# Parabolic Anderson model with rough initial condition: continuity in law of
the solution

Xiao Liang111University of Ottawa, Department of Mathematics and Statistics,
150 Louis Pasteur Private, Ottawa, Ontario, K1G 0P8, Canada. E-mail address:
<EMAIL_ADDRESS>
(December, 2023)

###### Abstract

The parabolic Anderson model (PAM) is one of the most interesting and
challenging SPDEs related to various physical phenomena, and can be described
mathematically as a stochastic heat equation driven by linear multiplicative
noise. In this paper, we consider PAM with initial condition given by a signed
Borel measure on $\mathbb{R}^{d}$. The forcing term under investigation is
examined in two cases: (i) the regular noise, with the spatial covariance
given by the Riesz kernel of order $\alpha\in(0,d)$ in spatial dimension
$d\geq 1$; (ii) the rough noise, which is a fractional noise in space with
Hurst index $H<1/2$ and $d=1$. In both cases, the noise is assumed to be
colored in time and we consider a general initial condition. The objective of
this article is to show that the solution is continuous in law with respect to
the spatial noise parameter. The similar problem for the constant initial
condition has been recently studied in [6].

MSC 2020: Primary 60H15; Secondary 60G60, 60H07

Keywords: stochastic partial differential equations, random fields, rough
noise, rough initial condition

##  1 Introduction and Main Results

In this article, we consider the following stochastic heat equation:

$\begin{cases}\dfrac{\partial u}{\partial
t}(t,x)=\dfrac{1}{2}\dfrac{\partial^{2}u}{\partial
x^{2}}(t,x)+u(t,x)\dot{W}(t,x),\quad t>0,x\in\mathbb{R}^{d}\\\\[8.00003pt]
u(0,\cdot)=u_{0}(\cdot),\end{cases}$ (1)

with initial condition given by a signed Borel measure $u_{0}$ on
$\mathbb{R}^{d}$ such that

$\int_{\mathbb{R}^{d}}e^{-a|x|^{2}}|u_{0}|(\mathrm{d}x)<\infty,\text{ for all
}a>0.$ (2)

This general initial condition, known as “rough initial condition”, was
introduced in [7], and has been considered in the analysis of the stochastic
heat equation in several other papers; see e.g. [3, 8, 9]. Here
$|u_{0}|:=u_{0,+}+u_{0,-}$, where $u_{0}=u_{0,+}-u_{0,-}$ is the Jordan
decomposition and $u_{0,\pm}$ are two non-negative Borel measure with disjoint
support.

We assume that $W$ is a space-time homogeneous Gaussian noise. More precisely,
$W=\\{W(\varphi);\varphi\in
C_{0}^{\infty}(\mathbb{R}_{+}\times\mathbb{R}^{d})\\}$ is a zero-mean Gaussian
process, defined on a complete probability space
$(\Omega,\mathcal{F},\mathbb{P})$, with the following covariance structure:

$\mathbb{E}[W(\varphi)W(\psi)]=\int_{\mathbb{R}_{+}^{2}\times\mathbb{R}^{d}}\mathcal{F}\varphi(t,\cdot)(\xi)\overline{\mathcal{F}\psi(s,\cdot)(\xi)}\gamma_{0}(t-s)dtds\mu(d\xi)=:\langle\varphi,\psi\rangle_{\mathcal{H}},$
(3)

where $\mathcal{F}\varphi(t,\cdot)(\xi)=\int_{\mathbb{R}^{d}}e^{-i\xi\cdot
x}\varphi(t,x)dx$ is the Fourier transform of the function $\varphi(t,\cdot)$,
$\xi\cdot x$ is the Euclidean product in $\mathbb{R}^{d}$, and $\gamma_{0}$ is
a non-negative and non-negative definite, i.e. for any
$\phi\in\mathcal{S}(\mathbb{R})$,

$\int_{\mathbb{R}}(\phi\ast\widetilde{\phi})(t)\gamma_{0}(t)dt\geq 0.$

We denote by $\mathcal{H}$ the Hilbert space defined as the completion of
$C_{0}^{\infty}(\mathbb{R}_{+}\times\mathbb{R}^{d})\\}$ with respect to the
inner product $\langle\cdot,\cdot\rangle_{\mathcal{H}}$. In this setting, the
noise $W$ can be extended to an isonormal Gaussian process
$\\{W(\varphi);\varphi\in\mathcal{H}\\}$, and the solution can be defined
using Malliavin calculus.

By the Bochner-Schwartz theorem, there exists a tempered measure $\mu_{0}$ on
$\mathbb{R}$ such that $\gamma_{0}$ is the Fourier transform of $\mu_{0}$ in
the space $\mathcal{S}_{\mathbb{C}}^{\prime}(\mathbb{R})$ of tempered
distributions on $\mathbb{R}$. Then,

$\int_{\mathbb{R}^{2}}\phi(t)\phi(s)\gamma_{0}(t-s)dtds=\int_{\mathbb{R}}|\mathcal{F}\phi(\tau)|^{2}\mu_{0}(d\tau),$
(4)

for any $\phi\in\mathcal{S}_{\mathbb{C}}(\mathbb{R})$, where
$\mathcal{S}_{\mathbb{C}}(\mathbb{R})$ is the space of $\mathbb{C}$-valued
rapidly decreasing $C^{\infty}$-functions on $\mathbb{R}$, and
$\mathcal{F}\phi$ is the Fourier transform of $\phi$, given by
$\mathcal{F}\phi(\tau)=\int_{\mathbb{R}}e^{-i\tau t}\phi(t)dt$ for all
$\tau\in\mathbb{R}$. We assume that there exists a function
$g_{0}:\mathbb{R}\to[0,\infty]$ such that

$\mu_{0}(d\tau)=g_{0}(\tau)d\tau.$ (5)

We will be interested in the temporal covariance function

$\gamma_{0}(t)=\alpha_{H_{0}}|t|^{2H_{0}-2}$ (6)

with $H_{0}\in(1/2,1)$, in which case
$g_{0}(\tau)=c_{H_{0}}|\tau|^{1-2H_{0}}$, with
$\alpha_{H_{0}}=H_{0}(2H_{0}-1)$ and

$c_{H_{0}}=\frac{\Gamma(2H_{0}+1)\sin(\pi H_{0})}{2\pi}.$ (7)

In this case, $\gamma_{0}$, we say that the noise $W$ is fractional in time
with index $H_{0}>1/2$.

Similarly, there exists a tempered measure $\mu$ on $\mathbb{R}^{d}$ such that
$\gamma$ is the Fourier transform of $\mu$ in
$\mathcal{S}^{\prime}_{\mathbb{C}}(\mathbb{R})$, i.e. for all
$\varphi\in\mathcal{S}_{\mathbb{C}}(\mathbb{R}^{d})$,

$\int_{\mathbb{R}^{2d}}\varphi(x)\varphi(y)\gamma(x-y)dxdy=\int_{\mathbb{R}^{d}}|\mathcal{F}\varphi(\xi)|^{2}\mu(d\xi).$
(8)

The spatial correlation pattern of the noise is defined by this spatial
spectral measure $\mu$. Typically, this measure is assumed to be tempered,
non-negative, and non-negative definite. We will consider two cases:

$\mu(d\xi)=\left\\{\begin{array}[]{ll}|\xi|^{-\alpha}d\xi&\mbox{with
$\alpha\in(0,d)$ and $d\geq 1$ ({\em Case I: the regular case})}\\\
c_{H}|\xi|^{1-2H}d\xi&\mbox{with $H\in(0,1/2)$ and $d=1$ ({\em Case II: the
rough case})}\end{array}\right.$

In Case II, the constant $c_{H}$ is defined in (7).

We are now ready to give the definition of the solution to equation (1).

######  Definition 1.1.

We say that a process
$u=\\{u(t,x),(t,x)\in\mathbb{R}_{+}\times\mathbb{R}^{d}\\}$ is a (mild
Skorohod) solution of equation (1) if for any
$(t,x)\in\mathbb{R}_{+}\times\mathbb{R}^{d}$, $u(t,x)$ is
$\mathcal{F}_{t}$-measurable, $\mathbb{E}|u(t,x)|^{2}<\infty$ and solves the
following equation:

$u(t,x)=w(t,x)+\int_{0}^{t}\int_{\mathbb{R}^{d}}G_{t-s}(x-y)u(s,y)W(\delta
s,\delta y),$

where the stochastic integral is interpreted in the Skorohod sense and
$w(t,x)$ is the solution of the homogeneous heat equation with initial data
$u_{0}$:

$w(t,x)=\int_{\mathbb{R}^{d}}G_{t}(x-y)u_{0}(\mathrm{d}y).$ (9)

Note that if $u_{0}\geq 0$, then (2) is the necessary and sufficient condition
for $w(t,x)$ to be finite. By Lemma B.2 of [3], we know that $w(t,x)$ is
continuous on $\mathbb{R}_{+}\times\mathbb{R}^{d}$. Hence $w(t,x)$ is
uniformly bounded in any compact subset of
$\mathbb{R}_{+}\times\mathbb{R}^{d}$. We recall that $G_{t}(x)$ is the
fundamental solution of the heat equation in $\mathbb{R}^{d}$ with

$G_{t}(x)=\frac{1}{(2\pi t)^{d/2}}\exp\left(-\frac{|x|^{2}}{2t}\right),$

and the Fourier transform of $G_{t}$ in any dimension $d$ is given by
$\mathcal{F}G_{t}(\xi)=e^{-t|\xi|^{2}/2}$. Moreover, we note that
$|w(t,x)|\leq w_{+}(t,x)$, where

$w_{+}(t,x)=\int_{\mathbb{R}^{d}}G_{t}(x-y)|u_{0}|(\mathrm{d}y),$ (10)

and it can be shown that condition (2) is equivalent to

$w_{+}(t,x)<\infty\quad\text{ for all }t>0\text{ and }x\in\mathbb{R}^{d}.$

The solution to equation (1) is understood in Skorohod sense. We recall that
$\delta:\text{Dom}(\delta)\subset L^{2}(\Omega;\mathcal{H})\to L^{2}(\Omega)$
is the divergence operator with respect to $W$ which is defined as the adjoint
of the Malliavin derivative $D$ with respect to $W$. If
$u\in\text{Dom}(\delta)$, we use the notation

$\delta(u)=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}u(t,x)W(\delta t,\delta x)$

and we say $\delta(u)$ is the Skorohod integral of $u$ with respect to $W$.
Further elaboration on the subject of Malliavin calculus is available in [13].

Using the methodology introduced in [11], we know that if it exists, the
solution of equation (1) has the series expansion:

$u(t,x)=w(t,x)+\sum_{n\geq 1}I_{n}\big{(}f_{t,x,n}\big{)}\quad\mbox{in}\quad
L^{2}(\Omega),$ (11)

the terms of these series being orthogonal in $L^{2}(\Omega)$. Here $I_{n}$ is
the multiple Wiener integral of order $n$ with respect to $W$, and $f_{t,x,n}$
is given by: for all $0<t_{1}<\ldots<t_{n}<t$,

$\displaystyle f_{t,x,n}(t_{1},x_{1},\ldots,t_{n},x_{n})$
$\displaystyle=G_{t-t_{n}}(x-x_{n})\ldots
G_{t_{2}-t_{1}}(x_{2}-x_{1})w(t_{1},x_{1})$
$\displaystyle=\int_{\mathbb{R}^{d}}G_{t-t_{n}}(x-x_{n})\cdots
G_{t_{2}-t_{1}}(x_{2}-x_{1})G_{t_{1}}(x_{1}-x_{0})u_{0}(\mathrm{d}x_{0})$

where the second expression of this kernel is obtained by (9). By the
orthogonality of the terms in this series, the necessary and sufficient
condition for the existence of solution is

$\sum_{n\geq 1}\mathbb{E}|I_{n}\big{(}f_{t,x,n}\big{)}|^{2}=\sum_{n\geq
1}n!\,\|\widetilde{f}_{t,x,n}\|_{\mathcal{H}^{\otimes n}}^{2}<\infty.$

where $\widetilde{f}$ is the symmetrization of the function $f$, defined by:

$\widetilde{f}(t_{1},x_{1},\ldots,t_{n},x_{n})=\frac{1}{n!}\sum_{\rho\in
S_{n}}f(t_{\rho(1)},x_{\rho(1)},\ldots,t_{\rho(n)},x_{\rho(n)}),$

and $S_{n}$ is the set of permutations of $\\{1,\ldots,n\\}$.

In what follows, we will need the following result which gives the explicit
expression of the Fourier transform of the kernel $f_{t,x,n}$ in the space
variables.

######  Lemma 1.2 (Lemma 2.5 of [3]).

For any $0<t_{1}<\ldots<t_{n}<t=t_{n+1}$ and for any
$\xi_{1},\ldots,\xi_{n}\in\mathbb{R}^{d}$, we have

$\displaystyle\mathcal{F}f_{t,x,n}(t_{1},\cdot,\ldots,t_{n},\cdot)(\xi_{1},\ldots,\xi_{n})$
$\displaystyle\qquad=\prod_{k=1}^{n}\exp\bigg{\\{}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\bigg{\\}}\exp\bigg{\\{}-\frac{i}{t}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot
x\bigg{\\}}$
$\displaystyle\qquad\quad\times\int_{\mathbb{R}^{d}}\exp\Big{\\{}-i\Big{[}\sum_{j=1}^{n}\Big{(}1-\frac{t_{j}}{t}\Big{)}\xi_{j}\Big{]}\cdot
x_{0}\Big{\\}}G_{t}(x-x_{0})u_{0}(\mathrm{d}x_{0}).$

In Case I with $\gamma_{0}$ arbitrary, a sufficient condition for the
existence of the solution the heat equation (1) is Dalang’s condition:

$\int_{\mathbb{R}^{d}}\frac{1}{1+|\xi|^{2}}\mu(d\xi)<\infty,$

which is equivalent to

$d-\alpha<2.$ (12)

(see [2]). In Case II with temporal covariance $\gamma_{0}$ given by (6), a
sufficient condition for the existence of the solution is $H_{0}+H>3/4$ (see
[10]). We note that the authors of [10] considered a slightly different
initial condition compared to the present paper.

The following theorems are the main results of the present article.

######  Theorem 1.3 (The Regular Case).

Let $\\{W^{\alpha}\\}_{\alpha\in(0,d)}$ be a family of zero-mean Gaussian
processes with covariance (3) in which $\gamma_{0}$ is chosen such that its
corresponding measure $\mu_{0}$ satisfies (5), and the measure $\mu$ is given
by:

$\mu(d\xi)=|\xi|^{-\alpha}d\xi\quad\mbox{with $\alpha\in(0,d)$ and $d\geq
1$}.$ (13)

Let $u^{\alpha}$ be a continuous modification of the Skorohod solution of
equation (1) with noise $W$ replaced by $W^{\alpha}$. Assume that the initial
condition $u_{0}$ is a signed measure which satisfies (2). For any
$\alpha\in\big{(}\max(d-2,0),d\big{)}$ and for any $t_{0}\in(0,T)$, if
$\alpha_{n}\to\alpha^{*}\in(\max(d-2,0),d)$, then

$u^{\alpha_{n}}\stackrel{{\scriptstyle
d}}{{\longrightarrow}}u^{\alpha^{*}}\quad\ \mbox{in
$C([t_{0},T]\times\mathbb{R}^{d})$},$

where $C([t_{0},T]\times\mathbb{R}^{d})$ is equipped with the topology of
uniform convergence on compact sets.

######  Theorem 1.4 (The Rough Case).

Let $\\{W^{H}\\}_{H\in(0,1/2)}$ be a family of zero-mean Gaussian processes
with covariance (3) in which $\gamma_{0}$ is given by (6) and the measure
$\mu$ is given by:

$\mu(d\xi)=c_{H}|\xi|^{1-2H}d\xi\quad\mbox{with $H\in(0,1/2)$ and $d=1$.}$
(14)

Fix $H_{0}\in(1/2,1)$. We define $\ell=\max(3/4-H_{0},0)$. Assume that the
initial condition $u_{0}$ is a signed measure which satisfies (2). For any
$H\in(\ell,1/2)$ and for any $t_{0}\in(0,T)$, let $u^{H}$ be a continuous
modification of the Skorohod solution of equation (1) with noise $W$ replaced
by $W^{H}$. If $H_{n}\to H^{*}\in(\ell,1/2)$, then

$u^{H_{n}}\stackrel{{\scriptstyle d}}{{\longrightarrow}}u^{H^{*}}\quad\
\mbox{in $C([t_{0},T]\times\mathbb{R})$},$

where $C([t_{0},T]\times\mathbb{R})$ is equipped with the topology of uniform
convergence on compact sets.

######  Remark 1.5.

We note that in Theorems 1.3 and 1.4, the convergence is only on compact sets
of the form $[t_{0},T]\times K$ with $0<t_{0}<T$ and $K\subset\mathbb{R}^{d}$
compact. This limitation is due to some uniform moment estimates, which are
needed for the proof of tightness.

The proof of Theorems 1.3 and 1.4 follow the same strategy. We first prove the
convergence of the corresponding finite-dimensional distributions and
tightness. In both cases, we need to guarantee that all noise processes are
defined on the same probability space, for all parameter values $\alpha$, or
$H$. To construct all noise processes $(W^{\alpha})_{\alpha\in(0,d)}$,
respectively $(W^{H})_{H\in(0,1/2)}$, on the same probability space, we need
to assume that $\mu_{0}$ has density (see (5)). We refer reader to [6] for
this construction of noise, based on a complex-valued Gaussian white noise. To
prove tightness, we apply Kolmogorov-Centsov theorem. For this, we derive some
moment estimates for the increments of the solution, ensuring that all bounds
remain uniformly bounded for parameter values ($\alpha$ or $H$) in a compact
set.

We conclude the introduction by providing a brief overview of the article’s
organization and introducing the notation that will be used throughout. The
proofs of Theorems 1.3 and 1.4 can be found in Section 2, respectively Section
3. We denote

$f(\boldsymbol{t_{n}},\boldsymbol{x_{n}})=f(t_{1},x_{1},\ldots,t_{n},x_{n}),$

where $\boldsymbol{t_{n}}=(t_{1},\ldots,t_{n})\in\mathbb{R}_{+}^{n}$ and
$\boldsymbol{x_{n}}=(x_{1},\ldots,x_{n})\in(\mathbb{R}^{d})^{n}$.
Occasionally, we will use
$\boldsymbol{\xi_{n}}=(\xi_{1},\ldots,\xi_{n})\in(\mathbb{R}^{d})^{n}$. We
will use the convention:

$G_{t}(x)=0\quad\mbox{for any $t<0$ and $x\in\mathbb{R}^{d}$.}$ (15)

We let
$T_{n}(t)=\\{\boldsymbol{t_{n}}=(t_{1},\ldots,t_{n});0<t_{1}<\ldots<t_{n}<t\\}$
be the $n$-dimensional simplex. For any $p\geq 1$, we denote by
$\|\cdot\|_{p}$ the norm in $L^{p}(\Omega)$. We let $\sigma$ be the surface
measure on the unit sphere $S_{1}(0)=\\{z\in\mathbb{R}^{d};|z|=1\\}$, and
$c_{d}$ be the area of $S_{1}(0)$, i.e.

$c_{d}=\int_{S_{1}(0)}\sigma(dz).$ (16)

##  2 Regular noise

In this section, we consider equation (1) driven by a Gaussian noise $W$ with
covariance (3) in which $\gamma_{0}$ is chosen such that its corresponding
measure $\mu_{0}$ satisfies (5), and the measure $\mu$ is given by (13). The
existence and uniqueness of the solution was proved in [3].

To emphasize the dependence on the parameter $\alpha$, we denote the noise,
the Hilbert space, and the solution, by
$W^{\alpha},\mathcal{H}^{\alpha},u^{\alpha}$, respectively. Denoting the
multiple integral of order $n$ with respect to $W^{\alpha}$ as
$I_{n}^{\alpha}$, the series expansion (11) takes the following form:

$u^{\alpha}(t,x)=1+\sum_{n\geq 1}I_{n}^{\alpha}(f_{t,x,n}).$

As mentioned above, we need the family $\\{W^{\alpha}\\}_{\alpha\in(0,d)}$ to
be defined on the same probability space. We omit the details and refer the
reader to [6] for this constructions, which involves a $\mathbb{C}$-valued
Gaussian measure $\widehat{W}$, defined as follows:

$\widehat{W}(A)=W_{1}(A)+iW_{2}(A)$ (17)

where $W_{1}$ and $W_{2}$ are independent space-time Gaussian white noise
processes on $\mathbb{R}^{d+1}$. We recall the following representation of the
multiple integral with respect to the process $W^{\alpha}$: for any
$t\mapsto\varphi(t,\cdot)\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$,

$I_{k}^{\alpha}(\varphi)=\int_{(\mathbb{R}\times\mathbb{R}^{d})^{k}}\mathcal{F}_{t}[\mathcal{F}_{x}\varphi(\boldsymbol{t_{k}},\bullet)(\boldsymbol{\xi_{k}})](\boldsymbol{\tau_{k}})\prod_{j=1}^{k}\sqrt{g_{0}(\tau_{j})}|\xi_{j}|^{-\alpha/2}\widehat{W}(d\tau_{1},d\xi_{1})\ldots\widehat{W}(d\tau_{k},d\xi_{k}),$
(18)

where $\mathcal{F}_{x}$ is the Fourier transform in the space variables and
$\mathcal{F}_{t}$ is the Fourier transform in the time variables, i.e. the
Fourier transform of the function
$\boldsymbol{t_{k}}\mapsto\mathcal{F}_{x}\varphi(\boldsymbol{t_{k}},\bullet)(\boldsymbol{\xi_{k}})$.
Whenever there is no risk of confusion, we drop the lower indices $t,x$ from
the Fourier transform notation.

The following three lemmas will play an important role below. The first one
can be found for instance in [1] and the rest are taken from [3].

######  Lemma 2.1.

For any $\alpha\in\big{(}\max(d-2,0),d\big{)}$ and $t>0$,

$\displaystyle\int_{\mathbb{R}^{d}}|\mathcal{F}G_{t}(\xi)|^{2}|\xi-\eta|^{-\alpha}d\xi\leq
K_{d,\alpha}\,t^{-(d-\alpha)/2},$

where

$K_{d,\alpha}:=\int_{\mathbb{R}^{d}}\frac{1}{1+|\xi|^{2}}|\xi|^{-\alpha}d\xi\leq
c_{d}\left(\frac{1}{d-\alpha}+\frac{1}{2-(d-\alpha)}\right),$ (19)

with $c_{d}$ is given by (16).

######  Lemma 2.2 (Lemma 3.4 of [3]).

Let $\mu$ be a tempered measure on $\mathbb{R}^{d}$ such that its Fourier
transform in $\mathcal{S}^{\prime}_{\mathbb{C}}(\mathbb{R}^{d})$ is a locally
integrable function $\gamma$. Assume that $\gamma$ is non-negative. Then for
any $\psi\in\mathcal{S}(\mathbb{R}^{d})$ such that $\psi\ast\widetilde{\psi}$
is non-negative, where $\widetilde{\psi}(x)=\psi(-x)$ for all
$x\in\mathbb{R}^{d}$, we have:

$\displaystyle\sup_{\eta\in\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\big{|}\mathcal{F}\psi(\xi+\eta)\big{|}^{2}\mu(\mathrm{d}\xi)=\int_{\mathbb{R}^{d}}\big{|}\mathcal{F}\psi(\xi)\big{|}^{2}\mu(\mathrm{d}\xi).$
(20)

In particular, for any $a>0$ and $t>0$,

$\displaystyle\sup_{\eta\in\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}e^{-a|t\xi+\eta|^{2}}\mu(\mathrm{d}\xi)=\int_{\mathbb{R}^{d}}e^{-a|t\xi|^{2}}\mu(\mathrm{d}\xi).$
(21)

######  Lemma 2.3 (Lemma 4.1 of [3]).

For any $h>-1$, we have

$\int_{0<t_{1}<\ldots<t_{n}<t}\big{[}t_{1}(t_{2}-t_{1})\ldots(t-t_{n})\big{]}^{h}\mathrm{d}t_{1}\ldots\mathrm{d}t_{n}=\frac{\Gamma(h+1)^{n+1}}{\Gamma\Big{(}(n+1)(h+1)\Big{)}}t^{n(h+1)+h}.$

The following result will be used in the proof of Theorem 1.3, in order to
show the finite dimensional convergence.

######  Lemma 2.4.

If $\alpha_{n}\to\alpha^{*}\in\big{(}\max(d-2,0),d\big{)}$, then for any
$t>0$, $x\in\mathbb{R}^{d}$ and $k\geq 1$,

$\mathbb{E}|I_{k}^{\alpha_{n}}(f_{t,x,k})-I_{k}^{\alpha^{*}}(f_{t,x,k})|^{2}\to
0,\quad\mbox{as}\quad n\to\infty.$

###### Proof.

By (18), we have

$\displaystyle
I_{k}^{\alpha_{n}}(f_{t,x,k})-I_{k}^{\alpha^{*}}(f_{t,x,k})=\int_{(\mathbb{R}\times\mathbb{R}^{d})^{n}}H_{k}^{(n)}(\boldsymbol{\tau_{k}},\boldsymbol{\xi_{k}})\widehat{W}(d\tau_{1},d\xi_{1})\ldots\widehat{W}(d\tau_{n},d\xi_{n}),$
(22)

where

$H_{k}^{(n)}(\boldsymbol{\tau_{k}},\boldsymbol{\xi_{k}})=\mathcal{F}\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{\tau_{k}})\prod_{j=1}^{k}\sqrt{g_{0}(\tau_{j})}\left(\prod_{j=1}^{k}|\xi_{j}|^{-\alpha_{n}/2}-\prod_{j=1}^{k}|\xi_{j}|^{-\alpha^{*}/2}\right)$

and

$\displaystyle\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{t_{k}})$
$\displaystyle:=\mathcal{F}f_{t,x,k}(\boldsymbol{t_{k}},\cdot)(\boldsymbol{\xi_{k}})$
$\displaystyle=\prod_{j=1}^{k}\exp\bigg{\\{}-\frac{1}{2}\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\Big{|}\sum_{i=1}^{j}t_{i}\xi_{i}\Big{|}^{2}\bigg{\\}}\exp\bigg{\\{}-\frac{i}{t}\Big{(}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{)}\cdot
x\bigg{\\}}$
$\displaystyle\qquad\quad\times\int_{\mathbb{R}^{d}}\exp\Big{\\{}-i\Big{[}\sum_{j=1}^{k}\Big{(}1-\frac{t_{j}}{t}\Big{)}\xi_{j}\Big{]}\cdot
x_{0}\Big{\\}}G_{t}(x-x_{0})u_{0}(\mathrm{d}x_{0}),$ (23)

with $t_{k+1}=t$. Note that (22) is the same as in the proof of Lemma 3.3 of
[6], but here we use a different expression for the Fourier transform
$\mathcal{F}f_{t,x,k}(\boldsymbol{t_{k}},\cdot)(\boldsymbol{\xi_{k}})$, due to
the general initial condition. This form is specific to the heat equation,
since it was developed using the semigroup property of the heat kernel.
Proceeding as in the proof of Lemma 3.3 of [6], we infer that

$\displaystyle
Q_{n}:=\mathbb{E}|I_{k}^{\alpha_{n}}(f_{t,x,k})-I_{k}^{\alpha^{*}}(f_{t,x,k})|^{2}\leq
k!\Gamma_{0,t}^{k}\int_{T_{k}(t)}A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{t_{k}})d\boldsymbol{t_{k}}$

where $T_{k}(t)=\\{0<t_{1}<\ldots<t_{k}<t\\}$,

$\displaystyle
A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{s_{k}})=\int_{(\mathbb{R}^{d})^{k}}\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{t_{k}})\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{s_{k}})\bigg{|}\prod_{j=1}^{k}|\xi_{j}|^{-\alpha_{n}/2}-\prod_{j=1}^{k}|\xi_{j}|^{-\alpha^{*}/2}\bigg{|}^{2}d\boldsymbol{\xi_{k}}$

and $\Gamma_{0,t}=\int_{-t}^{t}\gamma_{0}(s)ds$. It suffices to show that the
integral appearing in the upper bound above converges to $0$ as $n\to\infty$,
i.e.

$T_{n}:=\int_{T_{k}(t)}\int_{(\mathbb{R}^{d})^{k}}\big{|}\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{t_{k}})\big{|}^{2}\left|\prod_{j=1}^{k}|\xi_{j}|^{-\alpha_{n}/2}-\prod_{j=1}^{k}|\xi_{j}|^{-\alpha^{*}/2}\right|^{2}d\boldsymbol{\xi_{k}}d\boldsymbol{t_{k}}\to
0.$

Note that the integrand converges pointwise to $0$ on
$T_{k}(t)\times(\mathbb{R}^{d})^{k}$, as $n\to\infty$. To apply the Dominated
Convergence Theorem, it suffices to find an integrable bound for the term
depending on $\alpha_{n}$:

$g_{n}(\boldsymbol{t_{k}},\boldsymbol{\xi_{k}}):=\big{|}\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{t_{k}})\big{|}^{2}\prod_{j=1}^{k}|\xi_{j}|^{-\alpha_{n}},$

and we need to find a function $g$ on $T_{k}(t)\times(\mathbb{R}^{d})^{k}$,
such that

$g_{n}(\boldsymbol{t_{k}},\boldsymbol{\xi_{k}})\leq
g(\boldsymbol{t_{k}},\boldsymbol{\xi_{k}})\quad\mbox{for all
$n$},\quad\mbox{and}\quad\int_{T_{k}(t)}\int_{(\mathbb{R}^{d})^{k}}g(\boldsymbol{t_{k}},\boldsymbol{\xi_{k}})d\boldsymbol{\xi_{k}}d\boldsymbol{t_{k}}<\infty.$

Recall that $\max\\{d-2,0\\}<\alpha^{\ast}<d$. Fix numbers $a$ and $b$ such
that $\max\\{d-2,0\\}<a<\alpha^{\ast}<b<d$. Since
$\alpha_{n}\to\alpha^{\ast}$, there exists $N\in\mathbb{N}$ such that
$a\leq\alpha_{n}\leq b$ for all $n\geq N$. A natural candidate for $g$ is

$\displaystyle g(\boldsymbol{t_{k}},\boldsymbol{\xi_{k}})$
$\displaystyle:=\big{|}\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{t_{k}})\big{|}^{2}\prod_{j=1}^{k}\big{(}|\xi_{j}|^{-b}1_{\\{|\xi_{j}|\leq
1\\}}+|\xi_{j}|^{-a}1_{\\{|\xi_{j}|>1\\}}\big{)}$ $\displaystyle\leq
w_{+}^{2}(t,x)\prod_{j=1}^{k}\exp\bigg{\\{}-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\Big{|}\sum_{i=1}^{j}t_{i}\xi_{i}\Big{|}^{2}\bigg{\\}}\Big{(}|\xi_{j}|^{-b}1_{\\{|\xi_{j}|\leq
1\\}}+|\xi_{j}|^{-a}1_{\\{|\xi_{j}|>1\\}}\Big{)}.$ (24)

We fix $j=1,\ldots,k$. By Lemma 2.1, we have

$\displaystyle\int_{\mathbb{R}^{d}}e^{-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\big{|}\sum_{i=1}^{j}t_{i}\xi_{i}\big{|}^{2}}\Big{(}|\xi_{j}|^{-b}1_{\\{|\xi_{j}|\leq
1\\}}+|\xi_{j}|^{-a}1_{\\{|\xi_{j}|>1\\}}\Big{)}\mathrm{d}\xi_{j}$
$\displaystyle\qquad\leq\int_{|\xi_{j}|\leq
1}|\xi_{j}|^{-b}\mathrm{d}\xi_{j}+\sup_{\eta\in\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}e^{-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\big{|}t_{j}\xi_{j}+\eta\big{|}^{2}}|\xi_{j}|^{-a}\mathrm{d}\xi_{j}$
$\displaystyle\qquad=\frac{c_{d}}{d-b}+\int_{\mathbb{R}^{d}}e^{-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}|t_{j}\xi_{j}|^{2}}|\xi_{j}|^{-a}\mathrm{d}\xi_{j}=\frac{c_{d}}{d-b}+\int_{\mathbb{R}^{d}}e^{-\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\xi_{j}^{2}}|\xi_{j}|^{-a}\mathrm{d}\xi_{j}$
$\displaystyle\qquad\leq\frac{c_{d}}{d-b}+\bigg{(}\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\bigg{)}^{-\frac{d-a}{2}}K_{d,a}=:C\Big{[}1+\bigg{(}\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\bigg{)}^{-\frac{d-a}{2}}\Big{]}$
(25)

where $c_{d}$ given in (16) and $K_{d,a}$ are given in (19). Observe that for
any $0<t_{1}<\ldots<t_{k}<t_{k+1}=t$, we have
$\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\leq t_{j+1}-t_{j}\leq t$. Therefore we
have $\Big{(}\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\Big{)}^{-\frac{d-a}{2}}\geq
t^{-\frac{d-a}{2}},$ and multiplying by $t^{\frac{d-a}{2}}$ on both sides, we
obtain: $1\leq
t^{\frac{d-a}{2}}\Big{(}\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\Big{)}^{-\frac{d-a}{2}}$
which implies

$\displaystyle
1+\Big{(}\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\Big{)}^{-\frac{d-a}{2}}$
$\displaystyle\leq\Big{(}\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\Big{)}^{-\frac{d-a}{2}}\Big{(}t^{\frac{d-a}{2}}+1\Big{)}.$
(26)

Using relations (24), (25) and (26), we have

$\displaystyle\int_{T_{k}(t)}\int_{(\mathbb{R}^{d})^{k}}g(\boldsymbol{t_{k}},\boldsymbol{\xi_{k}})d\boldsymbol{\xi_{k}}d\boldsymbol{t_{k}}$
$\displaystyle\leq
w^{2}(t,x)C^{k}\int_{T_{k}(t)}\prod_{j=1}^{k}\Big{(}\frac{(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\Big{)}^{-\frac{d-a}{2}}\mathrm{d}\mathbb{t}$
$\displaystyle\leq
C\frac{\Gamma\big{(}\frac{2-(d-a)}{2}\big{)}^{n+1}}{\Gamma\Big{(}(n+1)(\frac{2-(d-a)}{2})\Big{)}}t^{n(\frac{2-(d-a)}{2})-\frac{d-a}{2}}$

where for the last line, we used Lemma 2.3. ∎

For the proof of tightness, we need some upper bounds for the moments of the
increments of $u_{\alpha}$. These bounds have been obtained in [3] and are
given by Theorem 2.5 below. However, for our problem, we will need the
explicit form of the constant $C$ appearing in this theorem, and this will be
described in Theorem 2.7. Note that Theorem 2.5 guarantees the existence of a
Hölder continuous modification of the solution $u_{\alpha}$. We work with this
modification, which we denote also by $u_{\alpha}$.

######  Theorem 2.5 (Theorem 1.3 of [3]).

Let $u^{\alpha}$ be the solution of equation (1) starting from an initial
measure $u_{0}$ that satisfying (2). Suppose that:

$\int_{\mathbb{R}^{d}}\Big{(}\frac{1}{1+|\xi|^{2}}\Big{)}^{\beta}\mu(d\xi)<\infty\quad\text{
for some }\beta\in(0,1).$

Then for any $p\geq 2$ and $a>1$ there exists a constant $C>0$ depending on
$p,a,\lambda$ and $\beta$ such that for any $(t,x)$,
$(t^{\prime},x^{\prime})\in K_{a}:=[1/a,a]\times[-a,a]^{d}$,

$\displaystyle\|u^{\alpha}(t,x)-u^{\alpha}(t^{\prime},x^{\prime})\|_{p}\leq
C\big{(}|t-t^{\prime}|^{\frac{1-\beta}{2}}+|x-x^{\prime}|^{1-\beta}\big{)},$
(27)

Consequently, for any $a>1$, the process $\\{u^{\alpha}(t,x);(t,x)\in
K_{a}\\}$ has a modification which is a.s. $\theta_{1}$-Hölder continuous in
time and a.s. $\theta_{2}$-Hölder continuous in space, for any
$\theta_{1}\in(0,(1-\beta)/2)$ and $\theta_{2}\in(0,1-\beta)$.

We will use the following constant:

$k_{\alpha}(t):=\int_{\mathbb{R}^{d}}e^{-\frac{t|\xi|^{2}}{2}}|\xi|^{-\alpha}d\xi=C^{(1)}_{d,\alpha}\,t^{-(d-\alpha)/2},$
(28)

where the last equation is obtained using the change of variables
$\xi^{\prime}=\sqrt{t}\xi$. The constant $C^{(1)}_{d,\alpha}$ is defined by:

$C^{(1)}_{d,\alpha}:=\int_{\mathbb{R}^{d}}e^{-|\xi|^{2}/2}|\xi|^{-\alpha}d\xi.$

Note that $C^{(1)}_{d,\alpha}$ can be uniformly bounded, for all
$\alpha\in[a,b]$ such that $\max\\{d-2,0\\}<a<b<d$.

The following result will be used several times in the proof of Theorem 2.7
below.

######  Lemma 2.6 (Lemma 3.5 of [3]).

For any $0<t_{1}<\ldots<t_{n}<t:=t_{n+1}$, we have that

$\int_{\mathbb{R}^{nd}}\exp\bigg{\\{}-\sum_{k=1}^{n}\bigg{(}\frac{t_{k+1}-t_{k}}{t_{k+1}t_{k}}\bigg{|}\sum_{j=1}^{k}t_{j}\xi_{j}\bigg{|}^{2}\bigg{)}\bigg{\\}}\prod_{k=1}^{n}|\xi_{k}|^{-\alpha}d\xi_{k}\leq\prod_{j=1}^{n}k_{\alpha}\bigg{(}\frac{2(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\bigg{)}.$

By (28), we have:

$\displaystyle\prod_{j=1}^{n}k_{\alpha}\bigg{(}\frac{2(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\bigg{)}$
$\displaystyle=C^{(1)}_{d,\alpha}\Big{(}\frac{2(t_{2}-t_{1})t_{1}}{t_{2}}\Big{)}^{-(d-\alpha)/2}\ldots
C^{(1)}_{d,\alpha}\Big{(}\frac{2(t-t_{n})t_{n}}{t}\Big{)}^{-(d-\alpha)/2}$
$\displaystyle=\big{(}C^{(1)}_{d,\alpha}\big{)}^{n}2^{-(d-\alpha)n/2}t^{(d-\alpha)/2}\big{[}t_{1}(t_{2}-t_{1})\ldots(t-t_{n})\big{]}^{-(d-\alpha)/2}.$
(29)

The next result gives the explicit form of the constant $C$ as a function of
$\alpha$, which allow us to bound it uniformly for all $\alpha$ in a compact
interval $[a,b]$. Then we can use it to demonstrate the tightness of
$(u^{\alpha})_{\alpha\in[a,b]}$.

######  Theorem 2.7.

Assume that $\max(d-2,0)<a<b<d$. We fix $T>t_{0}>0$. Let
$K\subset\mathbb{R}^{d}$ be a compact set. For any $p\geq 2$,
$\beta\in\big{(}\frac{d-a}{2},1\big{)}$, $t,t^{\prime}\in[t_{0},T]$ and
$x,x^{\prime}\in K$, we have:

$\sup_{\alpha\in[a,b]}\|u^{\alpha}(t,x)-u^{\alpha}(t^{\prime},x^{\prime})\|_{p}\leq
C\Big{(}|t-t^{\prime}|^{\frac{1-\beta}{2}}+|x-x^{\prime}|^{1-\beta}\Big{)}.$

Here $C$ is a constant that depends on $(d,p,T,t_{0},\beta,a,b)$.

###### Proof.

We proceed with a similar argument as in the proof of Theorem 1.3 of [3], with
the addition of an examination of a missing term that was overlooked in that
particular proof. To simplify the writing, we let $\theta=1-\beta$.

Step 1 (time increments). Let $t\in[t_{0},T]$ and $h>0$ be such that
$t+h\in[t_{0},T]$. By hypercontractivity,

$\displaystyle\big{\|}u^{\alpha}(t+h,x)-u^{\alpha}(t,x)\big{\|}_{p}$
$\displaystyle\leq\sum_{n\geq
1}(p-1)^{n/2}\big{\|}I_{n}^{\alpha}\big{(}f_{t+h,x,n}-f_{t,x,n}\big{)}\big{\|}_{2}$
$\displaystyle\leq\sum_{n\geq
1}(p-1)^{n/2}\left(\frac{2}{n!}\Big{(}A_{n}^{\alpha}(t,h)+B_{n}^{\alpha}(t,h)\Big{)}\right)^{1/2},$

where
$A_{n}^{\alpha}(t,h)=(n!)^{2}\|\widetilde{f}_{t+h,x,n}1_{[0,t]^{n}}-\widetilde{f}_{t,x,n}\|_{\mathcal{H}_{\alpha}^{\otimes
n}}^{2}$ and
$B_{n}^{\alpha}(t,h)=(n!)^{2}\|\widetilde{f}_{t+h,x,n}1_{[0,t+h]^{n}-[0,t]^{n}}\|_{\mathcal{H}_{\alpha}^{\otimes
n}}^{2}$.

We examine $A_{n}^{\alpha}(t,h)$. Note that

$\displaystyle
A_{n}^{\alpha}(t,h)\leq\Gamma_{0,t}^{n}\int_{[0,t]^{n}}\psi_{t,h,n}^{\alpha}(\boldsymbol{t_{n}})d\boldsymbol{t_{n}},$
(30)

where the function $\psi_{t,h,n}^{\alpha}(\boldsymbol{t_{n}})$ is defined as
follows: if $\boldsymbol{t_{n}}\in[0,t]^{n}$ is fixed and $\rho$ is the
permutation of $1,\ldots,n$ such that $t_{\rho(1)}<\ldots<t_{\rho(n)}$ with
$t_{\rho(n+1)}=t$, and

$\displaystyle\psi_{t,h,n}^{\alpha}(\boldsymbol{t_{n}})$
$\displaystyle:=\int_{(\mathbb{R}^{d})^{n}}\Big{|}\mathcal{F}\Big{(}f_{t+h,x,n}(t_{\rho(1)},\cdot,\ldots,t_{\rho(n)},\cdot)-f_{t,x,n}(t_{\rho(1)},\cdot,\ldots,t_{\rho(n)},\cdot)\Big{)}(\boldsymbol{\xi_{n}})\Big{|}^{2}\prod_{j=1}^{n}|\xi_{j}|^{-\alpha}d\boldsymbol{\xi_{n}}.$

Assume for simplicity that $t_{1}<\ldots<t_{n}$. By Lemma 1.2,

$\displaystyle\Big{|}\mathcal{F}\big{(}f_{t+h,x,n}(\boldsymbol{t_{n}},\cdot)-f_{t,x,n}(\boldsymbol{t_{n}},\cdot)\big{)}(\boldsymbol{\xi_{n}})\Big{|}=\prod_{k=1}^{n-1}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\Big{|}I^{(1)}_{1}I^{(2)}_{1}-I^{(1)}_{2}I^{(2)}_{2}\Big{|}$

where

$\displaystyle I^{(1)}_{1}$
$\displaystyle:=\exp\Big{(}-\frac{1}{2}\frac{t+h-t_{n}}{t_{n}(t+h)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}$
$\displaystyle I^{(2)}_{1}$
$\displaystyle:=\int_{\mathbb{R}^{d}}\exp\bigg{(}-i\Big{(}\sum_{j=1}^{n}\xi_{j}\Big{)}\cdot
x_{0}\bigg{)}\exp\bigg{(}-\frac{i}{t+h}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})\bigg{)}\,G_{t+h}(x-x_{0})u_{0}(\mathrm{d}x_{0})$
$\displaystyle I^{(1)}_{2}$
$\displaystyle:=\exp\Big{(}-\frac{1}{2}\frac{t-t_{n}}{t_{n}t}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}$
$\displaystyle I^{(2)}_{2}$
$\displaystyle:=\int_{\mathbb{R}^{d}}\exp\Big{(}-i\Big{(}\sum_{j=1}^{n}\xi_{j}\Big{)}\cdot
x_{0}\Big{)}\exp\bigg{(}-\frac{i}{t}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})\bigg{)}\,G_{t}(x-x_{0})u_{0}(\mathrm{d}x_{0}).$

Introducing the mixed term $I^{(1)}_{1}I^{(2)}_{2}$, we get:

$\displaystyle\Big{|}\mathcal{F}\big{(}f_{t+h,x,n}(\boldsymbol{t_{n}},\cdot)-f_{t,x,n}(\boldsymbol{t_{n}},\cdot)\big{)}(\boldsymbol{\xi_{n}})\Big{|}\leq
K_{1}+K_{2}$ (31)

where

$\displaystyle K_{1}$
$\displaystyle:=\prod_{k=1}^{n-1}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\times\Big{(}I^{(1)}_{1}\big{|}I^{(2)}_{1}-I^{(2)}_{2}\big{|}\Big{)},$
$\displaystyle K_{2}$
$\displaystyle:=\prod_{k=1}^{n-1}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\times\Big{(}\big{|}I^{(2)}_{2}\big{|}\big{|}I^{(1)}_{1}-I^{(1)}_{2}\big{|}\Big{)}.$

We first consider $K_{1}$. Note that

$\displaystyle\Big{|}I^{(2)}_{1}-I^{(2)}_{2}\Big{|}$
$\displaystyle\leq\int_{\mathbb{R}^{d}}\bigg{|}\exp\Big{(}-\frac{i}{t+h}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})\Big{)}\,G_{t+h}(x-x_{0})$
$\displaystyle\qquad\qquad-\exp\Big{(}-\frac{i}{t}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})\Big{)}\,G_{t}(x-x_{0})\bigg{|}u_{0}(\mathrm{d}x_{0})\leq
T_{1}+T_{2}$

where

$\displaystyle T_{1}$
$\displaystyle:=\int_{\mathbb{R}^{d}}\bigg{|}\exp\Big{(}-\frac{i}{t+h}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})\Big{)}\,G_{t+h}(x-x_{0})\bigg{|}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\bigg{|}G_{t+h}(x-x_{0})-G_{t}(x-x_{0})\bigg{|}u_{0}(\mathrm{d}x_{0})$
$\displaystyle T_{2}$
$\displaystyle:=\int_{\mathbb{R}^{d}}\bigg{|}\exp\Big{(}-\frac{i}{t+h}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})\Big{)}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad-\exp\Big{(}-\frac{i}{t}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})\Big{)}\bigg{|}G_{t}(x-x_{0})u_{0}(\mathrm{d}x_{0}).$

We study $T_{1}$ first. Using the inequality $1-e^{-x}\leq x^{\theta}$ for any
$x>0$ and for any $\theta\in[0,1]$, we have:

$\displaystyle T_{1}$
$\displaystyle\leq\int_{\mathbb{R}^{d}}\bigg{|}G_{t+h}(x-x_{0})-G_{t}(x-x_{0})\bigg{|}u_{0}(\mathrm{d}x_{0})\leq\frac{1}{(2\pi
t)^{d/2}}\int_{\mathbb{R}^{d}}e^{-\frac{|x-x_{0}|^{2}}{2(t+h)}}\Big{|}1-e^{-\frac{h|x-x_{0}|^{2}}{2t(t+h)}}\Big{|}u_{0}(\mathrm{d}x_{0})$
$\displaystyle\leq\frac{1}{(2\pi
t)^{d/2}}\int_{\mathbb{R}^{d}}e^{-\frac{|x-x_{0}|^{2}}{2(t+h)}}\Big{(}\frac{h|x-x_{0}|^{2}}{2t(t+h)}\Big{)}^{\theta}u_{0}(\mathrm{d}x_{0})\leq
h^{\theta}C_{t_{0},d,\theta,T}.$

For $T_{2}$, using the fact that
$\big{|}e^{-ia}-e^{-ib}\big{|}\leq|e^{-ia}||1-e^{-i(b-a)}|\leq|b-a|^{\theta}$,
for any $\theta\in[0,1]$, we have:

$\displaystyle T_{2}$
$\displaystyle\leq\int_{\mathbb{R}^{d}}\bigg{|}\frac{\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})}{t}-\frac{\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})}{t+h}\bigg{|}^{\theta}G_{t}(x-x_{0})u_{0}(\mathrm{d}x_{0})$
$\displaystyle=\int_{\mathbb{R}^{d}}\bigg{|}\frac{h\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x-x_{0})}{t(t+h)}\bigg{|}^{\theta}G_{t}(x-x_{0})u_{0}(\mathrm{d}x_{0})\leq
h^{\theta}C_{t_{0},\theta}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{\theta}.$

Therefore, there exist a constant $C^{(1)}$ depends on $t_{0}$, $T$, $\theta$,
$d$ such that

$\displaystyle K_{1}$ $\displaystyle\leq
h^{\theta}C^{(1)}\,\prod_{k=1}^{n-1}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}$
$\displaystyle\qquad\qquad\quad\times\exp\Big{(}-\frac{1}{2}\frac{t+h-t_{n}}{t_{n}(t+h)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\times\Big{(}1+\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{\theta}\Big{)}$
$\displaystyle\leq
h^{\theta}C^{(1)}\,\prod_{k=1}^{n-1}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}$
$\displaystyle\qquad\qquad\quad\times\exp\Big{(}-\frac{1}{2}\frac{t-t_{n}}{t_{n}t}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\times\Big{(}1+\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{\theta}\Big{)}.$

We need the following inequality (see relation 5.2 in [3]): for any $A>0$ and
$\theta>0$, there exists a constant $C_{\theta}>0$ depending on $\theta$ such
that

$\displaystyle e^{-Ax^{2}}x^{2\theta}\leq
C_{\theta}A^{-\theta}e^{-\frac{A}{2}x^{2}},\text{ for all }x\geq 0.$

Using above inequality with $A=\frac{t-t_{n}}{t_{n}t}$ and
$x=\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}$, we obtain:

$\displaystyle\exp\Big{(}-\frac{t-t_{n}}{t_{n}t}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2\theta}\leq
C_{\theta}\Big{(}\frac{t-t_{n}}{t_{n}t}\Big{)}^{-\theta}\exp\Big{(}-\frac{1}{2}\frac{t-t_{n}}{t_{n}t}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}.$
(32)

Note that for any $t\in[t_{0},T]$, we have

$\displaystyle\Big{(}\frac{t-t_{n}}{t_{n}t}\Big{)}^{-\theta}\leq
T^{2\theta}(t-t_{n})^{-\theta}.$ (33)

Hence, denoting $t_{n+1}=t$, there exist a constant $C^{(2)}$ depends on
$t_{0}$, $T$, $\theta$, $d$ such that

$\displaystyle K^{2}_{1}$ $\displaystyle\leq
h^{2\theta}\big{(}C^{(1)}\big{)}^{2}\bigg{\\{}\prod_{k=1}^{n}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}$
$\displaystyle\qquad+\prod_{k=1}^{n-1}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\times
C_{\theta}\Big{(}\frac{t-t_{n}}{t_{n}t}\Big{)}^{-\theta}\exp\Big{(}-\frac{1}{2}\frac{t-t_{n}}{t_{n}t}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\bigg{\\}}$
$\displaystyle\leq
h^{2\theta}C^{(2)}\prod_{k=1}^{n}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\big{[}1+(t-t_{n})^{-\theta}\big{]}.$
(34)

We consider next the term $K_{2}$. Note that using the fact that
$(1-e^{-x})^{2}\leq 1-e^{-x}\leq x^{\theta}$ for all $x>0$ and
$\theta\in[0,1]$, we have:

$\displaystyle\Big{|}I^{(1)}_{1}-I^{(1)}_{2}\Big{|}^{2}$
$\displaystyle=\exp\Big{(}-\frac{t-t_{n}}{t_{n}t}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\bigg{|}1-\exp\Big{(}-\frac{1}{2}\frac{h}{t(t+h)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\bigg{)}\bigg{|}^{2}$
$\displaystyle\leq\exp\Big{(}-\frac{t-t_{n}}{t_{n}t}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\Big{(}\frac{1}{2}\frac{h}{t(t+h)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}^{\theta}$
$\displaystyle\leq
h^{\theta}C_{t_{0},\theta,T}\Big{(}\frac{t-t_{n}}{t_{n}t}\Big{)}^{-\theta}\exp\Big{(}-\frac{1}{2}\frac{t-t_{n}}{t_{n}t}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)},$

where for the last inequality, we used the fact that $t(t+h)\geq t^{2}_{0}$,
followed by relation (32). Moreover, we see that $|I_{2}^{(2)}|\leq w(t,x)\leq
C$, using that fact that $w$ is a continuous function on $[t_{0},T]\times K$.
Therefore, we obtain:

$\displaystyle K^{2}_{2}$ $\displaystyle\leq
h^{\theta}C^{(3)}\,(t-t_{n})^{-\theta}\prod_{k=1}^{n}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)},$
(35)

where $C^{(3)}>0$ is a constant depends on $t_{0}$, $T$, $\theta$, $d$.

We now give an upper bound for $\psi_{t,h,n}^{\alpha}(\boldsymbol{t_{n}})$. By
relations (34), (35) and Lemma 2.6, we get:

$\displaystyle\psi_{t,h,n}^{\alpha}(\boldsymbol{t_{n}})$
$\displaystyle\leq\int_{(\mathbb{R}^{d})^{n}}2(K^{2}_{1}+K^{2}_{2})\prod_{k=1}^{n}|\xi_{k}|^{-\alpha}\mathrm{d}\boldsymbol{\xi}\leq
h^{\theta}C^{(4)}(t-t_{n})^{-\theta}\prod_{j=1}^{n}k_{\alpha}\Big{(}\frac{2(\frac{t_{j+1}}{2}-\frac{t_{j}}{2})(\frac{t_{j}}{2})}{\frac{t_{j+1}}{2}}\Big{)}.$

To evaluate the last expression, we use (29). We obtain that:

$\displaystyle\psi_{t,h,n}^{\alpha}(\boldsymbol{t_{n}})\leq
h^{\theta}C^{(5)}\,t^{(d-\alpha)/2}\big{[}t_{1}(t_{2}-t_{1})\ldots(t-t_{n})\big{]}^{-(d-\alpha)/2}(t-t_{n})^{-\theta},$

where $C^{(5)}$ is a constant that depends on $d$, $t_{0}$, $T$, $a$, $b$, and
$\beta$.

Returning to (30), we obtain:

$\displaystyle A^{(\alpha)}_{n}(t,h)$ $\displaystyle\leq
h^{\theta}\Gamma_{0,t}^{n}\,n!\,C^{(5)}t^{(d-\alpha)/2}$
$\displaystyle\qquad\quad\int_{0<t_{n}<t}\int_{T_{n-1}(t_{n})}\big{[}t_{1}(t_{2}-t_{1})\ldots(t_{n}-t_{n-1})\big{]}^{-\frac{d-\alpha}{2}}d\boldsymbol{t_{n-1}}(t-t_{n})^{-\frac{d-\alpha}{2}-\theta}dt_{n}$
$\displaystyle=h^{\theta}\Gamma_{0,t}^{n}\,n!\,C^{(5)}t^{(d-\alpha)/2}\frac{\Gamma\big{(}1-\frac{d-\alpha}{2}\big{)}^{n}}{\Gamma\big{(}n(1-\frac{d-\alpha}{2})\big{)}}\int_{0}^{t}t_{n}^{(n-1)(1-\frac{d-\alpha}{2})-\frac{d-\alpha}{2}}(t-t_{n})^{-\frac{d-\alpha}{2}-\theta}dt_{n}$
$\displaystyle\leq
h^{\theta}\Gamma_{0,t}^{n}\,n!\,C^{(5)}\frac{\Gamma\big{(}1-\frac{d-\alpha}{2}\big{)}^{n}}{\Gamma\big{(}n(1-\frac{d-\alpha}{2})\big{)}}t^{\frac{d-\alpha}{2}}t^{(n-1)(1-\frac{d-\alpha}{2})}\int_{0}^{t}t_{n}^{-\frac{d-\alpha}{2}}(t-t_{n})^{-\frac{d-\alpha}{2}-\theta}dt_{n}$
$\displaystyle=h^{\theta}\Gamma_{0,t}^{n}\,n!\,C^{(5)}\frac{\Gamma\big{(}1-\frac{d-\alpha}{2}\big{)}^{n}}{\Gamma\big{(}n(1-\frac{d-\alpha}{2})\big{)}}t^{n\big{(}1-\frac{d-\alpha}{2}\big{)}-\theta}B\Big{(}1-\frac{d-\alpha}{2},1-\frac{d-\alpha}{2}-\theta\Big{)},$
(36)

where we used Lemma 2.3 in the first equality and $B(a,b)$ denotes the beta
function. Note that $B(1-\frac{d-\alpha}{2},1-\frac{d-\alpha}{2}-\theta)$ is
finite since $1-\frac{d-\alpha}{2}-\theta>0$ for any
$\beta\in(\frac{d-a}{2},1)$ arbitrary. This expression can be bounded
uniformly in $\alpha\in[a,b]$, using monotonicity properties of the $\Gamma$
function. We have:

$\displaystyle\sup_{\alpha\in[a,b]}\sum_{n\geq
1}(p-1)^{n/2}\Big{(}\frac{1}{n!}A^{(\alpha)}_{n}(t,h)\Big{)}^{1/2}\leq
Ch^{(1-\beta)/2},$

where $C$ depends on ($d$, $t_{0}$, $T$, $p$, $a$, $b$, $\beta$).

Next, we examine $B_{n}^{\alpha}(t,h)$. Let $D_{t,h}=[0,t+h]^{n}\verb
2\2[0,t]^{n}$. Then

$\displaystyle B_{n}^{\alpha}(t,h)$
$\displaystyle\leq\Gamma_{0,t+h}^{n}\int_{[0,t+h]^{n}}\gamma_{t,h,n}^{\alpha}(\boldsymbol{t_{n}})1_{D_{t,h}}(\boldsymbol{t_{n}})d\boldsymbol{t_{n}},$
(37)

where

$\gamma_{t,h,n}^{\alpha}(\boldsymbol{t_{n}}):=(n!)^{2}\int_{(\mathbb{R}^{d})^{n}}\big{|}\mathcal{F}\widetilde{f}_{t+h,x,n}(\boldsymbol{t_{n}},\cdot)(\boldsymbol{\xi_{n}})\big{|}^{2}\prod_{j=1}^{n}|\xi_{j}|^{-\alpha}d\boldsymbol{\xi_{n}}.$

Note that for any $\mathbb{t}\in D_{t,h}$, there exists a permutation $\rho$
of $\\{1,\ldots,n\\}$ such that $t_{\rho(1)}<\ldots<t_{\rho(n)}<t+h$ and
$t<t_{\rho(n)}<t+h$. For the estimate below, we assume for simplicity that
$\rho$ is the identity, i.e. $0<t_{1}<\ldots<t_{n}$ and $t<t_{n}<t+h$. Then,
by Lemma 1.2 and using the fact that $w(t,x)\leq C$ for all $t\in[t_{0},T]$
and $x\in K$, we get:

$\displaystyle\big{|}\mathcal{F}f_{t+h,x,n}(\boldsymbol{t_{n}},\cdot)(\boldsymbol{\xi_{n}})\big{|}^{2}\leq
C\prod_{k=1}^{n-1}\exp\Big{(}-\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\exp\Big{(}-\frac{t+h-t_{n}}{t_{n}(t+h)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}.$

A similar relation holds for arbitrary $\rho$, with $t_{j}$, $\xi_{j}$
replaced by $t_{\rho(j)}$, $\xi_{\rho(j)}$ respectively.

Using Lemma 2.6, we have:

$\displaystyle\int_{(\mathbb{R}^{d})^{n}}\big{|}\mathcal{F}f_{t+h,x,n}(\boldsymbol{t_{n}},\cdot)(\boldsymbol{\xi_{n}})\big{|}^{2}\prod_{j=1}^{n}|\xi_{j}|^{-\alpha}d\boldsymbol{\xi_{n}}$
$\displaystyle\qquad\leq
C\int_{(\mathbb{R}^{d})^{n}}\prod_{k=1}^{n-1}\exp\Big{(}-\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\exp\Big{(}-\frac{t+h-t_{n}}{t_{n}(t+h)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\prod_{j=1}^{n}|\xi_{j}|^{-\alpha}d\boldsymbol{\xi_{n}}$
$\displaystyle\qquad\leq
C\prod_{j=1}^{n-1}k_{\alpha}\bigg{(}\frac{2(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\bigg{)}\times
k_{\alpha}\bigg{(}\frac{2(t+h-t_{n})t_{n}}{t+h}\bigg{)}$
$\displaystyle\qquad\leq
C_{1}\big{[}t_{1}(t_{2}-t_{1})\ldots(t_{n}-t_{n-1})\big{]}^{-(d-\alpha)/2}\Big{(}\frac{t+h-t_{n}}{t+h}\Big{)}^{-(d-\alpha)/2},$

where we used (29) in the last inequality and $C_{1}>0$ is a constant that
depends on $d$, $t_{0}$, $a$ and $b$.

We come back to (37). Using the change of variable $s=t+h-t_{n}$, we have:

$\displaystyle B_{n}^{\alpha}(t,h)$
$\displaystyle\leq\Gamma^{n}_{0,t+h}C_{1}\,n!\int_{t}^{t+h}\Big{(}\frac{t+h-t_{n}}{t+h}\Big{)}^{-(d-\alpha)/2}dt_{n}$
$\displaystyle\qquad\qquad\Big{(}\int_{0<t_{1}<\ldots<t_{n}}\big{[}t_{1}(t_{2}-t_{1})\ldots(t_{n}-t_{n-1})\big{]}^{-(d-\alpha)/2}dt_{1}\ldots
dt_{n-1}\Big{)}$
$\displaystyle\leq\Gamma^{n}_{0,t+h}C_{1}\,n!\frac{\Gamma\Big{(}1-\frac{d-\alpha}{2}\Big{)}^{n}}{\Gamma\Big{(}n\big{(}1-\frac{d-\alpha}{2}\big{)}\Big{)}}\int_{t}^{t+h}t_{n}^{(n-1)\big{(}1-\frac{d-\alpha}{2}\big{)}-\frac{d-\alpha}{2}}\Big{(}\frac{t+h-t_{n}}{t+h}\Big{)}^{-(d-\alpha)/2}dt_{n}$
$\displaystyle=\Gamma^{n}_{0,t+h}C_{1}\,n!\frac{\Gamma\Big{(}1-\frac{d-\alpha}{2}\Big{)}^{n}}{\Gamma\Big{(}n\big{(}1-\frac{d-\alpha}{2}\big{)}\Big{)}}\Big{(}\frac{1}{t+h}\Big{)}^{-(d-\alpha)/2}\int_{0}^{h}(t+h-s)^{n\big{(}1-\frac{d-\alpha}{2}\big{)}-1}s^{-(d-\alpha)/2}ds$
$\displaystyle\leq\Gamma^{n}_{0,t+h}C_{1}\,n!\frac{\Gamma\Big{(}1-\frac{d-\alpha}{2}\Big{)}^{n}}{\Gamma\Big{(}n\big{(}1-\frac{d-\alpha}{2}\big{)}\Big{)}}(t+h)^{(d-\alpha)/2}T^{n\big{(}1-\frac{d-\alpha}{2}\big{)}-1}\int_{0}^{h}s^{-(d-\alpha)/2}ds$
$\displaystyle\leq\Gamma^{n}_{0,t+h}C_{2}\,n!\frac{\Gamma\Big{(}1-\frac{d-\alpha}{2}\Big{)}^{n}}{\Gamma\Big{(}n\big{(}1-\frac{d-\alpha}{2}\big{)}\Big{)}}T^{(n-1)\big{(}1-\frac{d-b}{2}\big{)}}h^{1-\beta},$

where in the last inequality we used the fact that $h^{1-(d-\alpha)/2}\leq
h^{1-\beta}$ since $\beta>\frac{d-a}{2}>\frac{d-\alpha}{2}$. Note that the
preceding expression can be uniformly bounded for all $\alpha\in[a,b]$. Hence,
we obtain:

$\displaystyle\sup_{\alpha\in[a,b]}\sum_{n\geq
1}(p-1)^{n/2}\Big{(}\frac{1}{n!}B^{(\alpha)}_{n}(t,h)\Big{)}^{1/2}\leq
C\,h^{(1-\beta)/2},$

where $C$ is a constant depending on ($d$, $t_{0}$, $T$, $p$, $a$, $b$).

Step 2 (space increments). We denote $z=x^{\prime}-x$. By hypercontractivity,

$\|u^{\alpha}(t,x+z)-u^{\alpha}(t,x)\|_{p}\leq\sum_{n\geq
1}(p-1)^{n/2}\|I_{n}^{\alpha}(f_{t,x+z,n}-f_{t,x,n})\|_{p}\leq\sum_{n\geq
1}(p-1)^{n/2}\left(\frac{1}{n!}C_{n}^{\alpha}(t,z)\right)^{1/2},$

where
$C_{n}^{\alpha}(t,z)=(n!)^{2}\|\widetilde{f}_{t,x+z,n}-\widetilde{f}_{t,x,n}\|_{\mathcal{H}_{\alpha}^{\otimes
n}}^{2}$. Note that

$\displaystyle C_{n}^{\alpha}(t,z)$
$\displaystyle\leq\Gamma_{0,t}^{n}\int_{[0,t]^{n}}\psi_{t,z,n}^{\alpha}(\boldsymbol{t_{n}})d\boldsymbol{t_{n}},$
(38)

where we fixed $\boldsymbol{t_{n}}\in[0,t]^{n}$ and $\rho$ is the permutation
of $1,\ldots,n$ such that $t_{\rho(1)}<\ldots<t_{\rho(n)}$ with
$t_{\rho(n+1)}=t$ and

$\psi_{t,z,n}^{\alpha}(\boldsymbol{t_{n}}):=\int_{(\mathbb{R}^{d})^{n}}\Big{|}\mathcal{F}\Big{(}f_{t,x+z,n}(t_{\rho(1)},\cdot,\ldots,t_{\rho(1)},\cdot)-f_{t,x,n}(t_{\rho(1)},\cdot,\ldots,t_{\rho(1)},\cdot)\Big{)}(\boldsymbol{\xi_{n}})\Big{|}^{2}\prod_{j=1}^{n}|\xi_{j}|^{-\alpha}d\boldsymbol{\xi_{n}}.$

Assume for simplicity that $t_{1}<\ldots<t_{n}$. By Lemma 1.2, we have:

$\displaystyle\mathcal{F}\big{(}f_{t,x+z,n}(\boldsymbol{t_{n}},\cdot)-f_{t,x,n}(\boldsymbol{t_{n}},\cdot)\big{)}(\boldsymbol{\xi_{n}})=\prod_{k=1}^{n}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}(J_{1}+J_{2})$

where

$\displaystyle J_{1}$
$\displaystyle:=\bigg{(}\exp\Big{(}-\frac{i}{t}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot(x+z)\Big{)}-\exp\Big{(}-\frac{i}{t}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot
x\Big{)}\bigg{)}$
$\displaystyle\qquad\quad\int_{\mathbb{R}^{d}}\exp\Big{(}-i\Big{[}\sum_{j=1}^{n}\Big{(}1-\frac{t_{j}}{t}\Big{)}\xi_{j}\Big{]}\cdot
x_{0}\Big{)}\,G_{t}(x+z-x_{0})u_{0}(\mathrm{d}x_{0})$ $\displaystyle J_{2}$
$\displaystyle:=\exp\Big{(}-\frac{i}{t}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot
x\Big{)}$
$\displaystyle\qquad\quad\int_{\mathbb{R}^{d}}\exp\Big{(}-i\Big{[}\sum_{j=1}^{n}\Big{(}1-\frac{t_{j}}{t}\Big{)}\xi_{j}\Big{]}\cdot
x_{0}\Big{)}\,\Big{(}G_{t}(x+z-x_{0})-G_{t}(x-x_{0})\Big{)}u_{0}(\mathrm{d}x_{0}).$

For the term $J_{1}$, using the fact that $|1-e^{-ix}|^{2}\leq x^{2\theta}$,
for any $\theta\in[0,1]$, we get

$\displaystyle J^{2}_{1}$
$\displaystyle\leq\bigg{|}\exp\Big{(}-\frac{i}{t}\Big{(}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{)}\cdot
z\Big{)}-1\bigg{|}^{2}w_{+}^{2}(t,x+z)\leq
C_{1}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2\theta}|z|^{2\theta},$

where $C_{1}>0$ is a constant depends on $t_{0}$, $d$, $\theta$ and $T$. Note
that $|w_{+}(t,x+z)|=|w_{+}(t,x^{\prime})|\leq C$ for all $x^{\prime}\in K$.

We now turn to the analysis of $J_{2}$. We use the following inequality (see
Lemma 4.1 of [8]): for any $\alpha\in(0,1]$,

$\big{|}G_{t}(x)-G_{t}(y)\big{|}\leq\frac{C_{\alpha}}{t^{\alpha/2}}\big{[}G_{2t}(x)+G_{2t}(y)\big{]}|x-y|^{\alpha},$

for any $t>0$ and $x,y\in\mathbb{R}^{d}$, where $C_{\alpha}>0$ is a constant
depending on $\alpha$. Using the above inequality, we have

$\displaystyle J^{2}_{2}$
$\displaystyle\leq\Big{|}\int_{\mathbb{R}^{d}}\exp\Big{(}-i\Big{[}\sum_{j=1}^{n}\Big{(}1-\frac{t_{j}}{t}\Big{)}\xi_{j}\Big{]}\cdot
x_{0}\Big{)}\,\Big{(}G_{t}(x+z-x_{0})-G_{t}(x-x_{0})\Big{)}u_{0}(\mathrm{d}x_{0})\Big{|}^{2}$
$\displaystyle\leq\Big{(}\int_{\mathbb{R}^{d}}\frac{C_{\theta}}{t^{\theta/2}}\big{[}G_{2t}(x+z-x_{0})+G_{2t}(x-x_{0})\big{]}|z|^{\theta}u_{0}(\mathrm{d}x_{0})\Big{)}^{2}\leq
C_{2}|z|^{2\theta}.$

Here we used the fact that $w(2t,\cdot)$ is uniformly bounded on compact sets
and $t\geq t_{0}$.

Hence, there exists a constant $C$ that depends on $t_{0}$, $d$, $\theta$ and
$T$ such that

$\displaystyle\Big{|}\mathcal{F}\big{(}f_{t,x+z,n}(\boldsymbol{t_{n}},\cdot)-f_{t,x,n}(\boldsymbol{t_{n}},\cdot)\big{)}(\boldsymbol{\xi_{n}})\Big{|}^{2}\leq|z|^{2\theta}C(F_{1}+F_{2}),$
(39)

where

$F_{1}:=\prod_{k=1}^{n}\exp\Big{(}-\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2\theta}$

and

$F_{2}:=\prod_{k=1}^{n}\exp\Big{(}-\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}.$

We only need to study $F_{1}$ since the term involving $F_{2}$ is uniformly
bounded by Lemma 2.6. Using (32) and (33), we have:

$\displaystyle
F_{1}\leq\prod_{k=1}^{n}\exp\Big{(}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}C_{\theta}T^{\theta}(t-t_{n})^{-\theta}.$
(40)

By (39), (40) and Lemma 2.6, we get:

$\displaystyle\psi_{t,z,n}^{\alpha}(\boldsymbol{t_{n}})$
$\displaystyle\leq|z|^{2\theta}C_{3}\bigg{(}(t-t_{n})^{-\theta}\prod_{j=1}^{n}k_{\alpha}\Big{(}\frac{2(\frac{t_{j+1}}{2}-\frac{t_{j}}{2})(\frac{t_{j}}{2})}{\frac{t_{j+1}}{2}}\Big{)}+\prod_{j=1}^{n}k_{\alpha}\Big{(}\frac{2(t_{j+1}-t_{j})t_{j}}{t_{j+1}}\Big{)}\bigg{)}$
$\displaystyle\leq|z|^{2\theta}C_{4}\bigg{(}\big{[}t_{1}(t_{2}-t_{1})\ldots(t-t_{n})\big{]}^{-(d-\alpha)/2}(t-t_{n})^{-\theta}+\big{[}t_{1}(t_{2}-t_{1})\ldots(t-t_{n})\big{]}^{-(d-\alpha)/2}\bigg{)}$

where we used relation (29) in the last inequality and $C_{3}$, $C_{4}$ are
constants depending on $t_{0}$, $d$, $a$, $b$, $\theta$ and $T$.

We come back to (38). Similarly to (36), using the estimate of
$\psi_{t,z,n}^{\alpha}(\boldsymbol{t_{n}})$ above, we have:

$\displaystyle C^{(\alpha)}_{n}(t,z)$
$\displaystyle\leq|z|^{2\theta}\Gamma^{n}_{0,t}n!C_{5}\bigg{(}\frac{\Gamma\big{(}1-\frac{d-\alpha}{2}\big{)}^{n}}{\Gamma\big{(}n\big{(}1-\frac{d-\alpha}{2}\big{)}\big{)}}(T\vee
1)^{n(1-\frac{d-\alpha}{2})-\theta}+\frac{\Gamma\big{(}1-\frac{d-\alpha}{2}\big{)}^{n+1}}{\Gamma\big{(}(n+1)\big{(}1-\frac{d-\alpha}{2}\big{)}\big{)}}(T\vee
1)^{n(1-\frac{d-\alpha}{2})}\bigg{)}$

where requires again that $\beta\in(\frac{d-a}{2},1)$. Therefore,

$\displaystyle\sup_{\alpha\in[a,b]}\sum_{n\geq
1}(p-1)^{n/2}\Big{(}\frac{1}{n!}C^{(\alpha)}_{n}(t,z)\Big{)}^{1/2}\leq
C|z|^{1-\beta},$

where $C$ is a constant depending on $d$, $t_{0}$, $T$, $p$, $a$, $b$ and
$\beta$. ∎

Now we are ready to prove Theorem 1.3. Note that this proof employs similar
concepts as those found in the proof of Theorem 2.1 of [6].

Proof of Theorem 1.3: Step 1. (finite-dimensional distribution convergence) In
this step, we prove that: for any
$(t_{1},x_{1}),\ldots,(t_{k},x_{k})\in[t_{0},T]\times\mathbb{R}^{d}$,

$\big{(}u^{\alpha_{n}}(t_{1},x_{1}),\ldots,u^{\alpha_{n}}(t_{k},x_{k})\big{)}\stackrel{{\scriptstyle
d}}{{\to}}\big{(}u^{\alpha^{*}}(t_{1},x_{1}),\ldots,u^{\alpha^{*}}(t_{k},x_{k})\big{)}.$

It is enough to prove that $u^{\alpha_{n}}(t,x)\to u^{\alpha^{*}}(t,x)$ in
$L^{2}(\Omega)$ as $n\to\infty$, for any
$(t,x)\in[t_{0},T]\times\mathbb{R}^{d}$. To do this, we approximate
$u^{\alpha}(t,x)$ by the partial sum:

$u_{m}^{\alpha}(t,x)=w(t,x)+\sum_{k=1}^{m}I_{k}^{\alpha}(f_{t,x,k}).$ (41)

Note that for any $m\geq 1$, by Lemma 2.4,

$\displaystyle\mathbb{E}|u_{m}^{\alpha_{n}}(t,x)-u_{m}^{\alpha^{*}}(t,x)|^{2}\leq
m\sum_{k=1}^{m}\mathbb{E}|I_{k}^{\alpha_{n}}(f_{t,x,k})-I_{k}^{\alpha^{*}}(f_{t,x,k})|^{2}\quad\mbox{as
$n\to\infty$}.$

Moreover,
$\mathbb{E}|u_{m}^{\alpha^{\ast}}(t,x)-u^{\alpha^{\ast}}(t,x)|^{2}\to 0,\text{
as }m\to\infty$.

It remains to prove that

$\sup_{n\geq 1}\mathbb{E}|u_{m}^{\alpha_{n}}(t,x)-u^{\alpha_{n}}(t,x)|^{2}\to
0,\text{ as }m\to\infty.$

To see this, we choose specific values for $a$ and $b$ such that
$\max(d-2,0)<a<\alpha^{*}<b<d$. Since $\alpha_{n}\to\alpha^{*}$, there exists
$N\in\mathbb{N}$ such that $a<\alpha_{n}<b$ for any $n\geq N$. Therefore, it
suffices to prove that:

$\sup_{\alpha\in[a,b]}\mathbb{E}|u_{m}^{\alpha}(t,x)-u^{\alpha}(t,x)|^{2}=\sup_{\alpha\in[a,b]}\sum_{k\geq
m+1}\mathbb{E}|I_{k}^{\alpha}(f_{t,x,k})|^{2}\to 0\quad\mbox{as
$m\to\infty$.}$ (42)

Note that

$\mathbb{E}|I_{k}^{\alpha}(f_{t,x,k})|^{2}=k!\|\widetilde{f}_{t,x,k}\|_{\mathcal{H}_{\alpha}^{\otimes
k}}^{2}\leq
k!\Gamma_{0,t}^{k}\int_{[0,t]^{n}}A_{k}^{\alpha}(\boldsymbol{t_{k}})d\boldsymbol{t_{k}},$

where $\Gamma_{0,t}=\int_{-t}^{t}\gamma_{0}(s)ds$ and
$A_{k}^{\alpha}(\boldsymbol{t_{k}})=\int_{(\mathbb{R}^{d})^{k}}\big{|}\mathcal{F}\widetilde{f}_{t,x,k}(\boldsymbol{t_{k}},\bullet)(\boldsymbol{\xi_{k}})\big{|}^{2}\prod_{j=1}^{k}|\xi_{j}|^{-\alpha}d\boldsymbol{\xi_{k}}$.
We denote $\boldsymbol{t_{k}}=(t_{1},\ldots,t_{k})\in[0,t]^{k}$ be arbitrary
and $\rho$ be a permutation of $1,\ldots,k$ such that
$t_{\rho(1)}<\ldots<t_{\rho(k)}$, with $t_{\rho(k+1)}=t$. Then by Lemma 2.6,
we have:

$\displaystyle A_{k}^{\alpha}(\boldsymbol{t_{k}})$
$\displaystyle=\frac{1}{(k!)^{2}}\int_{(\mathbb{R}^{d})^{k}}\bigg{|}\prod_{j=1}^{k}\exp\bigg{\\{}-\frac{1}{2}\frac{t_{\rho(j+1)}-t_{\rho(j)}}{t_{\rho(j)}t_{\rho((j+1)}}\Big{|}\sum_{i=1}^{j}t_{\rho(i)}\xi_{\rho(i)}\Big{|}^{2}\bigg{\\}}\exp\bigg{\\{}-\frac{i}{t}\Big{(}\sum_{j=1}^{k}t_{\rho(j)}\xi_{\rho(j)}\Big{)}\cdot
x\bigg{\\}}$
$\displaystyle\qquad\quad\times\int_{\mathbb{R}^{d}}\exp\Big{\\{}-i\Big{[}\sum_{j=1}^{k}\Big{(}1-\frac{t_{\rho(j)}}{t}\Big{)}\xi_{\rho(j)}\Big{]}\cdot
x_{0}\Big{\\}}G_{t}(x-x_{0})u_{0}(\mathrm{d}x_{0})\bigg{|}^{2}\prod_{j=1}^{k}|\xi_{j}|^{-\alpha}d\boldsymbol{\xi_{k}}$
$\displaystyle\leq\frac{w_{+}^{2}(t,x)}{(k!)^{2}}\prod_{j=1}^{k}k_{\alpha}\Big{(}\frac{2(t_{\rho(j+1)}-t_{\rho(j)})t_{\rho(j)}}{t_{\rho(j+1)}}\Big{)}$
$\displaystyle=\frac{w_{+}^{2}(t,x)(C^{(1)}_{d,\alpha})^{k}2^{-(d-\alpha)k/2}}{(k!)^{2}}t^{\frac{d-\alpha}{2}}[t_{\rho(1)}(t_{\rho(2)}-t_{\rho(1)})\ldots(t-t_{\rho(k)})]^{-\frac{d-\alpha}{2}},$

where $k_{\alpha}$ and $C^{(1)}_{d,\alpha}$ are given in (28), respectively
(29). Using Lemma 2.3, we get:

$\displaystyle\mathbb{E}|I_{k}^{\alpha}(f_{t,x,k})|^{2}$ $\displaystyle\leq
Ck!\,\Gamma_{0,t}^{k}(C^{(1)}_{d,\alpha})^{k}2^{-(d-\alpha)k/2}t^{\frac{d-\alpha}{2}}$
$\displaystyle\qquad\qquad\qquad\frac{1}{(k!)^{2}}\sum_{\rho\in
S_{n}}\int_{t_{\rho(1)}<\ldots<t_{\rho(k)}}[t_{\rho(1)}(t_{\rho(2)}-t_{\rho(1)})\ldots(t-t_{\rho(k)})]^{-\frac{d-\alpha}{2}}d\boldsymbol{t_{k}}$
$\displaystyle=C\,\Gamma_{0,t}^{k}(C^{(1)}_{d,\alpha})^{k}2^{-(d-\alpha)k/2}t^{k(1-\frac{d-\alpha}{2})}\frac{\big{(}\Gamma(1-\frac{d-\alpha}{2})\big{)}^{k+1}}{\Gamma\Big{(}(k+1)\big{(}1-\frac{d-\alpha}{2}\big{)}\Big{)}}.$

The last thing we need to do is to uniformly bound above expression for all
$\alpha\in[a,b]$. Note that for any $\alpha\in[a,b]$,

$C^{(1)}_{d,\alpha}\leq
c_{d}\Big{(}\frac{1}{d-b}+\frac{2}{2-(d-a)}\Big{)}=:C_{d,a,b}$

and

$2^{-(d-\alpha)k/2}\leq 2^{-(d-b)k/2}.$

Moreover, $\Gamma(1-\frac{d-\alpha}{2})\leq\Gamma(1-\frac{d-a}{2})$,
$t^{k(1-\frac{d-\alpha}{2})}\leq(t\vee 1)^{k(1-\frac{d-b}{2})}$ and

$\Gamma\Big{(}(k+1)\big{(}1-\frac{d-\alpha}{2}\big{)}\Big{)}\geq\Gamma\Big{(}(k+1)\big{(}1-\frac{d-a}{2}\big{)}\Big{)}\quad\mbox{for
any}\ k\geq m_{0},$

where $m_{0}$ is chosen such that $m_{0}(1-\frac{d-a}{2})>x_{0}$, and
therefore for any $m\geq m_{0}$,

$\displaystyle\sup_{\alpha\in[a,b]}\sum_{k\geq
m+1}\mathbb{E}\big{|}I^{\alpha}_{k}(f_{t,x,k})\big{|}^{2}\leq\sum_{k\geq
m+1}C\Gamma^{k}_{0,T}C^{k}_{d,a,b}\frac{\big{(}\Gamma(1-\frac{d-a}{2})\big{)}^{k+1}}{\Gamma\Big{(}(k+1)\big{(}1-\frac{d-a}{2}\big{)}\Big{)}}(t\vee
1)^{k(1-\frac{d-b}{2})}\to 0.$ (43)

Relation (42) follows.

Step 2. (tightness) The fact that the sequence $(u^{\alpha_{n}})_{n\geq 1}$ is
tight in $C([0,T]\times\mathbb{R}^{d})$ follows by Proposition 2.3 of [14],
using Theorem 2.7 above.

∎

##  3 Rough noise

In this section, we consider equation (1) driven by a Gaussian noise $W$ where
the temporal covariance function $\gamma_{0}$ is given by (6), and the measure
$\mu$ is given by (14). The existence of the solution was proved in [4].

To emphasize the dependence on the parameter $H$, we denote the noise, the
Hilbert space, and the solution, by $W^{H},\mathcal{H}^{H},u^{H}$,
respectively. Denoting the multiple integral of order $n$ with respect to
$W^{H}$ as $I_{n}^{H}$, the series expansion (11) takes the following form:

$u^{H}(t,x)=1+\sum_{n\geq 1}I_{n}^{H}(f_{t,x,n}).$

Similarly to Section 4 of [6], all Gaussian processes
$\\{W^{H}\\}_{H\in(0,1/2)}$ can be constructed on the same probability space.
Let $\widehat{W}$ be the $\mathbb{C}$-valued Gaussian measure given by (17).
We recall the following representation of the multiple integral with respect
to the process $W^{H}$: for any
$t\mapsto\varphi(t,\cdot)\in\mathcal{S}^{\prime}(\mathbb{R}^{d})$,

$I_{k}^{H}(\varphi)=c_{H_{0}}^{k/2}c_{H}^{k/2}\int_{(\mathbb{R}\times\mathbb{R})^{k}}\mathcal{F}_{t}[\mathcal{F}_{x}\varphi(\boldsymbol{t_{k}},\bullet)(\boldsymbol{\xi_{k}})](\boldsymbol{\tau_{k}})\prod_{j=1}^{k}|\tau_{j}|^{\frac{1}{2}-H_{0}}|\xi_{j}|^{\frac{1}{2}-H}\widehat{W}(d\tau_{1},d\xi_{1})\ldots\widehat{W}(d\tau_{k},d\xi_{k}).$
(44)

We will use the following lemmas.

######  Lemma 3.1 (Lemma 3.1 of [5]).

Let $G$ be the fundamental solution of the heat equation, the integral
$\int_{\mathbb{R}}|\mathcal{F}G_{t}(\xi)|^{2}|\xi|^{\alpha}\mathrm{d}\xi$ is
finite if and only if $\alpha>-1$ and in this case,

$\int_{\mathbb{R}}|\mathcal{F}G_{t}(\xi)|^{2}|\xi|^{\alpha}\mathrm{d}\xi=\Gamma\Big{(}\frac{1+\alpha}{2}\Big{)}\
t^{-(1+\alpha)/2}.$

######  Lemma 3.2 (Lemma A.1 in [4]).

Suppose that $\alpha_{i}>-1$, $\beta_{i}>-1$ for any $i=1,\ldots,n$, and

$\sum_{i=1}^{k}(\alpha_{i}+\beta_{i})+k+1+\alpha_{k+1}>0\,\text{ for all
}k=1,\ldots,n-1.$ (45)

Then by setting $t_{n+1}=t$, $|\alpha|=\sum_{i=1}^{n}\alpha_{i}$ and
$|\beta|=\sum_{i=1}^{n}\beta_{i}$, we have that

$\displaystyle
I_{n}(t,\alpha_{1},\ldots,\alpha_{n},\beta_{1},\ldots,\beta_{n}):=\int_{0<t_{1}<\ldots<t_{n}<t}\prod_{i=1}^{n}t_{i}^{\alpha_{i}}(t_{i+1}-t_{i})^{\beta_{i}}\mathrm{d}\mathbb{t}$
$\displaystyle=\frac{\Gamma(\alpha_{1}+1)\prod_{i=1}^{n}\Gamma(\beta_{i}+1)}{\Gamma\big{(}|\alpha|+|\beta|+n+1\big{)}}\prod_{k=1}^{n-1}\frac{\Gamma\big{(}\sum_{i=1}^{k}(\alpha_{i}+\beta_{i})+k+1+\alpha_{k+1}\big{)}}{\Gamma\big{(}\sum_{i=1}^{k}(\alpha_{i}+\beta_{i})+k+1\big{)}}t^{|\alpha|+|\beta|+n}.$

We will use a similar approximation technique as in Section 2. More precisely,
we study first the $L^{2}(\Omega)$-continuity in $H$ of the multiple integral
of $I_{k}^{H}(f_{t,x,k})$ for fixed $k$. This is achieved by the following
lemma. Note that the proof is the same as in the proof of Lemma 4.2 of [6],
the difference being that here we use a different expression for the Fourier
transform
$\mathcal{F}f_{t,x,k}(\boldsymbol{t_{k}},\cdot)(\boldsymbol{\xi_{k}})$, due to
the general initial condition. The following inequality is also required,
which is a consequence of the Littlewood-Hardy inequality: for any $\varphi\in
L^{1/H_{0}}(\mathbb{R}_{+}^{k})$,

$\alpha_{H_{0}}^{k}\int_{\mathbb{R}_{+}^{k}}\int_{\mathbb{R}_{+}^{k}}\prod_{j=1}^{k}|t_{j}-s_{j}|^{2H_{0}-2}\varphi(\boldsymbol{t_{k}})\varphi(\boldsymbol{s_{k}})d\boldsymbol{t_{k}}d\boldsymbol{s_{k}}\leq
b_{H_{0}}^{k}\left(\int_{\mathbb{R}_{+}^{k}}|\varphi(\boldsymbol{t_{k}})|^{1/H_{0}}d\boldsymbol{t_{k}}\right)^{2H_{0}},$
(46)

where $b_{H_{0}}>0$ depends only on $H_{0}$.

######  Lemma 3.3.

Let $\ell=\max(3/4-H_{0},0)$. If $H_{n}\to H^{*}\in(\ell,1/2)$, then

$Q_{n}:=\mathbb{E}|I_{k}^{H_{n}}(f_{t,x,k})-I_{k}^{H^{*}}(f_{t,x,k})|^{2}\to
0,\quad\mbox{as $n\to\infty$},$

for any $t>0$, $x\in\mathbb{R}$ and $k\geq 1$.

###### Proof.

Proceeding as in the proof of Lemma 4.2 of [6], by (46), we infer that

$\displaystyle Q_{n}\leq
k!\,b_{H_{0}}^{k}\left(\int_{T_{k}(t)}A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{t_{k}})^{\frac{1}{2H_{0}}}d\boldsymbol{t_{k}}\right)^{2H_{0}}.$

where

$\displaystyle
A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{s_{k}})=\int_{\mathbb{R}^{k}}\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{t_{k}})\overline{\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{s_{k}})}\left|c^{k}_{H_{n}}\prod_{j=1}^{k}|\xi_{j}|^{\frac{1}{2}-H_{n}}-c^{k}_{H^{\ast}}\prod_{j=1}^{k}|\xi_{j}|^{\frac{1}{2}-H^{\ast}}\right|^{2}d\boldsymbol{\xi_{k}}$

and $\phi_{\boldsymbol{\xi_{k}}}(\boldsymbol{t_{k}})$ is defined by (23) with
$d=1$.

Therefore, it suffices to show that:

$\int_{T_{k}(t)}A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{t_{k}})^{\frac{1}{2H_{0}}}d\boldsymbol{t_{k}}\to
0,\quad\mbox{as $n\to\infty$}.$ (47)

For this, we apply the Dominated Convergence Theorem. We will show that

$A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{t_{k}})\to 0\quad\mbox{as
$n\to\infty$},$ (48)

and there exists a function $h_{k}(\boldsymbol{t_{k}})$ such that

$A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{t_{k}})\leq
h_{k}(\boldsymbol{t_{k}})\quad\mbox{and}\int_{T_{k}(t)}h_{k}(\boldsymbol{t_{k}})d\boldsymbol{t_{k}}<\infty.$
(49)

We first prove (48). By the change of variables $z_{i}=t_{i}\xi_{i}$ for
$i=1,\ldots,j$, followed by $\eta_{j}=\sum_{i=1}^{j}z_{i}$ for $j=1,\ldots,k$
(with $\eta_{0}=0$), we have

$\displaystyle A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{t_{k}})\leq
w_{+}^{2}(t,x)\int_{\mathbb{R}^{k}}\prod_{j=1}^{k}\exp\Big{(}-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\Big{|}\sum_{i=1}^{j}t_{i}\xi_{i}\Big{|}^{2}\Big{)}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\bigg{|}c^{k/2}_{H_{n}}\prod_{j=1}^{k}\big{|}\xi_{j}\big{|}^{\frac{1}{2}-H_{n}}-c^{k/2}_{H^{\ast}}\prod_{j=1}^{k}\big{|}\xi_{j}\big{|}^{\frac{1}{2}-H^{\ast}}\bigg{|}^{2}d\boldsymbol{\xi_{k}}$
$\displaystyle=w_{+}^{2}(t,x)\Big{(}\prod_{j=1}^{k}t_{j}^{-1}\Big{)}\int_{\mathbb{R}^{k}}\prod_{j=1}^{k}\exp\Big{(}-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\Big{|}\sum_{i=1}^{j}z_{i}\Big{|}^{2}\Big{)}$
$\displaystyle\qquad\qquad\qquad\bigg{|}c^{k/2}_{H_{n}}\prod_{j=1}^{k}t_{j}^{H_{n}-\frac{1}{2}}\prod_{j=1}^{k}\big{|}z_{j}\big{|}^{\frac{1}{2}-H_{n}}-c^{k/2}_{H^{\ast}}\prod_{j=1}^{k}t_{j}^{H^{\ast}-\frac{1}{2}}\prod_{j=1}^{k}\big{|}z_{j}\big{|}^{\frac{1}{2}-H^{\ast}}\bigg{|}^{2}d\boldsymbol{z_{k}}$
$\displaystyle=w_{+}^{2}(t,x)\int_{\mathbb{R}^{k}}\prod_{j=1}^{k}\exp\Big{(}-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\Big{|}\eta_{j}\Big{|}^{2}\Big{)}$
$\displaystyle\qquad\bigg{|}c^{k/2}_{H_{n}}\prod_{j=1}^{k}t_{j}^{H_{n}-1}\prod_{j=1}^{k}\big{|}\eta_{j}-\eta_{j-1}\big{|}^{\frac{1}{2}-H_{n}}-c^{k/2}_{H^{\ast}}\prod_{j=1}^{k}t_{j}^{H^{\ast}-1}\prod_{j=1}^{k}\big{|}\eta_{j}-\eta_{j-1}\big{|}^{\frac{1}{2}-H^{\ast}}\bigg{|}^{2}d\boldsymbol{\eta_{k}}.$
(50)

We denote by $B^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})$ the integrand
in the last integral above. Clearly,
$B^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})\to 0$ as $n\to\infty$. So
relation (48) will follow by the Dominated Convergence Theorem, as long as we
show that there exists a function
$B(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})$ such that

$B^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})\leq
B(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})\quad\mbox{and}\quad\int_{\mathbb{R}^{k}}B(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})d\boldsymbol{\eta_{k}}<\infty.$

Note that $B^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})\leq
2\big{(}B_{1}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})+B_{2}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})\big{)}$,
where

$B_{1}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}}):=\prod_{j=1}^{k}\exp\Big{(}-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\big{|}\eta_{j}\big{|}^{2}\Big{)}c^{k}_{H_{n}}\prod_{j=1}^{k}t_{j}^{2H_{n}-2}\prod_{j=1}^{k}\big{|}\eta_{j}-\eta_{j-1}\big{|}^{1-2H_{n}}$

and $B_{2}$ has a similar expression with $H_{n}$ replaced by $H^{*}$. It
suffices to consider $B_{1}^{(n)}$.

Fix numbers $a$ and $b$ such that

$\ell<a<H^{*}<b<\frac{1}{2}.$ (51)

Then $H_{n}\in[a,b]$ for $n$ large enough. Since the constant $c_{H}$ defines
a continuous function of $H$, $c_{H_{n}}$ is bounded by a constant $c$. We use
the inequality

$\prod_{j=1}^{k}|\eta_{j}-\eta_{j-1}|^{1-2H}\leq\sum_{\boldsymbol{a}\in
A_{k}}\prod_{j=1}^{k}|\eta_{j}|^{(1-2H)a_{j}},$ (52)

where $A_{k}$ is a set of multi-indices $\boldsymbol{a}=(a_{1},\ldots,a_{k})$
such that $a_{1}\in\\{1,2\\}$, $a_{n}\in\\{0,1\\}$, $a_{j}\in\\{0,1,2\\}$ for
$j=1,\ldots,k-1$, $\sum_{j=1}^{k}a_{j}=k$, and ${\rm card}(A_{k})=2^{k-1}$.
Note that $|\eta_{j}|^{(1-2H)a_{j}}\leq f_{a}(|\eta_{j}|)$ for any
$j=1,\ldots,n$, where $a$ is the constant from $\\{0,1,2\\}$ and the functions
$f_{0},f_{1},f_{2}$ are defined as follows: $f_{0}(r)=1$ for any $r>0$,

$f_{1}(r)=\left\\{\begin{array}[]{ll}r^{1-2a}&\mbox{if $r\geq 1$}\\\
1&\mbox{if $r\in(0,1)$}\end{array}\right.\quad\mbox{and}\quad
f_{2}(r)=\left\\{\begin{array}[]{ll}r^{2(1-2a)}&\mbox{if $r\geq 1$}\\\
1&\mbox{if $r\in(0,1)$}.\end{array}\right.$

Moreover, for any $H_{n}\in[a,b]$,

$\displaystyle\prod_{j=1}^{k}t^{2H_{n}-2}_{j}$
$\displaystyle=\Big{(}\prod_{j=1}^{k}t^{-1}_{j}\Big{)}^{2-2H_{n}}\leq\prod_{j=1}^{k}\big{(}t^{-1}_{j}\vee
1\big{)}^{2-2a}\leq\prod_{j=1}^{k}\big{(}t_{j}\wedge
1\big{)}^{2a-2}\leq\prod_{j=1}^{k}\big{(}t_{j}\vee 1\big{)}^{2a-2}.$

It follows that:

$\displaystyle B_{1}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})$
$\displaystyle\leq c^{k}\Big{(}\prod_{j=1}^{k}(t_{j}\vee
1)^{2a-2}\Big{)}\sum_{\mathbb{a}\in
A_{k}}\prod_{j=1}^{k}\exp\Big{(}-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}\Big{|}\eta_{j}\Big{|}^{2}\Big{)}\prod_{j=1}^{k}f_{a_{j}}(|\eta_{j}|)=:F(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}}).$

By the same argument,
$B_{2}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})\leq
F(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})$. Hence,
$B^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})\leq
2F(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})$ and so,

$A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{t_{k}})=\int_{\mathbb{R}^{k}}B^{(n)}(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})d\boldsymbol{\eta_{k}}\leq
4\int_{\mathbb{R}^{k}}F(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})d\boldsymbol{\eta_{k}}.$
(53)

It remains to prove that
$\int_{\mathbb{R}^{k}}F(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})d\boldsymbol{\eta_{k}}<\infty$.
We give below an estimate for this integral which will be used for the proof
of relation (49). Note that

$\int_{\mathbb{R}^{k}}F(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})d\boldsymbol{\eta_{k}}=c^{k}\prod_{j=1}^{k}(t_{j}\vee
1)^{2a-2}\sum_{\mathbb{a}\in A_{k}}\prod_{j=1}^{k}I(a_{j}),\ \mbox{with}\
I(a_{j})=\int_{\mathbb{R}}e^{-\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}|\eta_{j}|^{2}}f_{a_{j}}(|\eta_{j}|)d\eta_{j}.$

We use the same arguments for the study of the upper bound of $I(a_{j})$ in
the proof of Lemma 4.2 in [6] with $t_{j+1}-t_{j}$ replaced by
$\frac{t_{j+1}-t_{j}}{t_{j}t_{j+1}}$. Moreover, we have

$\begin{cases}(t_{j}\vee 1)^{2a-2}(t_{j}t_{j+1})^{\frac{1}{2}}\leq
t^{\frac{1}{2}}_{j+1}\frac{1}{(t_{j}\vee 1)^{2-2a}}(t_{j}\vee
1)^{\frac{1}{2}}\leq t^{\frac{1}{2}}_{j+1},&\text{ since
}\frac{1}{2}<2-2a,\\\\[8.00003pt] (t_{j}\vee 1)^{2a-2}(t_{j}t_{j+1})^{1-a}\leq
t^{1-a}_{j+1}\frac{1}{(t_{j}\vee 1)^{1-a}}(t_{j}\vee 1)^{1-a}\leq
t^{1-a}_{j+1},&\text{ since }1-a<2-2a,\\\\[8.00003pt] (t_{j}\vee
1)^{2a-2}(t_{j}t_{j+1})^{\frac{3-4a}{2}}\leq
t^{\frac{3-4a}{2}}_{j+1}\frac{1}{(t_{j}\vee 1)^{\frac{3-4a}{2}}}(t_{j}\vee
1)^{\frac{3-4a}{2}}\leq t^{\frac{3-4a}{2}}_{j+1},&\text{ since
}\frac{3-4a}{2}<2-2a.\end{cases}$

Since $\frac{1}{2}<1-a<\frac{3-4a}{2}$, $t^{\frac{1}{2}}_{j+1}$,
$t^{1-a}_{j+1}$ and $t^{\frac{3-4a}{2}}_{j+1}$ are all dominated by $(T\vee
1)^{\frac{3-4a}{2}}$. Hence,

$\displaystyle(t_{j}\vee 1)^{2a-2}I(a_{j})$ $\displaystyle\leq
c_{a}t^{\frac{3-4a}{2}}_{j+1}\Big{\\{}(t_{j+1}-t_{j})^{-\frac{1}{2}}+(t_{j+1}-t_{j})^{-(1-a)}+(t_{j+1}-t_{j})^{-\frac{3-4a}{2}}\Big{\\}}$
$\displaystyle\leq c_{a}c_{t,a}(T\vee
1)^{\frac{3-4a}{2}}\,(t_{j+1}-t_{j})^{-\frac{3-4a}{2}}.$

Since $\text{card}(A_{k})=2^{k-1}$, we obtain:

$\int_{\mathbb{R}^{k}}F(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})\mathrm{d}\boldsymbol{\eta}\leq
2^{k-1}c^{k}c^{k}_{a}c^{k}_{t,a}(T\vee
1)^{k\cdot\frac{3-4a}{2}}\prod_{j=1}^{k}(t_{j+1}-t_{j})^{-\frac{3-4a}{2}}.$
(54)

This proves the integrability of $F(\boldsymbol{t_{k}},\boldsymbol{\eta_{k}})$
and so, relation (48) follows by the Dominated Convergence Theorem.

Now we prove (49). By relations (53) and (54), we have

$\displaystyle\Big{(}A_{k}^{(n)}(\boldsymbol{t_{k}},\boldsymbol{t_{k}})\Big{)}^{\frac{1}{2H_{0}}}$
$\displaystyle\leq\Big{(}2^{k+1}c^{k}c_{a}^{k}c^{k}_{t,a}(T\vee
1)^{k\cdot\frac{3-4a}{2}}\prod_{j=1}^{k}(t_{j+1}-t_{j})^{-\frac{3-4a}{2}}\Big{)}^{\frac{1}{2H_{0}}}=:h_{k}(\mathbb{t}).$

Note that $h_{k}(\mathbb{t})$ is integrable over the symplex $T_{k}(t)$ since
$-\frac{3-4a}{4H_{0}}>-1$ which is equivalent to $a>\frac{3}{4}-H_{0}$. This
finishes the proof of (49) and the justification of the application of the
Dominated Convergence Theorem.

∎

We now prove some moment estimates for the increments of the solution $u^{H}$,
uniform in $H$, which will be used below for the proof of tightness of the
family $(u^{H})_{H\in[a,b]}$. In particular, Theorem 3.4 shows that the
process $u^{H}$ has a continuous modification. We will work with this
modification.

######  Theorem 3.4.

Let $\ell=\max(3/4-H_{0},0)$. Let $[a,b]$ be a compact set in $[0,1]$ such
that:

$\ell<a<b<\frac{1}{2}.$ (55)

Let $u^{H}$ be the solution of the heat equation (1) with noise $W^{H}$. For
any $p\geq 2$, $T>0$, $c_{0}\in(0,\frac{2H_{0}+a-1}{2H_{0}})$ and

$0<\theta<2H_{0}(1-c_{0})+a-1,$ (56)

there exist a constant $C>0$ such that for any $t,t^{\prime}\in[t_{0},T]$ and
$x,x^{\prime}\in K$,

$\sup_{H\in[a,b]}\mathbb{E}|u(t,x)-u(t^{\prime},x^{\prime})|^{p}\leq
C\Big{(}|t-t^{\prime}|^{p\theta/2}+|x-x^{\prime}|^{p\theta}\Big{)}.$

###### Proof.

Step 1 (time increments). Let $t\in[t_{0},T]$ and $h>0$ be such that
$t+h\in[t_{0},T]$. As in the proof of Theorem 2.7,

$\displaystyle\big{\|}u^{H}(t+h,x)-u^{H}(t,x)\big{\|}_{p}$
$\displaystyle\leq\sum_{n\geq
1}(p-1)^{n/2}\left(\frac{2}{n!}\Big{(}A_{n}^{H}(t,h)+B_{n}^{H}(t,h)\Big{)}\right)^{1/2},$

where
$A_{n}^{H}(t,h)=(n!)^{2}\|\widetilde{f}_{t+h,x,n}1_{[0,t]^{n}}-\widetilde{f}_{t,x,n}\|_{\mathcal{H}_{H}^{\otimes
n}}^{2}$ and
$B_{n}^{H}(t,h)=(n!)^{2}\|\widetilde{f}_{t+h,x,n}1_{[0,t+h]^{n}-[0,t]^{n}}\|_{\mathcal{H}_{H}^{\otimes
n}}^{2}$.

We examine $A_{n}^{H}(t,h)$. Note that by inequality (46),

$\displaystyle A^{H}_{n}(t,h)$ $\displaystyle\leq
b^{n}_{H_{0}}\bigg{(}\int_{[0,t]^{n}}\big{(}\psi_{t,h,n}^{H}(\boldsymbol{t_{n}})\big{)}^{\frac{1}{2H_{0}}}\mathrm{d}{\mathbb{t}}\bigg{)}^{2H_{0}},$
(57)

where the function $\psi_{t,h,n}^{H}(\boldsymbol{t_{n}})$ is defined as
follows: if $\boldsymbol{t_{n}}\in[0,t]^{n}$ is fixed and $\rho$ is the
permutation of $1,\ldots,n$ such that $t_{\rho(1)}<\ldots<t_{\rho(n)}$ with
$t_{\rho(n+1)}=t$, and

$\displaystyle\psi_{t,h,n}^{H}(\boldsymbol{t_{n}})$
$\displaystyle:=\int_{\mathbb{R}^{n}}\Big{|}\mathcal{F}\Big{(}f_{t+h,x,n}(t_{\rho(1)},\cdot,\ldots,t_{\rho(n)},\cdot)-f_{t,x,n}(t_{\rho(1)},\cdot,\ldots,t_{\rho(n)},\cdot)\Big{)}(\boldsymbol{\xi_{n}})\Big{|}^{2}\prod_{j=1}^{n}|\xi_{j}|^{1-2H}d\boldsymbol{\xi_{n}}.$

Assume for simplicity that $t_{1}<\ldots<t_{n}$. To estimate
$\psi_{t,h,n}^{H}(\boldsymbol{t_{n}})$, we use estimates (31), (34) and (35),
for the increments of the Fourier transform appearing above. Hence, there
exists a constant $C^{(1)}$ depending on $t_{0}$, $\theta$ and $T$ such that

$\displaystyle\psi_{t,h,n}^{H}(\boldsymbol{t_{n}})\leq
h^{\theta}C^{(1)}\big{(}t-t_{n}\big{)}^{-\theta}I^{(n)}_{t/2}(\boldsymbol{t_{n}}/2)$
(58)

where

$\displaystyle
I^{(n)}_{t}(\boldsymbol{t_{n}}):=c_{H}^{n}\int_{\mathbb{R}^{n}}\prod_{k=1}^{n}\exp\bigg{\\{}-\frac{t_{k+1}-t_{k}}{t_{k+1}t_{k}}\Big{|}\sum_{i=1}^{k}t_{i}\xi_{i}\Big{|}^{2}\bigg{\\}}|\xi_{k}|^{1-2H}d\boldsymbol{\xi_{n}}.$
(59)

and $t_{n+1}=t$.

We study $I^{(n)}_{t}(\boldsymbol{t_{n}})$. Using the change of variables
$z_{i}=t_{i}\xi_{i}$ for $i=1,\ldots,k$, followed by
$\eta_{k}=\sum_{i=1}^{k}z_{i}$ for $k=1,\ldots,n$, we have:

$\displaystyle I^{(n)}_{t}(\boldsymbol{t_{n}})$ $\displaystyle\leq
c_{H}^{n}\Big{(}\prod_{k=1}^{n}t_{k}\Big{)}^{2H-2}\int_{\mathbb{R}^{n}}\prod_{k=1}^{n}\exp\bigg{\\{}-\frac{t_{k+1}-t_{k}}{t_{k+1}t_{k}}|\eta_{k}|^{2}\bigg{\\}}|\eta_{1}|^{1-2H}\prod_{k=2}^{n}\big{(}|\eta_{k}|^{1-2H}+|\eta_{k-1}|^{1-2H}\big{)}\mathrm{d}\boldsymbol{\eta_{n}}$
$\displaystyle\leq
c_{H}^{n}\Big{(}\prod_{k=1}^{n}t_{k}\Big{)}^{2H-2}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}\prod_{k=1}^{n}\bigg{\\{}\int_{\mathbb{R}}\exp\Big{(}-\frac{t_{k+1}-t_{k}}{t_{k+1}t_{k}}|\eta_{k}|^{2}\Big{)}|\eta_{k}|^{\alpha_{k}}\mathrm{d}\eta_{k}\bigg{\\}},$

where for the last inequality, we used relation (52) and the set $D_{n}^{(H)}$
is the set of multi-indices
$\boldsymbol{\alpha_{n}}=(\alpha_{1},\ldots,\alpha_{n})$ with
$\alpha_{j}=(1-2H)a_{j}$ and $\boldsymbol{a_{n}}=(a_{1},\ldots,a_{n})\in
A_{n}$. Using Lemma 3.1 to compute the $d\eta_{k}$ integrals above with
$k=1,\ldots,n$, we have:

$\displaystyle I^{(n)}_{t}(\boldsymbol{t_{n}})$ $\displaystyle\leq
c_{H}^{n}\Big{(}\prod_{k=1}^{n}t_{k}\Big{)}^{2H-2}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}\prod_{k=1}^{n}\Gamma\Big{(}\frac{1+\alpha_{k}}{2}\Big{)}\Big{(}\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{)}^{-\frac{1+\alpha_{k}}{2}}$
$\displaystyle\quad\leq C_{H,1}^{n}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}t^{\frac{1+\alpha_{n}}{2}}t_{1}^{\frac{4H-3+\alpha_{1}}{2}}\Big{(}\prod_{k=2}^{n}t_{k}^{\frac{4H-2+\alpha_{k-1}+\alpha_{k}}{2}}\Big{)}\prod_{k=1}^{n}(t_{k+1}-t_{k})^{-\frac{1+\alpha_{k}}{2}}$
(60)

where for the second inequality we used the constant $C_{H,1}$ defined by

$C_{H,1}=c_{H}\max\bigg{\\{}\Gamma\Big{(}\frac{1}{2}\Big{)},\Gamma\Big{(}1-H\Big{)},\Gamma\Big{(}\frac{3-4H}{2}\Big{)}\bigg{\\}}.$
(61)

Both $c_{H}$ and $C_{H,1}$ can be uniformly bounded for all $H\in[a,b]$.
Hence, we obtain:

$\displaystyle\big{(}t-t_{n}\big{)}^{-\theta}I^{(n)}_{t/2}(\boldsymbol{t_{n}}/2)$
$\displaystyle\leq(C^{(1)})^{n}2^{n}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}t^{\frac{1+\alpha_{n}}{2}}t_{1}^{\frac{4H-3+\alpha_{1}}{2}}\Big{(}\prod_{k=2}^{n}t_{k}^{\frac{4H-2+\alpha_{k-1}+\alpha_{k}}{2}}\Big{)}\Big{(}\prod_{k=1}^{n-1}\big{(}t_{k+1}-t_{k}\big{)}^{-\frac{1+\alpha_{k}}{2}}\Big{)}(t-t_{n})^{-\frac{1+\alpha_{n}}{2}-\theta}.$
(62)

Taking power $\frac{1}{2H_{0}}$ on both sides of (58) above, we obtain:

$\displaystyle\Big{(}\psi^{H}_{t,z,n}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}\leq\Big{(}h^{\theta}C^{(1)}2^{n}\Big{)}^{\frac{1}{2H_{0}}}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}\prod_{k=1}^{n}t^{\widetilde{\alpha}_{k}}_{k}\prod_{k=1}^{n-1}\big{(}t_{k+1}-t_{k}\big{)}^{\widetilde{\beta}_{k}}\big{(}t-t_{n}\big{)}^{\widetilde{\beta}_{n,\theta}}$
(63)

where

$\displaystyle\widetilde{\alpha}_{k}=\begin{cases}\frac{4H-3+\alpha_{1}}{4H_{0}},&\quad
k=1\\\\[8.00003pt] \frac{4H-2+\alpha_{k-1}+\alpha_{k}}{4H_{0}},&\quad
k=2,\ldots,n\end{cases}$ (64)

and

$\displaystyle\widetilde{\beta}_{k}=-\frac{1+\alpha_{k}}{4H_{0}},\;k=1,\ldots,n-1$
(65)

and

$\displaystyle\widetilde{\beta}_{n,\theta}=-\frac{1+\alpha_{n}+2\theta}{4H_{0}}.$
(66)

A similar estimate holds for arbitrary $(t_{1},\ldots,t_{n})\in[0,t]^{n}$ with
$t_{\rho(1)}<\ldots<t_{\rho(n)}$ for same permutation $\rho\in S_{n}$, where
$S_{n}$ is the set of all permutation of $\\{1,\ldots,n\\}$. From here, we
derive that

$\displaystyle\int_{[0,t]^{n}}\left(\psi_{t,h,n}^{H}(\boldsymbol{t_{n}})\right)^{\frac{1}{2H_{0}}}d\boldsymbol{t_{n}}$
$\displaystyle\qquad\leq\Big{(}h^{\theta}C^{(1)}2^{n}\Big{)}^{\frac{1}{2H_{0}}}n!\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}\int_{T_{n}(t)}\prod_{k=1}^{n}t^{\widetilde{\alpha}_{k}}_{k}\prod_{k=1}^{n-1}\big{(}t_{k+1}-t_{k}\big{)}^{\widetilde{\beta}_{k}}\big{(}t_{n+1}-t_{n}\big{)}^{\widetilde{\beta}_{n,\theta}}d\boldsymbol{t_{n}}.$
(67)

To compute the integral

${\cal
I}(\boldsymbol{\alpha_{n}}):=\int_{T_{n}(t)}\prod_{k=1}^{n}t^{\widetilde{\alpha}_{k}}_{k}\prod_{k=1}^{n-1}\big{(}t_{k+1}-t_{k}\big{)}^{\widetilde{\beta}_{k}}\big{(}t_{n+1}-t_{n}\big{)}^{\widetilde{\beta}_{n,\theta}}d\boldsymbol{t_{n}},$

we use Lemma 3.2. To apply this Lemma, we first need
$\widetilde{\alpha}_{k}>-1$ for all $k=1,\ldots,n$ and
$\widetilde{\beta}_{k}>-1$ for all $k=1,\ldots,n-1$ which hold since
$\alpha_{k}>0$ for all $k=1,\ldots,n$. We further need to check
$\widetilde{\beta}_{n,\theta}>-1$ and condition (45) holds for
$k=1,\ldots,n-1$. When $\alpha_{n}=0$, we need $-\frac{1+2\theta}{4H_{0}}>-1$,
i.e. $\theta<2H_{0}-1/2$. When $\alpha_{n}=1-2H$, we need
$-\frac{2+2H+2\theta}{4H_{0}}>-1$, i.e. $\theta<2H_{0}+H-1$. Note that
$2H_{0}-1/2>2H_{0}+H-1$ since $H<1/2$. Hence, for all $H\in[a,b]$ arbitrary,
we encounter the condition

$0<\theta<2H_{0}+a-1.$ (68)

The verification of (45) has been done in [12]. Under condition (68), we can
apply Lemma 3.2 to deduce that:

$\displaystyle{\cal
I}(\boldsymbol{\alpha_{n}})=\frac{\Gamma(\widetilde{\alpha}_{1}+1)\prod_{k=1}^{n-1}\Gamma(\widetilde{\beta}_{k}+1)\Gamma(\widetilde{\beta}_{n,\theta}+1)}{\Gamma\big{(}|\widetilde{\alpha}|+|\widetilde{\beta}|+n+1\big{)}}\gamma_{n}t^{|\widetilde{\alpha}|+|\widetilde{\beta}|+n}$
(69)

with

$\gamma_{n}=\gamma_{n}(\boldsymbol{\alpha_{n}})=\prod_{k=1}^{n-1}\frac{\Gamma\big{(}\sum_{i=1}^{k}(\widetilde{\alpha}_{i}+\widetilde{\beta}_{i})+k+1+\widetilde{\alpha}_{k+1}\big{)}}{\Gamma\big{(}\sum_{i=1}^{k}(\widetilde{\alpha}_{i}+\widetilde{\beta}_{i})+k+1\big{)}}.$

Note that $\gamma_{n}$ does not depend on $\widetilde{\beta}_{n,\theta}$, and

$\gamma_{n}\leq 1.$

This proof require significant effort. We refer the reader to Step 4 of the
proof of Theorem 1.1 in [4].

To bound these remaining factors in ${\cal I}(\boldsymbol{\alpha_{n}})$, we
use some monotonicity properties of the Gamma function: $\Gamma$ is decreasing
on $(0,x_{0})$ and increasing on $(x_{0},\infty)$, where $x_{0}\approx 1.4$.
Therefore, for any $\alpha_{j}\in\\{0,1-2H,2(1-2H)\\}$ and $H\in[a,b]$, we
have
$\Gamma(\widetilde{\alpha}_{1}+1)\leq\Gamma\big{(}\frac{2H_{0}+a-1}{2H_{0}}\big{)}$
and
$\prod_{k=1}^{n-1}\Gamma(\widetilde{\beta}_{k}+1)\leq\big{(}\Gamma\big{(}1-\frac{3-4a}{4H_{0}}\big{)}\big{)}^{n-1}$.
Next, we observe that
$\widetilde{\beta}_{n,\theta}+1=1-\frac{1+\alpha_{n}+2\theta}{4H_{0}}$ takes
values in the interval $[1-\frac{1-a+\theta}{2H_{0}},1-\frac{1}{4H_{0}}]$
whose lower bound may be close to 0. Since $\lim_{x\to 0}\Gamma(x)=\infty$, we
control this term by fixing an arbitrary value
$c_{0}\in(0,\frac{2H_{0}+a-1}{2H_{0}})$, and then choosing

$0<\theta<2H_{0}(1-c_{0})+a-1$ (70)

With this choice of $\theta$, (68) holds, and more importantly
$1-\frac{1-a+\theta}{2H_{0}}>c_{0}$, so that for any
$\alpha_{n}\in\\{0,1-2H\\}$ and $H\in[a,b]$,

$\Gamma\Big{(}1-\frac{1+\alpha_{n}+2\theta}{4H_{0}}\Big{)}\leq\Gamma(c_{0}).$
(71)

For any $H\in[a,b]$, the upper bound of
$|\widetilde{\alpha}|+|\widetilde{\beta}|+n$ is

$\displaystyle|\widetilde{\alpha}|+|\widetilde{\beta}|+n$
$\displaystyle\leq\frac{n(2H_{0}+H-1)}{2H_{0}}-\frac{1+\alpha_{n}}{4H_{0}}\leq\frac{n(2H_{0}+H-1)}{2H_{0}}-\frac{1}{4H_{0}}$
$\displaystyle\leq\frac{n(2H_{0}+b-1)}{2H_{0}}-\frac{1}{4H_{0}}\leq\frac{n(2H_{0}+b-1)}{2H_{0}}$

and the lower bound of $|\widetilde{\alpha}|+|\widetilde{\beta}|+n$ is

$\displaystyle|\widetilde{\alpha}|+|\widetilde{\beta}|+n$
$\displaystyle\geq\frac{(n-1)(2H_{0}+H-1)}{2H_{0}}-\frac{1+\alpha_{n}}{4H_{0}}\geq\frac{(n-1)(2H_{0}+H-1)}{2H_{0}}-\frac{1-H}{2H_{0}}$
$\displaystyle\geq\frac{(n-1)(2H_{0}+a-1)}{2H_{0}}-\frac{1-a}{2H_{0}}.$

To bound the $\Gamma$-value appearing in the denominator, we pick an integer
$m_{0}$ such that
$\frac{(m_{0}-1)(2H_{0}+a-1)}{2H_{0}}-\frac{1-a}{2H_{0}}>x_{0}$. Then, for any
$n\geq m_{0}$ and for any $H\in[a,b]$,

$\displaystyle\Gamma\Big{(}|\widetilde{\alpha}|+|\widetilde{\beta}|+n+1\Big{)}>\Gamma\Big{(}\frac{(n-1)(2H_{0}+a-1)}{2H_{0}}-\frac{1-a}{2H_{0}}+1\Big{)}>c^{n-1}\big{[}(n-1)!\big{]}^{\frac{2H_{0}+a-1}{2H_{0}}}.$

Hence, combining all the estimates above, for any $\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}$ and for any $H\in[a,b]$,

$\displaystyle{\cal
I}(\boldsymbol{\alpha_{n}})\leq\frac{\big{(}C^{(2)}\big{)}^{n-1}}{[(n-1)!]^{\frac{2H_{0}+a-1}{2H_{0}}}}(t\vee
1)^{\frac{n(2H_{0}+b-1)}{2H_{0}}}.$

Returning to (67), we have:

$\int_{[0,t]^{n}}\Big{(}\psi_{t,h,n}^{H}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}d\boldsymbol{t_{n}}\leq
h^{\frac{\theta}{2H_{0}}}\frac{\big{(}C^{(3)}\big{)}^{n-1}}{[(n-1)!]^{\frac{a-1}{2H_{0}}}}.$

Finally, coming back to (57), we obtain:

$A_{n}^{H}(t,h)\leq
h^{\theta}\frac{\big{(}C^{(4)}\big{)}^{(n-1)2H_{0}}}{[(n-1)!]^{a-1}}.$

Consequently, for any $p\geq 2$, $t\in[t_{0},T]$, $H\in[a,b]$ and $\theta$ as
in (70),

$\sum_{n\geq 1}(p-1)^{n/2}\left(\frac{1}{n!}A_{n}^{H}(t,h)\right)^{1/2}\leq
Ch^{\theta/2}.$ (72)

where $C$ depends on $p$, $a$, $b$, $\theta$, $c_{0}$, $H_{0}$, and $T$.

We examine $B_{n}^{H}(t,h)$. Let $D_{t,h}=[0,t+h]^{n}\verb 2\2[0,t]^{n}$. By
inequality (46),

$B_{n}^{H}(t,h)\leq
b_{H_{0}}^{n}\left(\int_{[0,t+h]^{n}}\gamma_{t,h,n}^{H}(\boldsymbol{t_{n}})^{\frac{1}{2H_{0}}}1_{D_{t,h}}(\boldsymbol{t_{n}})d\boldsymbol{t_{n}},\right)^{2H_{0}}$
(73)

where

$\gamma_{t,h,n}^{H}(\boldsymbol{t_{n}}):=(n!)^{2}c_{H}^{n}\int_{\mathbb{R}^{n}}\big{|}\mathcal{F}\widetilde{f}_{t+h,x,n}(\boldsymbol{t_{n}},\cdot)(\boldsymbol{\xi_{n}})\big{|}^{2}\prod_{j=1}^{n}|\xi_{j}|^{1-2H}d\boldsymbol{\xi_{n}}.$

Note that for any $\mathbb{t}\in D_{t,h}$, there exists a permutation $\rho$
of $\\{1,\ldots,n\\}$ such that $t_{\rho(1)}<\ldots<t_{\rho(n)}<t+h$ and
$t<t_{\rho(n)}<t_{\rho(n+1)}=t+h$. For the estimate below, we assume for
simplicity that $\rho$ is the identity, i.e. $0<t_{1}<\ldots<t_{n}$ and
$t<t_{n}<t_{n+1}=t+h$. Then, by Lemma 1.2 and using the fact that $w(t,x)\leq
C$ for all $t\in[t_{0},T]$ and $x\in K$, we get:

$\displaystyle\big{|}\mathcal{F}f_{t+h,x,n}(\boldsymbol{t_{n}},\cdot)(\boldsymbol{\xi_{n}})\big{|}^{2}\leq
C\prod_{k=1}^{n-1}\exp\Big{(}-\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\Big{)}\exp\Big{(}-\frac{t+h-t_{n}}{t_{n}(t+h)}\Big{|}\sum_{j=1}^{n}t_{j}\xi_{j}\Big{|}^{2}\Big{)}.$

A similar relation holds for arbitrary $\rho$, with $t_{j}$, $\xi_{j}$
replaced by $t_{\rho(j)}$, $\xi_{\rho(j)}$ respectively. Hence, using the same
approach as in (59) and (60) and Lemma 3.1, we have:

$\displaystyle\gamma_{t,h,n}^{H}(\boldsymbol{t_{n}})\leq
C\;c_{H}^{n}\Big{(}\prod_{k=1}^{n}t_{k}\Big{)}^{2H-2}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}\Big{(}\prod_{k=1}^{n-1}\int_{\mathbb{R}}\exp\Big{(}-\frac{t_{k+1}-t_{k}}{t_{k}t_{k+1}}\Big{)}|\eta_{k}|^{\alpha_{k}}\mathrm{d}\eta_{k}\Big{)}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\Big{(}\int_{\mathbb{R}}\exp\Big{(}-\frac{t+h-t_{n}}{t_{n}(t+h)}\Big{)}|\eta_{n}|^{\alpha_{n}}\mathrm{d}\eta_{n}\Big{)}$
$\displaystyle=Cc_{H}^{n}\Big{(}\prod_{k=1}^{n}t_{k}\Big{)}^{2H-2}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}\prod_{k=1}^{n-1}\Gamma\Big{(}\frac{1+\alpha_{k}}{2}\Big{)}\Big{(}\frac{t_{k+1}-t_{k}}{t_{k+1}t_{k}}\Big{)}^{-\frac{1+\alpha_{k}}{2}}\Gamma\Big{(}\frac{1+\alpha_{n}}{2}\Big{)}\Big{(}\frac{t+h-t_{n}}{t_{n}(t+h)}\Big{)}^{-\frac{1+\alpha_{n}}{2}}$
$\displaystyle\leq CC_{H,1}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}t_{1}^{\frac{4H-3+\alpha_{1}}{2}}\Big{(}\prod_{k=2}^{n}t_{k}^{\frac{4H-2+\alpha_{k-1}+\alpha_{k}}{2}}\Big{)}(t+h)^{\frac{1+\alpha_{n}}{2}}\prod_{k=1}^{n-1}\big{(}t_{k+1}-t_{k}\big{)}^{-\frac{1+\alpha_{k}}{2}}\big{(}t+h-t_{n}\big{)}^{-\frac{1+\alpha_{n}}{2}}$
(74)

where $C_{H,1}$ is given by (61) and it can be uniformly bounded for all
$H\in[a,b]$.

Hence, taking power $\frac{1}{2H_{0}}$ on both sides of relation (74), we
obtain:

$\displaystyle\Big{(}\gamma_{t,h,n}^{H}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}$
$\displaystyle\leq\big{(}C^{(5)}\big{)}^{\frac{1}{2H_{0}}}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}\prod_{k=1}^{n}t_{k}^{\widetilde{\alpha}_{k}}\big{(}t_{k+1}-t_{k}\big{)}^{\widetilde{\beta}_{k}}(t+h)^{\frac{1+\alpha_{n}}{4H_{0}}},$
(75)

where $\widetilde{\alpha}_{1},\ldots,\widetilde{\alpha}_{n}$ are given by
(64); $\widetilde{\beta}_{1},\ldots,\widetilde{\beta}_{n-1}$ are given by (65)
and

$\widetilde{\beta}_{n}=-\frac{1+\alpha_{n}}{4H_{0}}$

Integrating over the set $D_{t,h}$ and using Lemma 3.2, we obtain:

$\displaystyle\int_{[0,t+h]^{n}}\Big{(}\gamma_{t,h,n}^{H}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}1_{D_{t,h}}({\mathbb{t}})d\mathbb{t}$
$\displaystyle\quad\leq\big{(}C^{(5)}\big{)}^{\frac{1}{2H_{0}}}n!\int_{t}^{t+h}\bigg{(}\int_{T_{n-1}(t_{n})}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}(t+h)^{\frac{1+\alpha_{n}}{4H_{0}}}\prod_{k=1}^{n-1}t^{\widetilde{\alpha}_{k}}_{k}(t_{k+1}-t_{k})^{\widetilde{\beta}_{k}}(t+h-t_{n})^{\widetilde{\beta}_{n}}d\boldsymbol{t_{n-1}}\bigg{)}dt_{n}$
$\displaystyle\quad\leq\big{(}C^{(5)}\big{)}^{\frac{1}{2H_{0}}}n!\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}(t+h)^{\frac{1+\alpha_{n}}{4H_{0}}}\int_{t}^{t+h}J(t_{n})\cdot(t+h-t_{n})^{\widetilde{\beta}_{n}}\mathrm{d}t_{n}$
(76)

where

$\displaystyle J(t_{n})$
$\displaystyle:=\int_{T_{n-1}(t_{n})}\prod_{k=1}^{n-1}t_{k}^{\widetilde{\alpha}_{k}}(t_{k+1}-t_{k})^{\widetilde{\beta}_{k}}d\boldsymbol{t_{n-1}}$
$\displaystyle=\frac{\Gamma(\widetilde{\alpha}_{1}+1)\prod_{k=1}^{n-1}\Gamma(\widetilde{\beta}_{k}+1)}{\Gamma\big{(}\sum_{k=1}^{n-1}\widetilde{\alpha}_{k}+\sum_{k=1}^{n-1}\widetilde{\beta}_{k}+n\big{)}}\gamma_{n-1}t^{\sum_{k=1}^{n-1}\widetilde{\alpha}_{k}+\sum_{k=1}^{n-1}\widetilde{\beta}_{k}+n-1}$
(77)

with

$\gamma_{n-1}=\gamma_{n-1}(\boldsymbol{\alpha_{n-1}})=\prod_{k=1}^{n-2}\frac{\Gamma\big{(}\sum_{i=1}^{k}(\widetilde{\alpha}_{i}+\widetilde{\beta}_{i})+k+1+\widetilde{\alpha}_{k+1}\big{)}}{\Gamma\big{(}\sum_{i=1}^{k}(\widetilde{\alpha}_{i}+\widetilde{\beta}_{i})+k+1\big{)}}.$

Note that $\gamma_{n-1}\leq 1$, by using the same arguments in Step 4 of the
proof of Theorem 1.1 in [4]. To bound these factors in (77), we use the same
approach as in the study of ${\cal I}(\boldsymbol{\alpha_{n}})$ above. It
follows that:

$\displaystyle
J(t_{n})\leq\frac{\big{(}C^{(6)}\big{)}^{n-1}}{\big{[}(n-1)!\big{]}^{\frac{2H_{0}+a-1}{2H_{0}}}}\;(t\vee
1)^{\frac{(n-1)(2H_{0}+b-1)}{2H_{0}}}.$

Using the change of variable $s=t+h-t_{n}$, we conclude that

$\displaystyle\int_{[0,t+h]^{n}}\gamma_{t,h,n}^{(H)}(\boldsymbol{t_{n}})^{\frac{1}{2H_{0}}}1_{D_{t,h}}(\boldsymbol{t_{n}})d\boldsymbol{t_{n}}\leq\frac{\big{(}C^{(7)}\big{)}^{n-1}}{\big{[}(n-1)!\big{]}^{\frac{2H_{0}+a-1}{2H_{0}}}}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}\frac{1}{1-\frac{1+\alpha_{n}}{4H_{0}}}h^{1-\frac{1+\alpha_{n}}{4H_{0}}}.$

Note that $(1-\frac{1+\alpha_{n}}{4H_{0}})^{-1}$ is bounded by
$\frac{2H_{0}}{2H_{0}+a-1}$ for any $\alpha_{n}\in\\{0,1-2H\\}$ and
$H\in[a,b]$. Moreover, due to condition (68),

$\frac{\theta}{2H_{0}}<\frac{2H_{0}+a-1}{2H_{0}}\leq
1-\frac{1+\alpha_{n}}{4H_{0}}\leq 1-\frac{1}{4H_{0}}.$

Therefore,

$\int_{[0,t+h]^{n}}\gamma_{t,h,n}^{(H)}(\boldsymbol{t_{n}})^{\frac{1}{2H_{0}}}1_{D_{t,h}}(\boldsymbol{t_{n}})d\boldsymbol{t_{n}}\leq
h^{\frac{\theta}{2H_{0}}}\frac{\big{(}C^{(8)}\big{)}^{n-1}}{\big{[}(n-1)!\big{]}^{\frac{2H_{0}+a-1}{2H_{0}}}}.$

Returning to (73), we infer that $B_{n}(t,h)\leq
h^{\theta}\frac{C^{n-1}}{[(n-1)!]^{a-1}}$, and hence, for any $p\geq 2$,
$t\in[t_{0},T]$ and $\theta$ satisfying (68) (and in particular, for any
$\theta$ chosen as in (70)),

$\sum_{n\geq 1}(p-1)^{n/2}\left(\frac{1}{n!}B_{n}(t,h)\right)^{1/2}\leq
Ch^{\theta/2}.$

Here $C$ is a constant depending on $p$, $a$, $b$, $c_{0}$, $H_{0}$, and $T$.

Step 2 (space increments). We denote $z=x^{\prime}-x$. As in the proof of
Theorem 2.7,

$\|u^{H}(t,x+z)-u^{H}(t,x)\|_{p}\leq\sum_{n\geq
1}(p-1)^{n/2}\|I_{n}^{H}(f_{t,x+z,n}-f_{t,x,n})\|_{p}\leq\sum_{n\geq
1}(p-1)^{n/2}\left(\frac{1}{n!}C_{n}^{H}(t,z)\right)^{1/2},$

where
$C_{n}^{H}(t,z)=(n!)^{2}\|\widetilde{f}_{t,x+z,n}-\widetilde{f}_{t,x,n}\|_{\mathcal{H}_{H}^{\otimes
n}}^{2}$. By (46), we have:

$\displaystyle C^{H}_{n}(t,z)$ $\displaystyle\leq
b^{n}_{H_{0}}\bigg{(}\int_{[0,t]^{n}}\Big{(}\Psi^{H}_{t,z,n}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}d\boldsymbol{t_{n}}\bigg{)}^{2H_{0}},$
(78)

where we fixed $\boldsymbol{t_{n}}\in[0,t]^{n}$ and $\rho$ is the permutation
of $1,\ldots,n$ such that $t_{\rho(1)}<\ldots<t_{\rho(n)}$ with
$t_{\rho(n+1)}=t$ and

$\Psi^{H}_{t,z,n}(\boldsymbol{t_{n}}):=\int_{\mathbb{R}^{n}}\Big{|}\mathcal{F}\Big{(}f_{t,x+z,n}(t_{\rho(1)},\cdot,\ldots,t_{\rho(1)},\cdot)-f_{t,x,n}(t_{\rho(1)},\cdot,\ldots,t_{\rho(1)},\cdot)\Big{)}(\boldsymbol{\xi_{n}})\Big{|}^{2}\prod_{j=1}^{n}|\xi_{j}|^{1-2H}d\boldsymbol{\xi_{n}}.$

Assume for simplicity that $t_{1}<\ldots<t_{n}$. To study
$\Psi^{H}_{t,z,n}(\boldsymbol{t_{n}})$, we can use the same estimates (39) and
(40), for the increments of the Fourier transform appearing above. Using the
definition of $I^{(n)}_{t}(\boldsymbol{t_{n}})$ in (59), there exist a
constant $C$ denepdning on $t_{0}$, $\theta$ and $T$ such that

$\displaystyle\Psi^{H}_{t,z,n}(\boldsymbol{t_{n}})\leq|z|^{2\theta}C\big{(}I_{1}+I_{2}\big{)}$
(79)

where

$\displaystyle I_{1}$
$\displaystyle:=C_{\theta}T^{\theta}(t-t_{n})^{-\theta}c_{H}^{n}\int_{\mathbb{R}^{n}}\prod_{k=1}^{n}\exp\bigg{\\{}-\frac{1}{2}\frac{t_{k+1}-t_{k}}{t_{k+1}t_{k}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\bigg{\\}}\big{|}\xi_{k}\big{|}^{1-2H}d\boldsymbol{\xi_{n}}$
$\displaystyle\leq
C_{\theta}T^{\theta}(t-t_{n})^{-\theta}I^{(n)}_{t/2}(\boldsymbol{t_{n}}/2)$

and

$\displaystyle I_{2}$
$\displaystyle:=c_{H}^{n}\int_{\mathbb{R}^{n}}\prod_{k=1}^{n}\exp\bigg{\\{}-\frac{t_{k+1}-t_{k}}{t_{k+1}t_{k}}\Big{|}\sum_{j=1}^{k}t_{j}\xi_{j}\Big{|}^{2}\bigg{\\}}\big{|}\xi_{k}\big{|}^{1-2H}d\boldsymbol{\xi_{n}}=I^{(n)}_{t}(\boldsymbol{t_{n}}).$

Taking power $1/(2H_{0})$ on (79), we obtain:

$\displaystyle\Big{(}\Psi^{H}_{t,z,n}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}\leq|z|^{\frac{\theta}{H_{0}}}C^{\frac{1}{2H_{0}}}\bigg{(}C^{\frac{1}{2H_{0}}}_{\theta}T^{\frac{\theta}{2H_{0}}}\Big{(}(t-t_{n})^{-\theta}I^{(n)}_{t/2}(\boldsymbol{t_{n}}/2)\Big{)}^{\frac{1}{2H_{0}}}+\Big{(}I^{(n)}_{t}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}\bigg{)}.$

We estimate separately the two terms in the inequality above. We start with
the first summand. Using (62), we have

$\displaystyle
C^{\frac{1}{2H_{0}}}_{\theta}T^{\frac{\theta}{2H_{0}}}\Big{(}(t-t_{n})^{-\theta}I^{(n)}_{t/2}(\boldsymbol{t_{n}}/2)\Big{)}^{\frac{1}{2H_{0}}}$
$\displaystyle\qquad\leq\Big{(}C_{\theta}T^{\theta}(C^{(2)})^{n}2^{n}\Big{)}^{\frac{1}{2H_{0}}}\sum_{\boldsymbol{\alpha}\in
D^{(H)}_{n}}t^{\frac{1+\alpha_{n}}{4H_{0}}}\prod_{k=1}^{n}t^{\widetilde{\alpha}_{k}}_{k}\prod_{k=1}^{n-1}\big{(}t_{k+1}-t_{k}\big{)}^{\widetilde{\beta}_{k}}\big{(}t-t_{n}\big{)}^{\widetilde{\beta}_{n,\theta}}$

where where $\widetilde{\alpha}_{1},\ldots,\widetilde{\alpha}_{n}$ are given
by (64); $\widetilde{\beta}_{1},\ldots,\widetilde{\beta}_{n-1}$ are given by
(65) and $\widetilde{\beta}_{n,\theta}$ is defined in (66). For the second
summand, we use (60) and take power $\frac{1}{2H_{0}}$, we obtain:

$\displaystyle\Big{(}I^{(n)}_{t}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}\leq
C_{H,1}^{\frac{n}{2H_{0}}}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}t^{\frac{1+\alpha_{n}}{4H_{0}}}\prod_{k=1}^{n}t_{k}^{\widetilde{\alpha}_{k}}(t_{k+1}-t_{k})^{\widetilde{\beta}_{k}}$

where $\widetilde{\alpha}_{k}$ and $\widetilde{\beta}_{k}$ are defined the
same as in (75).

Putting these estimates for the two terms together, we obtain that for any
$0<t_{1}<\ldots<t_{n}<t_{n+1}=t$, there exists a constant $C$ depending on
$a$, $b$, $t_{0}$, $\theta$ and $T$ such that

$\displaystyle\Big{(}\Psi^{H}_{t,z,n}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}\leq|z|^{\frac{\theta}{H_{0}}}C\bigg{(}2^{\frac{n}{2H_{0}}}\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}t^{\frac{1+\alpha_{n}}{4H_{0}}}\prod_{k=1}^{n}t^{\widetilde{\alpha}_{k}}_{k}\prod_{k=1}^{n-1}\big{(}t_{k+1}-t_{k}\big{)}^{\widetilde{\beta}_{k}}\big{(}t-t_{n}\big{)}^{\widetilde{\beta}_{n,\theta}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\sum_{\boldsymbol{\alpha_{n}}\in
D_{n}^{(H)}}t^{\frac{1+\alpha_{n}}{4H_{0}}}\prod_{k=1}^{n}t_{k}^{\widetilde{\alpha}_{k}}(t_{k+1}-t_{k})^{\widetilde{\beta}_{k}}\bigg{)}.$
(80)

A similar estimate holds for arbitrary $(t_{1},\ldots,t_{n})\in[0,t]^{n}$ with
$t_{\rho(1)}<\ldots<t_{\rho(n)}$ for same permutation $\rho\in S_{n}$.

We now integrate
$\big{(}\Psi^{H}_{t,z,n}(\boldsymbol{t_{n}})\big{)}^{\frac{1}{2H_{0}}}$ over
$[0,t]^{n}$. For this, we use Lemma 3.2. For the first integral, we use (69)
and for the second integral we use (77). Note that, the polynomial function of
$t$ in the right side of (69), respectively (77), is
$t^{|\widetilde{\alpha}|+|\widetilde{\beta}|+n}=t^{n\frac{2H_{0}+H-1}{2H_{0}}-\frac{1+\alpha_{n}}{4H_{0}}}$
and the factor $t^{-\frac{1+\alpha_{n}}{4H_{0}}}$ cancels when we multiplying
the integral by $t^{\frac{1+\alpha_{n}}{4H_{0}}}$, as required when
integrating the right hand side of (80) with respect to $\boldsymbol{t_{n}}$.
Using the same estimates above, we see that

$\int_{[0,t]^{n}}\Big{(}\Psi^{H}_{t,z,n}(\boldsymbol{t_{n}})\Big{)}^{\frac{1}{2H_{0}}}d\boldsymbol{t_{n}}\leq|z|^{\frac{\theta}{H_{0}}}\frac{\big{(}C^{(9)}\big{)}^{n-1}}{[(n-1)!]^{\frac{a-1}{2H_{0}}}}.$

We conclude that

$\sum_{n\geq 1}(p-1)^{n/2}\left(\frac{1}{n!}C_{n}^{H}(t,z)\right)^{1/2}\leq
C|z|^{\theta},$

where $C$ is a constant depending on $t_{0}$, $T$, $p$, $a$, $b$, $c_{0}$ and
$\theta$.

∎

Proof of Theorem 1.4: Step 1. (finite dimensional convergence) In this step,
we prove that:

$\big{(}u^{H_{n}}(t_{1},x_{1}),\ldots,u^{H_{n}}(t_{k},x_{k})\big{)}\stackrel{{\scriptstyle
d}}{{\to}}\big{(}u^{H^{*}}(t_{1},x_{1}),\ldots,u^{H^{*}}(t_{k},x_{k})\big{)},$

for any $(t_{1},x_{1}),\ldots,(t_{k},x_{k})\in[0,T]\times\mathbb{R}$. It is
enough to prove that for any $(t,x)\in[0,T]\times\mathbb{R}$,
$u^{H_{n}}(t,x)\to u^{H^{*}}(t,x)$ in $L^{2}(\Omega)$ as $n\to\infty$.

As in the proof of Theorem 1.3, we approximate $u^{H}(t,x)$ by the partial
sum:

$u_{m}^{H}(t,x)=1+\sum_{k=1}^{m}I_{n}^{H}(f_{t,x,k}).$

Then $u_{m}^{H^{\ast}}(t,x)\to u^{H^{\ast}}(t,x)$ in $L^{2}(\Omega)$ as
$m\to\infty$. Moreover, by Lemma 3.3, for any $m\geq 1$ fixed,
$u_{m}^{H_{n}}(t,x)\to u_{m}^{H*}(t,x)$ in $L^{2}(\Omega)$ as $n\to\infty$.
So, it remains to prove that

$\sup_{n\geq 1}\mathbb{E}|u_{m}^{H_{n}}(t,x)-u^{H_{n}}(t,x)|^{2}\to 0$

as $m\to\infty$. We choose values $a$ and $b$ such that (51) holds. Since
$H_{n}\to H^{*}$, there exists $N\in\mathbb{N}$ such that $a<H_{n}<b$ for all
$n\geq N$. Therefore, it suffices to prove that:

$\sup_{H\in[a,b]}\mathbb{E}|u_{m}^{H}(t,x)-u^{H}(t,x)|^{2}=\sup_{H\in[a,b]}\sum_{k\geq
m+1}\mathbb{E}|I_{k}^{H}(f_{t,x,k})|^{2}\to 0\quad\mbox{as $m\to\infty$}.$
(81)

To prove (81), we need to estimate $\mathbb{E}|I_{k}^{H}(f_{t,x,k})|^{2}$.
Those estimates have been done in [4], but we need to revisit them here since
we are interested in obtaining a uniform bound in $H$. First, note that by
Littlewood-Hardy inequality (46),

$\mathbb{E}|I_{k}^{H}(f_{t,x,k})|^{2}=k!\,\|\widetilde{f}_{t,x,k}\|_{\mathcal{H}_{H}^{\otimes
k}}^{2}\leq
k!\,b_{0}^{k}\left(\int_{[0,t]^{k}}A_{k}^{H}(\boldsymbol{t_{k}})^{\frac{1}{2H_{0}}}d\boldsymbol{t_{k}}\right)^{2H_{0}},$
(82)

where
$A_{k}^{H}(\boldsymbol{t_{k}})=c_{H}^{k}\int_{\mathbb{R}^{k}}|\mathcal{F}\widetilde{f}_{t,x,k}(\boldsymbol{t_{k}},\bullet)(\boldsymbol{\xi_{k}})|^{2}\prod_{j=1}^{k}|\xi_{j}|^{1-2H}d\boldsymbol{\xi_{k}}$.

Next, we would like to find an upper bound for
$A_{k}^{H}(\boldsymbol{t_{k}})$. For this, using the same arguments as in the
proof of Theorem 2.2 of [6] (the study of $A_{k}^{H}(\boldsymbol{t_{k}})$),
with $\mathcal{F}G_{t_{\rho(j+1)}-t_{\rho(j)}}(\eta_{j})$ replaced by
$\mathcal{F}G_{(t_{\rho(j+1)}-t_{\rho(j)})/t_{\rho(j+1)}t_{\rho(j)}}(\eta_{j})$,
we obtain:

$\displaystyle
A_{k}^{H}(\boldsymbol{t_{k}})\leq\frac{c_{H}^{k}}{(k!)^{2}}\Big{(}\prod_{j=1}^{k}t_{\rho(j)}\Big{)}^{2H-2}\sum_{\boldsymbol{\alpha}\in
D_{k}^{(H)}}\prod_{j=1}^{k}\bigg{\\{}\int_{\mathbb{R}}\exp\Big{(}-\frac{t_{\rho(j+1)}-t_{\rho(j)}}{t_{\rho(j+1)}t_{\rho(j)}}|\eta_{j}|^{2}\Big{)}|\eta_{j}|^{\alpha_{j}}d\eta_{j}\bigg{\\}}.$

where $D_{k}^{(H)}$ be the set of multi-indices
$\boldsymbol{\alpha}=(\alpha_{1},\ldots,\alpha_{k})$ with
$\alpha_{j}=(1-2H)a_{j}$ and $\boldsymbol{a}=(a_{1},\ldots,a_{k})\in A_{k}$
Using Lemma 3.1, we have:

$\displaystyle A_{k}^{H}(\boldsymbol{t_{k}})$
$\displaystyle\leq\frac{c_{H}^{k}}{(k!)^{2}}\Big{(}\prod_{j=1}^{k}t_{\rho(j)}\Big{)}^{2H-2}\sum_{\boldsymbol{\alpha}\in
D_{k}^{(H)}}\prod_{j=1}^{k}\Gamma\Big{(}\frac{1+\alpha_{j}}{2}\Big{)}\Big{(}\frac{t_{\rho(j+1)}-t_{\rho(j)}}{t_{\rho(j+1)}t_{\rho(j)}}\Big{)}^{-\frac{1+\alpha_{k}}{2}}$
$\displaystyle\leq\frac{C_{H,1}^{k}}{(k!)^{2}}\sum_{\boldsymbol{\alpha}\in
D_{k}^{(H)}}t^{\frac{1+\alpha_{n}}{2}}t_{\rho(1)}^{\frac{4H-3+\alpha_{1}}{2}}\Big{(}\prod_{j=2}^{k}t_{\rho(j)}^{\frac{4H-2+\alpha_{k-1}+\alpha_{k}}{2}}\Big{)}\prod_{j=1}^{k}(t_{\rho(j+1)}-t_{\rho(j)})^{-\frac{1+\alpha_{k}}{2}}$

where for the second inequality we used the constant $C_{H,1}$ defined by
(61).

Coming back to (82), we obtain that:

$\displaystyle\mathbb{E}|I_{k}^{H}(f_{t,x,k})|^{2}\leq
b^{2}_{H_{0}}w_{+}^{2}(t,x)C_{H,1}^{k}(k!)^{2H_{0}-1}\bigg{(}\sum_{\boldsymbol{\alpha}\in
D_{k}^{(H)}}t^{\frac{1+\alpha_{n}}{4H_{0}}}\int_{T_{k}(t)}\prod_{j=1}^{k}t_{j}^{\widetilde{\alpha}_{j}}(t_{j+1}-t_{j})^{\widetilde{\beta}_{j}}d\boldsymbol{t_{k}}\bigg{)}^{2H_{0}}$

where $\widetilde{\alpha}_{j}$ and $\widetilde{\beta}_{j}$ for all
$j=1,\ldots,k$ are the same as in (75). The last integral is calculated by
Lemma 3.2, the integral is

$\displaystyle\int_{T_{k}(t)}\prod_{j=1}^{k}t_{\rho(j)}^{\widetilde{\alpha}_{j}}(t_{\rho(j+1)}-t_{\rho(j)})^{\widetilde{\beta}_{j}}d\boldsymbol{t_{k}}$
$\displaystyle=\frac{\Gamma(\widetilde{\alpha}_{1}+1)\prod_{j=1}^{k}\Gamma(\widetilde{\beta}_{j}+1)}{\Gamma\big{(}\sum_{j=1}^{k}\widetilde{\alpha}_{j}+\sum_{j=1}^{k}\widetilde{\beta}_{j}+k+1\big{)}}\gamma_{k-1}t^{\sum_{j=1}^{k}\widetilde{\alpha}_{j}+\sum_{j=1}^{k}\widetilde{\beta}_{j}+k}$

Similarly as in (77), those factors appears in above can be bounded uniformly
in $H\in[a,b]$. There exists a constant $C>0$ such that

$\displaystyle\sup_{H\in[a,b]}\sum_{k\geq
m+1}\mathbb{E}\big{|}I^{H}_{k}(f_{k}(\cdot,t,x))\big{|}^{2}\leq\sum_{k\geq
m+1}C^{k}\frac{(t\vee 1)^{k(2H_{0}+b-1)}}{(k!)^{a}}.$

The last term clearly converges to $0$ as $m\to\infty$. This concludes the
proof of (81).

Step 2. (tightness) The fact that the sequence $(u^{H_{n}})_{n\geq 1}$ is
tight in $C([t_{0},T]\times\mathbb{R})$ follows by Proposition 2.3 of [14]
using Theorem 3.4 above.

∎

Acknowledgement. The author is grateful to Raluca Balan for insightful
discussions and for help in organizing and enhancing overall quality of this
work.

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