\end{align*}

Finally, our lower bound of interest, i.e. the right-hand side of (<ref>), is given by

\begin{align*}
&\sqrt{\frac{2}{2\beta+1} \log T} \left(\hspace{-0.05cm}\left( 1 + \frac{2\beta+1}{2\beta} \epsilon_T + \mathcal{O}(\epsilon_T^2)\right)\hspace{-0.05cm}\left( 1- \frac{3}{\sqrt{2}}\frac{a_T^{b_T}(y_T)}{ q_{b_T}(y_T)} \right) -1 \right) \\
&\hspace{1.5cm} + o_{\Pr_b,\textrm{unif}}(1) \\
&\hspace{1cm} = \sqrt{\frac{2}{2\beta+1} \log T} \left( \frac{2\beta+1}{2\beta}\epsilon_T ( 1+ \mathcal{O}(\epsilon_T)) \right. \\
&\hspace{3cm}\left. - \frac{3}{\sqrt{2}}\frac{a_T^{b_T}(y_T)}{q_{b_T}(y_T)} \left( 1+ \frac{2\beta +1}{2\beta} \epsilon_T + \mathcal{O}(\epsilon_T^2)\right) \right) + o_{\Pr_b,\textrm{unif}}(1).
\end{align*}

Now, both $1+ \mathcal{O}(\epsilon_T)$ and $1+ \frac{2\beta +1}{2\beta} \epsilon_T + \mathcal{O}(\epsilon_T^2)$ converge to one. Furthermore, by the choice of $\epsilon_T$ we have $\sqrt{\log T}\epsilon_T\rightarrow \infty$ and by Lemma <ref> we have $\sqrt{\log T} a_T^{b_T}(y_T)\rightarrow_{\Pr_b,\textrm{unif}} 0$ uniformly in $y_T$ and $b_T$. Hence, the whole expression diverges to $\infty$ and we are done with the case $\beta\leq 1$.

Now we turn to the case $\beta>1$:

Let $y_T$ maximize $|b_T(y) - b_0(y)|-\eta$ on the compact $J\subset (-A,A)$. As $b_T-b_0$ is bounded away from zero by $\eta$, by Lemma <ref> we find intervals $I_T$ with $\liminf_{T\to\infty}\lambda(I_T)>0$ on which $b_T-b_0$ is bounded away from zero. Thus, for $T$ large enough, $b_T-b_0$ has the same sign on $[y_T - h_T, y_T+h_T]$ and on this interval $|b_T-b_0|-\eta\in\mathcal{H}(\beta, L)$.

First, we assume additionally that

\begin{align}\label{eqp: upper_gamma_T}
\gamma_T:= \gamma_T(b_T) := |b_T(y_T)-b_0(y_T)| - \eta
\end{align}

tends to zero for $T\to\infty$. We have $\gamma_T\geq \tilde\delta_T$ and by Lemma <ref> there exists a constant $c'=c'(\beta,L)>0$ such that $|b-b_0|-\gamma\geq \frac{\gamma_T}{2}\geq\frac{\tilde\delta_T}{2}$ on $[y_T-c'\gamma_T^{1/\beta}, y_T+c'\gamma_T^{1/\beta}]$. Especially, the same is true on the subset

\[ \left[y_T-c'\tilde\delta_T^{1/\beta},y_T+c' \tilde\delta_T^{1/\beta}\right]  = \left[ y_T-c''h_T , y_T+c''h_T\right]\]

with $c'':=c''(\beta,L):= L^{1/\beta}c'(\beta,L)$. Then,

\begin{align}\label{eqp: E11}
\begin{split}
\int K_{y_T,c''h_T}(z)\left( |b_T(z) - b_0(z)|-\eta\right) dz &\geq \frac{\tilde{\delta}_T}{2}\int_{y_T-c''h_T}^{y_T+c''h_T} K_{y_T,h_T}(z) dz \\
&= \frac{1}{2} \tilde\delta_T c''h_T \int_{-1}^{1} K(z) dz.
\end{split}
\end{align}

As $|b-b_0|-\eta\geq 0$ on the interval $[y_T-c''h_T, y_T+c''h_T]$ we can follow the lines of the case $\beta\leq 1$ to see that

\begin{align*}
&\frac{\left| \frac{1}{\sqrt{T}} \int_0^T K_{y_T,c''h_T}(X_s) (b_T(X_s) - b_0(X_s)) ds\right| - \frac{\eta}{\sqrt{T}} \int_0^T K_{y_T,c''h_T}(X_s) ds}{\sigma \sqrt{\frac1T \int_0^T K_{y_T,c''h_T}(X_s)^2 ds}}\\
&\geq \frac{\sqrt{T/(c''h_T)}\left( q_{b_T}(y_T) - a_T^{b_T}(y_T)\right) \int_\R K_{y_T,c''h_T}(z) (|b_T(z)-b_0(z)|-\eta) dz}{\sigma \sqrt{(q_{b_T}(y_T) + a_T^{b_T}(y_T))\|K\|_{L^2}^2}} \\
&\hspace{0.1cm} \geq \frac{\sqrt{T/(c''h_T)}\left( q_{b_T}(y_T) - a_T^{b_T}(y_T)\right) \frac{1}{2} \tilde\delta_T c''h_T \|K\|_{L^1}}{\sigma \sqrt{(q_{b_T}(y_T) + a_T^{b_T}(y_T))\|K\|_{L^2}^2}} \\
& = \sqrt{T h_T} \tilde\delta_T \frac{\sqrt{c''} \|K\|_{L^1} }{2\sigma \|K\|_{L^2}^2} \frac{q_{b_T}(y_T) - a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T) + a_T^{b_T}(y_T)}}\\
&= c^{\frac{2\beta+1}{2\beta}} \hspace{-0.05cm}\underbrace{\frac{\sqrt{c''} \|K\|_{L^1} }{2\|K\|_{L^2}^2} L^{-\frac{1}{2\beta}}  }_{=:\tilde{c}(\beta,L,K)} \hspace{-0.1cm}\sqrt{\log T} (1+\epsilon_T)^{\frac{2\beta+1}{2\beta}}  q_{b_T}(y_T)^{-\frac12} \frac{q_{b_T}(y_T) - a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T) + a_T^{b_T}(y_T)}}.
\end{align*}

For the correction term $C(\cdot)$ we only have to use $c''h_T$ instead of $h_T$. Then (<ref>) turns to

\begin{align}\label{eqp: E12}
\begin{split}
\log(c''h_T) &= \frac{1}{2\beta +1}\left( \log\log T - \log T\right) + \mathcal{O}_T(1) + \log(c'')\\
&= \frac{1}{2\beta +1}\left( \log\log T - \log T\right) + \mathcal{O}_T(1),
\end{split}
\end{align}

which does not change the result

\[ C(\cdot) = \sqrt{2/(2\beta+1)\log T} + o_{\Pr_b,\textrm{unif}}(1)\]

in (<ref>), and we get the upper bound

\begin{align*}
&\frac{\left|\frac{1}{\sqrt{T}}\int_0^T K_{y_T,c''h_T}(X_s)\left(b_T(X_s) - b_0(X_s)\right)ds\right|-\frac{\eta}{\sqrt{T}}\int_0^T K_{y_T,c''h_T}(X_s) ds}{\sigma \sqrt{\frac1T\int_0^T K_{y_T,c''h_T}(X_s)^2 ds}} \\
&\hspace{5cm} -C\left(\frac{\frac1T\int_0^T K_{y_T,c''h_T}(X_s)^2 ds}{\frac1T \int_0^T \1_{[-A,A]}(X_s) ds}\right)\\
&\hspace{0.2cm}\geq c^{\frac{2\beta+1}{2\beta}} \tilde{c}(\beta, L, K) \sqrt{\log T} (1+\epsilon_T)^{\frac{2\beta+1}{2\beta}}  q_{b_T}(y_T)^{-\frac12}\frac{q_{b_T}(y_T) - a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T) + a_T^{b_T}(y_T)}}\\
&\hspace{5cm} - \sqrt{\frac{2}{2\beta+1} \log T} + o_{\Pr_b,\textrm{unif}}(1)\\
&\hspace{0.2cm} = \left( c^{\frac{2\beta+1}{2\beta}} \tilde{c}(\beta, L, K) (1+\epsilon_T)^{\frac{2\beta+1}{2\beta}}  q_{b_T}(y_T)^{-\frac12}\frac{q_b^{b_T}(y_T) - a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T)+ a_T^{b_T}(y_T)}} \right. \\
& \hspace{6.5cm}\left. - \sqrt{\frac{2}{2\beta+1}}\right)\sqrt{\log T} + o_{\Pr_b,\textrm{unif}}(1).
\end{align*}

Treating the term with $\epsilon_T$ and $a_T^{b_T}(y_T)$ in the same way as in the case $\beta\leq 1$, the bracket equals

\begin{align*}
&c^{\frac{2\beta+1}{2\beta}} \tilde{c}(\beta, L, K) \left( 1+ \frac{2\beta+1}{2\beta}\epsilon_T ( 1+ \mathcal{O}(\epsilon_T)) - \frac{3}{\sqrt{2}}\frac{a_T^{b_T}(y_T)}{ q_{b_T}(y_T)} \right. \\
&\hspace{5cm}\left. \cdot\left( 1+ \frac{2\beta +1}{2\beta} \epsilon_T + \mathcal{O}(\epsilon_T^2)\right)\right) -\sqrt{\frac{2}{2\beta+1}}
\end{align*}

and by the same arguments as above this term multiplied by $\sqrt{\log T}$ diverges if we choose $c$ large enough to suffice

\[ c^{\frac{2\beta+1}{2\beta}} \tilde{c}(\beta, L, K)\geq \sqrt{\frac{2}{2\beta+1}}. \]

To complete the proof, we assume there exists a sequence $(b_T)_{T\geq 0}$ with $\|b_T - b_0\|_J-\eta\geq \tilde\delta_T$ and

\[ \liminf_{T\to\infty} \Pr_{b_T}\left( T_{b_T}^\eta(X) > \kappa_{\eta,\alpha}\right) = c<1.\]

Then there exists a subsequence $(b_{T_n})_n$ with $\lim_n \Pr_{b_{T_n}}( T_{b_{T_n}}^\eta(X) >\kappa_{\eta,\alpha}) =c$. By the item shown above this implies that $\gamma_{T_n}(b_{T_n})$ does not converge to zero, whence there exists a subsubsequence $(b_{T_{n_k}})_k$ such that $\gamma_{T_{n_k}}(b_{T_{n_k}})$ is uniformly bounded away from zero, for $\gamma_T$ being given in (<ref>). If we can show in such a case that our test attains asymptotic power one, we have a contradiction and are done with the proof.

We again index the subsubsequence in $T$ for notational clarity.
By Lemma <ref> there exists a sequence of intervals $I_T$ with $\liminf_{T\to\infty}\lambda(I_T)>0$ and $|b_T-b_0|-\eta\geq \frac{\gamma_T}{2}$ on $I_T$. In particular, this holds true on each $[y_T-h_T, y+h_T]$ for $T$ large enough and we can repeat the argument from the first part, where

\[ \int K_{y_T,h_T}(z)\left( |b_T(z) - b_0(z)|-\eta\right) dz \geq \frac{\gamma_T}{2}\int_{y_T-h_T}^{y_T+h_T} K_{y_T,h_T}(z) dz\]

is of larger order than the right-hand side of (<ref>). Hence, the whole term

\[\frac{\left| \frac{1}{\sqrt{T}} \int_0^T K_{y_T,h_T}(X_s) (b_T(X_s) - b_0(X_s)) ds\right| - \frac{\eta}{\sqrt{T}} \int_0^T K_{y_T,h_T}(X_s) ds}{\sigma \sqrt{\frac1T \int_0^T K_{y_T,h_T}(X_s)^2 ds}}\]

dominates the correction term $C(\cdot)$ in order and therefore their difference tends to infinity.

The proof follows the lines of Theorem <ref>. For rate adaptivity in both $\beta$ and $L$ we can no longer choose the kernel $K_\beta$ that depends on $\beta$. Hence, the constant $c(\beta,L,K)$ has to be modified in the case $\beta\leq 1$, which is done in the same spirit as for $\beta>1$ in the proof of Theorem <ref>.
Then we note that

\begin{align}\label{eqp: E13}
\frac{1}{\beta}\log\left(\frac{c(\beta,L,K)}{L}\right) +\frac{1}{\beta}\left(1+\mathcal{O}(\epsilon_T)\right)  + \log(c''(\beta,L))
\end{align}

is uniformly bounded by Lemma <ref> over $(\beta,L)\in [\beta_1,\beta_2]\times [L_1,L_2]$ and hence (<ref>) and (<ref>) are still valid. The constant $c''(\beta,L)$ was introduced at the beginning of the case $\beta>1$ in the proof of Theorem <ref> and we have $c''(\beta,L)=1$ for $\beta\leq 1$. With boundedness of (<ref>) in $(\beta,L)$, the rest of the proof is then the same as for Theorem <ref>.

For sharp adaptivity in $L$, one uses that $\log(c_*/L)$ is uniformly bounded for $L\in [L_1,L_2]$ and again (<ref>) and (<ref>) remain valid.

The proof works as that for Theorem <ref> with $\eta=0$, i.e. we show that uniformly in probability

\begin{align}\label{eqp: E15}
\frac{\frac{1}{\sqrt{Th}} \left|\int_0^T K_{y,h}(X_s)\left(b(X_s)-b_0(X_s)\right) ds\right|}{\sigma \sqrt{\frac{1}{Th} \int_0^T K_{y,h}(X_s)^2 ds}} - C\left(\frac{\frac{1}{T} \int_0^T K_{y,h}(X_s)^2 ds}{\frac1T \int_0^T \1_{[-A,A]}(X_s) ds}\right)
\end{align}

tends to infinity for all $b_T-b_0\in H_{\neq} \cap\{ b-b_0\in \mathcal{H}(\beta, L)\}$ such that

\begin{align}\label{eqp: E14}
\|(b_T-b_0) (q_{b_T}/\sigma^2)^{\beta/(2\beta+1)}\|_J \geq (1+\epsilon_T)c_*\delta_T,
\end{align}

with a suitable choice of $(y,h)=(y_T,h_T)$.
Denote by $y_T$ a location where the supremum in (<ref>) is attained and define $\tilde\delta_T$ and $h_T$ by (<ref>) and (<ref>). This means, $\gamma_T:=|b_T(y_T) - b_0(y_T)|\geq \tilde\delta_T$. By the maximum condition of $y_T$ we have for all $z\in J$,

\begin{align*}
\left| b_T(y_T) - b_0(y_T)\right| q_{b_T}(y_T)^\frac{\beta}{2\beta+1} &\geq \left| b_T(z) - b_0(z)\right| q_{b_T}(z)^\frac{\beta}{2\beta+1} \\
& \geq \left| b_T(z) - b_0(z)\right| (L_*)^\frac{\beta}{2\beta+1},
\end{align*}

where the second step holds true by Lemma <ref>. In particular,

\begin{align}\label{eqp: E1}
\begin{split}
\left| b_T(z) - b_0(z)\right|&\leq (L_*)^{-\frac{\beta}{2\beta+1}} \left| b_T(y_T) - b_0(y_T)\right| q_{b_T}(y_T)^\frac{\beta}{2\beta+1}\\
& \leq \left(\frac{L^*}{L_*}\right)^\frac{\beta}{2\beta+1} \gamma_T .
\end{split}
\end{align}

With $d_{T,y_T}(z) := \frac{1}{\sigma^2 T} L_T^z(X)-q_{b_T}(y_T)$, we have as in the proof of Theorem <ref>,

\begin{align*}
&\frac{1}{\sqrt{Th_T}} \left|\int_0^T K_{y_T,h_T}(X_s)\left(b(X_s)-b_0(X_s)\right) ds\right| \\
&\hspace{1cm} = \sqrt{T/h_T} \left|\int_\R K_{y_T,h_T}(z) \left( b_T(z)-b_0(z)\right) \left( q_{b_T}(y_T) + d_{T,y_T}(z)\right) dz\right| \\
&\hspace{1cm}\geq \sqrt{T/h_T} q_{b_T}(y_T) \left| \int_\R K_{y_T,h_T}(z) \left( b_T(z) - b_0(z)\right) dz\right| \\
&\hspace{2cm} - \sqrt{T/h_T} a_T^{b_T}(y_T) \int_\R \left| K_{y_T,h_T}(z)\right| \left| b_T(z)- b_0(z)\right| dz.
\end{align*}

Again, $a_T^{b_T}(y_T) := \left\| \frac{1}{\sigma^2 T} L_T^\cdot - q_{b_T}(y_T)\right\|_{[y_T-h_T, y_T+h_T]} $.
From the bound on $|b_T-b_0|$ in (<ref>) we directly get

\begin{align*}
&- \sqrt{T/h_T} a_T^{b_T}(y_T) \int_\R \left| K_{y_T,h_T}(z)\right| \left| b_T(z)- b_0(z)\right| dz \\
&\hspace{3cm} \geq -\sqrt{Th_T} a_T^{b_T}(y_T) \gamma_T \left(\frac{L^*}{L_*}\right)^\frac{\beta}{2\beta+1} \|K_\beta\|_{L^1}.
\end{align*}

To find a lower bound for the other term, note that $\frac{\tilde\delta_T}{\gamma_T}|b_T(y_T) - b_0(y_T)|=\tilde\delta_T$ and as $\gamma_T\geq\tilde\delta_T$, by Lemma <ref> (i) we have

\[ \frac{\tilde\delta_T}{\gamma_T} \left( b_T(z) - b_0(z)\right) \in \mathcal{H}\left( \beta, (\tilde\delta_T/\gamma_T) L\right) \subset\mathcal{H}(\beta, L)\]

\[ \tilde\delta_T K_{y_T,h_T} \in \mathcal{H}\left(\beta, \tilde\delta_T h_T^{-\beta}\right) = \mathcal{H}(\beta, L).\]

Thus, by Lemma <ref> (iii),

\begin{align*}
&\left| \int_\R K_{y_T,h_T}(z) \left( b_T(z) - b_0(z)\right) dz\right|\\
&\hspace{2cm}= \frac{\gamma_T}{\tilde\delta_T^2} \left| \int_\R \tilde\delta_T K_{y_T,h_T}(z) \frac{\tilde\delta_T}{\gamma_T} \left( b_T(z) - b_0(z)\right) dz\right| \\
&\hspace{2cm}\geq \frac{\gamma_T}{\tilde\delta_T^2} \left\| \tilde\delta_T K_{y_T,h_T}\right\|_{L^2}^2\\
&\hspace{2cm}= h_T\gamma_T \|K_\beta\|_{L^2}^2.
\end{align*}

Together, we have

\begin{align*}
&\frac{1}{\sqrt{Th_T}} \left|\int_0^T K_{y_T,h_T}(X_s)\left(b(X_s)-b_0(X_s)\right) ds\right| \\
&\hspace{1cm} \geq \sqrt{Th_T}\gamma_T \|K_\beta\|_{L^2}^2 \underbrace{\left( q_{b_T}(y_T) - a_T^{b_T}(y_T) \left(\frac{L^*}{L_*}\right)^\frac{\beta}{2\beta+1} \frac{\|K_\beta\|_{L^1}}{\|K_\beta\|_{L^2}^2}\right)}_{=: q_{b_T}(y_T) - a_T^{b_T}(y_T) C'}.
\end{align*}

In the proof of Theorem <ref> we have seen in (<ref>) that

\[ \sqrt{\frac{1}{Th_T} \int_0^T K_{y_T,h_T}(X_s)^2 ds} \geq  \|K_\beta\|_{L^2} \sqrt{ q_{b_T}(y_T) + a_T^{b_T}(y_T)}\]

and in (<ref>) that

\[ C\left(\frac{\frac1T \int_0^T K_{y_T,h_T}(X_s)^2 ds}{\frac1T \int_0^T \1_{[-A,A]}(X_s) ds}\right) \leq \sqrt{\frac{2}{2\beta+1}\log T} + o_{\Pr_b, \textrm{unif}}(1).\]

Consequently, we have the following lower bound for (<ref>), which uses the evaluation of the term $\sqrt{Th_T}\tilde\delta_T$ in (<ref>):

\begin{align*}
&\sqrt{Th_T}\tilde\delta_T \frac{\gamma_T}{\tilde\delta_T} \frac{\|K_\beta\|_{L^2}}{\sigma} \frac{ q_{b_T}(y_T) - C' a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T) + a_T^{b_T}(y_T)}} - \sqrt{\frac{2}{2\beta+1}\log T} + o_{\Pr_b,\textrm{unif}}(1)\\
&\hspace{1cm} \geq \sqrt{\frac{2}{2\beta+1} \log T}\left( \frac{\gamma_T}{\tilde\delta_T} (1+\epsilon_T)^\frac{2\beta+1}{2\beta} q_{b_T}(y_T)^{-\frac12} \right. \\
&\hspace{5cm} \left. \cdot \frac{ q_{b_T}(y_T) - C' a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T) + a_T^{b_T}(y_T)}} - 1\right) + o_{\Pr_b,\textrm{unif}}(1).
\end{align*}

Again as in the proof of Theorem <ref>, we use a Taylor expansion to obtain

\[ (1+\epsilon_T)^\frac{2\beta+1}{2\beta} = 1+ \frac{2\beta+1}{2\beta} \epsilon_T + \mathcal{O}(\epsilon_T^2).\]

For $x\geq 0$ we have

\[ \frac{d}{dx}\frac{a-C'x}{\sqrt{a+x}} = -\frac{1}{\sqrt{a+x}}\left( C'+\frac12\frac{a-C'x}{a+x}\right) \geq -\frac{1}{\sqrt{a}} \left( C'+2^{-1}\right)\]

and conclude by a Taylor expansion that

\[ \frac{ q_{b_T}(y_T) - C' a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T) + a_T^{b_T}(y_T)}} \geq \sqrt{q_{b_T}(y_T)} - c a_T^{b_T}(y_T)\]

for some constant $c>0$. Our lower bound now turns into $o_{\Pr_b,\textrm{unif}}(1)$ plus

\begin{align*}
&\sqrt{\frac{2}{2\beta+1}\log T}\left( \frac{\gamma_T}{\tilde\delta_T}\left( 1+\frac{2\beta+1}{2\beta}\epsilon_T +\mathcal{O}(\epsilon_T^2)\right) \left( 1 -\frac{c a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T)}}\right) -1\right) \\
&\hspace{0.5cm}\geq\sqrt{\frac{2}{2\beta+1}\log T}\left( \frac{2\beta+1}{2\beta}\epsilon_T \left( 1+\mathcal{O}(\epsilon_T)\right) - \frac{c a_T^{b_T}(y_T)}{\sqrt{q_{b_T}(y_T)}} (1+\mathcal{O}(\epsilon_T))\right).
\end{align*}

This term tends to infinity, as by choice of $\epsilon_T$ we have $\epsilon_T\sqrt{\log T} \rightarrow\infty$ and $\sqrt{\log T} a_T^{b}(y) \rightarrow_{\Pr_{b},\textrm{unif}} 0$ by Lemma <ref>.

The results in Theorem <ref>, <ref> and <ref> and their auxiliary results (in particular Theorem <ref>) were derived under the assumption that the diffusion $X$ is started in the invariant density, i.e. $X_0\sim\mu_b$. As outlined in Remark <ref> and given in detail in Subsection <ref>, the moment inequality for the deviation of the normalized local time and invariant density given in Proposition <ref> remains valid under the assumption that $X_0=x_0\in [-A,A]$ is fixed. Our results used stationarity, i.e. $X_0\sim \mu_b$, only via Proposition <ref> and hence, Theorem <ref>, <ref> and <ref> remain valid if the diffusion $X$ is started at a fixed point $X_0=x_0\in [-A,A]$.

A closer look reveals that one only needs the uniform stochastic convergence

\[ \sup_{b\in\Sigma(C,A,\gamma,\sigma)} \Pr_b\left( \log T \left\| \frac{1}{\sigma^2 T} L_T^\cdot(X) - q_b\right\|_{[-A,A]} >\epsilon\right) \rightarrow 0 \]

to derive these theorems, which is a weaker statement than the moment inequality in Proposition <ref>.

§ PROOFS FOR SECTION <REF>

This Section contains all proofs of Section <ref>. In Subsection <ref> a proof of the continuity result in Theorem <ref> is given. In Subsection <ref> some details about fractional Brownian motion and the fractional diffusion model are given, in particular a Girsanov-type formula and some preliminaries on fractional calculus. Those results are used in the subsequent Subsection <ref>, <ref> and <ref> that contain the proofs of Proposition <ref>, Theorem <ref> and Theorem <ref>.

§.§ Proof of Theorem <ref>

We start with a few preliminary lemmas.

Let $\mathcal{S}$ be a metric space together with a function $f:\mathcal{S}\times\R\rightarrow \R$ such that

(a) for all $s\in\mathcal{S}$ we have $f(s,\cdot)\in L^1(\lambda)$,

(b) $f(\cdot, x):\mathcal{S}\rightarrow\R$ is continuous in $s_0\in\mathcal{S}$ for $\lambda$-almost every $x\in\R$, and

(c) there exists a neighborhood $U$ of $s_0$ and an integrable function $g$ such that for all $s\in U$ we have $|f(s,\cdot)|\leq g$ $\lambda$-almost everywhere.

Then the function

\[ s\mapsto \int_\R f(s,x) dx\]

is continuous in $s_0$.

Let $s\in \mathcal{S}$ be an element of a compact metric space $(\mathcal{S},d_\mathcal{S})$ and $f:\mathcal{S}\times\R\rightarrow\R$ be a function such that

(a) $|f|\leq c$ is bounded,

(b) $f(s,\cdot)$ is uniformly continuous with $\textrm{supp}(f(s,\cdot))\subset [-A,A]$ for every $s\in\mathcal{S}$, and

(c) $s\mapsto f(s,y)$ is continuous for almost all $y\in\R$.

Then for fixed $T>0$, the map

\begin{align*}
\mathcal{C}([0,T])\times \mathcal{S}\ \ &\rightarrow\ \ \ \ \ \R,\\
((x_u)_{0\leq u\leq T}, s)\ \  &\mapsto\ \int_0^T f(s, x_u) du,
\end{align*}

is continuous in $(x,s)$ for all $x\in\mathcal{C}([0,T])$ and $s\in\mathcal{S}$, where the left-hand side is equipped with the product topology on $\mathcal{C}([0,T])\times \mathcal{S}$, where $\mathcal{C}([0,T])$ is considered as a normed vector space with $\|\cdot\|_{[0,T]}$.

Let $\epsilon>0$. We have

\begin{align*}
&\left| \int_0^T f(s,x_u) du - \int_0^T f(s', x_u')du\right|\\
&\hspace{1.5cm} \leq \left| \int_0^T f(s,x_u) - f(s',x_u) du\right| + \left| \int_0^T f(s',x_u)-f(s',x_u') du \right|.
\end{align*}

For the first summand we want to apply Proposition <ref> to the function $g(s,u) := f(s,x_u)$ and check its conditions:

* We have $g(s,\cdot)\in L^1(\lambda)$ as

\[ \int_0^T |g(s,u)| du = \int_0^T |f(s,x_u)| du \leq cT. \]

* $g(\cdot, u)$ is continuous in each $s$ for all $u$ by condition (c).

* We have $|g(s,u)|\leq c$ by (a) and as $\textrm{supp}(f(s,\cdot))\subset [-A,A]$ for all $s\in\mathcal{S}$ by (b), we can choose $c\1_{[-A,A]}(u)$ as a majorant.

Then we can conclude that there exists a $\delta_1>0$ such that

\[ \left| \int_0^T f(s,x_u) - f(s',x_u) du\right| \leq \frac{\epsilon}{2} \]

for $d_\mathcal{S}(s,s')<\delta_1$.
By uniform continuity of $f(s,\cdot)$, for each $s\in\mathcal{S}$ there exists $\delta_s>0$ such that

\[ \left| f(s,y) - f(s,y')\right| <\frac{\epsilon}{2T} \quad \textrm{ for all }\ |y-y'|<\delta_s.\]

As $\mathcal{S}$ is compact, we have $\delta_2 := \inf_{s\in\mathcal{S}} \delta_s>0$ because the infimum is attained for some $s_0$. Then for $\|x-x'\|_{[0,T]}<\delta_2$,

\begin{align*}
\left| \int_0^T  f(s', x_u) - f(s', x_u') du\right| \leq T \sup_{u\in [0,T]}\left| f(s', x_u) - f(s', x_u')\right| <\frac{\epsilon}{2}.
\end{align*}

Putting together both estimates, the claim follows.

Let $(X, d_X)$ and $(Y,d_Y)$ be two metric spaces where $X$ is compact and $f:X\times Y\rightarrow \R$ a continuous function. Then the function $g:Y\rightarrow \R$ defined by $g(y) := \sup_{x\in X} f(x,y)$ is continuous as well.

Fix $y\in Y$ and choose $\epsilon>0$. As $f$ is continuous, we find for every $x\in X$ a $\delta_x(y)>0$ such that $|f(x,y) - f(x',y')|< \epsilon$ for all  $x'\in X$ and $y'\in Y$ with $d_X(x,x')\vee d_Y(y,y')\leq \delta_x(y)$. As $X$ is compact, there exists a finite cover

\[ B_{\delta_{x_1}(y)}(x_1), B_{\delta_{x_2}(y)}(x_2), \dots B_{\delta_{x_N}(y)}(x_N)\]

of $X$, where $B_\delta(x):=\{z\in X\mid d_X(x,z)<\delta\}$ denotes the open ball of radius $\delta$ around $x$. For $y'$ with $d_Y(y,y')< \min\{ \delta_{x_1}(y),\dots, \delta_{x_N}(y)\}$, we then have

\begin{align*}
\left| g(y) - g(y')\right| &= \left| \sup_{x\in X} f(x,y) - \sup_{x\in X} f(x,y')\right|\\
&\leq \sup_{x\in X} \left| f(x,y) - f(x,y')\right|\\
&= \max_{1\leq k\leq N} \sup_{x\in B_{\delta_{x_k}(y)}(x_k)}\left| f(x,y) - f(x,y')\right|
\leq\epsilon.
\end{align*}

\[ \overline{\mathcal{T}}_T:= \{(y,h)\in\R^2\mid h_{\min}(T)\leq h\leq A \textrm{ and } -A+h\leq y\leq A-h\}.  \]

With Lemma <ref>, we establish continuity of the mapping

\begin{align}\label{eqp: Gm1}
\begin{split}
\mathcal{C}([0,T])\times \overline{\mathcal{T}}_T\ \ &\longrightarrow\ \ \ \ \ \ \R,\\
( (x_s)_{s\in [0,T]}, (y,h))\  &\ \mapsto\ \  \int_0^T f_i((y,h), x_s) ds,
\end{split}
\end{align}

for the functions $f_1((y,h),z) := K_{y,h}(z)^2$, $f_2((y,h),z) := K_{y,h}(z) b_0(z)$ and $f_3((y,h),z) := \frac{1}{h} K'((y-z)/h)$ and additionally of

\begin{align}\label{eqp: Gm2}
\begin{split}
\mathcal{C}([0,T])\times \overline{\mathcal{T}}_T\ \ &\longrightarrow\ \ \ \ \ \ \R,\\
( (x_s)_{s\in [0,T]}, (y,h))\ \ &\ \mapsto\ \ \int_{x_0}^{x_T} K_{y,h}(z) dz.
\end{split}
\end{align}

Then we can conclude that

\begin{align*}
\mathcal{C}([0,T])\times \overline{\mathcal{T}}_T\ \ &\longrightarrow\ \ \ \ \R,\\
( (x_s)_{s\in [0,T]}, (y,h))\ \ &\ \mapsto\ \ \tilde\Psi_{T,y,h}^{b_0}(x),
\end{align*}

is continuous as a concatenation of continuous mappings and afterwards the claim follows by continuity of the map $C(\cdot)$, Lemma <ref> and the fact that $\overline{\mathcal{T}}_T$ is compact by Lemma <ref> and $\mathcal{T}_T\subset \overline{\mathcal{T}}_T$ is dense.

For continuity of (<ref>), we first note that for $(y,h), (y',h')\in\overline{\mathcal{T}}_T$ we have $K_{y,h}(x)\rightarrow K_{y',h'}(x)$ for all $x\in\R$ in the limit $(y',h')\to (y,h)$ by continuity of $K$ and the same holds true for $K'$. Therefore, for $f_i$, $i=1,2,3$, condition (c) of Lemma <ref> is fullfilled and condition (b) is clear, as we work with continuous functions having compact support, which is a subset of $[-A,A]$. Moreover, $f_1$ and $f_2$ are obviously bounded and $f_3$ is bounded, as $h$ is bounded away from zero on $\overline{\mathcal{T}}_T$, so condition (a) of Lemma <ref> holds as well. Consequently, Lemma <ref> is applicable and the mapping in (<ref>) is continuous for $f_i$, $i=1,2,3$.

Lastly, we show continuity of (<ref>) in $((x_s)_{s\in [0,T]}, (y,h))$ and pick some $((x_s')_{s\in [0,T]}, (y',h'))$. Suppose $x_0\leq x_T$ and $x_0'\leq x_T'$ (the other cases work the same), then

\begin{align*}
&\left| \int_{x_0}^{x_T} K_{y,h}(z) dz - \int_{x_0'}^{x_T'} K_{y',h'}(z) dz\right| \\
&\hspace{0.5cm} = \left|\int_{x_0}^{x_T} K_{y,h}(z) - K_{y',h'}(z) dz + \int_{x_0}^{x_T} K_{y',h'}(z) dz - \int_{x_0'}^{x_T'} K_{y',h'}(z) dz\right|\\
&\hspace{0.5cm} \leq \left( x_T - x_0\right) \|K_{y,h} - K_{y',h'}\|_{[-A,A]} \\
&\hspace{1.5cm} + \left| \int_{\min\{x_0,x_0'\}}^{\max\{x_0,x_0'\}} K_{y',h'}(z) dz + \int_{\min\{x_T,x_T'\}}^{\max\{x_T,x_T'\}} K_{y',h'}(z) dz\right|\\
&\hspace{0.5cm} \leq \left( x_T - x_0\right) \|K_{y,h} - K_{y',h'}\|_{[-A,A]}  + 2\|x-x'\|_{[0,T]} \|K\|_{[-1,1]}.
\end{align*}

The second summand vanishes for $\|x-x'\|_{[0,T]}\to 0$, the first one for $(y',h')\to (y,h)$ because

\begin{align*}
\| K_{y,h} - K_{y',h'}\|_{[-A,A]} &\leq \sup_{z \in [-A,A]} \|K'\|_{[-1,1]}\left| \frac{z-y}{h} - \frac{z-y'}{h'}\right|\\
& =\|K'\|_{[-1,1]}\cdot \sup_{z \in [-A,A]} \left| \frac{z(h-h') - yh'+y'h}{hh'}\right|
\end{align*}

tends to zero in this limiting scenario. Continuity of (<ref>) follows.

§.§ Preliminaries

Here, we first give some important preliminary results on the fractional calculus that are frequently used in the proofs of the results of Section <ref> in the following subsections. In the second part we introduce a change of measure formula for fractional Brownian motion and present how the density process can be expressed in terms of fractional integrals.

§.§.§ Fractional calculus and hypergeometric function

First, we give some elementary definitions and results about fractional integrals and derivatives. The standard reference on this subject is [56].

Let $\Gamma$ denote the Gamma function.
The (left-handed) Riemann-Liouville fractional integral $I_{0+}^\alpha$ of order $\alpha>0$ for any function $f\in L^1([0,T])$ is defined as

\begin{align}\label{eqp: I_+}
\left( I_{0+}^\alpha f\right)(x) := \frac{1}{\Gamma(\alpha)} \int_0^x \frac{f(y)}{(x-y)^{1-\alpha}} dy.
\end{align}

The fractional derivative can be introduced as its inverse operator. Assuming $0<\alpha<1$ and $p\geq1$, we denote by $I_{0+}^\alpha(L^p([0,T]))$ the image of $L^p([0,T])$ with respect to the operator $I_{0+}^\alpha$. Then for $p>1$ and for each $f\in I_{0+}^\alpha(L^p([0,T]))$, the function $g$ with $f=I_{0+}^\alpha g$ is unique in $L^p([0,T])$ and agrees with the (left-handed) Riemann-Liouville fractional derivative $D_{0+}^\alpha f$ of $f$ of order $\alpha$, which is for any $f\in L^1([0,T])$ given almost everywhere by

\[\left( D_{0+}^\alpha f\right)(x) :=  \frac{1}{\Gamma(1-\alpha)} \frac{d}{dx}\int_0^x \frac{f(y)}{(x-y)^\alpha} dy.\]

On $I_{0+}^\alpha(L^1([0,T]))$, the fractional derivative has the so-called Weyl representation (cf. Corollary subsequent to Theorem $13.1$ in [56])

\begin{align}\label{eq: Weil_rep}
\left( D_{0+}^\alpha f\right)(x) =  \frac{1}{\Gamma(1-\alpha)} \left( \frac{f(x)}{x^\alpha} + \alpha\int_0^x \frac{f(x)-f(y)}{(x-y)^{1+\alpha}} dy\right)\quad a.e.
\end{align}

Last but not least, we give two examples that are frequently used in the following subsections and can be found as $(2.26)$ and $(2.44)$ in [56]. For the function $f(x) = x^{-\mu}$ and $0<\mu<1$ we have

\begin{align}\label{eq: D_x^mu}
\left( D_{0+}^\alpha f\right)(x) = \frac{\Gamma(1-\mu)}{\Gamma(1-\mu-\alpha)} x^{-\mu-\alpha}
\end{align}

and for $f(x)= x^{\beta-1}$ with $\beta>0$ it holds that

\begin{align}\label{eq: I_x^beta}
\left( I_{0+}^\alpha f\right)(x) = \frac{\Gamma(\beta)}{\Gamma(\alpha+\beta)} x^{\alpha+\beta-1}.
\end{align}

The Gaussian hypergeometric function

\[ F:\C\times \C\times \left(\C\setminus\{0,-1,-2,\dots \}\right) \times \left(\C\setminus\{x\in\R: x\geq 1\}\right)\longrightarrow\C \]

is the analytic continuation of the power series

\begin{align}\label{eq: F_power}
(a,b,c;z)\ \mapsto\ \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n n!} z^n,
\end{align}

with the Pochhammer symbol $(z)_0=1$ and $(z)_n := z(z+1)\cdots (z+n-1)$, $z\in\C$, $n\in\N$ (cf. [14], Section $2$ and [48], $IV.20.3$).

We give some properties of the Gaussian hypergeometric function $F(a,b,c;z)$  with $(a,b,c,z)\in\C\times \C\times \left(\C\setminus\{0,-1,-2,\dots \}\right) \times \left(\C\setminus\{x\in\R: x\geq 1\}\right)$ that can be found in the literature and are used below.

(i) For $\Re(c)>\Re(b)>0$, we have Euler's formula

\[ F(a,b,c;z) = \frac{\Gamma(a)}{\Gamma(b)\Gamma(c-b)}\int_0^1 t^{b-1}(1-t)^{c-b-1} (1-tz)^{-a} dt, \]

cf. $(10)$ in [24], Section $2.1.3$, where $\Re(z)$ denotes the real part of $z\in\C$.

(ii) $F$ is commutative in its first two arguments i.e.

\[ F(a,b,c;z) = F(b,a,c;z). \]

This is given in $2.1.2$ in [24] for $|z|<1$ and follows for the analytic continuation by the identity theorem for analytic functions, cf. Theorem $III.3.2$ in [9], applied to $z\mapsto F(a,b,c;z)-F(b,a,c;z)$.

(iii) We have $F(a,0,c;z)=1$, which is immediate by the power series representation (<ref>) inside the convergence disk $\{z\in\C: |z|<1\}$ and follows for the analytic continuation by the identity theorem for analytic functions, cf. Theorem $III.3.2$ in [9].

(iv) We have $F(H-1/2, 1/2-H,H+1/2,x)\geq 0$ for real arguments $x\leq 0$. This follows for $H<\frac12$ by Euler's formula in (i) and for $H>\frac12$ correspondingly when first applying the commutativity relationship (ii). The case $H=\frac12$ is obtained from (iii).

§.§.§ Fractional Brownian motion

Let $\Gamma$ denote the Gamma function and define the kernel $K_H$ for $s\leq t$ and $H\in (0,1)$ by

\begin{align}\label{eq: K_H}
K_H(t,s) := \frac{(t-s)^{H-\frac12}}{\Gamma\left(H+\frac12\right)} F\left( H-\frac12, \frac12-H, H+\frac12; 1-\frac{t}{s}\right),
\end{align}

where $F$ denotes the Gaussian hypergeometric function, see Subsection <ref>.
The covariance function $R_H$ of a fractional Brownian motion in (<ref>) has a representation of the form (see [49], p. $106$ and [14], Lemma $3.1$)

\begin{align}\label{eqp: R_H}
R_H(t,s) = \int_0^{t\wedge s} K_H(t,u) K_H(s,u) du
\end{align}

and it follows from this that the process $W^H=(W_t^H)_{t\in [0,T]}$ with

\[ W_t^H := \int_0^t K_H(t,s) dW_s, \quad t\in [0,T],\]

is a fractional Brownian motion on the interval $[0,T]$. By $R_{1/2}(s,t) = s\wedge t$, we see that $W^H|_{H=1/2} = W$ is a standard Brownian motion.

To move on to a change of measure formula for the fractional Brownian motion, we introduce the operator $K_H$ associated with the kernel $K_H(\cdot, \cdot)$ on $L^2([0,T])$ that is given by

\[ \left( K_Hf\right)(t) := \int_0^t K_H(t,s) f(s) ds. \]

This operator is an isomorphism from $L^2([0,T])$ onto $I_{0+}^{H+\frac12}(L^2([0,T])$ (cf. [56], Theorem $10.4$).

It can be expressed for $f\in L^2([0,T])$ in terms of fractional integrals in the following way:

\begin{align}\label{eq: K_H_rep}
\left( K_H f\right) (t)  =\begin{cases}
I_{0+}^{2H}\left( (\cdot)^{\frac12-H} I_{0+}^{\frac12-H}\left( (\cdot)^{H-\frac12} f(\cdot)\right)(\cdot)\right)(t) & \quad \textrm{ for } H < \frac12, \\
I_{0+}^{1}\left( (\cdot)^{H-\frac12} I_{0+}^{H-\frac12}\left( (\cdot)^{\frac12-H} f(\cdot)\right)(\cdot)\right)(t) & \quad \textrm{ for } H > \frac12,
\end{cases}
\end{align}

see equation $(7)$ and $(8)$ in [49] and Theorem $10.4$ in [56]. For absolutely continuous $f\in I_{0+}^{H+1/2}\left(L^2([0,T])\right)$, its inverse operator $K_H^{-1}$ can be expressed as

\begin{align}\label{eq: K_H^-1_rep}
\left(K_H^{-1}f\right)(s) = \begin{cases}
s^{H-\frac12} I_{0+}^{\frac12-H} \left( (\cdot)^{\frac12-H} f'(\cdot)\right)(s) & \quad \textrm{ for } H<\frac12, \\
s^{H-\frac12} D_{0+}^{H-\frac12} \left( (\cdot)^{\frac12-H} f'(\cdot)\right)(s) & \quad \textrm{ for } H>\frac12,
\end{cases}
\end{align}

with weak derivative $f'$, see equation $(11)$ and $(13)$ in [49] where a derivation is given.

With the help of this inverse operator $K_H^{-1}$ we can now give a formulation of Girsanov's theorem for the fractional Brownian motion. Other sources on change of measure for fractional Brownian motion are [14] and [49].

Let $u=(u_t)_{t\in [0,T]}$ be an adapted process with integrable trajectories and set

\[ \tilde{W}_t^H := W_t^H + \int_0^t u_s ds.\]

Assume that

(a) $\int_0^\cdot u_s ds\in I_{0+}^{H+\frac12}(L^2([0,T]))$ almost surely,

(b) $\E[Z_T^H(u)]=1$, where

\begin{align*}
&Z_T^H(u) = \exp\left( -\int_0^T K_H^{-1}\left(\int_0^\cdot u_s ds\right)(s) dW_s \right. \\
&\hspace{5cm} \left. - \frac12\int_0^T K_H^{-1}\left(\int_0^\cdot u_s ds\right)^2 (s) ds\right).
\end{align*}

Then the shifted process $\tilde{W}^H$ is a fractional Brownian motion with Hurst parameter $H$ under the new probability measure $\tilde{\Pr}$ defined by $\frac{d\tilde{\Pr}}{d\Pr} = Z_T^H(u)$ and the process

\[ W_t + \int_0^t K_H^{-1}\left( \int_0^\cdot u_r dr\right)(s) ds \]

is a Brownian motion.

Note that from the representation (<ref>) and the form of the fractional operators in (<ref>) and (<ref>) it can be seen that the density process $Z_T^H(u)$ in Proposition <ref> is adapted for $u_s:= g(X_s^{H,b})$, where $X^{H,b}$ solves (<ref>) and $g$ is a measurable function. If $g$ is in particular Lipschitz continuous, then Proposition $1$ in [61] reveals that

\[ \left( K_H^{-1}\left( \int_0^\cdot g(X_s^{H,b}) ds\right)(t)\right)_{t\in [0,T]} \in L^2([0,T]). \]

Thus, the stochastic integral in Proposition <ref> is well-defined in the classical Itô sense.

In our paper we work under the condition $u_s = g(X^{H,b}_s)$ for a bounded and Lipschitz continuous function $g$ and need condition (a) in Proposition <ref> to be satisfied for those.

For $H<\frac12$, this condition (a) follows from $\int_0^T u_s^2 ds <\infty$, which was derived in [4] within the proof of Lemma $4.1$ on p. 313–314. Since $g$ is bounded, this follows trivially for $u=g(X_\cdot^{H,b})$. For $H>\frac12$, at the beginning of Section $5$ on p. 316–317 it has been derived that condition (a) is satisfied if $u$ has trajectories that are Hölder continuous of order $H-\frac12+\epsilon$ for some $\epsilon>0$. This is true for $u=g(X_\cdot^{H,b})$ due to the Hölder continuity of order $H-\epsilon$ with arbitrary $\epsilon\in (0,H)$ of the fractional Brownian motion $W^H$ and the Lipschitz assumption on $g$.

§.§ Proof of Proposition <ref>

Before we start with the main proof, we establish two preliminary lemmas.

Let $0\leq s<t\leq T$. Then

\begin{align*}
&\int_0^{s} \left( K_H(t,u) - K_H(s,u)\right)^2 du + \int_{s}^t\left( K_H(t,u)-1\right)^2 du  \\
&\hspace{2cm}= (t-s)^{2H} + (t-s) - 2\int_{s}^t K_H(t,u)du.
\end{align*}

We have

\begin{align*}
&\int_0^{s} \left( K_H(t,u) - K_H(s,u)\right)^2 du + \int_{s}^t\left( K_H(t,u)-1\right)^2 du \\
&\hspace{1cm} = \int_0^{s} K_H(t,u)^2 - 2K_H(t,u)K_H(s,u) + K_H(s,u)^2 du \\
&\hspace{2cm} + \int_{s}^t K_H(t,u)^2 - 2K_H(t,u) +1 du\\
&\hspace{1cm} = -2\int_0^{s} K_H(t,u)K_H(s,u) du + \int_0^t K_H(t,u)^2 du \\
&\hspace{2cm} + \int_0^{s} K_H(s,u)^2 du + (t-s) -2 \int_{s}^t K_H(t,u) du.
\end{align*}

The first three integrals can be easily evaluated using the formula

\[ R_H(s,t) = \int_0^{s\wedge t} K_H(t,u) K_H(s,u)du \]

given in (<ref>) where $R_H(s,t) =\frac12\left( |t|^{2H} + |s|^{2H} - |t-s|^{2H}\right)$ is the covariance function of fractional Brownian motion. Their sum can be rewritten as

\begin{align*}
-2\frac12\left( t^{2H} +s^{2H} - |t-s|^{2H}\right) + t^{2H} + s^{2H} = |t-s|^{2H}
\end{align*}

and the claim follows.

Let the metric $d_H$ on $[0,T]^2$ be defined as follows. For $s\leq t$,

\[ d_H(s,t)^2 :=\int_0^{s} \left( K_H(t,u) - K_H(s,u)\right)^2 du + \int_{s}^t\left( K_H(t,u)-1\right)^2 du \]

and for $s>t$ set $d_H(s,t) := d_H(t,s)$.
Then for every $r>0$, there exists $\delta=\delta(T,r)>0$ such that for $|\frac12-H|<\delta$ we have

\[ \sup_{s,t\in [0,T]} d_H(s,t) \leq r. \]

The idea of the proof is to split

\[ \sup_{s,t\in [0,T]} d_H(s,t) = \max_{i=1,2,3} \sup_{(s,t)\in A_i(\epsilon)} d_H(s,t),\]

where we define for some $0<\epsilon<1\wedge T$ the sets

\begin{align*}
A_1(\epsilon) := [0,\epsilon]^2,\quad A_2(\epsilon) := [0,\epsilon]\times [\epsilon,T],\quad \textrm{and } A_3(\epsilon) := [\epsilon,T]^2.
\end{align*}

Note that by the triangle inequality

\begin{align*}
\sup_{(s,t)\in A_2(\epsilon)} d_H(s,t) &\leq \sup_{\substack{ s\in [0,\epsilon], t\in [\epsilon,T]}} \left( d_H(s,\epsilon) + d_H(\epsilon,T)\right)\\
&\leq \sup_{(s,t)\in A_1(\epsilon)} d_H(s,t) + \sup_{(s,t)\in A_3(\epsilon)} d_H(s,t).
\end{align*}

We now show that for $|H-\frac12|<\delta$ and $\delta=\delta(T,\epsilon)>0$ sufficiently small,

\begin{align}\label{eqp: G10}
\sup_{(s,t)\in A_1(\epsilon)} d_H(s,t)^2< 2\sqrt{\epsilon}\quad \textrm{ and } \sup_{(s,t)\in A_3(\epsilon)} d_H(s,t)^2 <\epsilon.
\end{align}

The claim of this lemma follows from this if we take $0<\epsilon(r)<1$ small enough such that $4\sqrt{\epsilon} + 2\epsilon <r^2$ because in this case we have for $i=1,2,3$ that $\sup_{(s,t)\in A_i} d_H(s,t)<r$. We start by showing the first inequality in (<ref>). By  Remark <ref> (iv), $F(H-1/2,1/2-H,H+1/2;x)\geq 0$ for $x\leq 0$ such that $K_H$ in (<ref>) satisfies $K_H(t,s)\geq 0$ for $s\leq t$. Consequently, by Lemma <ref> for $H\in [\frac14,\frac34]$,

\begin{align*}
\sup_{(s,t)\in A_1(\epsilon)} d_H(s,t)^2 &\leq \sup_{(s,t)\in A_1(\epsilon)} \left( |t-s|^{2H} + |t-s|\right)\\
&\leq \sup_{(s,t)\in A_1(\epsilon)} 2|t-s|^\frac12 \leq 2\sqrt{\epsilon}.
\end{align*}

The second inequality in (<ref>) is more involved. Note that for $s,t\in [\epsilon,T]$ with $s\leq t$, we have

\[ 1-\frac{T}{\epsilon} \leq 1-\frac{t}{s} \leq 0.\]

We know the Gaussian hypergeometric function $F(a,b,c;z)$ to be analytic on $\C\times \C\times \left(\C\setminus\{0,-1,-2,\dots \}\right) \times \left(\C\setminus\{x\in\R: x\geq 1\}\right)$, see Subsection <ref>. Hence, its restriction on $[-\frac14,\frac14]\times [-\frac14,\frac14] \times [\frac12,\frac54]\times [1-T/\epsilon,0]$ is continuous. In particular, the function

\[ G(H,x) := \frac{1}{\Gamma(\frac12+H)} F\left( H-\frac12, \frac12-H, H+\frac12; x\right) \]

is continuous on $[\frac14,\frac34]\times [1-T/\epsilon,0]$. As $[1-T/\epsilon,0]$ is compact, by Lemma <ref> the map

\begin{align}\label{eqp: G11}
H\mapsto \sup_{x\in [1-T/\epsilon,0]} G(H,x) = \sup_{(s,t)\in A_3(\epsilon)} G\left(H, 1-\frac{t}{s}\right)
\end{align}

is continuous with $G(\frac12,x) =1$ for all $x\in [1-T/\epsilon,0]$, see Remark <ref> (iii). By direct evaluation,

\[ \int_s^t (t-u)^{H-\frac12} du = \left[ -\frac{1}{H+\frac12} (t-u)^{H+\frac12}\right]_s^t = \frac{2}{2H+1} (t-s)^{H+\frac12}.\]

Now, by Lemma <ref>, for $s\leq t$ and $(s,t)\in A_3(\epsilon)$,

\begin{align*}
d_H(s,t)^2 &= (t-s)^{2H} + (t-s) - 2\int_s^t K_H(t,u) du \\
&= (t-s)^{2H} + (t-s) - 2\int_s^t (t-u)^{H-\frac12} du \\
&\hspace{2cm}- 2\int_s^t (t-u)^{H-\frac12} \left( G(H,1-t/u) - 1\right) du \\
&\leq (t-s)^{2H} + (t-s) - \frac{4}{2H+1} (t-s)^{H+\frac12} \\
&\hspace{2cm} + \frac{4}{2H+1} (t-s)^{H+\frac12} \cdot \sup_{u\in [s,t]}\left| G(H,1-t/u) - 1\right|,
\end{align*}

and consequently,

\begin{align*}
\sup_{(s,t)\in A_3(\epsilon)} d_H(s,t)^2 &\leq \sup_{(s,t)\in A_3(\epsilon)} \left( |t-s|^{2H} + |t-s| - \frac{4}{2H+1} |t-s|^{H+\frac12} \right) \\
&\hspace{1cm} + \frac{4}{2H+1} T^{H+\frac12} \cdot \sup_{s,t\in [\epsilon,T]} \left| G\left(H,1-\frac{t}{s}\right) - 1\right|.
\end{align*}

By the continuity in (<ref>), there exists $\delta_1=\delta_1(T,\epsilon)>0$ such that the second summand is bounded from above by $\epsilon/2$ for $|H-\frac12|<\delta_1$. For the first summand, we apply the same reasoning: The map

\[ (s,t,H)\mapsto \left( |t-s|^{2H} + |t-s| - \frac{4}{2H+1} |t-s|^{H+\frac12} \right) \]

is clearly continuous on $[\epsilon,T]\times [\epsilon,T]\times [\frac14,\frac34]$ and by Lemma <ref> the same is true for

\[ H\mapsto \sup_{s,t\in [\epsilon,T]} \left( |t-s|^{2H} + |t-s| - \frac{4}{2H+1} |t-s|^{H+\frac12} \right).\]

As the supremum on the right-hand side is zero for $H=\frac12$, we conclude by this continuity that there exists $\delta_2=\delta_2(T,\epsilon)>0$ such that

\[ \sup_{(s,t)\in A_3(\epsilon)} \left( |t-s|^{2H} + |t-s| - \frac{4}{2H+1} |t-s|^{H+\frac12} \right)\leq \frac{\epsilon}{2} \]

for $|H-\frac12|<\delta_2$. Then the second inequality in (<ref>) follows for the choice $\delta=\delta_1\wedge \delta_2$.

From the SDE (<ref>) we directly get

\[ X_t^{H,b} - X_t^b = \int_0^t  \big( b(X_s^{H,b}) - b(X_s^b)\big) ds + \sigma( W_t^H - W_t).\]

As each $b\in\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)$ is Lipschitz with constant $L$, we derive

\[ |X_t^{H,b} - X_t^b| \leq L\int_0^t |X_s^{H,b} - X_s^b| ds + \sigma |W_t^H -W_t|.\]

From Gronwall's lemma (cf. $(2.11)$ in [33], Chapter $5$ with corresponding proof on p. 387–388) we then get

\[ |X_t^{H,b} - X_t^b| \leq \sigma |W_t^H - W_t| + L \int_0^t |W_s^H - W_s| e^{L (t-s)} ds\]

and in consequence

\[ \|X^{H,b} - X^b\|_{[0,T]} \leq \|W^H -W\|_{[0,T]} \left( \sigma + L\int_0^t e^{L (t-s)} ds\right).\]

Note that the right-hand side does not depend on $b$ anymore. Hence, the proof is finished if we show that $\|W^H -W\|_{[0,T]}$ converges to zero in probability. To do so, we use continuity of the centered Gaussian process $W^H-W$, $W_0^H=W_0=0$, and Dudley's theorem (cf. Theorem $11.17$ in [42]) that yields

\begin{align}\label{eqp: G8}
\begin{split}
\E\left[ \left\|W^H_t-W_t\right\|_{[0,T]}\right]  &\leq \E\left[|W_0^H - W_0|\right] + 2\E\left[ \sup_{t\in [0,T]} \left( W_t^H- W_t\right)\right] \\
&\leq 48\int_0^\infty \sqrt{\log( N(r,[0,T], d_{H})} dr
\end{split}
\end{align}

with metric $d_{H}(t,t') := \E\left[ (W_t^H - W_t - W_{t'}^H + W_{t'})^2\right]^\frac12$, $t,t'\in [0,T]$. The definition of the covering numbers $N(r,[0,T],d_H)$ is given in (<ref>). Subsequently, we prove that the right-hand side of this expression converges to zero for $H\to\frac12$ by dominated convergence. By the kernel representation $W_t^H := \int_0^t K_H(t,s)dW_s$ and Itô's isometry, for $t\geq t'$ the distance $d_H(t,t')^2 $ is given by

\begin{align*}
&\E\left[ \left( \int_0^t \left(K_H(t,s) -1\right) dW_s -\int_0^{t'} \left(K_H(t',s) -1\right) dW_s\right)^2 \right] \\
&\hspace{0.3cm}= \E\left[ \left( \int_0^{t'} \left(K_H(t,s) -K_H(t',s)\right) dW_s +\int_{t'}^t \left(K_H(t,s) -1\right) dW_s\right)^2 \right] \\
&\hspace{0.3cm}=\int_0^{t'} \left(K_H(t,s) -K_H(t',s)\right)^2 ds +\int_{t'}^t \left(K_H(t,s) -1\right)^2 ds,
\end{align*}

where we made use of the independent increments of Brownian motion and therefore independence of the stochastic integrals in the last step.

By Lemma <ref>, there exists $\delta>0$ such that

\[ \sup_{s,t\in [0,T]} d_H(s,t) \leq r \]

for $|\frac12-H|<\delta$. For those $H$, $N(r,[0,T],d_H)=1$ and hence we have pointwise (in $r$) convergence

\begin{align}\label{eqp: G9}
\lim_{H\to\frac12} \sqrt{\log(N(r,[0,T], d_H) )}=0.
\end{align}

Next, we construct an integrable majorant of

\[ \left(\sqrt{\log(N(r,[0,T], d_H))}\right)_{\frac13<H<1}.\]

To do so, we first consider $0<r\leq 1$ and define

\[ I_k :=\left[ (k-1)(r/\sqrt{2})^3, k (r/\sqrt{2})^3\right] \cap [0,T], \quad k\in\N. \]

Clearly, there are at most

\[ m\leq \frac{T}{(r/\sqrt{2})^3} + 1 = \frac{2^\frac32 T}{r^3} +1 \]

non-empty intervals $I_k$, $k=1,\dots, m$. Now we show that a minimal $d_H$-covering of $[0,T]$ for $\frac13<H<1$ cannot contain more than $m$ elements. This follows if $d_H(s,t)<r$ for $s,t\in I_k$. Indeed, by Lemma <ref> and Remark <ref> (iv),

\begin{align*}
d_H(s,t)^2 \leq |t-s|^{2H} + |t-s|\leq 2|t-s|^\frac23 \leq 2 \left(\frac{r}{\sqrt{2}}\right)^2 = r^2.
\end{align*}

Hence, we have for $0<r\leq 1$ that $N(r,[0,T], d_H) \leq 2^\frac32 T r^{-3} +1$. The case $r>1$ is easier. Here, it is enough to consider $r\leq 2\max\{ T^\frac23, T^{2}\}$ as $H\in (\frac13,1)$ and

\[ d_H(s,t)^2 \leq |t-s|^{2H} + |t-s| \leq 2\max\{ T^\frac23, T^{2}\}.   \]

We split $[0,T]$ into intervals of length $2^{-\frac32}$ with respect to the usual absolute value $|\cdot|$ and there are $m\leq 2^{\frac32}T+1$ of them. For $s,t$ in one of these intervals, $|t-s|\leq 2^{-\frac32}<\frac12$ and we find again by Lemma <ref> and Remark <ref> (iv),

\[ d_H(s,t)^2 \leq |t-s|^{2H} + |t-s| \leq |t-s|^\frac23 + |t-s| \leq \left( 2^{-\frac32}\right)^\frac23 + \frac12 =1<r.\]

We conclude, that for any $1<r\leq 2\max\{ T^\frac23, T^{2}\}$ we have the bound $N(r,[0,T],d_H) \leq 2^\frac32 T+1$ and hence

\begin{align*}
&\int_0^T \sqrt{\log( N(r,[0,T], d_H)} dr  \\
&\hspace{1cm} \leq \int_0^1 \sqrt{\log( 2^\frac32 T r^{-3} +1)} dr + \int_1^{2\max\{ T^{2/3}, T^{2}\}} \sqrt{\log(2^\frac32 T +1)} dr.
\end{align*}

The latter integral is clearly bounded and the first one is bounded by

\begin{align*}
\int_0^1 \sqrt{\log( 2^\frac32 T r^{-3} +1)} dr &\leq \int_0^1 \sqrt{\log( (2^\frac32T +1)r^{-3}) } dr\\
&= \int_0^1 \sqrt{\log(2^\frac32T +1) + 3\log(1/r)} dr\\
&\leq \int_0^1 \sqrt{\log(2^\frac32T+1)}  + \sqrt{3\log(1/r)} dr,
\end{align*}

where we used $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ for $a,b\geq 0$ in the last step. Here, the first integral is bounded and by substituting $r=e^{-u}$

\begin{align*}
\int_0^1 \sqrt{3\log(1/r)} dr &= \int_0^\infty \sqrt{3\log(e^{u})} e^{-u} du \\
&= \sqrt{3}\int_0^\infty \sqrt{u} e^{-u} du\\
&= \sqrt{3}\Gamma\left(\frac32\right) <\infty.
\end{align*}

Applying the dominated convergence theorem for $H\to\frac12$ in (<ref>) together with (<ref>) gives the result.

§.§ Proof of Theorem <ref>

First, we provide some helpful preliminary results. At the end of this subsection, the proof of Theorem <ref> and Remark <ref> are given.

Let $(E,d)$ be a metric space and $f:E\rightarrow\R$ a uniformly continuous function. Suppose for a parameter space $\Theta$ and $E$-valued random variables $X,X_1,X_2,\dots$ we have uniform stochastic convergence, i.e. for every $\epsilon>0$,

\[ \sup_{\theta\in\Theta} \Pr_\theta \left( d(X_n,X)>\epsilon \right) \stackrel{n\to\infty}{\longrightarrow} 0.\]

Then we also have for every $\epsilon>0$ that

\[ \sup_{\theta\in\Theta} \Pr_\theta \left( |f(X_n)-f(X)|>\epsilon \right) \stackrel{n\to\infty}{\longrightarrow} 0.\]

By the uniform continuity of $f$, for every $\epsilon>0$ we can find $\delta_\epsilon>0$ such that $|f(x)-f(y)|\leq\epsilon$ whenever $d(x,y)\leq\delta_\epsilon$. Then

\begin{align*}
\sup_{\theta\in\Theta} \Pr_\theta \left( |f(X_n)-f(X)|>\epsilon \right) \leq \sup_{\theta\in\Theta} \Pr_\theta\left( d(X_n,X) >\delta_\epsilon\right)
\end{align*}

and the right-hand side converges to zero by assumption.

From here on, we define for $c\in\R$ the set

\begin{align}\label{eqp: G21}
D_{c,T} := \{ f\in\mathcal{C}([0,T])\mid -c,c \in \textrm{im}(f)\}.
\end{align}

There exists $\epsilon>0$ such that

\[ \lim_{T\to\infty} \inf_{b\in \Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)} \Pr\left( X^b\in D_{A+\epsilon,T}\right) =1.\]

By $X_0^b=x_0\in [-A,A]$, continuity of paths and the occupation times formula <ref>, we have for any $\epsilon>0$

\begin{align*}
&\Pr\left( X^b\in D_{A+\epsilon,T}\right)\\
&\hspace{0.5cm} \geq \Pr\left( \int_0^T \1_{(A+\epsilon, A+2\epsilon]}(X_s^b) ds >0 \textrm{ and } \int_0^T \1_{[ -A-2\epsilon, -A-\epsilon)}(X_s^b) ds >0\right) \\
&\hspace{0.5cm} = \Pr\left( T \int_{A+\epsilon}^{A+2\epsilon} \frac{1}{T\sigma^2} L_T^z(X^b) dz >0 \textrm{ and } T \int_{-A-2\epsilon}^{-A-\epsilon} \frac{1}{T\sigma^2} L_T^z(X^b) dz >0\right) \\
&\hspace{0.5cm} =  \Pr\left( T \int_{A+\epsilon}^{A+2\epsilon} q_b(z) dz + T\int_{A+\epsilon}^{A+2\epsilon} \frac{1}{T\sigma^2}L_T^z(X^b) - q_b(z) dz >0 \right. \\
&\hspace{2cm} \left. \textrm{ and }T \int_{-A-2\epsilon}^{-A-\epsilon} q_b(z) dz + T\int_{-A-2\epsilon}^{-A-\epsilon} \frac{1}{T\sigma^2}L_T^z(X^b) - q_b(z) dz >0 \right).
\end{align*}

Now we specify $\epsilon := L_*/(4L^*)$. Lemma <ref> reveals that for all $A\leq y\leq A+2\epsilon$ and all $-A-2\epsilon\leq z\leq -A$,

\[ \left| q_b(A) - q_b(y)\right| \leq 2L^*\epsilon \leq \frac{L_*}{2}\ \textrm{ and }\ \left| q_b(-A) - q_b(z)\right| \leq 2L^*\epsilon \leq \frac{L_*}{2}. \]

Consequently, as $q_b(A),q_b(-A)\geq L_*$ by Lemma <ref>,

\[ \int_{A+\epsilon}^{A+2\epsilon} q_b(z) dz \geq \frac{(L_*)^2}{8L^*}\ \ \textrm{ and }\ \  \int_{-A-2\epsilon}^{-A-\epsilon} q_b(z) dz \geq \frac{(L_*)^2}{8L^*}\]

and we can conclude

\begin{align*}
\Pr\left( X^b\in D_{A+\epsilon,T}\right)  &\geq \Pr\left( \epsilon \left\|\frac{1}{T\sigma^2}L_T^\cdot(X) - q_b\right\|_\infty < \frac{(L_*)^2}{16L^*}\right)\\
& = \Pr\left( \left\|\frac{1}{T\sigma^2}L_T^\cdot(X) - q_b\right\|_\infty < \frac{L_*}{4}\right).
\end{align*}

In particular,

\begin{align*}
&\liminf_{T\to\infty} \inf_{b\in \Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)} \Pr\left( X^b\in D_{A+\epsilon,T}\right) \\
&\hspace{1cm}  \geq \liminf_{T\to\infty} \inf_{b\in \Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)} \Pr\left( \left\|\frac{1}{T\sigma^2}L_T^\cdot(X) - q_b\right\|_\infty < \frac{L_*}{4}\right)\\
& \hspace{1cm} = \liminf_{T\to\infty} \inf_{b\in \Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)} \left( 1- \Pr\left( \left\|\frac{1}{T\sigma^2}L_T^\cdot(X) - q_b\right\|_\infty \geq \frac{L_*}{4} \right)\right) \\
&\hspace{1cm} = 1 - \limsup_{T\to\infty}\sup_{b\in \Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)} \Pr\left( \left\|\frac{1}{T\sigma^2}L_T^\cdot(X) - q_b\right\|_\infty \geq \frac{L_*}{4}\right)
\end{align*}

and the right-hand side equals one by Corollary <ref>.

The set $D_{A,T}\subset\mathcal{C}([0,T])$ is closed with respect to the topology of uniform convergence.

Consider a sequence $(f_n)_{n\in\N}$ with $f_n\in D_{A,T}$ for all $n\in\N$, that converges to some function $f\in\mathcal{C}([0,T])$, i.e. $\|f_n-f\|_{[0,T]}\rightarrow 0$. Because of $f_n\in D_{A,T}$, there exists a sequence $(t_n)_{n\in\N}$ with $f_n(t_n)=A$. By compactness of $[0,T]$, we find a subsequence $(t_{n_k})_{k\in\N}$ with $t_{n_k}\rightarrow t$ for $k\to\infty$ and some $t\in [0,T]$. For this $t$, we have

\begin{align*}
| f(t) - A| = \left|f(t) - f_{n_k}(t_{n_k}) \right| \leq \left| f(t) - f(t_{n_k})\right| + \left| f(t_{n_k}) - f_{n_k}(t_{n_k})\right|.
\end{align*}

The first summand vanishes in the limit $k\to\infty$ by continuity of $f$ and the second one by $\|f_{n_k}-f\|_{[0,T]}\rightarrow 0$. Thus, $A\in \textrm{im}(f)$. By the same reasoning, we see that $-A\in \textrm{im}(f)$ and conclude $f\in D_{A,T}$.

Let $(a_n)_{n\in\N}$ be a sequence of real numbers, $a\in\R$ and $|a_n -a|\to 0$. Let $(b_n)_{n\in\N}$ be a sequence with $b_n\in \{ a_k : k\in\N\}$, $b\in\R$, and $|b_n-b|\to 0$. Then one of the following is true:

(i) $a=b$, or

(ii) there exists $n_0\in\N$ such that $b_n=b$ for all $n\geq n_0$.

Suppose (i) does not hold and choose $0<\epsilon<|b-a|$. As $b$ is the limit of $(b_n)_{n\in\N}$, there exists $n_0$ such that $|b-b_n|<\frac{\epsilon}{2}$ for all $n\geq n_0$, in particular $|b_n-a|>\frac{\epsilon}{2}$ for $n\geq n_0$. As there are only finitely many $a_{n_1},\dots, a_{n_l}$ with $|a_{n_i}-a|>\frac{\epsilon}{2}$, $i=1,\dots, l$, we have $b_{n}\in \{a_{n_1},\dots, a_{n_l}\}=:A$ for all $n\geq n_0$. We conclude that $(b_n)_{n\geq n_0}$ is a convergent $A$-valued sequence and as $A$ only contains isolated points, (ii) holds.

The following results is well-known, yet we did not find a version simplified for our context. For completeness, the proof is given.

The set $\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)$ is relatively compact with respect to the topology of uniform convergence on compact sets.

Let $(f_n)_{n\in\N}$ be a sequence in $\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)$. We have to prove that there exists a subsequence which is convergent in $\mathcal{C}(\R)$ with respect to the topology of uniform convergence on compact sets.
For any compact set $K\subset\R$, the restriction of $\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)$ to $K$ is relatively compact with respect to the topology of uniform convergence on $K$ by the Arzelà-Ascoli theorem. For any $n\in\N$, let $K_n:= [-n,n]$. Note that

\[ \bigcup_{n\in\N} K_n = \R. \]

For $K_1$ we find a subsequence $\big(n_k^{(1)}\big)_{k\in\N}$ and an $f_1\in \mathcal{C}(K_1)$ such that

\[ \left\| f_{n_k^{(1)}} -f_1\right\|_{K_1} \rightarrow 0\quad \textrm{ for } k\to\infty. \]

We proceed inductively as follows: If $\big( n_k^{(i)}\big)_{k\in\N}$ and $f_i$ are chosen, we find a subsequence $\big( n_k^{(i+1)}\big)_{k\in\N}$ of $\big( n_k^{(i)}\big)_{k\in\N}$ and $f_{i+1}$ such that

\[ \left\| f_{n_k^{(i+1)}} -f_{i+1}\right\|_{K_{i+1}} \rightarrow 0\quad \textrm{ for } k\to\infty. \]

Note that $f_i$ restricted to $K_j$ with $j<i$ coincides with $f_j$ on $K_j$ because $K_j\subset K_i$. Hence, there exists $f\in\mathcal{C}(\R)$ with $f$ restricted to $K_i$ equals $f_i$ for all $i\in\N$. Next, as every compact set $K\subset\R$ is (closed and) bounded, there exists $n\in\N$ such that $K\subset K_n$. Finally, an application of Cantor's diagonal subsequence argument reveals the diagonal sequence $(m_k)_{k\in\N}$ with $m_k := n_k^{(k)}$ such that for every compact set $K\subset\R$,

\[ \left\| f_{m_k} - f\right\|_{K} \rightarrow 0 \quad \textrm{ for } k\to\infty. \]

Here and subsequently, the image measure of a $\mathcal{C}([0,T])$-valued random variable $Y$ is denoted by $\Pr^Y$.

Let $L>0$ and $(h_n)_{n\in\N}$ be a sequence with $h_n\in (0,1)$ for all $n\in\N$ and $|h_n-\frac12|\to 0$. Then the family of measures

\[ \left\{\Pr^{X^{H,b}}: b\in\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L), H\in \{h_n: n\in\N\}\right\}\]

is tight.

Consider a sequence $(b_n,H_n)_{n\in\N}$ with $b_n\in \Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)$ and $H_n\in \{h_k:k\in\N\}$ for all $n\in\N$. As $\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L)$ is relatively compact with respect to the topology of uniform convergence on compact sets as a consequence of the Arzelà-Ascoli theorem (see Lemma <ref>) and $(H_n)_{n\in\N}$ contains a convergent subsequence by the Bolzano-Weierstrass theorem as it is a bounded sequence, we find a subsequence $(b_{n_k},H_{n_k})_{k\in\N}$ with $\|b_{n_k} -b\|_K\to 0$ for any compact $K\subset\R$ and $|H_{n_k}-H|\to 0$. We have

\begin{align}\label{eqp: G2}
\sup_{f\in\mathcal{F}_{BL}(\mathcal{C}([0,T]))}\hspace{-0.05cm} \E\left[ \left| f(X^{H_{n_k},b_{n_k}}) - f(X^{H,b})\right|\right]\hspace{-0.05cm} \leq \E\left[ \|X^{H_{n_k},b_{n_k}} - X^{H,b}\|_{[0,T]} \wedge 2\right]
\end{align}

by the definition (<ref>) of $\mathcal{F}_{BL}$ in Subsection <ref>. In order to conclude that the right-hand side converges to zero by the dominated convergence theorem, we have to show that $\|X^{H_{n_k},b_{n_k}} - X^{H,b}\|_{[0,T]} \wedge 2$ converges to zero pointwise, i.e. $\omega$-wise. For each $\omega\in\Omega$, there exists a compact set $K=K(\omega)\subset\R$ such that $X_t^{H,b}(\omega)\in K$ for all $t\in [0,T]$ by continuity of $X^{H,b}(\omega)$. Then for $t\leq T$,

\begin{align*}
&\left| X_t^{H_{n_k},b_{n_k}} - X_t^{H, b}\right|\\
&\hspace{1cm} = \left| \int_0^t b_{n_k}(X_s^{H_{n_k},b_{n_k}}) ds + \sigma W_t^{H_{n_k}} - \int_0^t b(X_s^{H,b}) ds - \sigma W_t^{H}\right| \\
&\hspace{1cm}\leq  \int_0^t \left| b_{n_k}(X_s^{H_{n_k},b_{n_k}}) - b( X_s^{H,b}) \right| ds + \sigma \left| W_t^{H_{n_k}} -W_t^H\right|\\
&\hspace{1cm}\leq \int_0^t \left|  b_{n_k}(X_s^{H_{n_k},b_{n_k}})  -b_{n_k}(X_s^{H,b}) \right|+\left| b_{n_k}( X_s^{H,b})- b( X_s^{H,b}) \right| ds \\
&\hspace{3cm} +  \sigma \left| W_t^{H_{n_k}} -W_t^H\right|\\
&\hspace{1cm}\leq  L \int_0^t  \left| X_s^{H_{n_k},b_{n_k}} - X_s^{H,b}\right| ds +t\|b_{n_k} - b\|_K + \sigma \left| W_t^{H_{n_k}} -W_t^H\right|.
\end{align*}

By Gronwall's lemma (cf. $(2.11)$ in [33], Chapter $5$ with corresponding proof on p. 387–388) we get

\begin{align*}
\left| X_t^{H_{n_k},b_{n_k}} - X_t^{H, b}\right|&\leq \left( t\|b_{n_k} -b\|_K + \sigma \left| W_t^{H_{n_k}} -W_t^H \right|\right) \\
&\hspace{0.5cm} + L\int_0^t \left( s\|b_{n_k} -b\|_K + \sigma \left| W_s^{H_{n_k}} -W_s^H \right|\right) e^{L(t-s)} ds
\end{align*}

and in consequence

\begin{align}\label{eqp: G6}
\begin{split}
&\left\| X_t^{H_{n_k},b_{n_k}} - X_t^{H, b}\right\|_{[0,T]} \\
&\hspace{1cm} \leq \left( T\|b_{n_k} -b\|_K + \sigma \left\| W^{H_{n_k}} -W^H\right\|_{[0,T]}\right) e^{LT}.
\end{split}
\end{align}

Now, we distinguish cases. By Lemma <ref> we either have

(i) $H_{n_k}=H$ for all $k\geq k_0$ for some $k_0\in\N$, or

(ii) $H=\frac12$.

First, suppose (i) holds true. Then for $k\geq k_0$, (<ref>) simplifies to

\[ \left\| X_t^{H_{n_k},b_{n_k}} - X_t^{H, b}\right\|_{[0,T]}\leq  T\|b_{n_k} -b\|_K\cdot e^{LT}, \]

which converges to zero as $k\to\infty$. Hence, the right-hand side of (<ref>) vanishes in the limit $k\to\infty$ by dominated convergence and we have

\[ \Pr^{X^{H_{n_k},b_{n_k}}}\Rightarrow \Pr^{X^{H,b}}. \]

Assume (ii) holds true. Then $\|W^{H_{n_k}} - W^H\|_{[0,T]}$ converges to zero in probability by Proposition <ref>. Therefore, we find a subsequence $(H_{n_{k_l}})_{l\in\N}$ with

\[ \|W^{H_{n_{k_l}}} - W^H\|_{[0,T]} \stackrel{l\to\infty}{\longrightarrow} 0\quad a.s. \]

(cf. Lemma $5.2$ in [32]). Then we repeat the above arguments with $n_k$ replaced by $n_{k_l}$, in particular (<ref>) tends to zero for $l\to\infty$ almost surely and consequently the right-hand side in (<ref>) converges to zero, which implies
\[ \Pr^{X^{H_{n_{k_l}},b_{n_{k_l}}}}\Rightarrow \Pr^{X^{H,b}}. \]

In either way, we have shown that any sequence $\{ \Pr^{X^{H_n,b_n}}:n\in\N \}$ contains a weakly convergent subsequence and hence the family

\[ \left\{\Pr^{X^{H,b}}: b\in\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(1,L), H\in \{ h_n: n\in\N\} \right\}\]

is relatively compact and by Prohorov's theorem ([8], Theorem $6.2$) it is tight.

Let $D_{c,T} := \{ f\in\mathcal{C}([0,T])\mid -c,c\in \textrm{im}(f)\}$ be given as in (<ref>). Unless stated otherwise, we denote $X^{H,b}=(X_t^{H,b})_{t\leq T}$ and $X^b=(X_t^b)_{t\leq T}$. First, we note that for any $\epsilon'>0$,

\begin{align*}
\Pr\left( X^{H,b}, X^b\in D_{A,T}\right) &\geq \Pr\left( X^b\in D_{A+\epsilon',T}\textrm{ and }\|X^{H,b}-X^b\|_{[0,T]} <\frac{\epsilon'}{2}\right) \\
&\geq \Pr\left( X^b\in D_{A+\epsilon', T}\right) - \Pr\left( \|X^{H,b}-X^b\|_{[0,T]} > \frac{\epsilon'}{2} \right)
\end{align*}

and in consequence

\begin{align}\label{eqp: G4}
\begin{split}
&\inf_b  \Pr\left( X^{H,b}, X^b \in D_{A,T}\right) \\
&\hspace{1cm} \geq \inf_b\Pr\left( X^b\in D_{A+\epsilon', T}\right)  -\sup_b \Pr\left( \|X^{H,b}-X^b\|_{[0,T]} > \frac{\epsilon'}{2} \right),
\end{split}
\end{align}

where $\inf_b$ and $\sup_b$ are taken over $\Sigma(C,A,\gamma,\sigma)\cap \mathcal{H}(1,L)$. By Lemma <ref> we find $T_0$ and $\epsilon'>0$ such that the first term is bounded away from zero for all $T\geq T_0$ and consequently by Proposition <ref>

\begin{align}\label{eqp: G3}
\liminf_{H\to\frac12} \inf_b\Pr\left( X^{H,b}, X^b \in D_{A,T}\right) >0
\end{align}

for $T\geq T_0$. As

\begin{align*}
&\Pr\left( \left. \left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^{b}) \right| >\epsilon \ \right| X^{H,b},X^{b}\in D_{A,T} \right) \\
&\hspace{1cm} = \frac{\Pr\left( \left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^{b}) \right| >\epsilon \textrm{ and } X^{H,b},X^{b}\in D_{A,T} \right)}{ \Pr\left(X^{H,b},X^b\in D_{A,T}\right)},
\end{align*}

the first claim of the proposition follows from (<ref>) once we have shown that

\[  \lim_{H\to\frac12} \sup_b \Pr\left( \left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^{b}) \right| >\epsilon \textrm{ and } X^{H,b},X^{b}\in D_{A,T} \right) = 0,\]

which is equivalent to showing that for any sequence $(H_n)_{n\in\N}$ with $H_n\rightarrow\frac12$

\begin{align}\label{eqp: G7}
\lim_{n\to\infty} \sup_b \Pr\left( \left| \tilde T_T^\eta(X^{H_n,b}) - \tilde T_T^\eta(X^{b}) \right| >\epsilon \textrm{ and } X^{H_n,b},X^{b}\in D_{A,T} \right) = 0.
\end{align}

To see this, let $(H_n)_{n\in\N}$ be such a sequence and $\epsilon''>0$. By Lemma <ref> there exist compact sets $K_1=K_1((H_n)_{n\in\N})$ and $K_2$ such that

\begin{align}\label{eqp: compacts}
\sup_{n\in\N}\sup_b\Pr\left( X^{H_n,b} \in K_1^c\right) \leq \epsilon'' \ \textrm{ and }\ \sup_b \Pr\left( X^{b} \in K_2^c\right) \leq \epsilon''.
\end{align}

With the compact set $K=K_1\cup K_2$,

\begin{align*}
&\sup_b \Pr\left( \left| \tilde T_T^\eta(X^{H_n,b}) - \tilde T_T^\eta(X^{b}) \right| >\epsilon \textrm{ and } X^{H_n,b},X^{b}\in D_{A,T} \right) \\
&\hspace{0.8cm} \leq \sup_b \Pr\left( \left| \tilde T_T^\eta(X^{H_n,b}) - \tilde T_T^\eta(X^{b}) \right| >\epsilon \textrm{ and } X^{H_n,b},X^{b}\in D_{A,T}\cap K \right) \\
&\hspace{2cm} + \sup_b  \Pr\left( \left\{ X^{H_n,b}, X^b \in K\right\}^c\right).
\end{align*}

By Lemma <ref>, the set $D_{A,T}\cap K$ is compact as a closed subset of a compact set. By Theorem <ref>, the map $\tilde T_T^\eta$ is continuous on $D_{A,T}$ and hence, it is uniformly continuous on the compact set $D_{A,T}\cap K$. Therefore, Lemma <ref> is applicable on this set and the limit $\lim_{n\to\infty}$ of the first term is zero by Proposition <ref>. By the union bound and (<ref>), the second term is bounded by $2\epsilon''$ and (<ref>) follows as $\epsilon''>0$ was arbitrary.

For the second claim of the theorem, we start with inequality (<ref>). As

\[  \liminf_{T\to\infty} \inf_b \Pr\left( X^b\in D_{A+\epsilon', T}\right) =1\]

by Lemma <ref> for a proper choice of $\epsilon'>0$, and

\[ \lim_{H\to\frac12} \sup_b \Pr\left( \|X^{H,b}-X^b\|_{[0,T]} > \frac{\epsilon'}{2} \right) =0 \]

by Proposition <ref>, we directly get from (<ref>) that

\[ \liminf_{T\to\infty}\liminf_{H\to\frac12} \inf_b \Pr\left( X^{H,b}, X^b\in D_{A,T}\right) =1.\]

For the other order of $\liminf$ we note that

\[ \left\{ (X_t^{H,b})_{t\leq T_0},(X_t^{b})_{t\leq T_0} \in D_{A,T}\right\} \subset  \left\{ (X_t^{H,b})_{t\leq T},(X_t^{b})_{t\leq T} \in D_{A,T}\right\} \]

for $T_0\leq T$ and hence for any $T_0>0$

\[ \liminf_{T\to\infty} \inf_{b} \Pr \left( X^{H,b}, X^b\in D_{A,T}\right) \geq \inf_{b}\Pr \left( (X_t^{H,b})_{t\leq T_0}, (X_t^b)_{t\leq T_0}\in D_{A,T_0}\right).  \]

Let $\epsilon>0$ be arbitrary. Choose $T_0$ large enough and $\delta>0$ such that $\inf_{b}\Pr\left( (X_t^b)_{t\leq T_0} \in D_{A+\delta,T_0}\right) \geq 1-\epsilon$, which is possible by Lemma <ref>. Then again by (<ref>) we find

\begin{align*}
&\liminf_{H\to\frac12} \liminf_{T\to\infty}\inf_{b\in \Sigma(C,A,\gamma,\sigma)\cap \mathcal{H}(1,L)} \Pr \left( X^{H,b}, X^b\in D_{A,T}\right)\\
&\hspace{1cm} \geq \liminf_{H\to\frac12} \inf_{b\in \Sigma(C,A,\gamma,\sigma)\cap \mathcal{H}(1,L)} \Pr \left( (X_t^{H,b})_{t\leq T_0}, (X_t^b)_{t\leq T_0}\in D_{A,T_0}\right) \\
&\hspace{1cm} \geq \inf_{b\in \Sigma(C,A,\gamma,\sigma)\cap \mathcal{H}(1,L)}\Pr\left( (X_t^b)_{t\leq T_0} \in D_{A+\delta,T_0}\right) \\
&\hspace{1.5cm} - \limsup_{H\to\frac12} \sup_{b\in \Sigma(C,A,\gamma,\sigma)\cap \mathcal{H}(1,L)} \Pr\left(\|X^{H,b} - X^b\|_{[0,T_0]} > \delta/2\right) \\
&\hspace{1cm} \geq 1-\epsilon,
\end{align*}

where we applied Proposition <ref> in the last step. The assertion follows by taking $\epsilon\searrow 0$.

As a first step, for any $\epsilon>0$,

\begin{align}\label{eqp: G5}
\begin{split}
\Pr\left( \tilde T_T^\eta(X^{H,b}) >\kappa\right) &\leq \Pr\left(  \tilde T_T^\eta(X^{H,b}) >\kappa, \left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^b)\right|\leq \epsilon \right)\\
&\hspace{1cm} +\Pr\left( \left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^b)\right|> \epsilon\right)\\
&\leq \Pr\left(  \tilde T_T^\eta(X^b) >\kappa-\epsilon\right)+\Pr\left( \left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^b)\right|> \epsilon\right).
\end{split}
\end{align}

To estimate the second summand, we denote $B_{H,b} := \{ X^{H,b}, X^b\in D_{A,T}\}$ and apply the law of total probability and get

\begin{align*}
&\Pr\left( \left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^b)\right|> \epsilon\right)\\
&\hspace{1cm} = \Pr\left( \left.\left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^b)\right|> \epsilon\ \right| B_{H,b}\right)\cdot\Pr\left( B_{H,b}\right)\\
&\hspace{2.5cm} + \Pr\left( \left.\left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^b)\right|> \epsilon\ \right| B_{H,b}^c\right)\cdot\Pr\left( B_{H,b}^c\right)\\
&\hspace{1cm} \leq \Pr\left( \left.\left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^b)\right|>\epsilon\ \right| B_{H,b}\right)+\Pr\left( B_{H,b}^c\right).
\end{align*}

By Theorem <ref>,

\[ \limsup_{T\to\infty}\limsup_{H\to\frac12} \sup_{b\in H_0(b_0,\eta)\cap\mathcal{H}(1,L)}\Pr\left( \left| \tilde T_T^\eta(X^{H,b}) - \tilde T_T^\eta(X^b)\right|> \epsilon\right) =0. \]

In conclusion, (<ref>) yields for any $\epsilon>0$ with $\kappa-\epsilon>\kappa_{\eta,\alpha}$

\begin{align*}
& \limsup_{T\to\infty}\limsup_{H\to\frac12} \sup_{b\in H_0(b_0,\eta)\cap\mathcal{H}(1,L)}\Pr\left( \tilde T_T^\eta(X^{H,b}) >\kappa\right)\\
&\hspace{2cm} \leq \limsup_{T\to\infty} \sup_{b\in H_0(b_0,\eta)\cap\mathcal{H}(1,L)}\Pr\left( \tilde T_T^\eta(X^{b}) >\kappa-\epsilon\right)\\
&\hspace{2cm} \leq \limsup_{T\to\infty} \sup_{b\in H_0(b_0,\eta)\cap\mathcal{H}(1,L)}\Pr\left( \tilde T_T^\eta(X^{b}) >\kappa_{\eta,\alpha}\right)\leq \alpha.
\end{align*}

The last step used that $T_T^\eta$ is uniformly asymptotically of level $\alpha$, see (<ref>), in combination with (<ref>) that implies that the restriction to $\mathcal{T}_T$ in the definition of $\tilde T_T^\eta$ compared to all of $\mathcal{T}$ is negligible in the limit.

§.§ Proof of Theorem <ref>

First, we present the following result that is used later in our proof.

Let $T>0$ and $\frac12<\beta<H<1$. Then we have the following moment estimate for any $k\geq 1$:

\[ \E\left[ \left(\sup_{s\neq r\in [0,T]}\frac{|W_s^H- W_r^H|}{|s-r|^\beta}\right)^{2k}\right] \leq 32^k (2T)^{2k(H-\beta)} \frac{(2k)!}{k!}.\]

Next, we establish $L^2(\Pr)$-convergence of the exponent of the density given by the fractional Girsanov theorem <ref> as $H\to\frac12$.

Let $b\in\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(\beta,L)$, $L>0$, and $g:\R\rightarrow\R$ be a function with $\|g\|_\infty <\infty$ that is Lipschitz continuous with constant $\|g\|_L$. Define

\[ v_H^g(s) := K_H^{-1}\left( \int_0^\cdot g(X_u^{H,b})du\right)(s).\]

Then we have

\[ \int_0^T v_H^g(s) dW_s - \frac12\int_0^T v_H^g(s)^2 ds \stackrel{H\to\frac12}{\longrightarrow}_{L^2(\Pr)} \int_0^T g(X_s^b) dW_s - \frac12\int_0^T g(X_s^b)^2 ds.\]

As the representation of $v_H^g$ in terms of fractional integrals and derivatives is different for $H<\frac12$ and $H>\frac12$ we distinguish between those two cases.

Case 1: $H<\frac12$. First of all, we show that $v_H^g(s)$ is bounded on $[0,T]$ uniformly in $H<\frac12$. This follows by

\begin{align}\label{eqp: G20}
\begin{split}
\left| v_H^g(s)\right| &= \left| s^{H-\frac12} I_{0+}^{\frac12 -H}\left( (\cdot)^{\frac12-H} g(X_\cdot^{H,b})\right)(s)\right|\\
&= s^{H-\frac12} \frac{1}{\Gamma(\frac12-H)} \left| \int_0^s \frac{u^{\frac12-H}g(X_u^{H,b})}{(s-u)^{\frac12 +H}} du \right|\\
&\leq s^{H-\frac12} \frac{\|g\|_\infty}{\Gamma(\frac12-H)} \int_0^s \frac{u^{\frac12-H}}{(s-u)^{\frac12+H}} du \\
&= s^{H-\frac12} \frac{\|g\|_\infty}{\Gamma(\frac12-H)}  \frac{\Gamma(\frac32-H)}{\Gamma(2-2H)} s^{1-2H}\\
&=\frac{\|g\|_\infty}{\Gamma(\frac12-H)}  \frac{\left(\frac12 -H\right)\Gamma(\frac12-H)}{\Gamma(2-2H)} s^{\frac12-H}\\
&\leq \|g\|_\infty \sqrt{T},
\end{split}
\end{align}

where we evaluated the integral in the forth step according to (<ref>). Furthermore, we used in the fifth stept that $x\Gamma(x) = \Gamma(x+1)$ for $x>0$ and in the last one that $s\in [0,T]$ and $(\frac12-H)(\Gamma(2-2H)^{-1}\leq 1$. This last property follows from $\Gamma(x)\geq\frac12$ for $x\in [1,2]$ which can be seen in the following way: For all those $x$ we have

\begin{align*}
\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} dt &= \int_0^1 t^{x-1} e^{-t} dt + \int_1^\infty t^{x-1} e^{-t} dt\\
&\geq \int_0^1 t e^{-t} dt + \int_1^\infty e^{-t} dt \\
&= \left[ -te^{-t}\right]_0^1 + \left[ -e^{-t}\right]_0^1 + \left[ -e^{-t}\right]_1^\infty \\
&= 1-e^{-1}
\end{align*}

and now we simply use that $e>2$.

To proceed with the proof, we will show $L^2(\Pr)$-convergence of the stochastic integral and the Lebesgue integral separately. For the stochastic integrals we have by Ito's isometry

\begin{align*}
\E\left[ \left( \int_0^T v_H^g(s) dW_s - \int_0^T g(X_s^b) dW_s\right)^2\right] &= \E\left[ \left( \int_0^T v_H^g(s) - g(X_s^b) dW_s\right)^2\right]\\
&= \E\left[ \int_0^T \left( v_H^g(s) -g(X_s^b) \right)^2 ds\right].
\end{align*}

By the uniform in $H$ boundedness of $v_H^g$ and $g$ we get uniform integrability of the inner integral process (indexed in $H$) and are done if we can show that it converges to zero in probability for $H\nearrow\frac12$ (cf. [32], Theorem $5.12$). This holds true, if $\| v_H^g -g(X_\cdot^b)\|_{L^2([0,T])}\rightarrow 0$ in probability.

Using again uniform integrability, we may show $L^2(\Pr)$-convergence of the Lebesgue integral if we establish that

\begin{align}\label{eqp: G15}
\|v_H^g\|_{L^2([0,T])}^2 = \int_0^T v_H^g(s)^2 ds\ \stackrel{H\nearrow\frac12}{\longrightarrow}\ \int_0^T g(X_s^b)^2 ds = \|g(X_\cdot^b)\|_{L^2([0,T])}^2
\end{align}

in probability. By the reverse triangle inequality we have

\begin{align}\label{eqp: G16}
\left| \|v_H^g\|_{L^2([0,T])} - \| g(X_\cdot^b)\|_{L^2([0,T])} \right| \leq \|v_H^g - g(X_\cdot^b)\|_{L^2([0,T])},
\end{align}

so convergence of $\|v_H^g - g(X_\cdot^b)\|_{L^2([0,T])}$ to zero in probability suffices to show $\|v_H^g\|_{L^2([0,T])} \rightarrow_\Pr \|g(X_\cdot^b)\|_{L^2([0,T])}$. By taking squares and applying the continuous mapping theorem we conclude that convergence in probability holds in (<ref>).
Thus, we are finished for both the Itô and the Lebesgue integral, if we show $\|v_H^g -g(X_\cdot^b)\|_{L^2([0,T])}\rightarrow_\Pr 0$. To this aim, we split in the following way:

\begin{align*}
v_H^g(s) - g(X_s^b) &= s^{H-\frac12} I_{0+}^{\frac12-H}\left( (\cdot)^{\frac12-H} g(X_\cdot^{H,b})\right) (s) - g(X_s^b) \\
&= \left( I_{0+}^{\frac12-H}\left( (\cdot)^{\frac12-H} g(X_\cdot^{H,b})\right)(s) - I_{0+}^{\frac12-H}\left( g(X_\cdot^b)\right)(s)\right)\\
&\hspace{1cm} + \left( I_{0+}^{\frac12-H}\left( g(X_\cdot^b)\right)(s) -g(X_s^b)\right) \\
&\hspace{1cm} + \left( s^{H-\frac12} -1\right) I_{0+}^{\frac12-H}\left( (\cdot)^{\frac12-H}g(X_\cdot^{H,b})\right)(s).
\end{align*}

By Minkowski's inequality for the $L^2([0,T])$-norm, it is enough to show  convergence in probability to zero for the $L^2([0,T])$-norm of each summand separately.

(1) For the first summand, we use the $L^2([0,T])$-estimate of equation $(2.72)$ in [56] to get

\begin{align*}
&\left\| I_{0+}^{\frac12-H}\left( (\cdot)^{\frac12-H} g(X_\cdot^{H,b})\right) - I_{0+}^{\frac12-H}\left( g(X_\cdot^b)\right)\right\|_{L^2([0,T])}\\
&\hspace{1cm}\leq \frac{T^{\frac12 -H}}{(\frac12-H)\Gamma(\frac12-H)} \left\| (\cdot)^{\frac12-H} g(X_\cdot^{H,b}) - g(X_\cdot^b)\right\|_{L^2([0,T])}.
\end{align*}

As $x\Gamma(x)=\Gamma(x+1)$ for $x>0$, we have for $H<\frac12$,

\[  \frac{T^{\frac12 -H}}{(\frac12-H)\Gamma(\frac12-H)} = \frac{T^{\frac12-H}}{\Gamma(\frac32 -H)} \stackrel{H\nearrow\frac12}{\longrightarrow}\ 1.\]

Next, we evaluate the squared $L^2([0,T])$-norm:

\begin{align*}
&\int_0^T  \left( u^{\frac12-H} g(X_u^{H,b}) - g(X_u^b)\right)^2 du \\
&\hspace{0.1cm} = \int_0^T \left( u^{\frac12-H} (g(X_u^{H,b}) - g(X_u^b)) - g(X_u^b)\left( 1- u^{\frac12-H} \right)\right)^2 du \\
&\hspace{0.1cm}\leq 2\int_0^T u^{1-2H}(g(X_u^{H,b}) - g(X_u^b))^2 du + 2\int_0^T g(X_u^b)^2 \left( u^{\frac12-H} -1\right)^2 du.
\end{align*}

The second summand is bounded by

\begin{align*}
2\|g\|_\infty^2 \int_0^T \left( u^{\frac12-H} -1\right)^2 du &= 2\|g\|_\infty^2 \int_0^T \left( u^{1-2H} - 2u^{\frac12-H} +1\right) du\\
&\hspace{-1.5cm}= 2\|g\|_\infty^2 \left( \frac{1}{2-2H} T^{2-2H} - \frac{2}{\frac32-H} T^{\frac32-H} + T\right),
\end{align*}

which converges to zero for $H\nearrow\frac12$. The first summand, on the other hand, can be bounded by

\begin{align*}
&2 \left\|g(X_\cdot^{H,b}) - g(X_\cdot^b)\right\|_{[0,T]} \int_0^T u^{1-2H} du \\
&\hspace{3cm}\leq 2\|g\|_L  \| X_\cdot^{H,b} - X_\cdot^b\|_{[0,T]}\cdot\frac{1}{2-2H}T^{2-2H}.
\end{align*}

This term converges to zero in probability for $H\nearrow\frac12$ because the term $\|X_\cdot^{H,b}-X_\cdot^b\|_{[0,T]}$ does by Proposition <ref> and the other factors are bounded.

(2) By Theorem $2.6$ in [56], we simply have for all $\omega\in\Omega$,

\[\left\| I_{0+}^{\frac12-H}\left( g(X_\cdot^b)\right) -g(X_\cdot^b)\right\|_{L^2([0,T])}\ \stackrel{H\nearrow\frac12}{\longrightarrow}\ 0.\]

(3) As in (2), we argue $\omega$-wise. We repeat the argument used for bounding $v_H^b$ in (<ref>) without the factor $s^{H-\frac12}$ and immediately see that

\[ \left| I_{0+}^{\frac12-H}\left( (\cdot)^{\frac12-H} g(X_\cdot^{H,b})\right)(s)\right| \leq \|g\|_\infty  T.\]

\begin{align*}
&\left\| \left( (\cdot)^{H-\frac12} -1\right) I_{0+}^{\frac12-H}\left( (\cdot)^{\frac12-H} g(X_\cdot^{H,b})\right) \right\|_{L^2([0,T])}\\
&\hspace{1cm}\leq \|g\|_\infty T \left( \int_0^T  \left( s^{H-\frac12} -1\right)^2 ds\right)^\frac12\\
&\hspace{1cm} = \|g\|_\infty T\left( \frac{T^{2H}}{2H} - 2\frac{2T^{\frac12+H}}{2H+1} + T\right)^\frac12\ \stackrel{H\nearrow\frac12}{\longrightarrow}\ 0.
\end{align*}

This finishes the proof in the case $H<\frac12$.

Case 2: $H>\frac12$. The idea is again to show $L^2(\Pr)$-convergence of the Itô integral and the Lebesgue integral separately. For the stochastic integral we have by Itô's isometry

\begin{align*}
\E\left[ \left( \int_0^T v_H^g(s) dW_s - \int_0^T g(X_s^b) dW_s\right)^2\right]  = \E\left[ \int_0^T \left( v_H^g(s) -g(X_s^b) \right)^2 ds\right],
\end{align*}

but this time we use (<ref>) to represent $v_H^g$ as

\begin{align*}
v_H^g(s) &= s^{H-\frac12} D_{0+}^{H-\frac12}\left( (\cdot)^{\frac12-H} g(X_\cdot^{H,b})\right)(s) \\
&= \frac{s^{H-\frac12}}{\Gamma\left(\frac32-H\right)}\left( \frac{s^{\frac12-H} g(X_s^{H,b})}{s^{H-\frac12}} \right. \\
&\hspace{2.5cm}\left. + \left( H-\frac12\right)\int_0^s \frac{s^{\frac12-H} g(X_s^{H,b}) - r^{\frac12-H} g(X_r^{H,b})}{(s-r)^{\frac12+H}} dr \right).
\end{align*}

Using repeatedly $(a+b)^2\leq 2a^2 +2b^2$, we see that

\begin{align}\label{eqp: G12}
\begin{split}
\E\left[ \int_0^T \left( v_H^g(s) - g(X_s^b)\right)^2 ds\right]  &\leq 2A_H+2B_H \\
&\leq 2A_H + 4B_H^1 + 4B_H^2,
\end{split}
\end{align}

\begin{align*}
A_H &= \E\left[ \int_0^T \left( \frac{s^{\frac12-H} g(X_s^{H,b})}{\Gamma\left(\frac32-H\right)} - g(X_s^b)   \right)^2 ds \right], \\
B_H &= \E\left[ \int_0^T \left( \frac{\left(H-\frac12\right)s^{H-\frac12}}{\Gamma\left(\frac32-H\right)} \int_0^s \frac{s^{\frac12-H} g(X_s^{H,b}) - r^{\frac12-H} g(X_r^{H,b})}{(s-r)^{\frac12+H}} dr\right)^2 ds\right],
\end{align*}

and we subsequently splitted $B_H$ in a similar way as in the proof of Theorem $3$ in [49] into

\begin{align*}
B_H^1 &= \E\left[\int_0^T \left( \frac{\left(H-\frac12\right)s^{H-\frac12}}{\Gamma\left(\frac32-H\right)} g(X_s^{H,b}) \int_0^s \frac{s^{\frac12-H} - r^{\frac12-H}}{(s-r)^{\frac12+H}} dr \right)^2  ds\right],\\
B_H^2 &= \E\left[\int_0^T  \left( \frac{\left(H-\frac12\right)s^{H-\frac12}}{\Gamma\left(\frac32-H\right)}  \int_0^s \frac{g(X_s^{H,b}) - g(X_r^{H,b})}{(s-r)^{\frac12+H}} r^{\frac12-H} dr \right)^2  ds\right].
\end{align*}

In the following we will show for each of $A_H, B_H^1$ and $B_H^{2}$ that it vanishes in the limit $H\searrow\frac12$.

$A_H$: For this term, we start by

\begin{align*}
&A_H = \E\left[\int_0^T \left( \left(\frac{s^{\frac12-H}}{\Gamma\left(\frac32-H\right)}-1\right) g(X_s^{H,b}) + \left( g(X_s^{H,b}) - g(X_s^b)\right) \right)^2 ds\right]\\
&\hspace{0.2cm}\leq 2\E\left[ \int_0^T \left(\frac{s^{\frac12-H}}{\Gamma\left(\frac32-H\right)}-1\right)^2 g(X_s^{H,b})^2 ds + \int_0^T \left( g(X_s^{H,b}) - g(X_s^b) \right)^2 ds\right] \\
&\hspace{0.2cm}\leq 2 \|g\|_\infty^2 \int_0^T \left(\frac{s^{\frac12-H}}{\Gamma\left(\frac32-H\right)}-1\right)^2 ds + T\E\left[ \|g(X_\cdot^{H,b}) - g(X_\cdot^b)\|_{[0,T]}^2 \right].
\end{align*}

For the first summand, we evaluate the integral and get

\begin{align*}
&\int_0^T \left(\frac{s^{\frac12-H}}{\Gamma\left(\frac32-H\right)}-1\right)^2 ds \\
&\hspace{1cm}= \int_0^T \frac{s^{1-2H}}{\Gamma\left(\frac32-H\right)^2} - \frac{2s^{\frac12-H}}{\Gamma\left(\frac32-H\right)} +1 \ ds\\
& \hspace{1cm} =\frac{T^{2-2H}}{(2-2H)\Gamma\left(\frac32-H\right)^2} - \frac{2 T^{\frac32-H}}{\left(\frac32-H\right)\Gamma\left(\frac32-H\right)} +T\stackrel{H\searrow\frac12}{\longrightarrow} 0.
\end{align*}

For the second summand, we note that the family of random variables

\[ \left\{\|g(X_\cdot^{H,b}) - g(X_\cdot^b)\|_{[0,T]}^2, \frac12\leq H<1 \right\} \]

is uniformly integrable as all of them are bounded by $4T\|g\|_\infty^2$. Thus, it suffices to establish convergence in probability to zero of this term instead (cf. [32], Theorem $5.12$). But this follows easily by the Lipschitz condition on $g$ and Proposition  <ref> as

\[ \|g(X_\cdot^{H,b}) - g(X_\cdot^b)\|_{[0,T]}^2 \leq \|g\|_L^2 \| X_\cdot^{H,b} - X_\cdot^b\|_{[0,T]}^2. \]

In conclusion, we have shown so far that $A_H\rightarrow 0$ for $H\searrow\frac12$.

$B_H^1$: Bounding $B_H^1$ essentially relies on finding the value of the inner integral. For this, we have

\begin{align}\label{eqp: G22}
\int_0^s \frac{s^{\frac12-H} - r^{\frac12-H}}{(s-r)^{\frac12+H}} dr  = \left(\frac{\Gamma\left(\frac32-H\right)^2}{\Gamma(2-2H)} -1\right)\left(H-\frac12\right)^{-1} s^{1-2H}
\end{align}

almost everywhere, which can be seen by evaluating the fractional derivative $D_{0+}^{H-1/2}\left((\cdot)^{1/2-H}\right)$ in two ways. First, by
direct evaluation via (<ref>),

\[ D_{0+}^{H-\frac12}\left((\cdot)^{\frac12-H}\right)(s) = \frac{\Gamma\left( \frac32-H\right)}{\Gamma(2-2H)} s^{1-2H}.\]

On the other hand, $s^{\frac12-H}\in I_{0+}^{H-1/2}(L^1([0,T])$ as $\int_0^T u^{1-2H}du <\infty$ and

\[ I_{0+}^{H-\frac12}\left( \frac{\Gamma(\frac32-H)}{\Gamma(2-2H)} (\cdot)^{1-2H}\right)(s) = s^{\frac12-H} \]

by (<ref>).
Then, by the Weyl representation (<ref>), $D_{0+}^{H-1/2}\left((\cdot)^{1/2-H}\right)$ is given by

\[ \frac{1}{\Gamma\left(\frac32-H\right)} \left( s^{1-2H} + \left(H-\frac12\right) \int_0^s \frac{s^{\frac12-H} - r^{\frac12-H}}{(s-r)^{\frac12+H}} dr\right) \quad a.e.\]

Combining these two results gives (<ref>) after rearranging. With this identity holding almost everywhere,

\begin{align*}
B_H^1 &\leq \frac{\left(H-\frac12\right)^2}{\Gamma\left(\frac32-H\right)^2} \|g\|_\infty^2 \left(\frac{\Gamma(\frac32-H)^2}{\Gamma(2-2H)}-1\right)^2 \left(H-\frac12\right)^{-2} \int_0^T s^{2H-1} s^{2-4H} ds\\
& = \frac{1}{(2-2H)\Gamma\left(\frac32-H\right)^2}\left(\frac{\Gamma(\frac32-H)^2}{\Gamma(2-2H)}-1\right)^2 \|g\|_\infty^2 T^{2-2H}
\end{align*}

which tends to zero for $H\searrow\frac12$.

$B_H^{2}$: Here, it is crucial to find a bound for the inner integral. For this, we first use the Lipschitz property of $g$ and get

\begin{align*}
\left|\int_0^s \frac{g(X_s^{H,b}) - g(X_r^{H,b})}{(s-r)^{\frac12+H}} r^{\frac12-H} dr\right| &\leq \int_0^s \frac{|g(X_s^{H,b}) - g(X_r^{H,b})|}{(s-r)^{\frac12+H}}  r^{\frac12-H} dr \\
&\leq \|g\|_L \int_0^s \frac{|X_s^{H,b} - X_r^{H,b}|}{(s-r)^{\frac12+H}} r^{\frac12-H} dr.
\end{align*}

By the definition of $X^{H,b}$ we furthermore have for $s\geq r$,

\[ X_s^{H,b} - X_r^{H,b} = \int_r^s b(X_u^{H,b}) du + \sigma (W_s^H - W_r^H). \]

By the at most linear growth condition on $b$,

\begin{align*}
\left| \int_r^s b(X_u^{H,b}) du\right| &\leq |s-r| \cdot \|b(X_\cdot^{H,b})\|_{[0,T]}\\
&\leq C|s-r| \left( 1+ \|X^{H,b}\|_{[0,T]}\right)\\
&\leq C|s-r| \left( 1+ CT + \|W^H\|_{[0,T]}\right)e^{CT},
\end{align*}

where the last step used Lemma $1$ in [61]. To proceed, we choose a number $ 0<\epsilon_H<\min\{H-\frac12,\frac14\} $
and estimate (note that $W_0=0$),

\begin{align*}
\sup_{0\leq t\leq T} |W_t^H| &= \sup_{0<t\leq T}\frac{|W_t^H|}{t^{H-\epsilon_H}} t^{H-\epsilon_H}\\
&\leq T^{H-\epsilon_H} \sup_{0<t\leq T}\frac{|W_t^H|}{t^{H-\epsilon_H}} \\
&\leq \left(T\vee T^{\frac14}\right) \sup_{s\neq t\in [0,T]} \frac{|W_t^H - W_s^H|}{|t-s|^{H-\epsilon_H}}.
\end{align*}

Introducing the random variable

\begin{align}\label{eqp: G13}
G_H := \sup_{s\neq r\in [0,T]} \frac{|W_s^H-W_r^H|}{|s-r|^{H-\epsilon_H}},
\end{align}

we have for a constant $C'=C'(T)>0$ not depending on $H$,

\[ |X_s^{H,b} - X_r^{H,b}| \leq C' |s-r| \left( 1+ G_H\right) + \sigma |W_s^H - W_r^H|.\]

Using $(a+b)^2\leq 2a^2 + 2b^2$ once again, we find $B_H^2\leq 2B_H^{2,1} + 2B_H^{2,2}$ with

\begin{align*}
B_H^{2,1} &= \E\left[ \|g\|_L^2 \int_0^T  \left( \frac{\left(H-\frac12\right)s^{H-\frac12}}{\Gamma\left(\frac32-H\right)} C' (1+G_H) \int_0^s \frac{(s-r)r^{\frac12-H} }{(s-r)^{\frac12+H}}  dr \right)^2  ds \right] , \\
B_H^{2,2} &= \E\left[\|g\|_L^2  \int_0^T  \left( \frac{\left(H-\frac12\right)s^{H-\frac12}}{\Gamma\left(\frac32-H\right)}  \sigma \int_0^s \frac{|W_s^H - W_r^H|}{(s-r)^{\frac12+H}} r^{\frac12-H} dr \right)^2  ds \right].
\end{align*}

For $B_H^{2,1}$ we first evaluate the inner integral and get

\begin{align*}
\int_0^s \frac{(s-r)}{(s-r)^{\frac12+H}} r^{\frac12-H} dr &=  \int_0^s \frac{r^{\frac12-H}}{(s-r)^{-\frac12+H}} dr = \frac{\Gamma\left(\frac32-H\right)}{\Gamma\left(3-2H\right)} s^{2-2H}
\end{align*}

using (<ref>). This gives the bound

\begin{align*}
B_H^{2,1} &\leq (C')^2\|g\|_L^2 \frac{\left(H-\frac12\right)^2 }{\Gamma(3-2H)^2} \E\left[ (1+G_H)^2\right]\int_0^T s^{3-2H} ds \\
&= (C')^2 \|g\|_L^2 \frac{\left(H-\frac12\right)^2}{\Gamma(3-2H)^2}  \frac{1}{4-2H} T^{4-2H} \E\left[ (1+G_H)^2\right]\\
&\leq (C')^2 \|g\|_L^2 \frac{\left(H-\frac12\right)^2}{\Gamma(3-2H)^2}  \frac{1}{4-2H} T^{4-2H} \left( 2+ 128 (2T)^{\epsilon_H} \right),
\end{align*}

where we used Lemma <ref> in the last step. An application of this lemma is possible since $H>H-\epsilon_H>\frac12$. It follows that $B_H^{2,1}\rightarrow 0$ for $H\searrow\frac12$.

A similar procedure will be applied for $B_H^{2,2}$. For $0<\epsilon_H< \min\{H-\frac12,\frac14\}$, we estimate

\begin{align*}
\int_0^s \frac{\sigma |W_s^H - W_r^H|}{(s-r)^{\frac12+H}} r^{\frac12-H} dr &\leq \sigma G_H \int_0^s \frac{r^{\frac12-H}}{(s-r)^{\frac12 + \epsilon_H}} dr \\
&= \sigma G_H \frac{\Gamma\left(\frac32-H\right)}{\Gamma(2-H-\epsilon_H)} s^{1-H-\epsilon_H}
\end{align*}

with (<ref>) and the random variable $G_H$ given in (<ref>).

This gives rise to the bound

\begin{align*}
B_H^{2,2} \leq \|g\|_L^2 \frac{\sigma^2\left(H-\frac12\right)^2}{\Gamma(2-H-\epsilon_H)^2} \E\left[ G_H^2\right] \int_0^T s^{1-2\epsilon_H} ds.
\end{align*}

Again, we can apply Lemma <ref> to bound the expectation whereas it is simple to evaluate the integral. This together gives

\[ B_H^{2,2}\leq \|g\|_L^2 \frac{\sigma^2\left(H-\frac12\right)^2}{\Gamma(2-H-\epsilon_H)^2}\left( \frac{1}{2-2\epsilon_H} T^{2-2\epsilon_H} \right) \left( 64 (2T)^{\epsilon_H}\right) \]

and it now can be seen that $B_H^{2,2}\rightarrow 0$ for $H\searrow\frac12$. This establishes the convergence of the Itô integrals of the statement as the right-hand side of (<ref>) converges to zero for $H\searrow \frac12$.

To show

\[ \int_0^T v_H^g(s)^2 ds \longrightarrow_{L^2(\Pr)} \int_0^T g(X_s^b)^2 ds \]

as $H\searrow \frac12$, we will prove that the family

\begin{align}\label{eqp: G17}
\left( \left(\int_0^T v_H^g(s)^2 ds\right)^2\right)_{\frac12< H\leq \frac34}
\end{align}

is uniformly integrable. Then it suffices to establish (cf. [32], Theorem $5.12$)

\[ \int_0^T v_H^g(s)^2 ds \longrightarrow_\Pr \int_0^T g(X_s^b)^2 ds. \]

By the reverse triangle inequality as in (<ref>) and the reasoning thereafter, it is enough to show that $\|v_H^g - g(X_\cdot^b)\|_{L^2([0,T])}$ converges to zero in probability. For any $\epsilon>0$,

\[ \Pr\left( \|v_H^g - g(X_\cdot^b)\|_{L^2([0,T])} >\epsilon\right) \leq \frac{1}{\epsilon^2}\E\left[ \int_0^T \left( v_H^g(s) - g(X_s^b)\right)^2 ds\right] \]

by an application of Markov's inequality and convergence of the right-hand side to zero for $H\searrow \frac12$ has been shown above, starting in (<ref>). Consequently, it remains to prove uniform integrability of the family in (<ref>). Without the summand $-g(X_s^b)$ inside the integral in the left-hand side of (<ref>),

\[ \E\left[ \int_0^T  v_H^g(s)^2 ds \right] \leq 2A_H' + 2B_H, \]

\[ A_H' = \|g\|_\infty^2 \int_0^T \left(\frac{s^{\frac12-H}}{\Gamma\left(\frac32-H\right)}\right)^2 ds\]

and $B_H$ just beyond (<ref>).

Proceeding as above with $G_H$ in (<ref>) and $0<\epsilon_H< \min\{H-\frac12,\frac14\}$ therein,

\begin{align*} \nonumber
\int_0^T v_H^g(s)^2 ds & \leq 2A_H' + 4B_H^1 + 8 B_H^{2,1} + 8B_H^{2,2} \\ \nonumber
&\leq 2\|g\|_\infty^2 \left( \frac{T^{2-2H}}{(2-2H)\Gamma\left(\frac32-H\right)^2} - \frac{2 T^{\frac32-H}}{\left(\frac32-H\right)\Gamma\left(\frac32-H\right)} +T\right)&\hspace{1cm}\\\nonumber
&\hspace{0.5cm} + 4\frac{1}{(2-2H) \Gamma\left(\frac32-H\right)^2} \left( \frac{\Gamma\left(\frac32-H\right)^2}{\Gamma(2-2H)} -1\right)^2 \|g\|_\infty^2 T^{2-2H}\\\nonumber
&\hspace{0.5cm} + 8(C')^2\|g\|_L^2  \frac{\left(H-\frac12\right)^2}{\Gamma(3-2H)^2} \frac{1}{4-2H} T^{4-2H} (1+G_H)^2\\\nonumber
&\hspace{0.5cm} +8\|g\|_L^2\sigma^2  \frac{\left(H-\frac12\right)^2}{\Gamma(2-H-\epsilon_H)} \frac{1}{2-2\epsilon_H} T^{2-2\epsilon_H} G_H^2 \\
&\leq C(g,T) \left( 1+ G_H^2\right)
\end{align*}

for some constant $C(T,g)>0$ depending only on $T$ and $g$, where we used that uniformly over $H\in (\frac12,\frac34]$, the terms involving the Gamma function are bounded away from zero and infinity and the $T$-dependent terms are bounded. Hence, by Lemma <ref>, for any integer $p>1$,

\begin{align}\label{eqp: G18}
\begin{split}
\E\left[ \left(\int_0^T v_H^g(s)^{2} ds\right)^{2p}\right]& \leq 2^{2p}C(g,T)^{2p} \E\left[1+ G_H^{2p}\right] \\
&\leq 2^{2p}C(g,T)^{2p} \left( 1+  (32)^{2p} (2(T\vee 1))^{2p} \frac{(4p)!}{(2p)!}\right),
\end{split}
\end{align}

where we used that $\epsilon_H$ in the definition of $G_H$ was chosen to be $\leq\frac14$. The bound is independent of $H$ and by Corollary $6.21$ in [35] the family in (<ref>) is uniformly integrable which completes the proof for $H>\frac12$.

Remember that in accordance with the fractional version of Girsanov's theorem, Proposition <ref>, we denote

\[ Z^{1/2}(u_s) = \exp\left( \int_0^T u_s dW_s - \frac12 \int_0^T u_s^2 ds\right) \]

for the corresponding likelihood in the case $H=\frac12$.

Let $b\in\Sigma(C,A,\gamma,\sigma)\cap\mathcal{H}(\beta,L)$, $L>0$. For any bounded and Lipschitz continuous function $g:\R\rightarrow\R$ we have

\[ \E\left[\left| Z_T^H(g(X_\cdot^{H,b})) - Z_T^{1/2}(g( X_\cdot^b))\right|\right] \stackrel{H\to\frac12}{\longrightarrow} 0.\]

As $L^2(\Pr)$-convergence implies convergence in probability, we have by the continuous mapping theorem and Lemma <ref> that

\begin{align*}
&\exp\left( \int_0^T v_H^g(s) dW_s -\frac12\int_0^T v_H^g(s)^2 ds\right)\\
&\hspace{4cm} \stackrel{H\to\frac12}{\longrightarrow}_\Pr \exp\left(\int_0^T g(X_s^b) dW_s -\frac12\int_0^T g(X_s^b)^2 ds\right)
\end{align*}

for any bounded and Lipschitz continuous function $g$. We can conclude convergence in $L^1(\Pr)$ if the left-hand side is a uniformly integrable $H$-indexed family. We prove this by showing that its $L^2(\Pr)$-norm is uniformly bounded (see Corollary $6.21$ in [35]), i.e.

\begin{align}\label{eqp: G19}
\E\left[ \exp\left( 2\int_0^T v_H^g(s) dW_s -\int_0^T v_H^g(s)^2 ds\right)\right]\leq c
\end{align}

for some $c$ independent of $H$.

As $v_H^g(s)^2\geq 0$, it is enough to bound the expectation

\[ \E\left[ \exp\left(2\int_0^T v_H^g(s) dW_s\right)\right] \]

and we use monotone convergence and the Burkholder-Davis-Gundy inequality (cf. [32], Theorem $20.12$) to bound the moments of the integral. The latter gives

\[ \E\left[ \left| \int_0^T v_H^g(s) dW_s\right|^p\right] \leq c_p \E\left[ \left(\int_0^T v_H^g(s)^2 ds\right)^\frac{p}{2}\right].\]

For the constant $c_p$ we have $c_p \leq C'p$ for some constant $C'>0$ (see [6], Proposition $4.2$).
Now we again distinguish the cases $H<\frac12$ and $H>\frac12$. In the first one, we have

\[ \E\left[ \left(\int_0^T v_H^g(s)^2 ds\right)^\frac{p}{2}\right] \leq \|g\|_\infty^p  T^p, \]

which follows by the bound (<ref>) established in the proof of Lemma <ref> and hence by monotone convergence,

\begin{align*}
\E\left[ \exp\left(2\int_0^T v_H^g(s) dW_s\right)\right] &\leq  \sum_{k=0}^\infty \frac{1}{k!} \E\left[ \left| 2\int_0^T v_H^g(s)dW_s\right|^k\right]\\
&\leq 1+ \sum_{k=1}^\infty \frac{c_k 2^k\|g\|_\infty^k  T^k}{k!}\\
&\leq 1 + C'\sum_{k=1}^\infty \frac{(2\|g\|_\infty T)^k}{(k-1)!}\\
&\leq 1+ C'(2\|g\|_\infty T) e^{2\|g\|_\infty T} <\infty.
\end{align*}

In the case $H>\frac12$ we restrict to the interval $\frac12<H<\frac34$, which is possible as we consider convergence $H\searrow \frac12$. First, as above,

\[ \E\left[ \exp\left(\int_0^T v_H^g(s) dW_s\right)\right]\leq 1+C'\sum_{k=1}^\infty  \frac{k 2^k}{k!} \E\left[ \left(\int_0^T v_H^g(s)^2 ds\right)^\frac{k}{2}\right]. \]

Next, using that for any random variable $X\in L^q(\Pr)$ and $1\leq p\leq q$,

\[ \E\left[ |X|^p\right] \leq \E\left[ |X|^q\right]^{p/q} \leq 1 + \E\left[ |X|^q\right],  \]

and (<ref>) subsequently,

\begin{align*}
&C'\sum_{k=1}^\infty  \frac{k 2^k}{k!} \E\left[ \left(\int_0^T v_H^g(s)^2 ds\right)^\frac{k}{2}\right] \\
&\hspace{0.5cm}\leq  4C'\sum_{k=1}^\infty \frac{4k 2^{4k}}{(4(k-1))!} \left( \E\left[ \left(\int_0^T v_H^g(s)^2 ds\right)^{2k}\right] +1\right) \\
&\hspace{0.5cm}\leq  4C'\sum_{k=1}^\infty \frac{4k 2^{4k}}{(4(k-1))!} \left( 2^{2k}C(T,g)^{2k}\left( 1+ (32)^{2k} (2(T\vee 1))^{2k}\frac{(4k)!}{(2k)!}\right) +1 \right) \\
&\hspace{0.5cm}\leq 4C'\sum_{k=1}^\infty \frac{(4k)^5}{(2k)!}\left( 2^5 C(g,T) 32 (T\vee 1) +2^2\right)^{2k}\\
&\hspace{0.5cm}\leq 4^6C'\sum_{k=1}^\infty \frac{k^5}{k!}\frac{1}{k!}\left(\left( 2^5 C(g,T) 32 (T\vee 1) +2^2\right)^2\right)^{k},
\end{align*}

where convergence of the series follows by a comparison with the exponential series as $k^5/k!$ is bounded.

In consequence, (<ref>) holds for both cases $H<\frac12$ and $H>\frac12$, which establishes the $L^1(\Pr)$-convergence of the Girsanov densities for a bounded and Lipschitz continuous function $g:\R\rightarrow\R$.

As in the proof of Theorem <ref>, we bound

\begin{align}\label{eqp: G25}
\begin{split}
&\sup_{\psi_T^H} \inf_{\substack{b\in H_1(b_0,\eta)\cap \{b-b_0\in\mathcal{H}(\beta,L)\}:\\ \Delta_J(b)\geq (1-\epsilon_T)c_*\delta_T}}\  \E \left[\psi_T^H(X^{H,b})\right] - \alpha \\
& \hspace{2cm}\leq  \E\left[\left(\frac{1}{N_T}\sum_{k=1}^{N_T} Z_T^H\left( (b_k-b_{0,\eta})(X^{H,b_{0,\eta}})\right) -1\right)\phi_T^H\right] \\
&\hspace{2cm} \leq \E\left[\left|\frac{1}{N_T}\sum_{k=1}^{N_T} Z_T^H\left( (b_k-b_{0,\eta})(X^{H,b_{0,\eta}})\right) -1\right|\right]
\end{split}
\end{align}

and in particular choose the same hypothesis $b_k$ as in this proof, which are bounded by one and Lipschitz continuous.
Now, by adding zero, this term is again bounded by the sum of

\[\E\left[\left|\frac{1}{N_T}\sum_{k=1}^{N_T} Z_T^{1/2}\left( (b_k-b_{0,\eta})(X^{b_{0,\eta}})\right) -1\right|\right] \]

\begin{align}\label{eqp: G23}
\begin{split}
& \E\left[ \left| \frac{1}{N_T} \sum_{k=1}^{N_T} \left( Z_T^H\left( (b_k-b_{0,\eta})(X^{H,b_{0,\eta}})\right)-Z_T^{1/2}\left( (b_k-b_{0,\eta})(X^{b_{0,\eta}})\right)\right)\right| \right]\\
&\hspace{1cm}\leq \frac{1}{N_T} \sum_{k=1}^{N_T} \E\left[ \left| Z_T^H\left( (b_k-b_{0,\eta})(X^{H,b_{0,\eta}})\right)-Z_T^{1/2}\left( (b_k-b_{0,\eta})(X^{b_{0,\eta}})\right)\right|\right].
\end{split}
\end{align}

Let $\epsilon>0$. Then, by the proof of Theorem <ref> in combination with Remark <ref>,

\begin{align}\label{eqp: G24}
\limsup_{T\to\infty} \E\left[\left|\frac{1}{N_T}\sum_{k=1}^{N_T} Z_T^{1/2}\left( (b_k-b_{0,\eta})(X^{b_{0,\eta}})\right) -1\right|\right] <  \frac{\epsilon}{2},
\end{align}

By Proposition <ref> there exists $\delta_k(T,\epsilon)>0$ such that for $|H-\frac12|<\delta_k(T,\epsilon)$,

\[ \E\left[\left| Z_T^H\left( (b_k-b_{0,\eta})(X^{H,b_{0,\eta}})\right)-Z_T^{1/2}\left( (b_k-b_{0,\eta})(X^{b_{0,\eta}})\right)\right|\right] <\frac{\epsilon}{2} \]

for every $k=1,\dots, N_T$.
Define $\delta(T) := \delta(T,\epsilon):=\min_{1\leq k\leq N}\delta_k(T,\epsilon)>0$. Then,

\[ \sup_{|H-\frac12|<\delta(T)} \max_{1\leq k\leq N_T} \E\left[\left| Z_T^H\hspace{-0.05cm} \left( (b_k-b_{0,\eta})(X^{H,b_{0,\eta}})\right)\hspace{-0.05cm}-\hspace{-0.05cm} Z_T^{1/2}\hspace{-0.05cm}\left( (b_k-b_{0,\eta})(X^{b_{0,\eta}})\right)\right|\right] \]

is bounded from above by $\frac{\epsilon}{2}$. In particular, $\sup_{|H-\frac12|<\delta(T)} $ of the right-hand side of (<ref>) is bounded from above by $\frac{\epsilon}{2}$ and the claim of the theorem follows with (<ref>) and (<ref>).

Institute of Mathematics

University of Freiburg

Ernst-Zermelo-Strasse 1

79104 Freiburg

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