$K^{n}_{t}(x)-P_{t-s}K^{n}_{s}(x)\leqslant
C[K_{t}(x)-P_{t-s}K_{s}(x)](t-s)^{\frac{3}{4}}+L^{n}_{s,t}(x)$ (5.39)

for every $(s,t)\in\Delta_{0,T_{0}}$ $x\in{D}$.

Fix $x\in{D}$, $T\in[0,T_{0}]$. We define

$\lambda^{T}_{s,t}(x)=P_{T-t}K_{t}(x)-P_{T-s}K_{s}(x),\quad(s,t)\in\Delta_{0,T}.$

From (5.37), we see that $0\leqslant\lambda^{T}_{s,t}(x)\leqslant
K_{T}(x)<\infty$. It is evident that $\lambda$ is additive, that is
$\lambda^{T}_{s,t}(x)=\lambda^{T}_{s,u}(x)+\lambda^{T}_{u,t}(x)$ for every
$s\leqslant u\leqslant t$. Hence, $(s,t)\mapsto\lambda^{T}_{s,t}(x)$ is a
random control per 4.6.

Define $A^{T,n}_{s,t}(x)$ as in the previous step and

$\mathcal{A}^{T,n}_{t}(x):=\int_{0}^{t}\int_{D}p_{T-r}(x,y)b^{n}(V_{r}(y)+\psi_{r}(y))\,dydr=P_{T-t}K^{n}_{t}(x).$

Then for $u:=(s+t)/2$ we have

$\delta
A^{T,n}_{s,u,t}(x)=\int_{u}^{t}\int_{D}p_{T-r}(x,y)[b^{n}(V_{r}(y)+P_{r-s}\psi_{s}(y))-b^{n}(V_{r}(y)+P_{r-u}\psi_{u}(y))]\,dydr.$

Applying consequently the Fubini theorem, (C.7), Lemma A.3(iv) and Lemma C.3,
we deduce that

$\displaystyle|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{u}\delta
A^{T,n}_{s,u,t}(x)|$
$\displaystyle=\Bigl{|}\int_{u}^{t}\int_{D}p_{T-r}(x,y)\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{u}[b^{n}(V_{r}(y)+P_{r-s}\psi_{s}(y))-b^{n}(V_{r}(y)+P_{r-u}\psi_{u}(y))]\,dydr\Bigr{|}$
$\displaystyle\leqslant\int_{u}^{t}\int_{D}p_{T-r}(x,y)\|G_{\rho_{r-u}(x)}b^{n}\|_{\mathcal{C}^{1}}|P_{r-s}\psi_{s}(y)-P_{r-u}\psi_{u}(y)|\,dydr$
$\displaystyle\leqslant
C\|b^{n}\|_{\mathcal{B}^{0}_{1}}\int_{u}^{t}\int_{D}p_{T-r}(x,y)|P_{r-u}\psi_{u}(y)-P_{r-s}\psi_{s}(y)|(r-u)^{-1/2}\,dydr,$

where we used the notation $\rho_{t}(x):=\mathop{\mbox{\rm Var}}(V_{t}(x))$.
Since $\psi=Pu_{0}+K$, we see that
$P_{r-u}\psi_{u}-P_{r-s}\psi_{s}=P_{r-u}K_{u}-P_{r-s}K_{s}$. Using the
elementary inequality

$|P_{r-u}K_{u}(y)-P_{r-s}K_{s}(y)|\leqslant
P_{r-u}|K_{u}(\cdot)-P_{u-s}K_{s}(\cdot)|(y)$

and (5.37), we get

$\displaystyle|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{u}\delta
A^{T,n}_{s,u,t}(x)|$ $\displaystyle\leqslant
C\|b^{n}\|_{\mathcal{B}^{0}_{1}}\int_{u}^{t}\int_{D}p_{T-u}(x,y)[K_{u}(y)-P_{u-s}K_{s}(y)](r-u)^{-1/2}\,dydr$
$\displaystyle\leqslant
C\|b^{n}\|_{\mathcal{B}^{0}_{1}}\lambda_{s,u}^{T}(x)(t-u)^{\frac{1}{2}}.$

As we have shown previously, $(s,t)\mapsto\lambda^{T}_{s,t}(x)$ is a random
control for every fixed $T,x$. Hence, the above estimate verifies condition
(4.13). The estimate (5.38) verifies condition (4.3) (with
$\alpha_{2}=\beta_{2}=0$ and $n=m$). It remains to verify condition (4.4). Let
$\Pi:=\\{0=t_{0},t_{1},...,t_{k}=T\\}$ be an arbitrary partition of $[0,T]$.
Denote by $|\Pi|$ its mesh size. Then we have

$\displaystyle\Bigl{|}\mathcal{A}_{t}-\sum_{i=0}^{k-1}A_{t_{i},t_{i+1}}\Bigr{|}$
$\displaystyle\quad\leqslant\sum_{i=0}^{k-1}\int_{t_{i}}^{t_{i+1}}\\!\\!\\!\int_{D}p_{T-r}(x,y)\bigl{|}b^{n}(V_{r}(y)+\psi_{r}(y))-b^{n}(V_{r}(y)+P_{r-{t_{i}}}\psi_{t_{i}}(y))\bigr{|}\,dydr.$

For each $i$, using the fact that $b^{n}$ is Lipschitz, we have

$\int_{t_{i}}^{t_{i+1}}\\!\\!\\!\int_{D}p_{T-r}(x,y)\bigl{|}b^{n}(V_{r}(y)+\psi_{r}(y))-b^{n}(V_{r}(y)+P_{r-{t_{i}}}\psi_{t_{i}}(y))\bigr{|}\,dydr\\\
\leqslant\|b^{n}\|_{\mathcal{C}^{1}}\int_{t_{i}}^{t_{i+1}}\\!\\!\\!\int_{D}p_{T-r}(x,y)|\psi_{r}(y)-P_{r-{t_{i}}}\psi_{t_{i}}(y)|\,dydr.$

Since $\psi=Pu_{0}+K$, we have
$|\psi_{r}(y)-P_{r-{t_{i}}}\psi_{t_{i}}(y)|=|K_{r}(y)-P_{r-{t_{i}}}K_{t_{i}}(y)|=K_{r}(y)-P_{r-{t_{i}}}K_{t_{i}}(y)$
where we have used (5.37). Hence, combining with the above estimate and
applying (5.37) once again yield

$\displaystyle\int_{t_{i}}^{t_{i+1}}\\!\\!\\!\int_{D}p_{T-r}(x,y)$
$\displaystyle\bigl{|}b^{n}(V_{r}(y)+\psi_{r}(y))-b^{n}(V_{r}(y)+P_{r-{t_{i}}}\psi_{t_{i}}(y))\bigr{|}\,dydr$
$\displaystyle\leqslant\|b^{n}\|_{\mathcal{C}^{1}}\int_{t_{i}}^{t_{i+1}}[P_{T-r}K_{r}(x)-P_{T-{t_{i}}}K_{t_{i}}(x)]dr$
$\displaystyle\leqslant\|b^{n}\|_{\mathcal{C}^{1}}[P_{T-t_{i+1}}K_{t_{i+1}}(x)-P_{T-{t_{i}}}K_{t_{i}}(x)](t_{i+1}-t_{i}).$

It follows that

$\displaystyle\Bigl{|}\mathcal{A}_{t}-\sum_{i=0}^{k-1}A_{t_{i},t_{i+1}}\Bigr{|}$
$\displaystyle\leqslant\|b^{n}\|_{\mathcal{C}^{1}}\sum_{i}[P_{T-t_{i+1}}K_{t_{i+1}}(x)-P_{T-{t_{i}}}K_{t_{i}}(x)](t_{i+1}-t_{i})$
$\displaystyle\leqslant\|b^{n}\|_{\mathcal{C}^{1}}(K_{T}(x)-P_{T}K_{0}(x))|\Pi|$

which converges to $0$ a.s. as $|\Pi|\to 0$. Thus, condition (4.4) holds.

Applying Theorem 4.7, we have

$\displaystyle|P_{T-t}K^{n}_{t}(x)-P_{T-s}K^{n}_{s}(x)|$
$\displaystyle\leqslant
C[P_{T-t}K_{t}(x)-P_{T-s}K_{s}(x)](t-s)^{\frac{1}{2}}+B^{T,n}_{s,t}(x)+|A^{T,n}_{s,t}(x)|$

where $\|B^{T,n}_{s,t}(x)\|_{{L_{m}}}\leqslant C|t-s|^{3/4}$ uniformly in
$T,n,x$. To obtain (5.39), it suffices to put $T=t$,
$L^{n}_{s,t}(x)=B^{t,n}_{s,t}(x)+|A^{t,n}_{s,t}(x)|$. Since $B^{T,n}$ is a
functional of $\delta A^{T,n}(x)$, measurability of $(x,\omega)\mapsto
L^{n}_{s,t}$ follows. The uniform moment estimate for $L^{n}_{s,t}(x)$ follows
from those of $B^{T,n}_{s,t}(x)$ and $A^{T,n}_{s,t}(x)$ in (5.38).

Step 4. We show that $\lim_{n\to\infty}P_{r}K^{n}_{t}(x)=P_{r}K_{t}(x)$ in
probability for every $t\in[0,T_{0}]$, $r\in[0,T_{0}-t]$ and $x\in{D}$.

The situation here is similar to Lemma B.5 except for a uniform moment bound
for $K^{n}_{t}(x)$. We replace this condition by estimate (5.39) and the
uniform moment bound for $L_{0,t}^{n}(x)$. Indeed, let $x\in{D}$ be fixed and
let $M>0$ be a positive number. We write

$P_{r}K^{n}_{t}(x)=\int_{|y|\leqslant
M}p_{r}(x,y)K^{n}_{t}(y)dy+\int_{|y|>M}p_{r}(x,y)K^{n}_{t}(y)dy.$

Since $K^{n}_{t}$ converges to $K_{t}$ uniformly on $\\{y\in{D}:|y|\leqslant
M\\}$, we see that $\int_{|y|\leqslant M}p_{r}(x,y)K^{n}_{t}(y)dy$ converges
to $\int_{|y|\leqslant M}p_{r}(x,y)K_{t}(y)dy$ a.s. To treat the second term,
we first set $s=0$ in (5.39) to obtain that

$|K^{n}_{t}(y)|\leqslant
Ct^{\frac{1}{2}}K_{t}(y)+L^{n}_{0,t}(y)\quad\textrm{a.s.}\quad\forall
y\in{D}.$

It follows that

$\displaystyle\int_{|y|>M}p_{r}(x,y)K^{n}_{t}(y)dy\leqslant
Ct^{\frac{1}{2}}\int_{|y|>M}p_{r}(x,y)K_{t}(y)dy+\int_{|y|>M}p_{r}(x,y)L^{n}_{0,t}(y)dy.$

By the Lebesgue monotone convergence theorem and (5.37), we see that a.s.

$\lim_{M\to\infty}\int_{|y|\leqslant
M}p_{r}(x,y)K_{t}(y)dy=\int_{D}p_{r}(x,y)K_{t}(y)dy\leqslant
K_{t+r}(x)<\infty$

so that a.s.

$\lim_{M\to\infty}\int_{|y|>M}p_{r}(x,y)K_{t}(y)dy=0.$

Lastly,

$\displaystyle\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt\int_{|y|>M}p_{r}(x,y)L^{n}_{0,t}(y)dy\leqslant\sup_{n\in\mathbb{Z}_{+},z\in{D}}\|L^{n}_{0,t}(z)\|_{L_{m}}\left(\int_{|y|>M}p_{r}(x,y)dy\right)$

which converges to $0$ as $M\to\infty$. These facts imply the claim.

Step 5. We define $L_{s,t}(x)=\liminf_{n}L^{n}_{s,t}(x)$ then by Fatou’s lemma
and Step 3, $\|L_{s,t}(x)\|_{{L_{m}}}\leqslant C|t-s|^{3/4}$ uniformly in $x$.
In (5.39), we send $n\to\infty$, applying the convergence in step 4 to obtain
that

$\displaystyle K_{t}(x)-P_{t-s}K_{s}(x)\leqslant
C(K_{t}(x)-P_{t-s}K_{s}(x))(t-s)^{\frac{1}{2}}+L_{s,t}(x).$

This implies that $K_{t}(x)-P_{t-s}K_{s}(x)\leqslant 2L_{s,t}(x)$ for
$t-s\leqslant\ell$ and $\ell$ is such that $C\ell^{1/2}\leqslant\frac{1}{2}$.
An application of $L_{m}$-norm yields

$\displaystyle\|K_{t}(x)-P_{t-s}K_{s}(x)\|_{L_{m}}\leqslant
2\|L_{s,t}(x)\|_{L_{m}}\leqslant 2C|t-s|^{3/4}$

for every $t-s\leqslant\ell$. Using the identity
$K_{t}-P_{t-s}K_{s}=\psi_{t}-P_{t-s}\psi_{s}$ once again and recalling that
$\psi=u-V$, the above estimate shows that $u-V$ belongs to
$\mathcal{C}^{3/4,0}L_{m}([S,T])$. This completes the proof of Proposition
3.8.

##  6 Proofs of regularity lemmas

In this section, we present the proofs of Lemmas 5.1, 5.2, 5.5, 5.6 and 5.7.
Throughout the section, we fix the filtration $(\mathcal{F}_{t})_{t\geqslant
0}$, which appears in the aforementioned lemmas.

###  6.1 Proof of Lemmas 5.1 and 5.2

We begin with the following auxiliary result, which will be crucial also for
the proof of Proposition 5.7. The proof relies on the stochastic sewing lemma
(Theorem 4.1) and the estimates in Lemmas C.3 and C.4.

######  Lemma 6.1.

Let $0\leqslant S\leqslant T$. Let $\gamma\in(-2,0)$, $m\in[2,\infty)$,
$n\in[m,\infty]$ and $p\in[n,\infty]$. Let
$h\colon\mathbb{R}\times[S,T]\times{D}\times\Omega\to\mathbb{R}$ be a globally
bounded measurable function. Let $X:[S,T]\times{D}\to\mathbb{R}$ be a
measurable function. Suppose further that the following conditions hold:

  1. 1)

for any fixed $(z,r,x)\in\mathbb{R}\times[S,T]\times{D}$ the random variable
$h(z,r,x)$ is $\mathcal{F}_{S}$-measurable;

  2. 2)

there exists a constant $\Gamma_{h}>0$ such that

$\sup_{\begin{subarray}{c}r\in[S,T]\\\
x\in{D}\end{subarray}}\|\,\|h(\cdot,r,x)\|_{\mathcal{B}^{\gamma}_{p}}\|_{L_{n}}\leqslant\Gamma_{h};$
(6.1)

  3. 3)

there exist constants $\Gamma_{X}>0$, $\nu\in[0,\frac{1}{2})$ such that for
every $r\in[S,T]$, $x\in{D}$

$\int_{D}|X_{r}(y)|dy\leqslant\Gamma_{X}(T-r)^{-\nu}.$ (6.2)

Then there exists a constant $C=C(\gamma,m,n,p)$ which does not depend on $S$,
$T$, $\Gamma_{h}$, $\Gamma_{X}$, $h$, $K$ such that for any $t\in[S,T]$

$\displaystyle\Bigl{\|}\Bigl{\|}\int_{S}^{t}\int_{D}X_{r}(y)h(V_{r}(y),r,y)\,dydr\Bigr{\|}_{{L_{m}}|\mathcal{F}_{S}}\Bigr{\|}_{{L_{n}}}\leqslant
C\Gamma_{h}\Gamma_{X}(T-S)^{-\nu}(t-S)^{1+\frac{\gamma}{4}-\frac{1}{4p}}.$
(6.3)

###### Proof.

The proof is based on the stochastic sewing lemma, Theorem 4.1. We put for
$S\leqslant s<t\leqslant T$,

$A_{s,t}:=\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\int_{s}^{t}\int_{D}X_{r}(y)h(V_{r}(y),r,y)\,dydr.$

Put also for $(t,x)\in[0,T]\times{D}$

$\mathcal{A}_{t}=\int_{S}^{t}\int_{D}X_{r}(y)h(V_{r}(y),r,y,\omega)\,dydr.$

Note that the integral in the left-hand side of inequality (6.3) is just
$\mathcal{A}_{t}-\mathcal{A}_{S}$.

Let us verify that all the conditions of Theorem 4.1 are satisfied. For any
$S\leqslant s<t\leqslant T$, $u:=(s+t)/2$ we have

$\displaystyle\delta A_{s,u,t}=\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\int_{u}^{t}\int_{D}X_{r}(y)h(V_{r}(y),r,y)\,dydr-\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{u}\int_{u}^{t}\int_{D}X_{r}(y)h(V_{r}(y),r,y)\,dydr.$

Therefore

$\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}\delta A_{s,u,t}=0$

and thus condition (4.2) holds.

Further, it is easy to see by the conditional Jensen’s inequality and
Minkowski’s integral inequality that for any $x\in{D}$, $S\leqslant
s<t\leqslant T$ and $u:=(s+t)/2$

$\displaystyle\|\delta A_{s,u,t}\|_{L_{n}}$ $\displaystyle\leqslant
2\Bigl{\|}\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{u}\int_{u}^{t}\int_{D}X_{r}(y)h(V_{r}(y),r,y)\,dydr\Bigr{\|}_{L_{n}}$
$\displaystyle\leqslant 2\int_{u}^{t}\int_{D}|X_{r}(y)|\|\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt^{u}h(V_{r}(y),r,y)\|_{L_{n}}\,dydr.$ (6.4)

Now for fixed $r\in[u,t]$, $y\in{D}$ we apply bound (C.8) to the function
$f\colon(z,\omega)\mapsto h(z,r,y,\omega)$ which is
$\mathscr{B}(\mathbb{R})\otimes\mathcal{F}_{S}$-measurable. We get

$\displaystyle\|\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{u}h(V_{r}(y),r,y)\|_{L_{n}}$ $\displaystyle\leqslant
C\|\,\|h(\cdot,r,y)\|_{\mathcal{B}^{\gamma}_{p}}\|_{L_{n}}(r-u)^{\frac{\gamma}{4}}(u-S)^{-\frac{1}{2p}}(r-S)^{\frac{1}{4p}}$
$\displaystyle\leqslant
C\|\,\|h(\cdot,r,y)\|_{\mathcal{B}^{\gamma}_{p}}\|_{L_{n}}(r-u)^{\frac{\gamma}{4}}(u-S)^{-\frac{1}{4p}},$
(6.5)

where the last inequality follows because $r-S\leqslant 2(u-S)$. Substituting
(6.5) into (6.1) and using (6.1), (6.2), we obtain that

$\displaystyle\|\delta A_{s,u,t}\|_{L_{n}}$ $\displaystyle\leqslant
C\Gamma_{h}\Gamma_{X}(u-S)^{-\frac{1}{4p}}(T-u)^{-\nu}(t-u)^{1+\frac{\gamma}{4}},$

where the last inequality follows from (B.7). From the assumptions
$\nu\in(0,1/2)$, $\gamma>-2$, $p\geqslant 2$, we see that condition (4.7)
holds with $\alpha_{2}=1/(4p)<1/2$, $\beta_{2}=\nu<1/2$,
$\varepsilon_{2}=1/2+\gamma/4>0$. Hence, by 4.3 inequality (4.3) is satisfied.

Finally, let us check condition (4.4). Let
$\Pi:=\\{S=t_{0},t_{1},...,t_{k}=t\\}$ be an arbitrary partition of $[S,t]$.
Denote by $|\Pi|$ its mesh size. Note that for any $i\in[0,k-1]$ we have
$\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt_{t_{i}}(\mathcal{A}_{t_{i+1}}-\mathcal{A}_{t_{i}}-A_{t_{i},t_{i+1}})=0$
and $\mathcal{A}_{t_{i+1}}-\mathcal{A}_{t_{i}}-A_{t_{i},t_{i+1}}$ is
$\mathcal{F}_{t_{i+1}}$–measurable. Therefore,
$\mathcal{A}_{t}-\sum_{i=0}^{k-1}A_{t_{i},t_{i+1}}$ is a sum of martingale
differences. Then using orthogonality,

$\displaystyle\Bigl{\|}\mathcal{A}_{t}-\sum_{i=0}^{k-1}A_{t_{i},t_{i+1}}\Bigr{\|}_{{L_{2}}}^{2}$
$\displaystyle=\sum_{i=0}^{k-1}\|\mathcal{A}_{t_{i+1}}-\mathcal{A}_{t_{i}}-A_{t_{i},t_{i+1}}\|_{{L_{2}}}^{2}$
$\displaystyle\leqslant
C\|h\|_{L_{\infty}}\Gamma_{X}\sum_{i=0}^{k-1}\Bigl{(}\int_{t_{i}}^{t_{i+1}}(T-r)^{-\nu}\,dr\Bigr{)}^{2}$
$\displaystyle\leqslant
C\|h\|_{L_{\infty}}\Gamma_{X}(T-S)^{1-\nu}\max_{i=0..(k-1)}\bigl{|}\int_{t_{i}}^{t_{i+1}}(T-r)^{-\nu}\,dr\bigr{|}$
$\displaystyle\leqslant
C\|h\|_{L_{\infty}}\Gamma_{X}(T-S)^{1-\nu}|\Pi|^{1-\nu}.$

Therefore, $\sum_{i=0}^{k-1}A_{t_{i},t_{i+1}}$ converges to $\mathcal{A}_{t}$
in probability as $|\Pi|\to\infty$ and thus condition (4.4) holds.

Thus, all the conditions of Theorem 4.1 are satisfied. Hence, by inequality
(4.5), taking into account (B.7) and (3.2) we have for any $t\in[S,T]$,
$x\in{D}$

$\|\|\mathcal{A}_{t}-\mathcal{A}_{S}\|_{{L_{m}}|\mathcal{F}_{S}}\|_{L_{n}}\leqslant\|A_{S,t}\|_{L_{n}}+C\Gamma_{h}\Gamma_{X}(T-S)^{-\nu}(t-S)^{1+\frac{\gamma}{4}-\frac{1}{4p}}.$
(6.6)

Applying (C.9) to the function $f\colon(z,\omega)\mapsto h(z,r,y,\omega)$, we
get

$\displaystyle\|A_{S,t}\|_{L_{n}}$
$\displaystyle\leqslant\int_{S}^{t}\int_{D}|X_{r}(y)|\,\|\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt^{S}h(V_{r}(y),r,y)\|_{L_{n}}\,dydr$
$\displaystyle\leqslant
C\int_{S}^{t}\int_{D}|X_{r}(y)|\,\|\,\|h(\cdot,r,y)\|_{\mathcal{B}^{\gamma}_{p}}\|_{L_{n}}(r-S)^{\frac{\gamma}{4}-\frac{1}{4p}}\,dydr$
$\displaystyle\leqslant
C\Gamma_{h}\Gamma_{X}(T-S)^{-\nu}(t-S)^{1+\frac{\gamma}{4}-\frac{1}{4p}},$

where in the last inequality we made use of (6.1), (6.2) and (B.7). Combining
this with (6.6), we finally obtain (6.3). ∎

An important consequence of Lemma 6.1 is the following statement.

######  Corollary 6.2.

Let $f\in\mathcal{B}^{\gamma}_{p}$ be a bounded function, $\gamma\in(-2,0)$,
$m\in[2,\infty)$, $p\in[m,\infty]$.

  1. $($i$)$

Let $X:[0,T]\times{D}\to\mathbb{R}$ be a measurable function satisfying (6.2)
for some $\Gamma_{X}>0$, $\nu\in[0,\frac{1}{2})$. Then there exists a constant
$C=C(\gamma,m,p,\nu)$ such that for any $0\leqslant s\leqslant t\leqslant T$
and any $\mathscr{B}(\mathbb{R})\otimes\mathcal{F}_{s}$-measurable function
$\kappa\in{\mathbf{B}L_{m}}$ one has

$\Bigl{\|}\Bigl{\|}\int_{s}^{t}\int_{D}X_{r}(y)f(V_{r}(y)+P_{r-s}\kappa(y))\,dydr\Bigr{\|}_{{L_{m}}|\mathcal{F}_{s}}\Bigr{\|}_{{L_{p}}}\\\
\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}(T-s)^{-\nu}(t-s)^{1+\frac{\gamma}{4}-\frac{1}{4p}}.$
(6.7)

  2. $($ii$)$

Let $\lambda\in(0,1]$ and assume that

$\gamma>-2+\lambda.$ (6.8)

Then there exists a constant $C=C(\gamma,\lambda,m,p)$ such that for any
$0\leqslant s\leqslant t\leqslant T_{1}$ and any
$\mathscr{B}(\mathbb{R})\otimes\mathcal{F}_{s}$-measurable functions
$\kappa_{1},\kappa_{2}\in{\mathbf{B}L_{m}}$ one has

$\displaystyle\sup_{x\in{D}}\Bigl{\|}\int_{s}^{t}\int_{D}p_{T-r}(x,y)\bigl{[}f(V_{r}(y)+P_{r-s}\kappa_{1}(y))-f(V_{r}(y)+P_{r-s}\kappa_{2}(y))\bigr{]}dr\Bigr{\|}_{{L_{m}}}$
$\displaystyle\qquad\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\|\kappa_{1}-\kappa_{2}\|^{\lambda}_{{\mathbf{B}L_{m}}}(t-s)^{1+\frac{\gamma-\lambda}{4}-\frac{1}{4p}}.$
(6.9)

  3. $($iii$)$

Let $\lambda,\lambda_{1},\lambda_{2}\in(0,1]$. Assume that (6.8) holds and
that

$\gamma>-2+\lambda_{1}+\lambda_{2}.$ (6.10)

Then there exists a constant $C=C(\gamma,\lambda,\lambda_{1},\lambda_{2},p)$
such that for any $0\leqslant s\leqslant u\leqslant t\leqslant 1$, any
$\mathscr{B}(\mathbb{R})\otimes\mathcal{F}_{s}$ measurable functions
$\kappa_{1},\kappa_{2}\in{\mathbf{B}L_{m}}$, any
$\mathscr{B}(\mathbb{R})\otimes\mathcal{F}_{u}$ measurable functions
$\kappa_{3},\kappa_{4}\in{\mathbf{B}L_{m}}$ one has

$\displaystyle\sup_{x\in{D}}\Bigl{\|}\int_{u}^{t}\int_{D}p_{T-r}(x,y)\bigl{[}f(V_{r}(y)+P_{r-u}\kappa_{1}(y))-f(V_{r}(y)+P_{r-u}\kappa_{2}(y))$
$\displaystyle\qquad\qquad\qquad-f(V_{r}(y)+P_{r-u}\kappa_{3}(y))+f(V_{r}(y)+P_{r-u}\kappa_{4}(y))\bigr{]}dr\Bigr{\|}_{{L_{m}}}$
$\displaystyle\qquad\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\sup_{y\in{D}}\|\|\kappa_{1}(y)-\kappa_{3}(y)\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{\infty}}}^{\lambda_{2}}\,\|\kappa_{1}-\kappa_{2}\|^{\lambda_{1}}_{{\mathbf{B}L_{m}}}(t-u)^{1+\frac{\gamma-\lambda_{1}-\lambda_{2}}{4}-\frac{1}{4p}}$
$\displaystyle\qquad\phantom{\leqslant}+C\|f\|_{\mathcal{B}^{\gamma}_{p}}\|\kappa_{1}-\kappa_{2}-\kappa_{3}+\kappa_{4}\|^{\lambda}_{{\mathbf{B}L_{m}}}(t-u)^{1+\frac{\gamma-\lambda}{4}-\frac{1}{4p}}.$
(6.11)

###### Proof.

The results are obtained by choosing suitable $n$, $K$ and $h$ in Lemma 6.1.
In what follows, we fix $0\leqslant s\leqslant u\leqslant t\leqslant T$ and
$x$ in ${D}$. We note that the expressions $P_{r}\kappa_{i}(y)$, $i=1,..,4$
are well defined thanks to Lemma B.4 and the assumption
$\kappa_{i}\in{\mathbf{B}L_{m}}$.

(i) We take $n=p$, $h(z,r,y,\omega):=f(z+P_{r-s}[\kappa(\cdot,\omega)](y))$,
where $z\in\mathbb{R}$, $r\in[s,t]$, $y\in{D}$, $\omega\in\Omega$. Let us
verify that $h$ satisfies all the assumptions of Lemma 6.1 with $s$ in place
of $S$. Since $\kappa$ is $\mathcal{F}_{s}$-measurable, we see that the random
variable $h(z,r,x)$ is $\mathcal{F}_{s}$ measurable and the first assumption
of Lemma 6.1 holds. It follows from (A.4) that

$\|f(\cdot+P_{r-s}\kappa(y))\|_{\mathcal{B}^{\gamma}_{p}}\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}.$

Hence (6.1) holds with $\Gamma_{h}:=\|f\|_{\mathcal{B}^{\gamma}_{p}}$. Thus
all the conditions of Lemma 6.1 are met. Now (6.7) follows directly from
inequality (6.3).

(ii) We fix $x\in{D}$, choose $n=m$ and $X_{r}(y):=p_{T-r}(x,y)$. Let us apply
Lemma 6.1 for the function

$h\colon(z,r,y,\omega)\mapsto
f(z+P_{r-s}\kappa_{1}(y))-f(z+P_{r-s}\kappa_{2}(y)),\quad
z\in\mathbb{R},r\in[s,t],y\in{D},\omega\in\Omega.$

It is easy to see that (6.2) is satisfied with $\Gamma_{X}=1$ and $\nu=0$. Let
us verify that $h$ satisfies all the assumptions of Lemma 6.1 with $s$ in
place of $S$. The first assumption clearly holds. To check the second
assumption, we note that (A.5) implies

$\|f(\cdot+P_{r-s}\kappa_{1}(y))-f(\cdot+P_{r-s}\kappa_{2}(y))\|_{\mathcal{B}^{\gamma-\lambda}_{p}}\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}|P_{r-s}\kappa_{1}(y)-P_{r-s}\kappa_{2}(y)|^{\lambda}.$

Therefore,

$\displaystyle\|\,\|h(\cdot,r,y)\|_{\mathcal{B}^{\gamma-\lambda}_{p}}\|_{{L_{m}}}$
$\displaystyle\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}\|P_{r-s}\kappa_{1}(y)-P_{r-s}\kappa_{2}(y)\|_{{L_{m}}}^{\lambda}$
$\displaystyle\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}\sup_{z\in{D}}\|\kappa_{1}(z)-\kappa_{2}(z)\|_{{L_{m}}}^{\lambda}$
$\displaystyle=\|f\|_{\mathcal{B}^{\gamma}_{p}}\|\kappa_{1}-\kappa_{2}\|_{{\mathbf{B}L_{m}}}^{\lambda},$

where in the first inequality we used the fact that
$\||\xi|^{\lambda}\|_{{L_{m}}}\leqslant\|\xi\|_{{L_{m}}}^{\lambda}$ for any
random variable $\xi$ and $\lambda\in(0,1]$; the second inequality follows
from (B.8); the third inequality follows from the definition of the norm
$\|\cdot\|_{{\mathbf{B}L_{m}}}$, see (3.5). Thus, inequality (6.1) is
satisfied with
$\Gamma_{h}:=\|f\|_{\mathcal{B}^{\gamma}_{p}}\|\kappa_{1}-\kappa_{2}\|_{{\mathbf{B}L_{m}}}^{\lambda}$.
Thus all the conditions of Lemma 6.1 are met. Bound (ii) follows now from
(6.3) and (3.2).

(iii). We fix $x\in{D}$, choose $n=m$, $X_{r}(y):=p_{T-r}(x,y)$ and

$h(z,r,y,\omega):=f(z+P_{r-u}\kappa_{1}(y))-f(z+P_{r-u}\kappa_{2}(y))\\\
\quad-f(z+P_{r-u}\kappa_{3}(y))+f(z+[P_{r-u}(\kappa_{3}+\kappa_{2}-\kappa_{1})](y)),$

where $z\in\mathbb{R}$, $r\in[u,t]$, $y\in{D}$, $\omega\in\Omega$. Let us
verify $h$ satisfies the assumptions of Lemma 6.1 with $u$ in place of $S$.

Again, it is easy to see that the first and the third conditions of Lemma 6.1
are satisfied with $\Gamma_{X}=1$ and $\nu=0$. To check the second condition,
we note that (A.6) yields

$\|h(\cdot,r,y)\|_{\mathcal{B}^{\gamma-\lambda_{1}-\lambda_{2}}_{p}}\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}|P_{r-u}\kappa_{1}(y)-P_{r-u}\kappa_{2}(y)|^{\lambda_{1}}\,|P_{r-u}\kappa_{1}(y)-P_{r-u}\kappa_{3}(y)|^{\lambda_{2}}.$

Using the fact that the functions $\kappa_{1}$ and $\kappa_{2}$ are
$\mathcal{F}_{s}$-measurable, we derive from the above bound

$\displaystyle\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt\|h(\cdot,r,y)\|_{\mathcal{B}^{\gamma-\lambda_{1}-\lambda_{2}}_{p}}^{m}=\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\|h(\cdot,r,y)\|_{\mathcal{B}^{\gamma-\lambda_{1}-\lambda_{2}}_{p}}^{m}$
$\displaystyle\quad\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}^{m}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt\Bigl{[}|P_{r-u}\kappa_{1}(y)-P_{r-u}\kappa_{2}(y)|^{m\lambda_{1}}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}|P_{r-u}(\kappa_{1}(y))-P_{r-u}\kappa_{3}(y)|^{m\lambda_{2}}\Bigr{]}$
$\displaystyle\quad\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}^{m}\|\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}|P_{r-u}\kappa_{1}(y)-P_{r-u}\kappa_{3}(y)|^{m\lambda_{2}}\|_{{L_{\infty}}}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt\bigl{[}|P_{r-u}\kappa_{1}(y)-P_{r-u}\kappa_{2}(y)|^{m\lambda_{1}}\bigr{]}$
$\displaystyle\quad\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}^{m}\|\,\|P_{r-u}[\kappa_{1}-\kappa_{3}](y)\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{\infty}}}^{m\lambda_{2}}\,\,\|P_{r-u}[\kappa_{1}-\kappa_{2}](y)\|_{L_{m}}^{m\lambda_{1}}$
$\displaystyle\quad\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}^{m}\sup_{y\in{D}}\|\|\kappa_{1}(y)-\kappa_{3}(y)\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{\infty}}}^{m\lambda_{2}}\,\|\kappa_{1}-\kappa_{2}\|^{m\lambda_{1}}_{{\mathbf{B}L_{m}}}.$

Here in the penultimate inequality we used the fact that $\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt^{s}[|\xi|^{\lambda}]\leqslant(\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt^{s}|\xi|)^{\lambda}$ for any random
variable $\xi$ and $\lambda\in(0,1]$, $s\geqslant 0$; in the last inequality
we applied bound (B.8).

Thus, inequality (6.1) is satisfied with

$\Gamma_{h}:=\|f\|_{\mathcal{B}^{\gamma}_{p}}\sup_{y\in{D}}\|\|\kappa_{1}(y)-\kappa_{3}(y)\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{\infty}}}^{\lambda_{2}}\,\|\kappa_{1}-\kappa_{2}\|^{\lambda_{1}}_{{\mathbf{B}L_{m}}}.$

Bound (iii) follows now from Lemma 6.1 and part ii of the Corollary. ∎

###### Proof of Lemma 5.1.

In 6.2i, we choose $X_{r}(y)=P_{T-r}(x,y)$. In this case, condition (6.2) is
satisfied with $\Gamma_{X}=1$ and $\nu=0$. Hence the estimates (6.7) and (3.2)
imply (5.1). ∎

To establish Lemma 5.2, we need the following result.

######  Lemma 6.3.

Assume that all the conditions of Lemma 5.2 are satisfied. Let
$X:[0,T]\times{D}\to\mathbb{R}$ be a measurable function satisfying (6.2) for
some $\Gamma_{X}>0$, $\nu\in[0,\frac{1}{2})$. Then there exists a constant
$C=C(\gamma,p,\tau,m,n)>0$ such that for any $S\in[0,T]$

$\displaystyle\Bigl{\|}\,\Bigl{\|}\int_{S}^{T}\int_{D}X_{r}(y)f(V_{r}(y)+\psi_{r}(y))\,dydr\Bigr{\|}_{{L_{m}}|\mathcal{F}_{S}}\Bigr{\|}_{{L_{n}}}$
$\displaystyle\quad\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}(T-S)^{1+\frac{1}{4}(\gamma-\frac{1}{p})-\nu}+C\|f\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}[\psi]_{{\mathcal{C}^{\tau,0}L_{m,n}({[S,T]})}}(T-S)^{\frac{3}{4}+\frac{1}{4}(\gamma-\frac{1}{p})+\tau-\nu}.$
(6.12)

###### Proof.

The proof of the lemma is based on the Stochastic Sewing lemma (Theorem 4.1),
6.2i, and dominated convergence arguments.

We can assume without loss of generality that
$[\psi]_{{\mathcal{C}^{\tau,0}L_{m,n}({[S,T]})}}<\infty$. Indeed, otherwise
the right-hand side of (6.3) is infinite. Therefore, recalling (3.2), we have
$[\psi]_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}}<\infty$. Thus for any
$s\in[S,T]$ we have

$\psi_{s}\in{\mathbf{B}L_{m}}.$ (6.13)

Step 1. Assume first that $f\in\mathcal{C}^{1}_{b}$. Fix $0\leqslant
S\leqslant T$ and put for $(s,t)\in\Delta_{S,T}$

$A_{s,t}:=\int_{s}^{t}\int_{D}X_{r}(y)f(V_{r}(y)+P_{r-s}\psi_{s}(y))\,dydr.$

Note that the expression $P_{r-s}\psi_{s}(y)$ is well-defined thanks to Lemma
B.4 and (6.13). Set now

$\mathcal{A}_{t}=\int_{S}^{t}\int_{D}X_{r}(y)f(V_{r}(y)+\psi_{r}(y))dydr.$

It is easy to see that the integral in the left-hand side of inequality (6.3)
is $\mathcal{A}_{T}-\mathcal{A}_{S}$. Let us verify that all the conditions of
Theorem 4.1 are satisfied.

Clearly, for any $S\leqslant s<u<t\leqslant T$ we get

$\delta
A_{s,u,t}=\int_{u}^{t}\int_{D}X_{r}(y)[f(V_{r}(y)+P_{r-s}\psi_{s}(y))-f(V_{r}(y)+P_{r-u}\psi_{u}(y))]\,dydr.$

For fixed $S\leqslant s<u<r\leqslant T$, $y\in{D}$ introduce a function
$h_{r,y}\colon\mathbb{R}\times\Omega\to\mathbb{R}$

$h_{r,y}\colon(z,\omega)\mapsto
f(z+P_{r-s}\psi_{s}(y))-f(z+P_{r-u}\psi_{u}(y)).$

Note that for fixed non-random parameters the random variable $h_{r,y}(z)$ is
$\mathcal{F}_{u}$-measurable. Hence, applying (C.9) with $u,r$ in place of
$s,t$, respectively, we deduce

$\displaystyle|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}\delta A_{s,u,t}|$
$\displaystyle=\Bigl{|}\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\int_{u}^{t}\int_{D}X_{r}(y)\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{u}h_{r,y}(V_{r}(y))\,dydr\Bigr{|}$ $\displaystyle\leqslant
C\int_{u}^{t}\int_{D}|X_{r}(y)|(r-u)^{-\frac{1}{4}+\frac{\gamma}{4}-\frac{1}{4p}}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\|h_{r,y}\|_{\mathcal{B}^{\gamma-1}_{p}}\,dydr$
$\displaystyle\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\int_{u}^{t}\int_{D}|X_{r}(y)|(r-u)^{-\frac{1}{4}+\frac{\gamma}{4}-\frac{1}{4p}}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}|P_{r-s}\psi_{s}(y)-P_{r-u}\psi_{u}(y)|\,dydr,$

where the last inequality follows from (A.5) with $\lambda=1$. Applying the
integral Minkowski inequality, we derive

$\\!\|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}\delta
A_{s,u,t}\|_{{L_{n}}}\\!\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\\!\int_{u}^{t}\\!\\!\int_{D}\\!\\!\\!|X_{r}(y)|(r-u)^{-\frac{1}{4}+\frac{\gamma}{4}-\frac{1}{4p}}\bigl{\|}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}|P_{r-s}\psi_{s}(y)-P_{r-u}\psi_{u}(y)|\bigr{\|}_{{L_{n}}}\,dydr.$
(6.14)

Using the fact that for any $s\geqslant 0$ and random variable $\xi$ one has
$\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}|\xi|\leqslant\|\xi\|_{{L_{m}}|\mathcal{F}_{s}}$ we deduce for
any $y\in{D}$,

$\displaystyle\|\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}|P_{r-s}\psi_{s}(y)-P_{r-u}\psi_{u}(y)|\|_{{L_{n}}}$
$\displaystyle=\|\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}|[P_{r-u}(P_{u-s}\psi_{s}-\psi_{u})](y)|\|_{{L_{n}}}$
$\displaystyle\leqslant\|\|[P_{r-u}(P_{u-s}\psi_{s}-\psi_{u})](y)\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{n}}}$
$\displaystyle\leqslant\sup_{z\in{D}}\|\|P_{u-s}\psi_{s}(z)-\psi_{u}(z)|\mathcal{F}_{s}\|_{{L_{m}}}\|_{{L_{n}}}$
$\displaystyle\leqslant[\psi]_{{\mathcal{C}^{\tau,0}L_{m,n}({[S,T]})}}(u-s)^{\tau},$

where the penultimate inequality follows from (B.8), and in the last
inequality we used the definition of the seminorm
$[\cdot]_{{\mathcal{C}^{\tau,0}L_{m,n}({[S,T]})}}$ given in (3.3).
Substituting this into (6.14), we obtain

$\displaystyle\|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}\delta
A_{s,u,t}\|_{{L_{n}}}$ $\displaystyle\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}[\psi]_{{\mathcal{C}^{\tau,0}L_{m,n}({[S,T]})}}(u-s)^{\tau}\int_{u}^{t}\int_{D}|X_{r}(y)|(r-u)^{-\frac{1}{4}+\frac{\gamma}{4}-\frac{1}{4p}}\,dydr$
$\displaystyle\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}[\psi]_{{\mathcal{C}^{\tau,0}L_{m,n}({[S,T]})}}(T-u)^{-\nu}(t-s)^{\frac{3}{4}+\frac{\gamma}{4}-\frac{1}{4p}+\tau},$

where the last inequality follows from assumption (6.2) and inequality (B.7).
Since, by assumption, $\gamma/4-1/(4p)+\tau>1/4$, we see that condition (4.2)
of Theorem 4.1 holds with $\alpha_{1}=0$, $\beta_{1}=\nu<1/2$,
$\varepsilon_{1}=\gamma/4-1/(4p)+\tau-1/4>0$.

Now we move on to the verification of condition (4.3). Applying 6.2i to the
function $k:=\psi_{s}$, we deduce that for any $(s,t)\in\Delta_{S,T}$

$\|\|A_{s,t}\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{n}}}\leqslant\|\|A_{s,t}\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{p}}}\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}(T-s)^{-\nu}(t-s)^{1+\frac{\gamma}{4}-\frac{1}{4p}}.$
(6.15)

This induces that for any $(s,t)\in\Delta_{S,T}$ and $u\in[s,t]$ one has

$\|\|\delta A_{s,u,t}\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{n}}}\leqslant
C\|f\|_{\mathcal{B}^{\alpha}_{p}}\Gamma_{X}(T-u)^{-\nu}(t-s)^{1+\frac{\gamma}{4}-\frac{1}{4p}}.$

Since $\gamma-\frac{1}{p}>-2$ by assumption, we see that condition (4.3) holds
with $\alpha_{2}=0$, $\beta_{2}=\nu<1/2$,
$\varepsilon_{2}=1/2+\gamma/4-1/(4p)>0$.

Finally, let us check condition (4.4). Let
$\Pi:=\\{S=t_{0},t_{1},...,t_{k}=t\\}$ be an arbitrary partition of $[S,t]$.
Denote by $|\Pi|$ its mesh size. Note that contrary to the proof of Lemma 6.1,
we cannot use here the orthogonality because the sum
$\sum_{i=0}^{k-1}(\mathcal{A}_{t_{i+1}}-\mathcal{A}_{t_{i}}-A_{t_{i},t_{i+1}})$
is not a sum of martingale differences. Indeed, in contrast to the proof of
Lemma 6.1, $\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt_{t_{i}}(\mathcal{A}_{t_{i+1}}-\mathcal{A}_{t_{i}})\neq
A_{t_{i},t_{i+1}}$. Nevertheless, we have

$\displaystyle\Bigl{\|}\mathcal{A}_{t}-\sum_{i=0}^{k-1}A_{t_{i},t_{i+1}}\Bigr{\|}_{{L_{m}}}$
$\displaystyle\quad\leqslant\sum_{i=0}^{k-1}\int_{t_{i}}^{t_{i+1}}\\!\\!\\!\int_{D}|X_{r}(y)|\bigl{\|}f(V_{r}(y)+\psi_{r}(y))-f(V_{r}(y)+P_{r-{t_{i}}}\psi_{t_{i}}(y))\bigr{\|}_{{L_{m}}}\,dydr$
$\displaystyle\quad\leqslant
C\Gamma_{X}(1+T)\|f\|_{\mathcal{C}^{1}}[\psi]_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}}|\Pi|^{\tau}.$

This verifies condition (4.4).

We see that for $f\in\mathcal{C}_{b}^{1}$ all the conditions of Theorem 4.1
are satisfied. Thus, by (4.5), taking into account (4.1), we have

$\displaystyle\|\|\mathcal{A}_{T}-\mathcal{A}_{S}\|_{{L_{m}}|\mathcal{F}_{S}}\|_{{L_{n}}}\leqslant$
$\displaystyle\|\|A_{S,T}\|_{{L_{m}}|\mathcal{F}_{S}}\|_{{L_{n}}}+C\|f\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}(T-S)^{1+\frac{1}{4}(\gamma-\frac{1}{p})-\nu}$
$\displaystyle+C\|f\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}[\psi]_{{\mathcal{C}^{\tau,0}L_{m,n}({[S,T]})}}(T-S)^{\frac{3}{4}+\frac{\gamma}{4}-\frac{1}{4p}+\tau-\nu}\,,$

which together with (6.15) implies (6.3).

Step 2. Let $f$ be a bounded continuous function. For each integer $k\geqslant
1$, define $f^{k}=G_{1/k}f$. Then $\\{f^{k}\\}$ is a sequence of bounded
continuous functions such that
$\sup_{k}\|f^{k}\|_{\mathcal{B}^{\gamma}_{p}}\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}$
(see Lemma A.3) and $\lim_{k}f^{k}(z)=f(z)$ for every $z\in\mathbb{R}$. We
apply (6.3) for $f^{k}$ to get

$\displaystyle\sup_{x\in{D}}\Bigl{\|}\Bigl{\|}\int_{S}^{T}\int_{D}X_{r}(y)f^{k}(V_{r}(y)+\psi_{r}(y))dydr\Bigr{\|}_{{L_{m}}|\mathcal{F}_{S}}\Bigr{\|}_{L_{n}}$
$\displaystyle\quad\leqslant
C\|f^{k}\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}(T-S)^{1+\frac{1}{4}(\gamma-\frac{1}{p})-\nu}+C\|f^{k}\|_{\mathcal{B}^{\gamma}_{p}}\Gamma_{X}[\psi]_{{\mathcal{C}^{\tau,0}L_{m,n}({[S,T]})}}(T-S)^{\frac{3}{4}+\frac{1}{4}(\gamma-\frac{1}{p})+\tau-\nu}.$

We now send $k\to\infty$, using the dominated convergence theorem and the
properties of the sequence $\\{f^{k}\\}$ described previously to obtain (6.3)
for the function $f$. ∎

Now we can finally give the proof of Lemma 5.2.

###### Proof of Lemma 5.2.

The result follows immediately from Lemma 6.3 and (3.2) by choosing
appropriate kernel $X$.

In part (i), we choose $X_{r}(y)=p_{T-r}(x,y)$ which satisfies condition (6.2)
with $\nu=0$ and $\Gamma_{X}=1$.

In part (ii), we choose $X_{r}(y)=p_{T-r}(x_{1},y)-p_{T-r}(x_{2},y)$, $n=m$.
In this case, thanks to (C.2), condition (6.2) holds with $\nu=\delta/2$ and
$\Gamma_{X}=C|x_{1}-x_{2}|^{\delta}$.

In part (iii), we choose $X_{r}(y)=p_{T+\bar{T}-r}(x,y)-p_{T-r}(x,y)$, $n=m$.
In this case, thanks to (C.4), condition (6.2) holds with $\nu=\delta/2$ and
$\Gamma_{X}=C\bar{T}^{\delta/2}$. ∎

###  6.2 Proof of Lemmas 5.5, 5.6 and 5.7

Now let us move to the proof of the key lemmas needed for the uniqueness
proof.

###### Proof of Lemma 5.5.

Fix $x\in{D}$, $(s,t)\in\Delta_{0,T_{0}}$, $\beta^{\prime}\in(-2,\beta)$. For
any $k,l\in\mathbb{Z}_{+}$ we apply 6.2i with $f=b_{k}-b_{l}$,
$\gamma=\beta^{\prime}$, $p=q$, $T=t$, $X_{r}(y)=p_{t-r}(x,y)$,
$\Gamma_{X}=1$, $\nu=0$, $\kappa=\psi_{s}$. Note that
$\psi_{s}\in{\mathbf{B}L_{m}}$ by (5.28). Recalling (3.2), we get that there
exists a constant $C>0$ independent of $k,l$ such that

$\|H^{l,\psi}_{s,t}(x)-H^{k,\psi}_{s,t}(x)\|_{{L_{m}}}\leqslant
C(1+T_{0})\|b_{l}-b_{k}\|_{\mathcal{B}^{\beta^{\prime}}_{q}}.$

Since the sequence $(b_{n})$ converges in $\mathcal{B}^{\beta^{\prime}}_{q}$,
we see that the right-hand side of the above inequality tends to $0$ as
$n,k\to\infty$. Hence the sequence of random variables
$(H^{k,\psi}_{s,t}(x))_{k\in\mathbb{Z}_{+}}$ is Cauchy in ${L_{m}}$. Thus this
sequence converges in ${L_{m}}$ (and hence in probability). We denote its
limit by $H^{\psi}_{s,t}(x)$.

By exactly the same argument, we see that the sequence
$(H^{k,\varphi}_{s,t}(x))_{k\in\mathbb{Z}_{+}}$ converges in probability to a
limit which we denote by $H^{\varphi}_{s,t}(x)$. Now, applying the Fatou’s
lemma we derive for each $x\in{D}$, $(s,t)\in\Delta_{0,T_{0}}$,
$\lambda\in[0,1]$

$\displaystyle\|H^{\psi}_{s,t}(x)-H^{\varphi}_{s,t}(x)\|_{{L_{m}}}$
$\displaystyle\leqslant\liminf_{k\to\infty}\|H^{k,\psi}_{s,t}(x)-H^{k,\varphi}_{s,t}(x)\|_{{L_{m}}}$
$\displaystyle\leqslant
C\sup_{k\in\mathbb{Z}_{+}}\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}\|\psi_{s}-\varphi_{s}\|_{{\mathbf{B}L_{m}}}^{\lambda}(t-s)^{1-\frac{\lambda}{4}+\frac{\beta}{4}-\frac{1}{4q}},$

where the second inequality follows from 6.2ii with $f=b_{k}$, $\gamma=\beta$,
$p=q$, $\kappa_{1}=\varphi_{s}$, $\kappa_{2}=\psi_{s}$. Taking into account
that by assumption
$\sup_{k\in\mathbb{Z}_{+}}\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}<\infty$ and
$\beta-1/q\geqslant-1$, we obtain

$\displaystyle\|H^{\psi}_{s,t}(x)-H^{\varphi}_{s,t}(x)\|_{{L_{m}}}\leqslant
C(1+T_{0})\|\psi_{s}-\varphi_{s}\|_{{\mathbf{B}L_{m}}}^{\lambda}(t-s)^{\frac{3}{4}-\frac{\lambda}{4}}.$

Now by taking $\lambda=1$ in this bound we get (5.30), and by taking
$\lambda=0$ we get (5.31). ∎

###### Proof of Lemma 5.6.

Since $u$ belongs to $\mathcal{V}(3/4)$ and
$\sup_{n}\|b_{n}\|_{\mathcal{B}^{\beta}_{q}}<\infty$, we obtain from Lemma
5.2(i) that

$\sup_{x\in{D}}\|K^{b_{n};u}_{s}(x)\|_{{L_{m}}}\leqslant
C(1+[\psi]_{{\mathcal{C}^{3/4,0}L_{m}({[0,T_{0}]})}})(1+T_{0}).$

Therefore for any fixed $x\in{D}$ the sequence
$(K^{b_{n};u}_{s}(x))_{n\in\mathbb{Z}_{+}}$ is uniformly integrable. Recalling
that $K^{b_{n};u}_{s}(x)$ converges to $K^{u}_{s}(x)$ in probability, we get

$\|K^{b_{n};u}_{s}(x)-K^{u}_{s}(x)\|_{L_{1}}\to 0\,\,\,\text{as
$n\to\infty$}.$

Thus,

$\displaystyle\|P_{t-s}K^{b_{n};u}_{s}(x)-P_{t-s}K^{u}_{s}(x)\|_{L_{1}}$
$\displaystyle=\Bigl{\|}\int_{{D}}p_{t-s}(x,y)(K^{b_{n};u}_{s}(y)-K^{u}_{s}(y))\,dy\Bigr{\|}_{L_{1}}$
$\displaystyle\leqslant\int_{{D}}p_{t-s}(x,y)\|K^{b_{n};u}_{s}(y)-K^{u}_{s}(y)\|_{L_{1}}\,dy\to
0,$

by the dominated convergence theorem (here we once again made use of (6.2)).
Thus, $P_{t-s}K^{b_{n};u}_{s}(x)\to P_{t-s}K^{u}_{s}(x)$ in probability as
$n\to\infty$. The convergence of $P_{t-s}K^{b_{n};v}_{s}(x)$ to
$P_{t-s}K^{v}_{s}(x)$ in probability is obtained by exactly the same argument.
∎

###### Proof of Lemma 5.7.

The proof is based on the stochastic sewing lemma with critical exponent,
Theorem 4.5. Let us verify that all the conditions of this theorem are
satisfied. A key tool for the verification will be 6.2. Fix $0\leqslant
S\leqslant T$, $x\in{D}$, $\tau>1/2$. Recall that we are given a sequence of
smooth functions $(b_{k})_{k\in\mathbb{Z}_{+}}$ such that $b_{k}\to b$ in
$\mathcal{B}^{\beta-}_{q}$ and
$\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}\leqslant\|b\|_{\mathcal{B}^{\beta}_{q}}$
for any $k\in\mathbb{Z}_{+}$.

We put for $(s,t)\in\Delta_{S,T}$, $x\in{D}$, $k\in\mathbb{Z}_{+}$

$A^{k}_{s,t}:=\int_{s}^{t}\int_{D}p_{T-r}(x,y)\bigl{[}b_{k}(V_{r}(y)+P_{r-s}\psi_{s}(y))-b_{k}(V_{r}(y)+P_{r-s}\varphi_{s}(y))\bigr{]}\,dydr.$

We note that $P_{r-s}\psi_{s}(y)$ and $P_{r-s}\varphi_{s}(y)$ are well-defined
thanks to Lemma B.4 and (5.28).

Put for $(t,x)\in[S,T]\times{D}$

$\mathcal{A}^{k}_{t}=\int_{S}^{t}\int_{D}p_{T-r}(x,y)\bigl{[}b_{k}(V_{r}(y)+\psi_{r}(y))-b_{k}(V_{r}(y)+\varphi_{r}(y))\bigr{]}\,dydr.$

From now on we fix $k\in\mathbb{Z}_{+}$, and will drop the superindex $k$ in
$A^{k}$ and $\mathcal{A}^{k}$.

Let us verify that all the conditions of Theorem 4.5 are satisfied. We get for
any $S\leqslant s<u<t\leqslant T$

$\displaystyle\delta A_{s,u,t}=\int_{u}^{t}\int_{D}$ $\displaystyle
p_{T-r}(x,y)\Bigl{[}b_{k}(V_{r}(y)+P_{r-s}\psi_{s}(y))-b_{k}(V_{r}(y)+P_{r-s}\varphi_{s}(y))$
$\displaystyle-
b_{k}(V_{r}(y)+P_{r-u}\psi_{u}(y))+b_{k}(V_{r}(y)+P_{r-u}\varphi_{u}(y))\Bigr{]}\,dydr.$

We begin by verifying (4.2). For fixed $S\leqslant s<u<r\leqslant T$,
$y\in{D}$ introduce functions
$h_{r,y},l_{r,y}\colon\mathbb{R}\times\Omega\to\mathbb{R}$

$\displaystyle h_{r,y}\colon(z,\omega)\mapsto
b_{k}(z+P_{r-s}\psi_{s}(y))-b_{k}(z+P_{r-s}\varphi_{s}(y))$
$\displaystyle\qquad\qquad\qquad-
b_{k}(z+P_{r-u}\psi_{u}(y))+b_{k}(z+P_{r-u}\psi_{u}(y)+P_{r-s}\varphi_{s}(y)-P_{r-s}\psi_{s}(y));$
$\displaystyle l_{r,y}\colon(z,\omega)\mapsto
b_{k}(z+P_{r-u}\varphi_{u}(y))-b_{k}(z+P_{r-u}\psi_{u}(y)+P_{r-s}\varphi_{s}(y)-P_{r-s}\psi_{s}(y)).$

Clearly, for fixed non-random parameters the random variables $h_{r,y}(z)$,
$l_{r,y}(z)$ are $\mathcal{F}_{u}$-measurable. Hence, by (C.9) with $u,r$ in
place of $s,t$, respectively, we deduce for any $\lambda\in[0,1]$

$\displaystyle|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}\delta A_{s,u,t}|$
$\displaystyle\,\,=\Bigl{|}\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\int_{u}^{t}\\!\\!\int_{D}p_{T-r}(x,y)\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{u}[h_{r,y}(V_{r}(y))+l_{r,y}(V_{r}(y))]\,dydr\Bigr{|}$
$\displaystyle\,\,\leqslant
C\int_{u}^{t}\\!\\!\int_{D}p_{T-r}(x,y)[(r-u)^{\frac{\beta-1-\lambda}{4}-\frac{1}{4q}}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\|h_{r,y}\|_{\mathcal{B}^{\beta-1-\lambda}_{q}}+(r-u)^{\frac{\beta-1}{4}-\frac{1}{4q}}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\|l_{r,y}\|_{\mathcal{B}^{\beta-1}_{q}}]\,dydr.$ (6.16)

We see that by (A.6) and (B.8)

$\displaystyle\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\|h_{r,y}\|_{\mathcal{B}^{\beta-1-\lambda}_{q}}$
$\displaystyle\leqslant\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}|P_{r-s}\psi_{s}(y)-P_{r-s}\varphi_{s}(y)|^{\lambda}\,\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt^{s}|P_{r-s}\psi_{s}(y)-P_{r-u}\psi_{u}(y)|$
$\displaystyle\leqslant\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}|P_{r-s}z_{s}(y)|^{\lambda}\,\|\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}|P_{r-s}\psi_{s}(y)-P_{r-u}\psi_{u}(y)|\|_{{L_{\infty}}}$
$\displaystyle\leqslant\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}|P_{r-s}z_{s}(y)|^{\lambda}\,\sup_{z\in{D}}\|\|P_{u-s}\psi_{s}(z)-\psi_{u}(z)\|_{{L_{1}}|\mathcal{F}_{s}}\|_{{L_{\infty}}}$
$\displaystyle\leqslant\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}}|P_{r-s}z_{s}(y)|^{\lambda}(t-s)^{\frac{3}{4}}.$

This, (B.8) and the fact that
$\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}\leqslant\|b\|_{\mathcal{B}^{\beta}_{q}}$
imply

$\|\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\|h_{r,y}\|_{\mathcal{B}^{\beta-1-\lambda}_{q}}\|_{{L_{m}}}\leqslant
C\|b\|_{\mathcal{B}^{\beta}_{q}}[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,1]})}}\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}^{\lambda}(t-s)^{\frac{3}{4}}.$
(6.17)

In a similar manner, by (A.5) and (B.8) we see that

$\displaystyle\|\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\|l_{r,y}\|_{\mathcal{B}^{\beta-1}_{q}}\|_{{L_{m}}}$
$\displaystyle\leqslant\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}\|P_{r-u}z_{u}(y)-P_{r-s}z_{s}(y)\|_{{L_{m}}}$
$\displaystyle\leqslant\|b\|_{\mathcal{B}^{\beta}_{q}}\sup_{y\in{D}}\|z_{u}(y)-P_{u-s}z_{s}(y)\|_{{L_{m}}}.$
(6.18)

Thus,

$\displaystyle\|\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\|l_{r,y}\|_{\mathcal{B}^{\beta-1}_{q}}\|_{{L_{m}}}$
$\displaystyle\leqslant\|b\|_{\mathcal{B}^{\beta}_{q}}(\sup_{y\in{D}}\|\varphi_{u}(y)-P_{u-s}\varphi_{s}(y)\|_{{L_{m}}}+\sup_{y\in{D}}\|\psi_{u}(y)-P_{u-s}\psi_{s}(y)\|_{{L_{m}}})$
$\displaystyle\leqslant\|b\|_{\mathcal{B}^{\beta}_{q}}([\psi]_{{\mathcal{C}^{3/4,0}L_{m}({[S,T]})}}+[\varphi]_{{\mathcal{C}^{3/4,0}L_{m}({[S,T]})}})(t-s)^{\frac{3}{4}}$
$\displaystyle\leqslant\|b\|_{\mathcal{B}^{\beta}_{q}}([\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}}+[\varphi]_{{\mathcal{C}^{3/4,0}L_{m}({[S,T]})}})(t-s)^{\frac{3}{4}}.$

Substituting this together with (6.17) into (6.2) and taking $\lambda=0$, we
obtain

$\|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}\delta
A_{s,u,t}\|_{{L_{m}}}\leqslant
C(T_{0},\|b\|_{\mathcal{B}^{\beta}_{q}},[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}},[\varphi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}})(t-s)^{\frac{5}{4}},$

where we used the fact that, by assumptions, $\beta\geqslant-1+1/q$. Thus,
condition (4.2) holds.

Now let us verify (4.3). Fix $\varepsilon\in(0,1+\beta)$. Let us apply
6.2(iii) with $f=b_{k}$, $\gamma=\beta$, $p=q$, $\lambda_{1}=\lambda=1$ and
$\lambda_{2}=\beta+1-\varepsilon$ and

$\displaystyle\kappa_{1}:=P_{u-s}\psi_{s},$
$\displaystyle\kappa_{2}:=P_{u-s}\varphi_{s},$
$\displaystyle\kappa_{3}:=\psi_{u},$ $\displaystyle\kappa_{4}:=\varphi_{u}.$

One can see that the functions $\kappa_{1}$ and $\kappa_{2}$ are
$\mathscr{B}(\mathbb{R})\times\mathcal{F}_{s}$ measurable, and the functions
$\kappa_{3}$ and $\kappa_{4}$ are
$\mathscr{B}(\mathbb{R})\times\mathcal{F}_{u}$ measurable, exactly as required
by the conditions of 6.2(iii). Further, we see that
$\kappa_{i}\in{\mathbf{B}L_{m}}$, $i=1,..,4$, thanks to (5.28) and (B.8).
Finally, conditions (6.8) and (6.10) are satisfied thanks to our choice of
$\lambda$, $\lambda_{1}$, $\lambda_{2}$. Therefore all the conditions of
6.2(iii) are met. Recall the notation $z:=\psi-\varphi$ and note that by (B.8)

$\displaystyle\|\kappa_{1}-\kappa_{2}\|_{{\mathbf{B}L_{m}}}$
$\displaystyle\leqslant\|z_{s}\|_{{\mathbf{B}L_{m}}}\leqslant\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}},$
$\displaystyle\sup_{y\in{D}}\|\|\kappa_{1}(y)-\kappa_{3}(y)\|_{{L_{m}}|\mathcal{F}_{s}}\|_{{L_{\infty}}}$
$\displaystyle\leqslant[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}}(t-u)^{3/4},$
$\displaystyle\|\kappa_{1}-\kappa_{2}-\kappa_{3}+\kappa_{4}\|_{{\mathbf{B}L_{m}}}$
$\displaystyle\leqslant\|z_{u}-P_{u-s}z_{s}\|_{{\mathbf{B}L_{m}}}$
$\displaystyle\leqslant\|R_{s,u}\|_{{\mathbf{B}L_{m}}}+\|H^{\psi}_{s,u}-H^{\varphi}_{s,u}\|_{{\mathbf{B}L_{m}}}$
$\displaystyle\leqslant\|R\|_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}}(t-u)^{\tau}+C\|b\|_{\mathcal{B}^{\gamma}_{p}}\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}(t-u)^{\frac{1}{2}},$

where the last inequality follows from (5.30) and the definition of the norm
$\|\cdot\|_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}}$ given in (5.27).
Substituting all this into (iii) and using the fact that
$\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}\leqslant\|b\|_{\mathcal{B}^{\beta}_{q}}$,
one gets

$\displaystyle\|\delta A_{s,u,t}\|_{{L_{m}}}\leqslant$ $\displaystyle
C\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}}^{\gamma+1-\varepsilon}(t-u)^{\frac{5}{4}+\frac{3\beta}{4}-\frac{1}{4q}-\frac{\varepsilon}{2}}$
$\displaystyle\phantom{\leqslant}+C\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}\|R\|_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}}(t-u)^{\frac{1}{2}+\tau}$
$\displaystyle\phantom{\leqslant}+C\|b_{k}\|_{\mathcal{B}^{\beta}_{q}}\|b\|_{\mathcal{B}^{\beta}_{q}}\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}(t-u)$
$\displaystyle\leqslant$ $\displaystyle
C(T_{0},\|b\|_{\mathcal{B}^{\beta}_{q}},[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}})\bigl{(}\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}+\|R\|_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}}\bigr{)}(t-s)^{\frac{1}{2}+\delta},$
(6.19)

where we have also used that $\beta-1/q\geqslant-1$ and we put
$\delta:=(\frac{3}{4}+\frac{3\beta}{4}-\frac{1}{4q}-\frac{\varepsilon}{2})\wedge\frac{1}{2}$.
Since $\beta>-1+1/(3q)$, we see that there exists
$\varepsilon=\varepsilon(\beta,q)>0$ small enough such that $\delta>0$. From
now on till the end of the proof we fix such $\varepsilon$. Then (6.2) implies
that condition (4.3) is satisfied.

Now let us check (4.4). We have

$\displaystyle\Bigl{\|}\mathcal{A}_{t}-\sum_{i=0}^{k-1}A_{t_{i},t_{i+1}}\Bigr{\|}_{{L_{1}}}$
$\displaystyle\quad\leqslant\sum_{i=0}^{k-1}\int_{t_{i}}^{t_{i+1}}\\!\\!\\!\int_{D}p_{T-r}(x,y)\bigl{\|}b_{k}(V_{r}(y)+\psi_{r}(y))-b_{k}(V_{r}(y)+P_{r-{t_{i}}}\psi_{t_{i}}(y))\bigr{\|}_{{L_{1}}}\,dydr$
$\displaystyle\quad\phantom{\leqslant}+\sum_{i=0}^{k-1}\int_{t_{i}}^{t_{i+1}}\\!\\!\\!\int_{D}p_{T-r}(x,y)\bigl{\|}b_{k}(V_{r}(y)+\varphi_{r}(y))-b_{k}(V_{r}(y)+P_{r-{t_{i}}}\varphi_{t_{i}}(y))\bigr{\|}_{{L_{1}}}\,dydr$
$\displaystyle\quad\leqslant\|b_{k}\|_{\mathcal{C}^{1}}\sum_{i=0}^{k-1}\int_{t_{i}}^{t_{i+1}}\\!\\!\\!\int_{D}p_{T-r}(x,y)(\|\psi_{r}(y)-P_{r-{t_{i}}}\psi_{t_{i}}(y)\|_{{L_{1}}}+\|\varphi_{r}(y)-P_{r-{t_{i}}}\varphi_{t_{i}}(y)\|_{{L_{1}}})\,dydr$
$\displaystyle\quad\leqslant
CT_{0}\|b_{k}\|_{\mathcal{C}^{1}}([\psi]_{{\mathcal{C}^{3/4,0}L_{m}({[0,T_{0}]})}}+[\varphi]_{{\mathcal{C}^{3/4,0}L_{m}({[0,T_{0}]})}})|\Pi|^{3/4}.$

This implies that $|\mathcal{A}_{t}-\sum_{i=0}^{k-1}A_{t_{i},t_{i+1}}|$
converges to $0$ in ${L_{1}}$ as $k\to\infty$. Hence (4.4) holds.

Finally, let us check (4.11). Substituting (6.17) and (6.2) into (6.2) and
taking $\lambda=1$, we obtain

$\displaystyle\|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}\delta
A_{s,u,t}\|_{{L_{m}}}$ $\displaystyle\leqslant
C(1+T_{0})\|b\|_{\mathcal{B}^{\beta}_{q}}[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}}\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}(t-s)$
$\displaystyle\phantom{\leqslant}+C(1+T_{0})\|b\|_{\mathcal{B}^{\beta}_{q}}(t-s)^{\frac{1}{2}}(\sup_{y\in{D}}\|R_{s,u}(y)\|+\sup_{y\in{D}}\|H^{\psi}_{s,u}(y)-H^{\varphi}_{s,u}(y)\|)$
$\displaystyle\leqslant
C(1+T_{0})\|b\|_{\mathcal{B}^{\beta}_{q}}(1+[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}})\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}(t-s)$
$\displaystyle\phantom{\leqslant}+C(1+T_{0})\|b\|_{\mathcal{B}^{\beta}_{q}}\|R\|_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}}(t-s)^{\frac{1}{2}+\tau},$

where the last inequality follows from (5.30). Recalling that, by assumptions,
we have $\tau>1/2$, we see that (4.11) holds.

Thus all the conditions of Theorem 4.5 are met. We deduce from (4.12)

$\displaystyle\|\mathcal{A}^{k}_{T}-\mathcal{A}^{k}_{S}-A^{k}_{S,T}\|_{L_{m}}$
$\displaystyle\leqslant
C\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}(1+|\log{\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}}|)(T-S)$
$\displaystyle\phantom{\leqslant}+C(\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}+\|R\|_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}})(T-S)^{\frac{1}{2}+\delta}$
$\displaystyle\phantom{\leqslant}+C\|R\|_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}}(T-S)^{\frac{1}{2}+\tau}$
$\displaystyle\leqslant
C\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}|\log{\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}}|(T-S)$
$\displaystyle\phantom{\leqslant}+C(\|z\|_{{\mathcal{C}^{0,0}L_{m}({[S,T]})}}+\|R\|_{{\mathcal{C}^{\tau,0}L_{m}({[S,T]})}})(T-S)^{\frac{1}{2}+\delta},$
(6.20)

where
$C=C(T_{0},\|b\|_{\mathcal{B}^{\beta}_{q}},[\psi]_{{\mathcal{C}^{3/4,0}L_{m,\infty}({[0,T_{0}]})}},\tau,\delta)$
is independent of $x$, $S$, $T$.

Now let us pass to the limit in (6.2) as $k\to\infty$. Recall the notation
$K^{h;\sigma}$ introduced in (5.12). Then it is easy to see that

$\mathcal{A}^{k}_{T}-\mathcal{A}^{k}_{S}=K^{b_{k};u}_{T}(x)-K^{b_{k};v}_{T}(x)-P_{T-S}[K^{b_{k};u}_{S}-K^{b_{k};v}_{S}](x).$

Applying Lemmas 5.6 and 5.5 we can conclude that

$\lim_{k\to\infty}\mathcal{A}^{k}_{T}-\mathcal{A}^{k}_{S}-A^{k}_{S,T}=z_{T}(x)-P_{T-S}z_{S}(x)-(H^{\psi}_{S,T}(x)-H^{\varphi}_{S,T}(x))=R_{S,T}(x),$

in probability. Inequality (5.34) follows now from (6.2) by Fatou’s lemma.
Inequality (5.7) follows immediately from (5.34) and (5.30). ∎

##  Appendix A Useful results on Besov spaces

We give a brief summary on nonhomogeneous Besov space which is sufficient for
our purpose. For a more detailed account on the topic, we refer to [BCD11,
Chapter 2]. Let $\varsigma,\varpi$ be the radial functions which are given by
[BCD11, Proposition 2.10]. We note that $\varsigma$ is supported on a ball
while $\varpi$ is supported on an annulus. Let $h_{-1}$ and $h$ respectively
be the inverse Fourier transform of $\varsigma$ and $\varpi$. The
nonhomogeneous dyadic blocks $\Delta_{j}$ are defined by

$\Delta_{-1}f=\int_{\mathbb{R}}h_{-1}(y)f(\cdot-y)dy\quad\textrm{and}\quad\Delta_{j}f=\int_{\mathbb{R}}h_{j}(y)f(\cdot-y)dy\quad\textrm{for}\quad
j\geqslant 0,$

where $h_{j}(y)=2^{j}h(2^{j}y)$, $j\geqslant 0$.

######  Definition A.1.

Let $\gamma\in\mathbb{R}$ and $1\leqslant p\leqslant\infty$. The
nonhomogeneous Besov space
$\mathcal{B}^{\gamma}_{p}=\mathcal{B}^{\gamma}_{p,\infty}(\mathbb{R})$
consists of all tempered distributions $f$ such that

$\|f\|_{\mathcal{B}^{\gamma}_{p}}:=\sup_{j\geqslant-1}2^{j\gamma}\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}<\infty.$

For a distribution $f$ in $\mathcal{B}^{\gamma}_{p}$, we note that
$\Delta_{j}f$ is a smooth function for each $j\geqslant-1$. In addition, the
Fourier transform of $\Delta_{-1}f$ is supported on a ball $\mathcal{B}$ while
for $j\geqslant 1$ the Fourier transform of $\Delta_{j}f$ is supported on the
annulus $2^{j}\mathcal{N}\subset 2^{j}\mathcal{B}$ for some annulus
$\mathcal{N}$.

To obtain various properties of Besov spaces, we will make use of the
following Bernstein’s inequalities. Let $f$ be a function in
$L_{p}(\mathbb{R})$. For every integer $k\geqslant 0$, every $\lambda>0$ and
$t>0$ we have

$\displaystyle\operatorname*{Supp}Ff\subset\lambda\mathcal{B}\Rightarrow\|\nabla^{k}f\|_{L_{p}(\mathbb{R})}\leqslant
C^{k+1}\lambda^{k}\|f\|_{L_{p}(\mathbb{R})},$ (A.1)
$\displaystyle\operatorname*{Supp}Ff\subset\lambda\mathcal{N}\Rightarrow\|G_{t}f\|_{L_{p}(\mathbb{R})}\leqslant
Ce^{-ct\lambda^{2}}\|f\|_{L_{p}(\mathbb{R})},$ (A.2)
$\displaystyle\operatorname*{Supp}Ff\subset\lambda\mathcal{B}\Rightarrow\|G_{t}f-f\|_{L_{p}(\mathbb{R})}\leqslant
C(t\lambda^{2}\wedge 1)\|f\|_{L_{p}(\mathbb{R})},$ (A.3)

where $Ff$ denotes the Fourier transform of $f$. We refer to [BCD11, Lemmas
2.1 and 2.4] for proofs of (A.1) and (A.2). For a proof of (A.3), we refer to
[MW17, Lemma 4].

######  Lemma A.2.

Let $f$ be a tempered distribution on $\mathbb{R}$, $\gamma\in\mathbb{R}$,
$p\in[1,\infty]$. Then for any $a,a_{1},a_{2},a_{3}\in\mathbb{R}$,
$\alpha,\alpha_{1},\alpha_{2}\in[0,1]$ one has

$\displaystyle\|f(a+\cdot)\|_{\mathcal{B}^{\gamma}_{p}}$
$\displaystyle=\|f\|_{\mathcal{B}^{\gamma}_{p}},$ (A.4)
$\displaystyle\|f(a_{1}+\cdot)-f(a_{2}+\cdot)\|_{\mathcal{B}^{\gamma}_{p}}$
$\displaystyle\leqslant
C|a_{1}-a_{2}|^{\alpha}\|f\|_{\mathcal{B}^{\gamma+\alpha}_{p}},$ (A.5)

and

$\|f(a_{1}+\cdot)-f(a_{2}+\cdot)-f(a_{3}+\cdot)+f(a_{3}+a_{2}-a_{1}+\cdot)\|_{\mathcal{B}^{\gamma}_{p}}\\\
\leqslant
C|a_{1}-a_{2}|^{\alpha_{1}}|a_{1}-a_{3}|^{\alpha_{2}}\|f\|_{\mathcal{B}^{\gamma+\alpha_{1}+\alpha_{2}}_{p}}.$
(A.6)

###### Proof.

Since
$\|\Delta_{j}f(a+\cdot)\|_{L_{p}(\mathbb{R})}=\|\Delta_{j}f(\cdot)\|_{L_{p}(\mathbb{R})}$,
(A.4) follows from Definition A.1.

For each $j\geqslant-1$, applying mean value theorem and (A.1), we have

$\displaystyle\|\Delta_{j}(f(a_{1}+\cdot)-f(a_{2}+\cdot))\|_{L_{p}(\mathbb{R})}\leqslant
C|a_{1}-a_{2}|2^{j}\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}.$

By triangle inequality, it is evident that

$\displaystyle\|\Delta_{j}(f(a_{1}+\cdot)-f(a_{2}+\cdot))\|_{L_{p}(\mathbb{R})}\leqslant
2\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}.$

Hence, by interpolation, we obtain

$\|\Delta_{j}(f(a_{1}+\cdot)-f(a_{2}+\cdot))\|_{L_{p}(\mathbb{R})}\leqslant
C|a_{1}-a_{2}|^{\alpha}2^{j\alpha}\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}.$ (A.7)

In view of Definition A.1, we multiply both sides with $2^{j\gamma}$ and take
supremum over $j\geqslant-1$ to obtain (A.5).

Similarly, using mean value theorem and (A.1), we have

$\|\Delta_{j}(f(a_{1}+\cdot)-f(a_{2}+\cdot)-f(a_{3}+\cdot)+f(a_{3}+a_{2}-a_{1}+\cdot))\|_{L_{p}(\mathbb{R})}\\\
\leqslant
C|a_{1}-a_{2}||a_{1}-a_{3}|2^{2j}\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}$

and

$\|\Delta_{j}(f(a_{1}+\cdot)-f(a_{2}+\cdot)-f(a_{3}+\cdot)+f(a_{3}+a_{2}-a_{1}+\cdot))\|_{L_{p}(\mathbb{R})}\\\
\leqslant
C\min(1,|a_{1}-a_{2}|2^{j},|a_{1}-a_{3}|2^{j})\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}.$

Interpolating between these inequalities, we obtain

$\|\Delta_{j}(f(a_{1}+\cdot)-f(a_{2}+\cdot)-f(a_{3}+\cdot)+f(a_{3}+a_{2}-a_{1}+\cdot))\|_{L_{p}(\mathbb{R})}\\\
\leqslant
C|a_{1}-a_{2}|^{\alpha_{1}}|a_{1}-a_{3}|^{\alpha_{2}}2^{j(\alpha_{1}+\alpha_{2})}\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}.$

We multiply both sides with $2^{j\gamma}$ and take supremum over
$j\geqslant-1$ to obtain (A.6). ∎

######  Lemma A.3.

For every $f$ in $\mathcal{B}^{\gamma}_{p}$, $\gamma\in\mathbb{R}$,
$p\in[1,\infty]$, we have

  1. (i)

provided that $\gamma<0$, $\|G_{t}f\|_{L_{p}(\mathbb{R})}\leqslant
C_{\gamma}\|f\|_{\mathcal{B}^{\gamma}_{p}}t^{\frac{\gamma}{2}}$ for all $t>0$;

  2. (ii)

$\lim_{t\to 0}G_{t}f=f$ in $\mathcal{B}^{\bar{\gamma}}_{p}$ for every
$\bar{\gamma}<\gamma$;

  3. (iii)

$\sup_{t>0}\|G_{t}f\|_{\mathcal{B}^{\gamma}_{p}}\leqslant\|f\|_{\mathcal{B}^{\gamma}_{p}}$
and

  4. (iv)

provided that $\gamma-\frac{1}{p}<0$, $\|G_{t}f\|_{\mathcal{C}^{1}}\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}t^{\frac{1}{2}(\gamma-\frac{1}{p}-1)}$ for
all $t>0$.

###### Proof.

(i) Fix $\gamma<0$. We have by (A.2) that there exists some constant $c>0$
such that for every $j\geqslant 0$

$\displaystyle\|G_{t}(\Delta_{j}f)\|_{L_{p}(\mathbb{R})}\leqslant
Ce^{-ct2^{2j}}\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}\leqslant
Ce^{-ct2^{2j}}2^{-j\gamma}\|f\|_{\mathcal{B}^{\gamma}_{p}}.$

For $j=-1$, we use the trivial bounds

$\displaystyle\|G_{t}(\Delta_{-1}f)\|_{L_{p}(\mathbb{R})}\leqslant\|\Delta_{-1}f\|_{L_{p}(\mathbb{R})}\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}.$

Since $G_{t}(\Delta_{j}f)=\Delta_{j}(G_{t}f)$, we have

$\sum_{j\geqslant-1}\|\Delta_{j}(G_{t}f)\|_{L_{p}(\mathbb{R})}\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}\sum_{j\geqslant-1}e^{-ct2^{2j}}2^{-j\gamma}.$
(A.8)

Clearly, if $j\leqslant x\leqslant j+1$, $j\geqslant-1$, then
$e^{-ct2^{2j}}2^{-j\gamma}\leqslant e^{-ct2^{2x}}2^{-(x+1)\gamma}$. Therefore,

$\sum_{j\geqslant-1}e^{-ct2^{2j}}2^{-j\gamma}\leqslant\int_{-1}^{\infty}e^{-ct2^{2x}}2^{-x\gamma}\,dx\leqslant
Ct^{\frac{\gamma}{2}}\int_{0}^{\infty}e^{-cy}y^{-\frac{\gamma}{2}-1}\,dy\leqslant
Ct^{\frac{\gamma}{2}},$

thanks to the assumption $\gamma<0$. Using this and the fact that
$\Delta_{j}(G_{t}f)=G_{t}(\Delta_{j}f)$, we derive from (A.8)

$\displaystyle\sum_{j\geqslant-1}\|\Delta_{j}(G_{t}f)\|_{L_{p}(\mathbb{R})}\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}t^{\frac{\gamma}{2}}.$

On the other hand, the identity $Id=\sum_{j\geqslant-1}\Delta_{j}$ holds in
distribution ([BCD11, Proposition 2.12]). The above estimate shows that it
also holds in $L_{p}(\mathbb{R})$ and hence,

$\displaystyle\|G_{t}f\|_{L_{p}(\mathbb{R})}\leqslant\sum_{j\geqslant-1}\|\Delta_{j}(G_{t}f)\|_{L_{p}(\mathbb{R})}\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}t^{\frac{\gamma}{2}}$

which implies (i).

(ii) From (A.3), we derive that for every $j\geqslant-1$ and
$\epsilon\in(0,1]$

$\|G_{t}(\Delta_{j}f)-\Delta_{j}f\|_{L_{p}(\mathbb{R})}\leqslant
c(t2^{2j})^{\varepsilon}\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}.$

This yields

$\displaystyle
2^{j(\gamma-2\varepsilon)}\|\Delta_{j}(G_{t}f-f)\|_{L_{p}(\mathbb{R})}\leqslant
ct^{\varepsilon}2^{j\gamma}\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}\leqslant
ct^{\varepsilon}\|f\|_{\mathcal{B}^{\gamma}_{p}}.$

By taking supremum over all $j\geqslant-1$ in the above inequality, we see
that $\lim_{t\downarrow 0}G_{t}f=f$ in
$\mathcal{B}^{\gamma-2\varepsilon}_{p}$. Since $\varepsilon$ is arbitrary in
$(0,1]$, this implies (ii) when $\bar{\gamma}\in[\gamma-2,\gamma)$. If
$\bar{\gamma}<\gamma-2$, then

$\|G_{t}f-f\|_{\mathcal{B}^{\bar{\gamma}}_{p}}\leqslant\|G_{t}f-f\|_{\mathcal{B}^{\gamma-2}_{p}}\to
0\,\,\text{as $t\to 0$}.$

This proves (ii).

(iii) follows immediately from the inequality
$\|G_{t}(\Delta_{j}f)\|_{L_{p}(\mathbb{R})}\leqslant\|\Delta_{j}f\|_{L_{p}(\mathbb{R})}$
and Definition A.1.

(iv) We write $G_{t}f=G_{t/2}G_{t/2}f$, apply standard heat kernel bounds and
part (i) to get

$\displaystyle\|G_{t}f\|_{\mathcal{C}^{1}}\leqslant
Ct^{-\frac{1}{2}}\|G_{t/2}f\|_{L_{\infty}(\mathbb{R})}\leqslant
Ct^{-\frac{1}{2}}t^{\frac{1}{2}(\gamma-\frac{1}{p})}\|f\|_{\mathcal{B}_{\infty}^{\gamma-1/p}}\leqslant
Ct^{-\frac{1}{2}}t^{\frac{1}{2}(\gamma-\frac{1}{p})}\|f\|_{\mathcal{B}_{p}^{\gamma}},$

where the last inequality follows from the embedding
$\mathcal{B}^{\gamma}_{p}\hookrightarrow\mathcal{B}^{\gamma-1/p}_{\infty}$.
This proves (iv). ∎

Recall that Schwarz distribution $\zeta^{-1}$ is defined in (2.9) and
$\zeta^{\alpha}_{+}$, $\zeta^{\alpha}_{-}$ are defined in (2.10).

######  Lemma A.4.

Let $\alpha\in(-1,0)$. Then we have

$\zeta^{-1}\in\mathcal{B}^{-1+\frac{1}{p}}_{p}(\mathbb{R})\,\,\,\forall
p\in(1,\infty]\quad\textrm{and}\quad\zeta^{\alpha}_{+},\zeta^{\alpha}_{-}\in\mathcal{B}^{\alpha+\frac{1}{p}}_{p}(\mathbb{R})\,\,\,\forall
p\in(|\alpha|^{-1},\infty].$

###### Proof.

By homogeneity, for every $j\geqslant 0$,
$\Delta_{j}\zeta^{-1}(x)=2^{j}\Delta_{0}\zeta^{-1}(2^{j}x)$ so that
$\|\Delta_{j}\zeta^{-1}\|_{L_{p}(\mathbb{R})}=2^{j(1-\frac{1}{p})}\|\Delta_{0}\zeta^{-1}\|_{L_{p}(\mathbb{R})}$.
Note that

$\displaystyle\Delta_{0}\zeta^{-1}(x)=\int_{0}^{\infty}\frac{h(x+y)-h(x-y)}{y}dy$

which is the Hilbert transform of $h$. It is known that Hilbert transform is
bounded on $L_{p}(\mathbb{R})$ for every $p\in(1,\infty)$. Hence, we have
$\|\Delta_{0}\zeta^{-1}\|_{L_{p}(\mathbb{R})}\leqslant
C\|h\|_{L_{p}(\mathbb{R})}$ which is finite. Similarly, we see that
$\|\Delta_{-1}\zeta^{-1}\|_{L_{p}(\mathbb{R})}$ is finite. This shows that
$\zeta^{-1}$ belong to $\mathcal{B}^{-1+\frac{1}{p}}_{p}(\mathbb{R})$ for
every $p\in(1,\infty)$. That $\zeta^{-1}$ belongs to
$\mathcal{B}^{-1}_{\infty}(\mathbb{R})$ follows from the Besov embedding
$\mathcal{B}^{-1+\frac{1}{p}}_{p}(\mathbb{R})\hookrightarrow\mathcal{B}^{-1}_{\infty}(\mathbb{R})$.

Similarly, we see that
$\|\Delta_{j}\zeta^{\alpha}_{\pm}\|_{L_{p}(\mathbb{R})}=2^{-j(\alpha+\frac{1}{p})}\|\Delta_{0}\zeta^{\alpha}_{\pm}\|_{L_{p}(\mathbb{R})}$.
Let $q\in[1,\infty]$ be such that $1+\frac{1}{p}=\gamma+\frac{1}{q}$. Applying
the refined Young convolution inequality ([BCD11, Theorem 1.5]), we see that
$\|\Delta_{0}\zeta^{\alpha}_{\pm}\|_{L_{p}(\mathbb{R})}\leqslant
C\|h\|_{L_{q}(\mathbb{R})}$, which is finite. That
$\|\Delta_{-1}\zeta^{\alpha}_{\pm}\|_{L_{p}(\mathbb{R})}$ is finite follows
from the same argument. ∎

######  Lemma A.5.

Let $f$ and $c,c_{0}$ be as in Theorem 2.12(i). For each $\lambda>0$, define
$f_{\lambda}(x)=\lambda f(\lambda x)$. Then $f_{\lambda}$ converges in
$\mathcal{B}^{(-1+\frac{1}{p})_{-}}_{p}(\mathbb{R})$ to the distribution
$c\zeta^{-1}+c_{0}\delta_{0}$ for every $p\in(1,\infty]$.

###### Proof.

By Besov embedding, it suffices to obtain the convergence in
$\mathcal{B}^{(-1+\frac{1}{p})_{-}}_{p}(\mathbb{R})$ for finite $p$. Fix
$p\in(1,\infty)$, $\beta>0$. Define a distribution

$\chi:=c\zeta^{-1}+c_{0}\delta_{0}$

and put $R(x):=f(x)-c/x$. Fix arbitrary $\varepsilon>0$. Choose $N>0$ large
enough so that

$\int_{|x|>N}|R(x)|\,dx<\varepsilon\quad\textrm{and}\quad\Bigl{|}c_{0}-\int_{|x|<N}f(x)\,dx\Bigr{|}<\varepsilon.$
(A.9)

This is possible thanks to the assumptions of the theorem.

For each $j\geqslant-1$, we write

$\displaystyle\Delta_{j}(f_{\lambda}-\chi)(y)$
$\displaystyle=\int_{\mathbb{R}}f_{\lambda}(x)h_{j}(y-x)dx-c\int_{\mathbb{R}_{+}}\frac{h_{j}(y-x)-h_{j}(y+x)}{x}\,dx-
c_{0}h_{j}(y)$ $\displaystyle=\lambda\int_{|x|>N/\lambda}R(\lambda
x)h_{j}(y-x)\,dx-c\int_{0}^{N/\lambda}\frac{h_{j}(y-x)-h_{j}(y+x)}{x}\,dx$
$\displaystyle\phantom{=}+\lambda\int_{|x|<N/\lambda}f(\lambda
x)(h_{j}(y-x)-h_{j}(y))\,dx-\Bigl{(}c_{0}-\int_{|x|<N}f(x)\,dx\Bigr{)}h_{j}(y)$
$\displaystyle=:I_{1,j}(y)+I_{2,j}(y)+I_{3,j}(y)+I_{4,j}(y).$ (A.10)

We begin with $I_{1,j}$. Applying the integral Minkowski inequality, the
change of variables $x^{\prime}=\lambda x$, and recalling (A.9), we deduce

$\displaystyle\|I_{1,j}\|_{L_{p}(\mathbb{R})}$
$\displaystyle\leqslant\lambda\|h_{j}\|_{L_{p}(\mathbb{R})}\int_{|x|>N/\lambda}|R(\lambda
x)|\,dx=\|h_{j}\|_{L_{p}(\mathbb{R})}\int_{|x|>N}|R(x)|\,dx$
$\displaystyle\leqslant
2^{j(1-\frac{1}{p})}(\|h\|_{L_{p}(\mathbb{R})}+\|h_{-1}\|_{L_{p}(\mathbb{R})})\varepsilon.$
(A.11)

To bound $I_{2,j}$, we note that for $j\geqslant 1$

$\|I_{2,j}\|_{L_{p}(\mathbb{R})}^{p}=2^{j(p-1)}\int_{\mathbb{R}}\Bigl{(}\int_{0}^{2^{j}N/\lambda}\frac{h(y-x)-h(y+x)}{x}\,dx\Bigr{)}^{p}\,dy.$
(A.12)

Introduce the following notation. For fixed $\delta>0$ define the truncated
Hilbert, Hilbert, and maximal Hilbert operators $H_{\delta}$, $H$, and
$H^{*}$, correspondingly, by

$\displaystyle
H_{\delta}(\varphi)(y)=\int_{\delta}^{+\infty}\frac{\varphi(y-x)-\varphi(y-x)}{x}dx,\quad
H(\varphi)=H_{0}(\varphi),$ $\displaystyle
H^{*}(\varphi)(y)=\sup_{\delta>0}|H_{\delta}(\varphi)(y)|,$

for each Schwarz function $\varphi$. With this notation in hand, we continue
(A.12) in the following way

$\|I_{2,j}\|_{L_{p}(\mathbb{R})}=2^{j(1-1/p)}\|H_{0}h-H_{2^{j}N/\lambda}h\|_{L_{p}(\mathbb{R})}.$
(A.13)

Note also that the following bound holds:

$\displaystyle\|I_{2,j}\|_{L_{p}(\mathbb{R})}$
$\displaystyle=2^{j(1-1/p)}(\|H_{0}h\|_{L_{p}(\mathbb{R})}+\|H_{2^{j}N/\lambda}h\|_{L_{p}(\mathbb{R})})$
$\displaystyle\leqslant
2^{j(1-1/p)}(\|H_{0}h\|_{L_{p}(\mathbb{R})}+\|H^{*}(h)\|_{L_{p}(\mathbb{R})}).$
(A.14)

Similarly,

$\|I_{2,-1}\|_{L_{p}(\mathbb{R})}=\|H_{0}h_{-1}-H_{N/\lambda}h_{-1}\|_{L_{p}(\mathbb{R})}.$
(A.15)

To treat $I_{3,j}$ we apply the integral Minkowski inequality and (A.7) with
$f=\delta_{0}$. We get for any $\beta>0$

$\displaystyle\|I_{3,j}\|_{L_{p}(\mathbb{R})}$
$\displaystyle\leqslant\lambda\int_{|x|\leqslant N/\lambda}|f(\lambda
x)|\|h_{j}(\cdot-x)-h_{j}(\cdot)\|_{L_{p}(\mathbb{R})}dx$
$\displaystyle\leqslant\lambda\int_{|x|\leqslant N/\lambda}|f(\lambda
x)||x|^{\beta}2^{j\beta}\|h_{j}\|_{L_{p}(\mathbb{R})}\,dx$
$\displaystyle=\lambda^{-\beta}2^{j\beta}\|h_{j}\|_{L_{p}(\mathbb{R})}\int_{|x|\leqslant
N}|f(x)||x|^{\beta}\,dx$
$\displaystyle\leqslant\lambda^{-\beta}2^{j(\beta+1-1/p)}(\|h\|_{L_{p}(\mathbb{R})}+\|h_{-1}\|_{L_{p}(\mathbb{R})})C_{N},$
(A.16)

where $C_{N}:=N^{\beta}\int_{|x|\leqslant N}|f(x)|\,dx$. Finally, it is easy
to treat $I_{4,j}$. Using (A.9), we get

$\displaystyle\|I_{4,j}\|_{L_{p}(\mathbb{R})}$ $\displaystyle\leqslant
2^{j(1-1/p)}(\|h\|_{L_{p}(\mathbb{R})}+\|h_{-1}\|_{L_{p}(\mathbb{R})})\varepsilon.$

Now we substitute this, (A) and (A) into (A). To bound $I_{2,j}$ we use (A.15)
for $j=-1$, (A.13) for $j\in[1,J]$, and (A) for $j\geqslant J$; here $J$ is a
parameter to be fixed later. We finally obtain

$\displaystyle\sup_{j\geqslant-1}2^{j(-1+\frac{1}{p}-\beta)}\|\Delta_{j}(f_{\lambda}-\chi)\|_{L_{p}(\mathbb{R})}$
$\displaystyle\qquad\leqslant
C(\|h\|_{L_{p}(\mathbb{R})}+\|h_{-1}\|_{L_{p}(\mathbb{R})})(\varepsilon+\lambda^{-\beta}C_{N})+\|H_{0}h_{-1}-H_{N/\lambda}h_{-1}\|_{L_{p}(\mathbb{R})}$
$\displaystyle\qquad+\phantom{\leqslant}\sup_{\delta\leqslant
2^{J}N/\lambda}\|H_{0}h-H_{\delta}h\|_{L_{p}(\mathbb{R})}+2^{-J\beta}(\|H_{0}h\|_{L_{p}(\mathbb{R})}+\|H^{*}(h)\|_{L_{p}(\mathbb{R})}).$

Sending first $\lambda\to\infty$, then $J\to\infty$, then
$\varepsilon\downarrow 0$, we see that the left-hand side of the above
inequality converges to $0$. Here we used the facts that
$\|H_{0}h\|_{L_{p}(\mathbb{R})}<\infty$,
$\|H^{*}(h)\|_{L_{p}(\mathbb{R})}<\infty$,
$\|H_{0}h-H_{\delta}h\|_{L_{p}(\mathbb{R})}\to 0$,
$\|H_{0}h_{-1}-H_{\delta}h_{-1}\|_{L_{p}(\mathbb{R})}\to 0$ as $\delta\to 0$;
all of them are established in [Gra14, Theorem 5.1.12].

Hence, $\lim_{\lambda\to\infty}f_{\lambda}=c\zeta^{-1}+c_{0}\delta_{0}$ in
$\mathcal{B}^{-1+\frac{1}{p}-\beta}_{p}$. Since $\beta$ is arbitrary in
$(0,1)$, this implies the convergence in
$\mathcal{B}^{(-1+\frac{1}{p})_{-}}_{p}$. ∎

######  Lemma A.6.

Let $f$ be a continuous function $\mathbb{R}\to\mathbb{R}$. Suppose that for
some $\alpha\in(0,1)$, $c_{+},c_{-}\in\mathbb{R}$ one has

$\lim_{x\to+\infty}f(x)x^{\alpha}=c_{+}\quad\textrm{and}\quad\lim_{x\to-\infty}f(x)|x|^{\alpha}=c_{-}.$
(A.17)

Let $f_{\lambda}(x):=\lambda^{\alpha}f(\lambda x)$, where $\lambda>0$,
$x\in\mathbb{R}$. Then $f_{\lambda}$ converges to
$c_{-}\zeta^{-\alpha}_{-}+c_{+}\zeta^{-\alpha}_{+}$ as $\lambda\to\infty$ in
$\mathcal{B}^{(-\alpha+\frac{1}{p})-}_{p}(\mathbb{R})$ for every
$p\in(\alpha^{-1},\infty]$.

###### Proof.

The proof is in the same spirit as the proof of Lemma A.5. Again, by the Besov
embedding it is sufficient to consider the case of finite $p$. Fix
$p\in(\alpha^{-1},\infty)$. Denote

$\chi(x):=c_{-}\zeta^{-\alpha}_{-}+c_{+}\zeta^{-\alpha}_{+},\quad
x\in\mathbb{R}$

and put $R(x):=f(x)-\chi(x)$.

Take arbitrary $\varepsilon>0$. It follows from (A.17), that there exists
$N>0$ such that

$|R(x)||x|^{\alpha}<\varepsilon\text{ whenever }|x|>N.$ (A.18)

For each $j\geqslant-1$, we write

$\displaystyle\Delta_{j}(f_{\lambda}-\chi)(y)$
$\displaystyle=\int_{\mathbb{R}}(f_{\lambda}(x)-\chi(x))h_{j}(y-x)\,dx$
$\displaystyle=\lambda^{\alpha}\int_{\mathbb{R}}R(\lambda x)h_{j}(y-x)\,dx$
$\displaystyle=\lambda^{\alpha}\Bigl{(}\int_{|x|<N/\lambda}+\int_{|x|>N/\lambda}\Bigr{)}R(\lambda
x)h_{j}(y-x)dx$ $\displaystyle=:I_{1}(y)+I_{2}(y).$ (A.19)

Fix $\beta\in(0,1-\alpha)$. To bound $I_{1}$ we apply Young inequality with
$p^{\prime},q^{\prime}>1$ such that $1/p^{\prime}+1/q^{\prime}=1+1/p$ and the
change of variables $x^{\prime}:=\lambda x$. We get

$\displaystyle\|I_{1}\|_{L_{p}(\mathbb{R})}$
$\displaystyle\leqslant\lambda^{\alpha}\|h_{j}\|_{L_{p^{\prime}}(\mathbb{R})}\Bigl{(}\int_{|x|<N/\lambda}|R(\lambda
x)|^{q^{\prime}}\,dx\Bigr{)}^{1/q^{\prime}}$
$\displaystyle=\lambda^{-\frac{1}{q^{\prime}}+\alpha}\|h_{j}\|_{L_{p^{\prime}}(\mathbb{R})}\Bigl{(}\int_{|x|<N}|R(x)|^{q^{\prime}}\,dx\Bigr{)}^{1/q^{\prime}}$
$\displaystyle\leqslant
C_{N}\lambda^{-\frac{1}{q^{\prime}}+\alpha}2^{j(1-\frac{1}{p^{\prime}})}(\|h\|_{L_{p^{\prime}}(\mathbb{R})}+\|h_{-1}\|_{L_{p^{\prime}}(\mathbb{R})})$

where $C_{N}:=CN(1+\sup_{|x|\leqslant N}|f(x)|)$. By choosing now
$1/q^{\prime}=\alpha+\beta<1$, we get $1/p^{\prime}=1+1/p-\alpha-\beta<1$ and
thus

$\|I_{1}\|_{L_{p}(\mathbb{R})}\leqslant
C_{N}\lambda^{-\beta}2^{j(\alpha+\beta-1/p)}(\|h\|_{L_{p^{\prime}}(\mathbb{R})}+\|h_{-1}\|_{L_{p^{\prime}}(\mathbb{R})})$
(A.20)

To deal with $I_{2}$ we apply (A.18) to deduce

$|I_{2}(y)|\leqslant\varepsilon\int_{|x|>N/\lambda}|x|^{-\alpha}|h_{j}(y-x)|\,dx\leqslant\varepsilon\int_{\mathbb{R}}|x|^{-\alpha}|h_{j}(y-x)|\,dx.$

Thus, the Hardy–Littlewood–Sobolev inequality on $\mathbb{R}$ with
$1/p^{\prime\prime}=1+1/p-\alpha$ [BCD11, Theorem 1.7] implies

$\|I_{2}\|_{L_{p}(\mathbb{R})}\leqslant\varepsilon\|h_{j}\|_{L_{p^{\prime\prime}}(\mathbb{R})}\leqslant\varepsilon
2^{j(\alpha-1/p)}(\|h\|_{L_{p^{\prime\prime}}(\mathbb{R})}+\|h_{-1}\|_{L_{p^{\prime\prime}}(\mathbb{R})}).$

Combining this with (A) and (A.20), we finally get

$\|f_{\lambda}-\chi\|_{\mathcal{B}^{-\alpha-\beta+1/p}_{p}}\leqslant
C_{N}\lambda^{-\beta}(\|h\|_{L_{p^{\prime}}(\mathbb{R})}+\|h_{-1}\|_{L_{p^{\prime}}(\mathbb{R})})+\varepsilon(\|h\|_{L_{p^{\prime\prime}}(\mathbb{R})}+\|h_{-1}\|_{L_{p^{\prime\prime}}(\mathbb{R})}).$

Taking now first $\lambda\to\infty$, and recalling then that $\varepsilon$ was
arbitrary completes the proof. ∎

##  Appendix B Other auxiliary results

######  Proposition B.1.

Let $\Lambda$ be a set and let
$(X_{n,\lambda})_{n\in\mathbb{Z}_{+},\lambda\in\Lambda}$ be a collection of
random elements taking values in a metric space $E$. Let
$(Y_{n})_{n\in\mathbb{Z}_{+}}$ be a collection of random elements taking
values in a metric space $\widetilde{E}$. Suppose that for each fixed $n$ the
random element $Y_{n}$ is independent of
$(X_{n,\lambda})_{\lambda\in\Lambda}$. Furthermore, assume that for each fixed
$\lambda\in\Lambda$ one has $X_{n,\lambda}\to X_{\lambda}$ and $Y_{n}\to Y$ in
probability as $n\to\infty$. Then $Y$ is independent of
$(X_{\lambda})_{\lambda\in\Lambda}$.

###### Proof.

Consider a collection of $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ for some
$n\geq 1$. Then we can construct a common subsequence such that

$X_{n_{k},\lambda_{j}}\to X_{\lambda_{j}}\mbox{ and }Y_{n_{k}}\to Y\mbox{
almost surely as $k\to\infty$ for all $1\leq j\leq n$.}$

Let $h_{1},h_{2}\ldots h_{n}:E\rightarrow\mathbb{R}$,
$g:\tilde{E}\rightarrow\mathbb{R}$ be bounded continuous functions. Then by
the Lebesgue dominated convergence theorem,

$\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt(\prod_{j=1}^{n}h_{j}(X_{n_{k},\lambda_{j}})g(Y_{n_{k}}))\to\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt(\prod_{j=1}^{n}h_{j}(X_{\lambda_{j}})g(Y))\mbox{ as $k\to\infty$ }$

and also

$\prod_{j=1}^{n}\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt(h_{j}(X_{n_{k},\lambda_{j}}))\hskip 0.6458pt\mathsf{E}\hskip
0.43057ptg(Y_{n_{k}})\to\prod_{j=1}^{n}\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt(h_{j}(X_{\lambda_{j}}))\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt(g(Y))\mbox{ as $k\to\infty$ }.$

We have assumed for each $k$ the random element $Y_{n_{k}}$ is independent of
$(X_{n_{k},\lambda_{j}})_{1\leq j\leq n}$. Therefore from the above it is
immediate that

$\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt\prod_{j=1}^{n}(h_{j}(X_{\lambda_{j}}))g(Y))=\prod_{j=1}^{n}\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt(h_{j}(X_{\lambda_{j}}))\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt(g(Y))$

As, $n\geq 1$, $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ and
$h_{1},h_{2}\ldots h_{n},g$ were arbitrary, the result follows. ∎

######  Lemma B.2 (Gaussian process representation).

Let $(\widetilde{\Omega},\widetilde{\mathcal{F}},\widetilde{\mathsf{P}})$ be a
filtered probability space. Let
$\widetilde{V}\colon[0,T_{0}]\times{D}\times\widetilde{\Omega}\to\mathbb{R}$
be a measurable function with the same law as $V$. Then on the same space
there exists a white noise $\widetilde{W}$ such that identity (2.4) holds with
$\widetilde{V}$ in place of $V$ and $\widetilde{W}$ in place of $W$.
Furthermore,

  1. (i)

$\mathcal{F}_{t}^{\widetilde{W}}=\mathcal{F}_{t}^{\widetilde{V}}$ for any
$t\in[0,T_{0}]$;

  2. (ii)

suppose additionally that there exists a filtration
$(\widetilde{\mathcal{F}}_{t})_{t\in[0,T_{0}]}$ such that
$\mathcal{F}_{t}^{\widetilde{V}}\subset\widetilde{\mathcal{F}}_{t}$ and for
any $(s,t)\in\Delta_{0,T_{0}}$, $\varphi\in\mathcal{C}_{c}^{\infty}$ the
random variable
$\int_{{D}}(\widetilde{V}_{t}(x)-P_{t-s}\widetilde{V}_{s}(x))\varphi(x)\,dx$
is independent of $\widetilde{\mathcal{F}}_{s}$. Then $\widetilde{W}$ is
$(\widetilde{\mathcal{F}}_{t})$-white noise.

###### Proof.

The result is probably well-known. However, we give a proof for the sake of
completeness. In what follows we will use the following notation:

$\langle f,g\rangle:=\int_{D}f(y)g(y)\,dy,$

for measurable functions $f,g\colon{D}\to\mathbb{R}$ for which the above
integral is well-defined. It is well-known that (2.4) is equivalent (see,
e.g., [Shi94, Theorem 2.1]) to representing $V$ as a solution to the additive
stochastic heat equation in a distributional form:

$\langle V_{t},\varphi\rangle=\frac{1}{2}\int_{0}^{t}\left\langle
V_{s},\partial^{2}_{yy}\varphi\right\rangle\,ds+W_{t}(\varphi),\quad\text{for
any $t\geq 0$, $\varphi\in\mathcal{C}_{c}^{\infty}$}.$ (B.1)

Since $\widetilde{V}$ has the same law as $V$, we immediately get that the
functional

$\widetilde{W}_{t}(\varphi):=\langle\widetilde{V}_{t},\varphi\rangle-\frac{1}{2}\int_{0}^{t}\left\langle\widetilde{V}_{s},\partial^{2}_{yy}\varphi\right\rangle\,ds,\quad
t\geq 0,\varphi\in\mathcal{C}_{c}^{\infty}$ (B.2)

has the same distributional properties as $W$. That is, for any
$\varphi\in\mathcal{C}_{c}^{\infty}$, the process
$(\widetilde{W}_{t}(\varphi))_{t\in[0,T_{0}]}$ is an
$(\mathcal{F}^{\widetilde{V}}_{t})$–Brownian motion with $\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt\widetilde{W}_{1}(\varphi)^{2}=\|\varphi\|^{2}_{L_{2}({D},dx)}$ and
clearly this also holds for any $\varphi\in{L_{2}}({D},dx)$ since
$\mathcal{C}_{c}^{\infty}$ is dense in $L^{2}({D},dx)$. Also
$\widetilde{W}_{t}(\varphi)$ and $\widetilde{W}_{t}(\psi)$ are independent
whenever $\varphi,\psi\in\mathcal{C}_{c}^{\infty}$ with
$\int_{D}\varphi(x)\psi(x)\,dx=0$, and again since $\mathcal{C}_{c}^{\infty}$
is dense in $L^{2}({D},dx)$, this holds for any
$\varphi,\psi\in{L_{2}}({D},dx)$ with $\int_{D}\varphi(x)\psi(x)\,dx=0$. This
immediately implies that $\widetilde{W}$ is an
$(\mathcal{F}^{\widetilde{V}}_{t})$-white noise. Thus, $\widetilde{V}$ is a
solution to

$\langle\widetilde{V}_{t},\varphi\rangle=\frac{1}{2}\int_{0}^{t}\left\langle\widetilde{V}_{s},\partial^{2}_{yy}\varphi\right\rangle\,ds+\widetilde{W}_{t}(\varphi),\quad\text{for
any $t\geq 0$, $\varphi\in\mathcal{C}_{c}^{\infty}$},$ (B.3)

with white noise $\widetilde{W}$. Using again [Shi94, Theorem 2.1], we get
that (2.4) holds with $\widetilde{V}$ in place of $V$ and $\widetilde{W}$ in
place of $W$.

To prove (i) note that

$\widetilde{V}_{t}(x)=\int_{0}^{t}\int_{{D}}p_{t-r}(x,y)\widetilde{W}(dr,dy),\quad
t\geqslant 0,\,x\in{D},$ (B.4)

and thus for any $t\in[0,T_{0}]$,
$\mathcal{F}_{t}^{\widetilde{V}}\subset\mathcal{F}_{t}^{\widetilde{W}}$. On
the other hand, from (B.2) we also immediately get that
$\mathcal{F}_{t}^{\widetilde{W}}\subset\mathcal{F}_{t}^{\widetilde{V}}$ for
any $t\in[0,T_{0}],$ and (i) follows.

Let us prove part (ii) of the proposition. We need just to show that for any
$(s,t)\in\Delta_{0,T_{0}}$, $\varphi\in L^{2}({D},dx)$

$\widetilde{W}_{t}(\varphi)-\widetilde{W}_{s}(\varphi)\;\text{is independent
of}\;\widetilde{\mathcal{F}}_{s}\,.$

Fix arbitrary $(s,t)\in\Delta_{0,T_{0}}$ and
$\varphi\in\mathcal{C}_{c}^{\infty}$. By (2.4) and stochastic Fubini theorem
we get that

$\int_{{D}}(\widetilde{V}_{t}(x)-P_{t-s}\widetilde{V}_{s}(x))\varphi(x)\,dx=\int_{s}^{t}\int_{D}P_{t-r}\varphi(y)\widetilde{W}(dr,dy).$

By our assumptions, we get that the above stochastic integral is independent
of $\widetilde{\mathcal{F}}_{s}$. Clearly, since
$\widetilde{\mathcal{F}}_{s}\subset\widetilde{\mathcal{F}}_{r}$ for any
$s\leqslant r$, we get that for any $(s_{1},s_{2})\in\Delta_{s,t}$,

$Y_{s_{1},s_{2}}:=\int_{s_{1}}^{s_{2}}\int_{D}P_{s_{2}-r}\varphi(y)\widetilde{W}(dr,dy)\;\text{is
independent of $\widetilde{\mathcal{F}}_{s}$.}$ (B.5)

For any $n\geq 1$ define $s^{n}_{k}:=s+\frac{(t-s)k}{n}$ for $k=0,\ldots,n$
and a function $f^{n}$ on $[s,t]\times{D}$ such that

$f^{n}(r,y)=\sum_{k=1}^{n}\operatorname{\mathbbm{1}}_{s^{n}_{k-1}\leq
r<s^{n}_{k}}P_{s^{n}_{k}-r}\varphi(y),\;s\leq r\leq t,y\in{D}.$

Now we are ready to define the sequence of random variables

$Y^{n}:=\sum_{k=1}^{n}Y_{s^{n}_{k-1},s^{n}_{k}}=\int_{s}^{t}\int_{D}f^{n}(r,y)\widetilde{W}(dr,dy).$

It is trivial to check that

$f^{n}\rightarrow\varphi,\;\text{in}\;L^{2}([s,t]\times{D}),\;\text{as
$n\rightarrow\infty$},$

and hence

$Y^{n}\rightarrow\int_{s}^{t}\int_{D}\varphi(y)\widetilde{W}(dr,dy)=\widetilde{W}_{t}(\varphi)-\widetilde{W}_{s}(\varphi),\;\text{as
$n\rightarrow\infty$},$

where convergence is in $L^{2}(\Omega)$. By (B.5) and properties of the white
noise we get that $Y^{n}$ is independent of $\widetilde{\mathcal{F}}_{s}$ for
all $n$, and hence the limit in $L^{2}(\Omega)$ of this sequence, is also
independent of $\widetilde{\mathcal{F}}_{s}$. Thus we get that
$\widetilde{W}_{t}(\varphi)-\widetilde{W}_{s}(\varphi)$ is independent of
$\widetilde{\mathcal{F}}_{s}$, for any $\varphi\in\mathcal{C}_{c}^{\infty}.$
We can easily get the same property for any $\varphi\in L^{2}({D},dx)$, by
approximating such $\varphi$ in $L^{2}({D},dx)$ by a sequence of functions
$\varphi^{n}\in\mathcal{C}_{c}^{\infty}$ and again passing to the limit of
corresponding sequence of random variables
$\widetilde{W}_{t}(\varphi_{n})-\widetilde{W}_{s}(\varphi_{n})$. ∎

######  Lemma B.3.

For every $\alpha,\beta\in[0,1)$, $(s,t)\in\Delta_{S,T}$ and $u=(s+t)/2$, we
have

$(u-S)^{-\alpha}(T-u)^{-\beta}(t-s)\leqslant
2^{1+\alpha}\int_{s}^{t}(r-S)^{-\alpha}(T-r)^{-\beta}dr$ (B.6)

and

$\int_{s}^{t}(T-r)^{-\beta}(r-s)^{-\alpha}\,dr\leqslant
C(T-s)^{-\beta}(t-s)^{1-\alpha}.$ (B.7)

###### Proof.

We have

$\displaystyle\int_{s}^{t}(r-S)^{-\alpha}(T-r)^{-\beta}dr\geqslant\int_{u}^{t}(r-S)^{-\alpha}(T-r)^{-\beta}dr\geqslant(t-S)^{-\alpha}(T-u)^{-\beta}(t-u).$

Since $u=(s+t)/2$, $t-S\leqslant 2(u-S)$, we obtain (B.6) from the above
inequalities.

For fixed $s\leqslant t$, the function
$q(T):=\int_{s}^{t}\left(\frac{T-s}{T-r}\right)^{\beta}(r-s)^{-\alpha}dr$ is
decreasing on $T\in[t,\infty)$. Hence we have $q(T)\leqslant
q(t)=\left(\int_{0}^{1}(1-r)^{-\beta}r^{-\alpha}dr\right)(t-s)^{1-\alpha}$.
This yields (B.7). ∎

######  Lemma B.4.

Let $f:\Omega\times{D}\to\mathbb{R}$ be a measurable function. Then for any
$s,t\geqslant 0$, $P_{t}f:\Omega\times{D}\to\mathbb{R}$ is measurable. Further
for any $x\in{D}$, $m\in[1,\infty]$, $n\in[1,\infty]$, and a $\sigma$-algebra
$\mathscr{G}$ we have

$\displaystyle\|\|P_{t}f(x)\|_{{L_{m}}|\mathscr{G}}\|_{{L_{n}}}\leqslant\sup_{y\in{D}}\|\,\|f(y)\|_{{L_{m}}|\mathscr{G}}\|_{{L_{n}}}\quad\textrm{and}\quad\|P_{t}f(x)\|_{{L_{m}}}\leqslant\sup_{y\in{D}}\|f(y)\|_{{L_{m}}}.$
(B.8)

In addition, if $f\in{\mathbf{B}L_{m}}$, there exists a set
$\Omega^{\prime}\subset\Omega$ of full measure such that for any
$\omega\in\Omega^{\prime}$

$P_{t}f(x,\omega)<\infty\quad\text{for Lebesgue almost every
$(t,x)\in[0,T_{0}]\times{D}$}.$ (B.9)

###### Proof.

It suffices to show the result assuming that $f$ is non-negative. In such
case, it is evident that $P_{t}f:\Omega\times\mathbb{R}\to\mathbb{R}$ is
measurable. Applying the conditional integral Minkowski inequality, we obtain
that

$\|P_{t}f(x)\|_{{L_{m}}|\mathscr{G}}\leqslant\int_{D}p_{t}(x,y)\|f(y)\|_{{L_{m}}|\mathscr{G}}\,dy.$

We then apply Minkowski inequality, to get

$\|\|P_{t}f(x)\|_{{L_{m}}|\mathscr{G}}\|_{L_{n}}\leqslant\int_{D}p_{t}(x,y)\|\|f(y)\|_{{L_{m}}|\mathscr{G}}\|_{L_{n}}\leqslant\sup_{y\in{D}}\|\|f(y)\|_{{L_{m}}|\mathscr{G}}\|_{L_{n}}\,dy$

which implies the former estimate in (B.8). The later one in (B.8) follows
from the former by setting $n=m$.

Finally, (B.9) follows from (B.8) and the Fubini theorem. ∎

######  Lemma B.5.

Let $(\varphi^{n})_{n\in\mathbb{Z}_{+}}$ be a sequence of continuous random
fields on ${D}$ such that $\varphi^{n}$ converges to $0$ in
$\mathcal{C}_{\textrm{uc}}({D})$ in probability and that
$\sup_{n\in\mathbb{Z}_{+},x\in{D}}\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt|\varphi^{n}(x)|<\infty$. Then for every
$(t,x)\in[0,T_{0}]\times{D}$, $P_{t}\varphi^{n}(x)$ converges to $0$ in
probability as $n\to\infty$.

###### Proof.

For every $M>0$, we have

$\displaystyle|P_{t}\varphi^{n}(x)|$ $\displaystyle\leqslant\int_{|y|\leqslant
M}p_{t}(x,y)|\varphi^{n}(y)|dy+\int_{|y|>M}p_{t}(x,y)|\varphi^{n}(y)|dy$
$\displaystyle\leqslant\sup_{|y|\leqslant
M}|\varphi^{n}(y)|+\int_{|y|>M}p_{t}(x,y)|\varphi^{n}(y)|dy.$

Hence, for every $\varepsilon>0$, we have by Chebyshev inequality,

$\displaystyle\mathsf{P}\left(|P_{t}\varphi^{n}(x)|>2\varepsilon\right)\leqslant\mathsf{P}(\sup_{|y|\leqslant
M}|\varphi^{n}(y)|>\varepsilon)+\varepsilon^{-1}\sup_{n,z}\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt|\varphi^{n}(z)|\left(\int_{|y|>M}p_{t}(x,y)dy\right).$

We send $n\to\infty$ then $M\to\infty$ to see that
$\lim_{n\to\infty}\mathsf{P}\left(|P_{t}\varphi^{n}(x)|>2\varepsilon\right)=0$.
This implies the result. ∎

##  Appendix C Heat kernel estimates

In this section we provide a number of standard simple statements
characterizing smoothing properties of the heat kernel. Recall that $G$ and
$P$ are the heat semigroups on $\mathbb{R}$ and ${D}$ defined in (2.1) and
2.1, correspondingly.

######  Lemma C.1.

For $(D,p)\in\\{(\mathbb{R},g),({[0,1]},p^{per}),({[0,1]},p^{Neu})\\}$, any
$t\in[0,T_{0}]$, $x\in{D}$ we have

$\frac{1}{\sqrt{2\pi}}t^{-1/2}\leqslant p_{t}(x,x)\leqslant
C(T_{0})+\frac{2}{\sqrt{2\pi}}t^{-1/2}.$ (C.1)

###### Proof.

Since $g_{t}(0)=\frac{1}{\sqrt{2\pi t}}$, and $p_{t}(x,x)\geqslant g_{t}(0)$
for $p\in\\{g,p^{per},p^{Neu}\\}$, the lower bound in (C.1) is trivial. We
apply the elementary inequality $e^{-|z|}\leqslant c|z|^{-1}$ to get

$\displaystyle p^{per}_{t}(x,x)$
$\displaystyle=g_{t}(0)+\sum_{n\in\mathbb{Z}:|n|\geqslant 1}g_{t}(n)\leqslant
g_{t}(0)+c\sum_{n\in\mathbb{Z}:|n|\geqslant 1}t^{\frac{1}{2}}n^{-2}$

which shows the upper bound in (C.1) for $p=p^{per}$. For $p^{Neu}$, using the
estimate $g_{t}(2x+2n)\leqslant g_{t}(2n)$ for $x\in[0,1]$, we have

$\displaystyle p^{Neu}_{t}(x,x)\leqslant
2g_{t}(0)+2\sum_{n\in\mathbb{Z}:|n|\geqslant 1}g_{t}(2n).$

From here, using the same argument as above, we obtain the upper bound in
(C.1) for $p=p^{Neu}$. ∎

######  Lemma C.2.

For every $\alpha\in[0,1]$, there exists a constant $C=C(\alpha,T_{0})>0$ such
that for any $s,t\in(0,T_{0}]$, $s\leqslant t$, $x,x_{1},x_{2}\in{D}$ we have

$\displaystyle\int_{{D}}|p_{t}(x_{1},y)-p_{t}(x_{2},y)|\,dy\leqslant
C|x_{1}-x_{2}|^{\alpha}t^{-\alpha/2},$ (C.2)
$\displaystyle\int_{{D}}p_{t}(x,y)|y-x|^{\alpha}\,dy\leqslant Ct^{\alpha/2},$
(C.3) $\displaystyle\int_{D}|p_{t}(x,y)-p_{s}(x,y)|dy\leqslant
Cs^{-\alpha/2}(t-s)^{\alpha/2}.$ (C.4)

###### Proof.

From the elementary estimate

$|g_{t}(x_{1}-y)-g_{t}(x_{2}-y)|\leqslant
C|x_{1}-x_{2}|^{\alpha}t^{-\alpha/2}(g_{2t}(x_{1}-y)+g_{2t}(x_{2}-y))$

and (2.2) and (2.3), we obtain that

$\displaystyle|p_{t}(x_{1},y)-p_{t}(x_{2},y)|\leqslant
C|x_{1}-x_{2}|^{\alpha}t^{-\alpha/2}(p_{2t}(x_{1},y)+p_{2t}(x_{2},y))$

for $p\in\\{g,p^{per},p^{Neu}\\}$. Integrating over $y\in{D}$ and note that
$\int_{D}p_{2t}(x,y)dy=1$ for each $x\in{D}$, we obtain (C.2) for
$p\in\\{g,p^{per},p^{Neu}\\}$.

The estimate (C.3) for $p_{t}(x,y)=g_{t}(x-y)$ follows easily by a change of
variable. For the other cases, we note that $g_{t}(z+k)\leqslant
cg_{t}(z)e^{-2\frac{|k|}{t}}$ for every $|z|\leqslant 2$ and every
$k\in\mathbb{R}$. Since $\sum_{n\in\mathbb{Z}}e^{-2\frac{|n|}{t}}\leqslant
C(T_{0})$, we see that from (2.2) and (2.3) that

$\displaystyle\int_{D}p_{t}(x,y)|y-x|^{\alpha}dy\leqslant
C(T_{0})\int_{\mathbb{R}}g_{t}(x-y)|x-y|^{\alpha}dy.$

From here, we obtain (C.3) for $p\in\\{p^{per},p^{Neu}\\}$ by a change of
variable.

To show (C.4), we use the estimate

$\displaystyle\int_{D}|p_{t}(x,y)-p_{s}(x,y)|dy\leqslant\int_{D}\int_{D}p_{t-s}(x,z)|p_{s}(z,y)-p_{s}(x,y)|dydz$

and (C.2) to obtain that

$\displaystyle\int_{D}|p_{t}(x,y)-p_{s}(x,y)|dy\leqslant
Cs^{-\alpha/2}\int_{D}p_{t-s}(x,z)|z-x|^{\alpha}dz.$

Taking into account (C.3), we obtain (C.4) from the above inequality. ∎

Recall that $V$ is defined in (2.4). For $t>0$, $x\in{D}$ put

$\rho_{t}(x):=\mathop{\mbox{\rm Var}}(V_{t}(x)).$ (C.5)

######  Lemma C.3.

For any $t\in[0,T_{0}]$, $x\in{D}$ we have

$\frac{\sqrt{t}}{\sqrt{\pi}}\leqslant{\rho_{t}}(x)\leqslant C(T_{0})\sqrt{t}.$
(C.6)

###### Proof.

We have from (C.1)

${\rho_{t}}(x)=\mathop{\mbox{\rm
Var}}(V(t,x))=\int_{0}^{t}p_{2(t-r)}(x,x)\,dr\geqslant\int_{0}^{t}\frac{1}{2\sqrt{\pi(t-r)}}\,dr=\frac{\sqrt{t}}{\sqrt{\pi}}.$

The upper bound on ${\rho_{t}}$ is established in exactly the same way. ∎

######  Lemma C.4.

Let $0\leqslant s\leqslant u\leqslant t\leqslant T_{0}$. Let
$f\colon\mathbb{R}\times\Omega\to\mathbb{R}$ be a bounded
$\mathscr{B}(\mathbb{R})\otimes\mathcal{F}_{s}$-measurable function. Then

$\displaystyle\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{u}f(V_{t}(x))=G_{\rho_{t-u}(x)}f(P_{t-u}V_{u}(x)).$ (C.7)

In addition, there exists a universal constant $C=C(T_{0})>0$ such that for
every $x\in{D}$, $\gamma<0$, $n\in[1,\infty]$ and $p\in[n,\infty]$

$\displaystyle\|\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{u}f(V_{t}(x))\|_{{L_{n}}}\leqslant
C\|\,\|f\|_{\mathcal{B}^{\gamma}_{p}}\|_{{L_{n}}}(t-u)^{\frac{\gamma}{4}}(u-s)^{-\frac{1}{2p}}(t-s)^{\frac{1}{4p}},$
(C.8) $\displaystyle|\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{s}f(V_{t}(x))|\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}(t-s)^{\frac{\gamma}{4}-\frac{1}{4p}}.$ (C.9)

###### Proof.

For $s\leqslant t$ introduce the process

$Z_{s,t}(x):=V_{t}(x)-P_{t-s}V_{s}(x)=\int_{s}^{t}\int_{D}p_{t-r}(x,y)W(dr,dy),\quad
x\in{D},\,0\leqslant s\leqslant t.$

By definition of $Z$, we have for any $0\leqslant s\leqslant u\leqslant t$,
$x\in{D}$

$V_{t}(x)=P_{t-u}V_{u}(x)+Z_{u,t}(x).$ (C.10)

It is immediate to see that $Z_{u,t}(x)$ is independent of $\mathcal{F}_{s}$
and is Gaussian with zero mean and variance

$\mathop{\mbox{\rm
Var}}(Z_{u,t}(x))=\int_{u}^{t}p_{2(t-r)}(x,x)dr=\rho_{t-u}(x).$ (C.11)

Using (C.10), this yields

$\hskip 0.6458pt\mathsf{E}\hskip
0.43057pt^{u}f(V_{t}(x))(\omega)=[G_{\rho_{t-u}(x)}f(\cdot,\omega)](P_{t-u}V_{u}(x)),$

which is (C.7).

Next, we show (C.8). To proceed, we further decompose
$P_{t-u}V_{u}(x)=P_{t-s}V_{s}(x)+P_{t-u}Z_{s,u}(x)$. The random variable
$P_{t-u}Z_{s,u}(x)=\int_{s}^{u}\int_{D}p_{t-r}(x,y)W(dr,dy)$ is independent of
$\mathcal{F}_{s}$ and has a Gaussian law with mean zero and variance

$\displaystyle\rho_{s,u,t}(x)$ $\displaystyle:=\mathop{\mbox{\rm
Var}}[P_{t-u}Z_{s,u}(x)]=\int_{s}^{u}p_{2(t-r)}(x,x)\,dr\geqslant
C(\sqrt{t-s}-\sqrt{t-u})$ $\displaystyle\geqslant C(u-s)(t-s)^{-1/2},$ (C.12)

where the first inequality follows from (C.1). Hence, using (C.7) and the fact
that $f$ is $\mathcal{F}_{s}$-measurable, we have

$\displaystyle\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}|\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt^{u}f(V_{t}(x))|^{n}$ $\displaystyle=\hskip
0.6458pt\mathsf{E}\hskip
0.43057pt^{s}\bigl{[}|G_{\rho_{t-u}(x)}f(P_{t-u}V_{u}(x))|^{n}\bigr{]}$
$\displaystyle=\int_{\mathbb{R}}g_{\rho_{s,u,t}(x)}(z)\left|G_{\rho_{t-u}(x)}f(P_{t-s}V_{s}(x)+z)\right|^{n}\,dz,$
(C.13)

where $g$ is the standard Gaussian density (2.1). Now we put
$q=\frac{p}{n}\geqslant 1$, $\frac{1}{q}+\frac{1}{q^{\prime}}=1$ and apply
Hölder inequality and (C) to estimate the above integral from above by

$\Bigl{[}\int_{\mathbb{R}}\bigl{|}G_{\rho_{t-u}(x)}f(P_{t-s}V_{s}(x)+z)\bigr{|}^{p}\,dz\Bigr{]}^{n/p}\,\,\|g_{\rho_{s,u,t}(x)}(\cdot)\|_{L^{q^{\prime}}},$
(C.14)

The first factor in the above expression equals
$\|G_{\rho_{t-u}(x)}f(\cdot,\omega)\|^{n}_{{L_{p}}(\mathbb{R})}$ and thus is
bounded above by

$C\|f(\cdot,\omega)\|_{\mathcal{B}^{\gamma}_{p}}^{n}\rho_{t-u}(x)^{\frac{\gamma
n}{2}}$

by Lemma A.3(i). The second factor in (C.14) is bounded by
$C\rho_{s,u,t}(x)^{-\frac{n}{2p}}$ for some constant $C>0$, where we again
used Lemma A.3(i) for Dirac delta. Taking into account (C.6), the lower bound
for $\rho_{s,u,t}(x)$ in (C), we continue (C) as follows:

$\displaystyle\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}|\hskip
0.6458pt\mathsf{E}\hskip 0.43057pt^{u}f(V_{t}(x))|^{n}$
$\displaystyle\leqslant
C\|f(\cdot,\omega)\|_{\mathcal{B}^{\gamma}_{p}}^{n}\rho_{t-u}(x)^{\frac{\gamma
n}{2}}\rho_{s,u,t}(x)^{-\frac{n}{2p}}$ $\displaystyle\leqslant
C(T_{0})\|f(\cdot,\omega)\|_{\mathcal{B}^{\gamma}_{p}}^{n}(t-u)^{\frac{\gamma
n}{4}}(u-s)^{-\frac{n}{2p}}(t-s)^{\frac{n}{4p}}.$

Taking expectation, we get (C.8).

To show (C.9), we use (C.7) with $u=s$. Applying Lemma A.3(i) and the Besov
embedding
$\mathcal{B}^{\gamma}_{p}\hookrightarrow\mathcal{B}^{\gamma-1/p}_{\infty}$
(see (3.6)), we deduce

$|\hskip 0.6458pt\mathsf{E}\hskip 0.43057pt^{s}f(V_{t}(x))|\leqslant
C\|f\|_{\mathcal{B}^{\gamma-1/p}_{\infty}}\rho_{t-s}(x)^{\frac{\gamma}{2}-\frac{1}{2p}}\leqslant
C\|f\|_{\mathcal{B}^{\gamma}_{p}}(t-s)^{\frac{\gamma}{4}-\frac{1}{4p}},$

where we also used (C.6). This implies (C.9). ∎

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