\rho^{-2})^{1/2}}{\rho^{2}}~{}d\rho$
(5.11) $\displaystyle\leq$ $\displaystyle
N\mathbb{M}_{x}|f|_{H}(s,\bar{y})a_{s-\kappa}^{-1}\int_{a_{s-\kappa}}^{\infty}\frac{\phi(\rho^{-2})^{1/2}}{\rho}~{}d\rho$
$\displaystyle\leq$ $\displaystyle
N\mathbb{M}_{x}|f|_{H}(s,\bar{y})a_{s-\kappa}^{-1}\phi(a_{s-\kappa}^{-2})^{1/2}$
$\displaystyle=$ $\displaystyle
N\mathbb{M}_{x}|f|_{H}(s,\bar{y})(s-\kappa)^{-1/2}\phi^{-1}((s-\kappa)^{-1})^{1/2}.$

Therefore, using (5.10) and (5.11), and coming back to (5.8), we get

$\displaystyle
c^{2d}\int_{0}^{r-2\phi(c^{-2})^{-1}}\mathbb{M}_{x}|f|^{2}_{H}(\kappa,\bar{y})(s-\kappa)^{-1}\phi^{-1}((s-\kappa)^{-1})d\kappa$
(5.12) $\displaystyle\leq$ $\displaystyle
Nc^{2d}\int_{-\infty}^{r-2\phi(c^{-2})^{-1}}\mathbb{M}_{x}|f|^{2}_{H}(\kappa,\bar{y})(r-\phi(c^{-2})^{-1}-\kappa)^{-1}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\,\phi^{-1}((r-\phi(c^{-2})^{-1}-\kappa)^{-1})\,\,d\kappa$
$\displaystyle\leq$ $\displaystyle
Nc^{2d}\int_{\phi(c^{-2})^{-1}}^{\infty}\mathbb{M}_{x}|f|^{2}_{H}(r-\phi(c^{-2})^{-1}-\kappa,\bar{y})\kappa^{-1}\phi^{-1}(\kappa^{-1})d\kappa$
$\displaystyle\leq$ $\displaystyle
Nc^{2d}\int_{1}^{\infty}\mathbb{M}_{x}|f|^{2}_{H}(r-\phi(c^{-2})^{-1}-\phi(c^{-2})^{-1}\kappa,\bar{y})\kappa^{-1}\phi^{-1}(\phi(c^{-2})\kappa^{-1})d\kappa$
$\displaystyle\leq$ $\displaystyle
Nc^{2(d-1)}\int_{1}^{\infty}\mathbb{M}_{x}|f|^{2}_{H}(r-\phi(c^{-2})^{-1}-\phi(c^{-2})^{-1}\kappa,\bar{y})\kappa^{-2}d\kappa.$

For the last inequality above, we used Lemma 3.1. Indeed, for $\kappa\geq 1$
and $t>0$

$\phi(t)\kappa^{-1}=\phi(\kappa
t\kappa^{-1})\kappa^{-1}\leq\phi(t\kappa^{-1}),\quad\phi^{-1}(\phi(t)\kappa^{-1})\leq
t\kappa^{-1}.$

Note that $|f|^{2}_{H}(r-\phi(c^{-2})^{-1}-\phi(c^{-2})^{-1}\kappa,\bar{y})$
in (5.12) can be replaced by $I_{\kappa>1}$ times of it. Therefore, by
integration by parts

$\displaystyle
c^{2(d-1)}\int_{1}^{\infty}\int_{0}^{\kappa}\mathbb{M}_{x}|f|^{2}_{H}(r-\phi(c^{-2})^{-1}-\phi(c^{-2})^{-1}\nu,\bar{y})d\nu~{}\kappa^{-3}d\kappa$
$\displaystyle\leq$ $\displaystyle
Nc^{2(d-1)}\phi(c^{-2})\int_{1}^{\infty}\int_{r-\phi(c^{-2})^{-1}-\phi(c^{-2})^{-1}\kappa}^{r+\phi(c^{-2})^{-1}}\mathbb{M}_{x}|f|^{2}_{H}(\nu,\bar{y})d\nu~{}\kappa^{-3}d\kappa$
$\displaystyle\leq$ $\displaystyle
Nc^{2(d-1)}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,\bar{y})\phi(c^{-2})\int_{1}^{\infty}(2\phi(c^{-2})^{-1}+\phi(c^{-2})^{-1}\kappa)\kappa^{-3}d\kappa$
$\displaystyle\leq$ $\displaystyle
Nc^{2(d-1)}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,\bar{y})\leq
N\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,\bar{y}),$

where (5.9) is used for the last inequality. Thus,

$I(s,y_{1},y_{2})\leq
N\int_{0}^{1}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,\eta
y_{1}+(1-\eta)y_{2})d\eta,$

where $N$ depends only on $d,T,a_{i},\delta_{i}$ $(i=1,2,3)$. Finally, we
conclude that

$\displaystyle\int_{r-\phi(c^{-2})^{-1}}^{r+\phi(c^{-2})^{-1}}\int_{\hat{B}_{c}}\int_{\hat{B}_{c}}I(s,y_{1},y_{2})\,dy_{1}dy_{2}ds$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-1}\int_{0}^{1}\int_{\hat{B}_{c}}\int_{\hat{B}_{c}}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,\eta
y_{1}+(1-\eta)y_{2})dy_{1}dy_{2}d\eta$ $\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-1}\int_{0}^{1}\int_{\hat{B}_{c}}\int_{\eta\hat{B}_{c}+(1-\eta)y_{2}}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,y)dydy_{2}d\eta$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-1}\int_{\hat{B}_{c}}\int_{\hat{B}_{c}}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,y)dydy_{2}$
$\displaystyle\leq$ $\displaystyle
N\phi(c^{-2})^{-1}c^{2d}\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t,x).$

The lemma is proved. $\Box$

For a measurable function $h(t,x)$ on $\mathbb{R}^{d+1}$ define the sharp
function

$h^{\\#}(t,x):=\sup_{Q_{c}(r,z)\ni(t,x)}\frac{1}{|Q_{c}(r,z)|}\int_{Q_{c}(r,z)}|h(s,y)-h_{Q_{c}(r,z)}|~{}dsdy,$

where

$h_{Q_{c}(r,z)}(x)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q_{c}(r,z)}h(s,y)~{}dsdy:=\frac{1}{|Q_{c}(r,z)|}\int_{Q_{c}(r,z)}h(s,y)~{}dsdy.$

The following two theorems are classical results and can be found in [23].

######  Theorem 5.6.

(Hardy-Littlewood) For $1<p<\infty$ and $f\in L_{p}(\mathbb{R}^{d})$, we have

$\|\mathcal{M}_{x}f\|_{L_{p}(\mathbb{R}^{d})}+\|\mathbb{M}_{x}f\|_{L_{p}(\mathbb{R}^{d})}\leq
N(d,p,\phi)\|f\|_{L_{p}(\mathbb{R}^{d})}.$

######  Theorem 5.7.

(Fefferman-Stein). For any $1<p<\infty$ and $h\in L_{p}(\mathbb{R}^{d+1})$,

$\|h\|_{L_{p}(\mathbb{R}^{d+1})}\leq
N(d,p,\phi)\|h^{\\#}\|_{L_{p}(\mathbb{R}^{d+1})}.$

###### Proof.

We can get this result from Theorem IV.2.2 in [23]. Indeed, due to (3.27) we
can easily check that the balls $Q_{c}(s,y)$ satisfy the conditions (i)-(iv)
in section 1.1 of [23] :
(i) $Q_{c}(t,x)\cap Q_{c}(s,y)\neq\emptyset$ implies $Q_{c}(s,y)\subset
Q_{N_{1}c}(t,x)$ ;
(ii) $|Q_{N_{1}c}(t,x)|\leq N_{2}|Q_{c}(t,x)|$ ;
(iii) $\cap_{c>0}\overline{Q}_{c}(t,x)=\\{(t,x)\\}$ and
$\cup_{c}Q_{c}(t,x)=\mathbb{R}^{d+1}$ ;
(iv) for each open set $U$ and $c>0$, the function $(t,x)\to|Q_{c}(t,x)\cap
U|$ is continuous. $\Box$

Proof of Theorem 1.1.

First assume $f(t,x)=0$ if $t\not\in[0,T]$.

Since the theorem is already proved if $p=2$ in Lemma 5.1, we assume $p>2$.

First, we prove

$((\hat{\mathcal{G}}f)^{\\#})^{2}(t,x)\leq N(G(t,x)+G(-t,x)),$ (5.13)

where

$G(t,x):=\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,x)+\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H}(t,x)+\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2}(t-T,x).$

Because of Jensen’s inequality, to prove (5.13) it suffices to prove that for
each $Q_{c}(r,z)\in\mathcal{F}$ and $(t,x)\in Q_{c}(r,z)$,

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q_{c}(r,z)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q_{c}(r,z)}|\hat{\mathcal{G}}f(s_{1},y_{1})-\hat{\mathcal{G}}f(s_{2},y_{2})|^{2}~{}ds_{1}dy_{1}ds_{2}dy_{2}$
(5.14) $\displaystyle\leq$ $\displaystyle N(G(t,x)+G(-t,x)).$

To prove this we use translation and apply Lemma 5.3, 5.4, and 5.5.

By the definition of $\hat{\mathcal{G}}f$ and the fact
$\hat{T}_{t}g(y+z)=\hat{T}_{t}g(z+\cdot)(y)$, we see for $s\geq 0$

$\displaystyle\hat{\mathcal{G}}f(s,y+z)$ $\displaystyle=$
$\displaystyle[\int_{0}^{s}|\phi(\Delta)^{1/2}\hat{T}_{s-\rho}f(\rho,\cdot)(y+z)|_{H}^{2}d\rho]^{1/2}$
(5.15) $\displaystyle=$
$\displaystyle[\int_{0}^{s}|\phi(\Delta)^{1/2}\hat{T}_{s-\rho}f(\rho,z+\cdot)(y)|_{H}^{2}d\rho]^{1/2}$
$\displaystyle=$
$\displaystyle[\int_{0}^{s}|\phi(\Delta)^{1/2}\hat{T}_{s-\rho}f(\rho,z+\cdot)(y)|_{H}^{2}d\rho]^{1/2}$
$\displaystyle=$ $\displaystyle\hat{\mathcal{G}}f(\cdot,z+\cdot)(s,y),$

and

$\hat{\mathcal{G}}f(-s,y+z)=\hat{\mathcal{G}}f(s,y+z)=\hat{\mathcal{G}}f(\cdot,z+\cdot)(s,y)=\hat{\mathcal{G}}f(\cdot,z+\cdot)(-s,y).$

Therefore we get

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q_{c}(r,z)}|\hat{\mathcal{G}}\\{f(\cdot,\cdot)\\}(s,y)|^{2}~{}dyds$
$\displaystyle=$
$\displaystyle\frac{1}{|Q_{c}(r)|}\int_{Q_{c}(r)}|\hat{\mathcal{G}}\\{f(\cdot,\cdot)\\}(s,y+z)|^{2}~{}dyds$
$\displaystyle=$
$\displaystyle\frac{1}{|Q_{c}(r)|}\int_{Q_{c}(r)}|\hat{\mathcal{G}}\\{f(\cdot,z+\cdot)\\}(s,y)|^{2}~{}dyds.$

This shows that we may assume that $z=0$ and $Q_{c}(r,z)=Q_{c}(r)$.

If $|r|\leq 2\phi(c^{-2})^{-1}$, then (5.14) follows from Lemma 5.3. Also if
$r>2\phi(c^{-2})^{-1}$ then (5.14) follows from Lemmas 5.4 and 5.5. Therefore
it only remains to consider the case $r<-2\phi(c^{-2})^{-1}$. In this case,
(5.14) follows from the identity

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q_{c}(r)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q_{c}(r)}|\hat{\mathcal{G}}f(s_{1},y_{1})-\hat{\mathcal{G}}f(s_{2},y_{2})|^{2}~{}ds_{1}dy_{1}ds_{2}dy_{2}$
$\displaystyle=$
$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q_{c}(-r)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q_{c}(-r)}|\hat{\mathcal{G}}f(s_{1},y_{1})-\hat{\mathcal{G}}f(s_{2},y_{2})|^{2}~{}ds_{1}dy_{1}ds_{2}dy_{2}.$

This is because, for $r^{\prime}:=-r$, we have $r^{\prime}>2\phi(c^{-2})^{-1}$
and this case is already proved above. Thus we have proved (5.13).

Recall that $\mathcal{G}f(t,x)=\hat{\mathcal{G}}f(t,x)$ for $t\in[0,T]$. Thus
by Theorem 5.7 and (5.13)

$\displaystyle\|\mathcal{G}f\|_{L_{p}([0,T]\times\mathbb{R}^{d})}^{p}=\|I_{[0,T]}\hat{\mathcal{G}}f\|_{L_{p}(\mathbb{R}^{d+1})}^{p}\leq
N\|(\hat{\mathcal{G}}f)^{\\#}\|_{L_{p}(\mathbb{R}^{d+1})}^{p}$
$\displaystyle\leq$ $\displaystyle
N\left(\|(\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H})^{1/2}(t,x)\|_{L_{p}(\mathbb{R}^{d+1})}^{p}+\|(\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|^{2}_{H})^{1/2}(t,x)\|_{L_{p}(\mathbb{R}^{d+1})}^{p}\right),$

where for the last inequality we use (5.13) and the fact that the $L_{p}$-norm
is invariant under reflection and translation. Now we use Theorem 5.6 to get

$\displaystyle\int_{\mathbb{R}^{d}}\int_{-\infty}^{\infty}(\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2})^{p/2}~{}dtdx+\int_{-\infty}^{\infty}\int_{\mathbb{R}^{d}}(\mathcal{M}_{x}\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2})^{p/2}~{}dtdx$
$\displaystyle\leq$ $\displaystyle
N\int_{-\infty}^{\infty}\int_{\mathbb{R}^{d}}(\mathbb{M}_{x}|f|_{H}^{2})^{p/2}~{}dtdx+N\int_{\mathbb{R}^{d}}\int_{-\infty}^{\infty}(\mathbb{M}_{t}\mathbb{M}_{x}|f|_{H}^{2})^{p/2}~{}dtdx$
$\displaystyle\leq$ $\displaystyle
N\int_{-\infty}^{\infty}\int_{\mathbb{R}^{d}}(|f|_{H}^{2})^{p/2}~{}dtdx+N\int_{-\infty}^{\infty}\int_{\mathbb{R}^{d}}(\mathbb{M}_{x}|f|_{H}^{2})^{p/2}~{}dtdx$
$\displaystyle\leq$ $\displaystyle
N\|f\|_{L_{p}(\mathbb{R}^{d+1},H)}^{p}=N\|f\|_{L_{p}([0,T]\times\mathbb{R}^{d},H)}^{p}.$

Finally, for the general case, choose $\zeta_{n}\in C_{0}^{\infty}(0,T)$ such
that $|\zeta_{n}|\leq 1$, $\zeta_{n}=1$ on $[1/n,T-1/n]$. Then from the above
inequality and Lebesgue dominated convergence theorem we conclude that

$\displaystyle\|\mathcal{G}f\|_{L_{p}([0,T]\times\mathbb{R}^{d})}\leq\limsup_{n\to\infty}\|\mathcal{G}(f\zeta_{n})\|_{L_{p}([0,T]\times\mathbb{R}^{d})}$
$\displaystyle\leq\limsup_{n\to\infty}\|f\zeta_{n}\|_{L_{p}(\mathbb{R}^{d+1},H)}\leq\|f\|_{L_{p}([0,T]\times\mathbb{R}^{d},H)}.$

Hence the theorem is proved. $\Box$

##  6\. An application

In this section we construct an $L_{p}$-theory for the stochastic integro-
differential equation

$du=(\phi(\Delta)u+f)dt+g^{k}dw^{k}_{t},\quad u(0)=0.$ (6.1)

The condition $u(0)=0$ is not necessary and is assumed only for the simplicity
of the presentation.

Let $(\Omega,\mathcal{I},P)$ be a complete probability space, and
$\\{\mathcal{I}_{t},t\geq 0\\}$ be an increasing filtration of $\sigma$-fields
$\mathcal{I}_{t}\subset\mathcal{I}$, each of which contains all
$(\mathcal{I},P)$-null sets. By $\mathcal{P}$ we denote the predictable
$\sigma$-field generated by $\\{\mathcal{I}_{t},t\geq 0\\}$ and we assume that
on $\Omega$ we are given independent one-dimensional Wiener processes
$w^{1}_{t},w^{2}_{t},...$, each of which is a Wiener process relative to
$\\{\mathcal{I}_{t},t\geq 0\\}$.

For $\gamma\in\mathbb{R}$, denote
$H^{\phi,\gamma}_{p}=(1-\phi(\Delta))^{-\gamma/2}L_{p}$, that is $u\in
H^{\phi,\gamma}_{p}$ if

$\|u\|_{H^{\phi,\gamma}_{p}}=\|(1-\phi(\Delta))^{\gamma/2}u\|_{p}:=\|\mathcal{F}^{-1}\\{(1+\phi(|\xi|^{2}))^{\gamma/2}\mathcal{F}(u)(\xi)\\}\|_{p}<\infty,$
(6.2)

where $\mathcal{F}$ is the Fourier transform and $\mathcal{F}^{-1}$ is the
inverse Fourier transform. Similarly, for a $\ell_{2}$-valued function
$u=(u^{1},u^{2},\cdots)$ we define

$\|u\|_{H^{\phi,\gamma}_{p}(\ell_{2})}=\||\mathcal{F}^{-1}\\{(1+\phi(|\xi|^{2}))^{\gamma/2}\mathcal{F}(u)(\xi)\\}|_{\ell_{2}}\|_{p}.$

The following result can be found, for instance, in [6].

######  Lemma 6.1.

(i) For any $\mu,\gamma\in\mathbb{R}$, the map
$(1-\phi(\Delta))^{\mu/2}:H^{\phi,\gamma}_{p}\to H^{\phi,\gamma-\mu}_{p}$ is
an isometry.

(ii) For any $\gamma\in\mathbb{R}$, $H^{\phi,\gamma}_{p}$ is a Banach space.

(iii) If $\gamma_{1}\leq\gamma_{2}$, then $H^{\phi,\gamma_{2}}_{p}\subset
H^{\phi,\gamma_{1}}_{p}$ and

$\|u\|_{H^{\phi,\gamma_{1}}_{p}}\leq c\|u\|_{H^{\phi,\gamma_{2}}_{p}}.$

(iv) Let $\gamma\geq 0$. Then there is a constant $c>1$ so that

$c^{-1}\|u\|_{H^{\phi,\gamma}_{p}}\leq(\|u\|_{p}+\|\phi(\Delta)^{\gamma/2}u\|_{p})\leq
c\|u\|_{H^{\phi,\gamma}_{p}}.$

###### Proof.

(i) follows from definition (6.2). For (ii), it suffices to prove the
completeness. Let $\\{u_{n}:n=1,2,\cdots\\}$ be a Cauchy sequence in
$H^{\phi,\gamma}_{p}$. Then $f_{n}:=(1-\phi(\Delta))^{\gamma/2}u_{n}$ is a
Cauchy sequence in $L_{p}$, and there exists $f\in L_{p}$ so that $f_{n}\to f$
in $L_{p}$. Define $u:=(1-\phi(\Delta))^{-\gamma/2}f$. Then $u\in
H^{\phi,\gamma}_{p}$ and

$\|u_{n}-u\|_{H^{\phi,\gamma}_{p}}=\|f_{n}-f\|_{p}\to 0.$

Finally (iii) and (iv) are consequences of a Fourier multiplier theorem.
Indeed, due to Theorem 0.2.6 in [22], we only need to show that for any
$\gamma_{1}\leq 0$ and $\gamma_{2}\in\mathbb{R}$

$|D^{n}[1+\phi(|\xi|^{2})^{\gamma_{1}}]|+|D^{n}[\frac{(1+\phi(|\xi|^{2}))^{\gamma_{2}}}{1+\phi(|\xi|^{2})^{\gamma_{2}}}]|+|D^{n}[\frac{(\phi(|\xi|^{2}))^{\gamma_{2}}}{(1+\phi(|\xi|^{2}))^{\gamma_{2}}}]|\leq
N(n)|\xi|^{-n}.$

This comes from Lemma 3.2. The lemma is proved. $\Box$

Denote

$\mathbb{H}^{\phi,\gamma}_{p}(T)=L_{p}(\Omega\times[0,T],\mathcal{P},H^{\phi,\gamma}_{p}),\quad\mathbb{L}_{p}(T):=\mathbb{H}^{\phi,0}_{p}(T),$
$\mathbb{H}^{\phi,\gamma}_{p}(T,\ell_{2})=L_{p}(\Omega\times[0,T],\mathcal{P},H^{\phi,\gamma}_{p}(\ell_{2})).$

######  Definition 6.2.

We write $u\in\mathcal{H}^{\phi,\gamma+2}_{p,\theta}(T)$ if
$u\in\mathbb{H}^{\phi,\gamma+2}_{p,\theta}(T)$, $u(0)=0$ and for some
$f\in\mathbb{H}^{\phi,\gamma}_{p,\theta}(T)$ and
$g=(g^{1},g^{2},\cdots)\in\mathbb{H}^{\phi,\gamma+1}_{p}(T,\ell_{2})$, it
holds that

$du=fdt+g^{k}dw^{k}_{t}$

in the sense of distributions, that is for any $\varphi\in
C^{\infty}_{0}(\mathbb{R}^{d})$, the equality

$(u(t),\varphi)=\int^{t}_{0}(f(s),\varphi)ds+\sum_{k}\int^{t}_{0}(g^{k}(s),\varphi)dw^{k}_{s}$
(6.3)

holds for all $t\leq T$ (a.s.). In this case we write

$f=\mathbb{D}u,\quad g=\mathbb{S}u.$

The norm in $\mathcal{H}^{\phi,\gamma+2}_{p}(T)$ is given by

$\|u\|_{\mathcal{H}^{\phi,\gamma+2}_{p}(T)}=\|u\|_{\mathbb{H}^{\phi,\gamma+2}_{p}(T)}+\|\mathbb{D}u\|_{\mathbb{H}^{\phi,\gamma}_{p}(T)}+\|\mathbb{S}u\|_{\mathbb{H}^{\phi,\gamma+1}_{p}(T,\ell_{2})}.$

######  Remark 6.3.

As explained in [16, Remark 3.2], for any
$g\in\mathbb{H}^{\phi,\gamma+1}_{p}(T,\ell_{2})$ and $\varphi\in
C^{\infty}_{0}(\mathbb{R}^{d})$, the series of stochastic integral
$\sum_{k}(g^{k},\varphi)dw^{k}_{t}$ converges in probability uniformly in
$[0,T]$ and defines a continuous square integrable martingale on $[0,T]$.

######  Theorem 6.4.

For any $\gamma$ and $p\geq 2$, $\mathcal{H}^{\phi,\gamma+2}_{p}(T)$ is a
Banach space. Also, for any $u\in\mathcal{H}^{\phi,\gamma+2}_{p}(T)$, we have
$u\in C([0,T],H^{\phi,\gamma}_{p})$ (a.s.) and

${\mathbb{E}}\sup_{t\leq T}\|u(t,\cdot)\|^{p}_{H^{\phi,\gamma}_{p}}\leq
N\left(\|\mathbb{D}u\|^{p}_{\mathbb{H}^{\phi,\gamma}_{p}(T)}+\|\mathbb{S}u\|^{p}_{\mathbb{H}^{\phi,\gamma}_{p}(T,\ell_{2})}\right).\mathrm{}$
(6.4)

In particular, for $t\leq T$,

$\|u\|^{p}_{\mathbb{H}^{\phi,\gamma}_{p}(t)}\leq
N\int^{t}_{0}\|u\|^{p}_{\mathcal{H}^{\phi,\gamma+2}_{p}(s)}\,ds.$

###### Proof.

Since the operator
$(1-\phi(\Delta))^{\gamma/2}:\mathcal{H}^{\phi,\gamma+2}_{p}(T)\to\mathcal{H}^{\phi,2}_{p}(T)$
is an isometry, it suffices to prove the case $\gamma=0$. In this case
$H^{\phi,\gamma}_{p}=L_{p}$, and therefore (6.4) is proved, for instance, in
Theorem 3.4 of [9]. Also, the completeness of the space
$\mathcal{H}^{\phi,\gamma+2}_{p}(T)$ can be proved using (6.4) as in the proof
of Theorem 3.4 of [9]. $\Box$

The following maximal principle will be used to prove the uniqueness result of
equation (6.1).

######  Lemma 6.5.

Let $\lambda>0$ be a constant. Suppose that $u$ is continuous in
$[0,T]\times\mathbb{R}^{d}$, $u(t,\cdot)\in C^{2}_{b}(\mathbb{R}^{d})$ for
each $t>0$, $u_{t},\phi(\Delta)u$ are continuous in
$(0,T]\times\mathbb{R}^{d})$, $u_{t}-\phi(\Delta)u+\lambda u=0$ for
$t\in(0,T]$, $u(t_{n},x)\to u(t,x)$ as $t_{n}\to t$ uniformly for
$x\in\mathbb{R}^{d}$, $u(0,x)=0$ for all $x\in\mathbb{R}^{d}$, and for each
$t$ $u(t,x)\to 0$ as $|x|\to\infty$. Then $u\equiv 0$ in
$[0,T]\times\mathbb{R}^{d}$.

###### Proof.

Suppose $\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}u(t,x)>0$. Then we claim that
there exists a $(t_{0},x_{0})\in[0,T]\times\mathbb{R}^{d}$ such that
$u(t_{0},x_{0})=\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}u(t,x)$. The
explanation is as follows. Since there exists a sequence $(t_{n},x_{n})$ such
that $u(t_{n},x_{n})\to\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}u(t,x)$, one
can choose a subsequence $t_{n_{k}}$ such that $t_{n_{k}}\to t_{0}$ for some
$t_{0}\in[0,T]$ as $k\to\infty$. If $\\{x_{n}\\}$ is unbounded, then there
exists a subsequence $x_{n_{k}}$ such that $|x_{n_{k}}|\to\infty$. Due to the
assumption : $u(t_{n},x)\to u(t,x)$ as $t_{n}\to t$ uniformly for
$x\in\mathbb{R}^{d}$, we have $u(t_{n_{k}},x_{n_{k}})\to u(t_{0},x_{n_{k}})$
as $k\to\infty$. But since $u(t_{0},x_{n_{k}})\to 0$ as $k\to\infty$ this is
contradiction to the fact
$u(t_{n_{k}},x_{n_{k}})\to\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}u(t,x)>0$.
Therefore, $\\{x_{n}\\}$ is bounded and this means that $x_{n}$ also has a
subsequence $x_{n_{k}}$ such that $x_{n_{k}}\to x_{0}$ for some
$x_{0}\in\mathbb{R}^{d}$. So we know that our claim is true. Note that since
$u(0,x)=0~{}\forall x\in\mathbb{R}^{d}$, $t_{0}>0$ and if $t_{0}\in(0,T)$ then
$u_{t}(t_{0},x_{0})=0$ otherwise $u_{t}(T,x_{0})\geq 0$. Recall that

$\displaystyle\phi(\Delta)u(t,x)=\lim_{\varepsilon\downarrow
0}\int_{\\{y\in\mathbb{R}^{d}:\,|y|>\varepsilon\\}}(u(t,x+y)-u(t,x))j(|y|)\,dy$

and $j$ is strictly positive. Therefore, we get $(u_{t}-\phi(\Delta)u+\lambda
u)(t_{0},x_{0})>0$ and this is contradiction to our assumption. So we have
$\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}u(t,x)\leq 0$. Similarly, we can
easily show that $\inf_{(t,x)\in[0,T]\times\mathbb{R}^{d}}u(t,x)\geq 0$. The
lemma is proved. $\Box$

The following result will be used to estimate the deterministic part of (6.1).

######  Lemma 6.6.

Let $m(\tau,\xi):=\frac{\phi(|\xi|^{2})}{i\tau+\phi(|\xi|^{2})}$. Then, $m$ is
a $L_{p}(\mathbb{R}^{d+1})$-multiplier. In other words,

$\|\mathcal{F}^{-1}(m\mathcal{F}f)\|_{L_{p}(\mathbb{R}^{d+1})}\leq
N\|f\|_{L_{p}(\mathbb{R}^{d+1})},\quad\forall f\in L_{p}(\mathbb{R}^{d+1}),$

where $N$ depends only on $d$ and $p$.

###### Proof.

First we estimate derivatives of $m$. Let
$\alpha=(\alpha_{1},\cdots,\alpha_{d})\neq 0$ be a $d$-dimensional multi-index
with $\alpha_{i}=0$ or $1$ for $i=1,\ldots,d$. Assume $\beta,\gamma$ are
multi-indices so that $\beta+\gamma=\alpha$. Then from Lemma 3.2 we can easily
get

$\displaystyle|D^{\beta}(\phi(|\xi|^{2}))|\leq
N\phi(|\xi|^{2})|\xi|^{-|\beta|}.$ (6.5)

Suppose $\gamma\neq 0$. Without loss of generality assume $\gamma_{1}=1$. Then
by Leibniz’s rule and (6.5), we get

$\displaystyle|D^{\gamma}(i\tau+\phi(|\xi|^{2}))^{-1}|$ (6.6) $\displaystyle=$
$\displaystyle|D^{\gamma^{\prime}}D^{\gamma_{1}}(i\tau+\phi(|\xi|^{2}))^{-1}|$
$\displaystyle=$
$\displaystyle|D^{\gamma^{\prime}}(i\tau+\phi(|\xi|^{2}))^{-2}D_{\xi^{1}}(\phi(|\xi|^{2}))|$
$\displaystyle\leq$ $\displaystyle
N\sum_{\bar{\gamma}^{\prime}+\hat{\gamma}^{\prime}=\gamma^{\prime}}|(i\tau+\phi(|\xi|^{2}))^{-(|\bar{\gamma}^{\prime}|+2)}$
$\displaystyle\quad\quad\quad\quad\quad\times[D_{\xi^{1}}(\phi(|\xi|^{2}))]^{\bar{\gamma}_{1}^{\prime}}\cdots[D_{\xi^{d}}(\phi(|\xi|^{2}))]^{\bar{\gamma}_{d}^{\prime}}D^{\hat{\gamma}^{\prime}}D_{\xi^{1}}(\phi(|\xi|^{2}))|$
$\displaystyle\leq$ $\displaystyle
N\sum_{\bar{\gamma}^{\prime}+\hat{\gamma}^{\prime}=\gamma^{\prime}}|\tau^{2}+\phi(|\xi|^{2})^{2}|^{-(|\bar{\gamma}^{\prime}|+2)/2}\phi(|\xi|^{2})^{|\bar{\gamma}^{\prime}|}|\xi|^{-|\bar{\gamma}^{\prime}|}\phi(|\xi|^{2})|\xi|^{-(|\hat{\gamma}^{\prime}|+1)}$
$\displaystyle\leq$ $\displaystyle N\
|\tau^{2}+\phi(|\xi|^{2})^{2}|^{-1/2}|\xi|^{-|\gamma|}.$

Obviously even if $\gamma=0$, (6.6) holds. Therefore from (6.5) and (6.6), we
get

$\displaystyle|D^{\alpha}m(\tau,\xi)|\leq
N\frac{\phi(|\xi|^{2})}{|\tau|+\phi(|\xi|^{2})}|\xi|^{-|\alpha|}.$ (6.7)

Next let
$\hat{\alpha}=(\alpha_{0},\alpha_{1},\ldots,\alpha_{d})=(\alpha_{0},\alpha)$
be a $(d+1)$-dimensional multi-index with $\alpha_{0}\neq 0$ and
$\alpha_{i}=0$ or $1$ for $i=1,\ldots,d$. Then from (6.7) we get

$\displaystyle|D^{\hat{\alpha}}m(\tau,\xi)|$ $\displaystyle\leq$
$\displaystyle N(|\tau|+\phi(|\xi|^{2}))^{-\alpha_{0}}|D^{\alpha}m(\tau,\xi)|$
(6.8) $\displaystyle\leq$ $\displaystyle
N\frac{\phi(|\xi|^{2})}{(|\tau|+\phi(|\xi|^{2}))^{\alpha_{0}+1}}|\xi|^{-|\alpha|}.$

Now to conclude that $m$ is a multiplier, we use Theorem 4.6’ in p 109 of
[24]. Due to (6.8), we see that for each $0<k\leq d+1$

$\displaystyle|\frac{\partial^{k}m}{\partial\tau\partial\xi_{1}\cdots\partial\xi_{k-1}}|\leq
N\frac{\phi(|\xi|^{2})}{(|\tau|+\phi(|\xi|^{2}))^{2}}|\xi|^{-|k-1|}.$

Therefore, for any dyadic rectangles $A=\Pi_{1\leq i\leq
k}[2^{k_{i}},2^{k_{i}+1}]$, we have

$\displaystyle\int_{\Pi}|\frac{\partial^{k}m}{\partial x_{1}\partial
x_{2}\cdots\partial x_{k}}|d\tau d\xi_{1}\cdots\xi_{k-1}\leq N.$

We can easily check that from (6.7) and (6.8), the above statement is also
valid for every one of the $n!$ permutations of the variables
$\tau,\xi_{1},\ldots,\xi_{d}$. The lemma is proved. $\Box$

Here is our $L_{p}$-theory.

######  Theorem 6.7.

For any $f\in\mathbb{H}^{\phi,\gamma}_{p}(T)$ and
$g=(g^{1},g^{2},\cdots)\in\mathbb{H}^{\phi,\gamma+1}_{p}(T,\ell_{2})$,
equation (6.1) has a unique solution $u\in\mathcal{H}^{\phi,\gamma+2}_{p}(T)$,
and for this solution

$\|u\|_{\mathcal{H}^{\phi,\gamma+2}_{p}(T)}\leq
N\|f\|_{\mathbb{H}^{\phi,\gamma}_{p}(T)}+N\|g\|_{\mathbb{H}^{\phi,\gamma+1}_{p}(T,\ell_{2})}.$
(6.9)

###### Proof.

Due to the isometry, we may assume $\gamma=0$. Note that if $u$ is a solution
of (6.1) in $\mathcal{H}^{\phi,2}_{p}(T)$, then we have $u\in C([0,T],L_{p})$
(a.s.) by Theorem 6.4.

Step 1. First we prove the uniqueness of Equation (6.1). Let $u_{1}$, $u_{2}$
be solutions of equation (6.1). Then putting $v:=u_{1}-u_{2}$, we see that $v$
satisfies (6.1) with $f=g^{k}=0$.

Take a non-negative smooth function $\varphi\in C^{\infty}_{0}$ with unit
integral. For $\varepsilon>0$, define
$\varphi_{\varepsilon}(x)=\varepsilon^{-d}\varphi(x/\varepsilon)$. Also denote
$w(t,x):=e^{-\lambda t}u(t,x)$ and $w^{\varepsilon}=w*\varphi_{\varepsilon}$.
Then by plugging $\varphi_{\varepsilon}(\cdot-x)$ in (6.3) in place of
$\varphi$, we have $w^{\varepsilon}_{t}-\phi(\Delta)w^{\varepsilon}+\lambda
w^{\varepsilon}=0$. Also one can easily check that $w^{\varepsilon}$ satisfies
the conditions in Lemma 6.5 and concludes that $w^{\varepsilon}\equiv 0$. This
certainly proves $v\equiv 0$ (a.s.).

Step 2. We consider the case $g=0$. By approximation argument (see the next
step for the detail), to prove the existence and (6.9), we may assume that $f$
is sufficiently smooth in $x$ and vanishes if $|x|$ is sufficiently large. In
this case, one can easily check that for each $\omega$,

$u(t,x)=\int^{t}_{0}T_{t-s}f(s,x)ds$ (6.10)

satisfies $u_{t}=\phi(\Delta)u+f$. In addition to it, denoting
$\bar{p}(t,x)=I_{0\leq t}p(t,x)$ and $\bar{f}=I_{0\leq t\leq T}f(t,x)$, we see
that for $(t,x)\in[0,T]\times\mathbb{R}^{d}$

$\displaystyle u(t,x)=\bar{p}(\cdot,\cdot)\ast\bar{f}(\cdot,\cdot)(t,x).$
(6.11)

We use notation $\mathcal{F}_{d}$ and $\mathcal{F}_{d+1}$ to denote the
Fourier transform for $x$ and $(t,x)$, respectively. Moreover for convenience
we put $\mathcal{F}_{d}u(t,x)=\mathcal{F}_{d}(u(t,\cdot))(x)$. Under this
setting, observe that

$\displaystyle\mathcal{F}_{d+1}(\bar{p})(\tau,\xi)=\int_{\mathbb{R}}e^{-i\tau
t}\mathcal{F}_{d}(\bar{p})(t,\xi)dt=\int_{0}^{\infty}e^{-i\tau
t}e^{-t\phi(|\xi|^{2})}dt=\frac{1}{i\tau+\phi(|\xi|^{2})}.$

So denoting $\bar{u}=\bar{p}(\cdot,\cdot)\ast\bar{f}(\cdot,\cdot)(t,x)$ we see

$\displaystyle\mathcal{F}_{d+1}^{-1}[(1+\phi(|\xi|^{2}))\mathcal{F}_{d+1}\bar{u}](t,x)$
$\displaystyle=$
$\displaystyle\bar{u}+\mathcal{F}_{d+1}^{-1}[\phi(|\xi|^{2})\mathcal{F}_{d+1}(\bar{p})\mathcal{F}_{d+1}(\bar{f})]$
(6.12) $\displaystyle=$
$\displaystyle\bar{u}+\mathcal{F}_{d+1}^{-1}[\frac{\phi(|\xi|^{2})}{i\tau+\phi(|\xi|^{2})}\mathcal{F}_{d+1}(\bar{f})].$

Due to generalized Minkowski’s inequality, we can easily check that
$\|\bar{u}\|_{L_{p}(\mathbb{R}^{d+1})}\leq
N\|f\|_{L_{p}([0,T]\times\mathbb{R}^{d})}$. Moreover we know that
$\frac{\phi(|\xi|^{2})}{i\tau+\phi(|\xi|^{2})}$ is a
$L_{p}(\mathbb{R}^{d+1})$-multiplier from Lemma 6.6. Therefore, from (6.11) we
conclude that

$\displaystyle\|u\|_{\mathbb{H}^{\phi,2}_{p}(T)}\leq\|f\|_{\mathbb{L}_{p}(T)}.$

Step 3. We consider the case $f=0$. First, assume that $g^{k}=0$ for all
sufficiently large $k$ (say for all $k\geq N_{0}$), and each $g^{k}$ is of the
type

$g^{k}(t,x)=\sum_{i=0}^{m_{k}}{\bf
1}_{(\tau^{k}_{i},\tau^{k}_{i+1}]}(t)g^{k_{i}}(x)\qquad\text{for }k\leq
N_{0},$ (6.13)

where $\tau^{k}_{i}$ are bounded stopping times and $g^{k_{i}}(x)\in
C^{\infty}_{0}(\mathbb{R}^{d})$. Define

$v(t,x):=\sum_{k=1}^{N_{0}}\int^{t}_{0}g^{k}(s,x)dw^{k}_{s}=\sum_{k=1}^{N_{0}}\sum_{i=1}^{m_{k}}g^{k_{i}}(x)(w^{k}_{t\wedge\tau^{k}_{i+1}}-w^{k}_{t\wedge\tau^{k}_{i}})$

and

$u(t,x):=v(t,x)+\int^{t}_{0}\phi(\Delta)T_{t-s}v(s,x)\,ds=v(t,x)+\int^{t}_{0}T_{t-s}\phi(\Delta)v(s,x)\,ds.$
(6.14)

Then $u-v=\int^{t}_{0}T_{t-s}\phi(\Delta)v(s,x)ds$, and therefore (see (6.10))
we have

$(u-v)_{t}=\phi(\Delta)(u-v)+\phi(\Delta)v=\phi(\Delta)u,$

and

$du=d(u-v)+dv=\phi(\Delta)udt+\sum_{k=1}^{N_{0}}g^{k}dw^{k}_{t}.$

Also by (6.14) and stochastic Fubini theorem ([18, Theorem 64]), almost
surely,

$\displaystyle u(t,x)$ $\displaystyle=$ $\displaystyle
v(t,x)+\sum_{k=1}^{N_{0}}\int^{t}_{0}\int^{s}_{0}\phi(\Delta)T_{t-s}g^{k}(r,x)dw^{k}_{r}ds$
$\displaystyle=$ $\displaystyle
v(t,x)-\sum_{k=1}^{N_{0}}\int^{t}_{0}\int^{t}_{r}\frac{\partial}{\partial
s}T_{t-s}g^{k}(r,x)dsdw^{k}_{r}$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{N_{0}}\int^{t}_{0}T_{t-s}g^{k}(s,x)dw^{k}_{s}.$

Hence,

$\phi(\Delta)u(t,x)=\sum_{k=1}^{N_{0}}\int^{t}_{0}\phi(\Delta)^{1/2}T_{t-s}\phi(\Delta)^{1/2}g^{k}(s,\cdot)(x)dw^{k}_{s},$

and by Burkholder-Davis-Gundy’s inequality, we have

${\mathbb{E}}\left[\big{|}\phi(\Delta)u(t,x)\big{|}^{p}\right]\leq
c(p){\mathbb{E}}\left[\left(\int^{t}_{0}\sum_{k=1}^{N_{0}}|\phi(\Delta)^{1/2}T_{t-s}\phi(\Delta)^{1/2}g^{k}(s,\cdot)(x)|^{2}ds\right)^{p/2}\right].$

Also, similarly we get

${\mathbb{E}}\left[\big{|}u(t,x)\big{|}^{p}\right]\leq
c(p){\mathbb{E}}\left[\left(\int^{t}_{0}\sum_{k=1}^{N_{0}}|T_{t-s}g^{k}(s,\cdot)(x)|^{2}ds\right)^{p/2}\right].$

Now it is enough to use Theorem 1.1 and Lemma 6.1 to conclude

$\|u\|_{\mathbb{H}^{\phi,2}_{p}(T)}\leq
N\|g\|_{\mathbb{H}^{\phi,1}_{p}(T,\ell_{2})}.$ (6.15)

For general $g$, take a sequence $g_{n}$ so that $g_{n}\to g$ in
$\mathbb{H}^{\phi,1}_{p}(T,\ell_{2})$ and each $g_{n}$ satisfies above
described conditions. Then, by the above result, for
$u_{n}:=\int^{t}_{0}T_{t-s}g^{k}_{n}dw^{k}_{s}$, we have
$du_{n}=\phi(\Delta)u_{n}dt+g^{k}_{n}dw^{k}_{t}$, and

$\|u_{n}\|_{\mathbb{H}^{\phi,2}_{p}(T)}\leq
N\|g_{n}\|_{\mathbb{H}^{\phi,1}_{p}(T,\ell_{2})},$
$\|u_{n}-u_{m}\|_{\mathbb{H}^{\phi,2}_{p}(T)}\leq
N\|g_{n}-g_{m}\|_{\mathbb{H}^{\phi,1}_{p}(T,\ell_{2})}.$

Thus $u_{n}$ is a Cauchy sequence in $\mathcal{H}^{\phi,2}_{p}(T)$ and
converges to a certain function $u\in\mathcal{H}^{\phi,2}_{2}(T)$. One easily
gets (6.15) by letting $n\to\infty$, and by Theorem 6.4 it also follows
${\mathbb{E}}\sup_{t\leq T}\|u_{n}-u_{m}\|^{p}_{L_{p}}\to 0$ as
$n,m\to\infty$. Finally by taking the limit from

$(u_{n}(t),\varphi)=\int^{t}_{0}(\phi(\Delta)u_{n},\varphi)ds+\sum_{k}\int^{t}_{0}(g^{k}_{n},\varphi)dw^{k}_{s},\quad\forall\,t\leq
T\,(a.s.)$

and remembering ${\mathbb{E}}\sup_{t\leq T}\|u_{n}-u\|^{p}_{L_{p}}\to 0$, we
prove that $u$ satisfies

$(u(t),\varphi)=\int^{t}_{0}(\phi(\Delta)u,\varphi)ds+\sum_{k}\int^{t}_{0}(g^{k},\varphi)dw^{k}_{s},\quad\forall\,t\leq
T\,(a.s.)$

Step 4. General case. The uniqueness follows from Step 1. For the existence
and the estimate it is enough to add the solutions in Steps 2 and 3. The
theorem is proved.

$\Box$

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