&\hspace{1cm} =  \sup_{b\in\Sigma(C,A,\gamma,\sigma)}\Pr_b\left( \sup_{f\in\mathcal{F}}\left|\int_\R f(z) \left( \frac{1}{\sigma^2 T}L_T^z(X) -q_b(z)\right) dz \right| >\epsilon\right) \\
&\hspace{1cm} \leq \sup_{b\in\Sigma(C,A,\gamma,\sigma)}\Pr_b\left(  \left\| \frac{1}{\sigma^2 T}L_T^\cdot(X) -q_b\right\|_{[-A,A]}\ \sup_{f\in\mathcal{F}}\int_\R |f(z)| dz  >\epsilon\right) \\
&\hspace{1cm} \leq \sup_{b\in\Sigma(C,A,\gamma,\sigma)}\Pr_b\left(  \left\| \frac{1}{\sigma^2 T}L_T^\cdot (X) -q_b\right\|_{[-A,A]}  >\frac{\epsilon}{2A C_\mathcal{F}}\right),
\end{align*}

where we used that by assumption $\int_\R |f(z)|dz \leq 2A C_{\mathcal{F}}$ uniformly in $f\in\mathcal{F}$. Now, the last probability converges to zero for $T\to\infty$ due to Corollary <ref> and the claim follows.

With $b_0 \in \textrm{Lip}_\textrm{loc}(\R)$, we have $b_0\pm\eta\in \textrm{Lip}_\textrm{loc}(\R)$. Next,

\begin{align*}
|b_0(x)\pm\eta| \leq |b_0(x)| + \eta \leq \left(\frac{C}{2} -\eta\right) (1+|x|) + \eta \leq \frac{C}{2}(1+|x|).
\end{align*}

Finally, for $|x|\geq A$, we have $\sigma^{-2}b_0(x)\textrm{sign}(x)\leq -(\gamma+\frac{\eta}{\sigma^2})$ and hence for those $x$,

\begin{align*}
\sigma^{-2}(b_0(x)\pm\eta)\textrm{sign}(x) &= \sigma^{-2}b_0(x)\textrm{sign}(x) \pm\sigma^{-2}\textrm{sign}(x)\eta \\
&\leq -\gamma - \sigma^{-2}\eta +\sigma^{-2}\eta = -\gamma.
\end{align*}

§.§ Generalization of Proposition <ref>

In this subsection we will generalize Proposition $\ref{moments_local-invariant}$ to diffusions that are not started in the invariant distribution. To this aim we consider a diffusion $X^{x_0} = (X_t^{x_0})_{t\geq 0}$ solving

\[ dX_t^{x_0} = b(X_t^{x_0}) dt + \sigma dW_t, \quad X_0^{x_0}= x_0, \]

for some fixed initial value $x_0\in [-A,A]$. We denote by $\xi$ a random variable that is distributed under the invariant law $\mu_b$ as in the preceeding subsections. Furthermore, we set $\tilde{G}:= \sup_{b\in\Sigma(C,A,\gamma,\sigma)} C_{b,\sigma^2}$ which is finite as seen in (<ref>) and define

\begin{align}\label{eqp: constant_G}
G:= 2\tilde G \sup_{b\in\Sigma(C,A,\gamma,\sigma)} \int_{-A}^A \exp\left( -\int_0^x 2\sigma^{-2} b(u) du \right) dx,
\end{align}

where we see that the integral is uniformly bounded over $b\in\Sigma(C,A,\gamma,\sigma)$ in the same way as in (<ref>).

For the proof of Proposition <ref>, the stationarity assumption is used in the form of the moment bound

\[ \E_b\left[ |X_t|^p\right] = \E_b\left[ |\xi|^p\right] \leq c_{mo}^p p^p \]

for some constant $c_{mo}\geq 0$, see Lemma $17$ in [1]. In order to derive the result for $X^{x_0}$, we first show that a moment inequality of the form

\[ \E_b\left[ |X_t^{x_0}|^p\right] \leq \C_{mo}^p p^p \]

holds for a different constant $\C_{mo}>0$, see Lemma <ref>. Then the result of Proposition <ref> can be transferred to $X^{x_0}$ as this result only uses the order $p^p$ of the moment bound which is the same for $X_0=x_0$ and $X_0=\xi\sim\mu_b$. This is described in detail at the end of the subsection.

Let $f:\R\rightarrow\R$ be a measurable function and $\xi\sim\mu_b$. Then for the constant $G$ given in (<ref>), we have

\[ \left| \E_b\left[ \int_0^T f(X_t^{x_0}) dt\right] - T\E_b\left[ f(\xi)\right]\right| \leq G\E_b\left[ |f(\xi)|\right]. \]

We know that the claim is true for $f\geq 0$, which was shown in Lemma $3.4.8$ in [40]. It is important to note that we used the exact constant in equation $(4.21)$ in the reference, by which

\[ \E_b[\tau]\leq 2(a-c)\tilde G = 2 \tilde G\int_{-A}^A \exp\left( -\int_0^x 2\sigma^{-2} b(u) du\right) dx\]

in the proof of Lemma $3.4.8$, where this upper bound of $\E_b[\tau]$ was used and yields the constant $G$ within the claim.

For the general case, we split $f$ into its positive and negative part, i.e. $f = f^+ - f^-$. As $f^+, f^-\geq 0$, we get

\begin{align*}
&\left| \E_b\left[ \int_0^T f(X_t^{x_0}) dt\right] - T\E_b\left[ f(\xi)\right]\right| \\
&\hspace{1cm} = \left| \E_b\left[ \int_0^T f^+(X_t^{x_0}) - f^-(X_t^{x_0}) dt\right] - T\E_b\left[ f^+(\xi) - f^-(\xi)\right]\right|  \\
&\hspace{1cm} \leq \left| \E_b\left[ \int_0^T f^+(X_t^{x_0}) dt\right] - T\E_b\left[ f^+(\xi)\right]\right|\\
&\hspace{3cm} + \left| \E_b\left[ \int_0^T f^-(X_t^{x_0}) dt\right] - T\E_b\left[ f^-(\xi)\right]\right| \\
&\hspace{1cm} \leq G\E_b\left[ f^+(\xi)\right] + G\E_b\left[ f^-(\xi)\right] \\
&\hspace{1cm} = G\E_b\left[ |f(\xi)|\right].
\end{align*}

We assume the following lemma to be commonly known, but as we did not find a reference for it, a proof is given.

For the absolute value function $|\cdot|:\R\rightarrow\R$ the following holds true:

(i) The function $x|x|$ is continuously differentiable.

(ii) For $n$ odd and $n\geq 3$, the function $|x|^n$ is twice continuously differentiable with

\begin{align*}
\frac{d}{dx} |x|^n = nx|x|^{n-2}, \quad \textrm{ and }\quad
\frac{d^2}{dx^2} |x|^n = n(n-1)|x|^{n-2}.
\end{align*}

(i) It is clear that $x|x|$ is differentiable for $x\neq 0$ with

\begin{align*}
\frac{d}{dx} (x|x|) = \begin{cases}
2x & \textrm{ for } x>0, \\
-2x & \textrm{ for } x<0.
\end{cases}
\end{align*}

For $x=0$, the left- and right-hand limits of the difference quotient have to be identical. This is the case, as

\[ \lim_{h\searrow 0} \frac{h(-h)}{h} = 0 = \lim_{h\nearrow 0}\frac{h^2}{h}.\]

In particular, we see that the derivative is continuous in $x=0$.

(ii) This follows by usual differentiation rules and the fact that $\frac{d}{dx}|x|=\frac{x}{|x|}$ for $x\neq 0$. In $x=0$ we consider the limit of the difference quotient as in (i), which exists and equals zero.

Let $\xi\sim \mu_b$. Then for $n\geq 2$,

\[ n\E_b\left[ |\xi|^{n-2}\xi b(\xi)\right] + \frac{n(n-1)}{2} \sigma^2 \E_b\left[ |\xi|^{n-2}\right] = 0.\]

The function $|x|^{n-2}x$ is differentiable with derivative

\begin{align*}
\frac{d}{dx}\left( |x|^{n-2}x\right) &= |x|^{n-2} + x (n-2) |x|^{n-3}\frac{x}{|x|}\\
&= |x|^{n-2} + (n-2) |x|^{n-3} \frac{|x|^2}{|x|}\\
&= (n-1)|x|^{n-2}.
\end{align*}

This is clear for $n$ even and can be seen by Lemma <ref> otherwise. By the relation $q_b'(x) = 2\sigma^{-2}b(x)q_b(x)$ and partial integration,

\begin{align*}
n\E_b\left[ |\xi|^{n-2}\xi b(\xi)\right]  &= n\int_\R |x|^{n-2}x b(x) q_b(x) dx\\
&= \frac{n\sigma^2}{2} \int_\R  x|x|^{n-2} q_b'(x) dx \\
&= \frac{n\sigma^2}{2}\left( \left[ x|x|^{n-2}q_b(x)\right]_{-\infty}^\infty - \int_\R (n-1)|x|^{n-2} q_b(x) dx\right) \\
&= -\frac{\sigma^2}{2} n(n-1) \E_b\left[ |\xi|^{n-2}\right].
\end{align*}

In the last step we used that $\lim_{x\to\pm\infty} x|x|^{n-2}q_b(x) =0$, which follows from the fact that by (<ref>) for some constant $\tilde c$,

\begin{align*}
q_b(x) &= \frac{1}{C_{b,\sigma}\sigma^2} \exp\left( \int_0^x \frac{2b(y)}{\sigma^2} dy\right)\\
& \leq \frac{1}{C_{b,\sigma}\sigma^2} \exp\left( \tilde c -2\gamma (|x|-A)_+\right).
\end{align*}

There exists a constant $\C_{mo}>0$ such that for all $p\geq 1$,

\[ \E_b\left[ |X_t^{x_0}|^p\right]^\frac1p \leq \C_{mo} p \quad \textrm{ for all } t\geq 0.\]

First, let $p\geq 2$. Then we have by the Itô formula and Lemma <ref>,

\begin{align*}
|X_t^{x_0}|^p &= |x_0|^p + p \int_0^t X_s^{x_0} |X_s^{x_0}|^{p-2} dX_s^{x_0} + \frac{p(p-1)}{2}\int_0^t |X_s^{x_0}|^{p-2} d\langle X^{x_0}\rangle_s \\
&= |x_0|^p + p\sigma \int_0^t X_s^{x_0} |X_s^{x_0}|^{p-2} dW_s + p\int_0^t X_s^{x_0}|X_s^{x_0}|^{p-2} b(X_s^{x_0}) ds\\
&\hspace{2cm} + \frac{p(p-1)}{2}\int_0^t \sigma^2 |X_s^{x_0}|^{p-2} ds.
\end{align*}

Consequently, the following is true for the expectation, where we apply Lemma <ref> for adding zero in the second step and Lemma <ref> in the last one:

\begin{align*}
&\E_b\left[|X_t^{x_0}|^p\right] \\
&\hspace{0.1cm}= |x_0|^p + p \E_b\left[ \int_0^t X_s^{x_0}|X_s^{x_0}|^{p-2} b(X_s^{x_0}) ds\right] + \frac{p(p-1)}{2} \E_b\left[ \int_0^t \sigma^2 |X_s^{x_0}|^{p-2} ds\right] \\
&\hspace{0.1cm}= |x_0|^p - t\left( p\E_b\left[ |\xi|^{p-2}\xi b(\xi)\right] + \frac{p(p-1)}{2}\E_b\left[ \sigma^2 |\xi|^{p-2}\right]\right) \\
&\hspace{1cm} +  p \E_b\left[ \int_0^t X_s^{x_0}|X_s^{x_0}|^{p-2} b(X_s^{x_0}) ds\right] + \frac{p(p-1)}{2} \E_b\left[ \int_0^t \sigma^2 |X_s^{x_0}|^{p-2} ds\right]  \\
&\hspace{0.1cm}\leq  |x_0|^p + p\left| \E_b\left[ \int_0^t X_s^{x_0}|X_s^{x_0}|^{p-2} b(X_s^{x_0}) ds \right] - t\E_b\left[ |\xi|^{p-2}\xi b(\xi)\right]\right| \\
&\hspace{1cm} + \frac{p(p-1)}{2}\left| \E_b\left[\int_0^t \sigma^2 |X_s^{x_0}|^{p-2}ds\right] - t\E_b\left[ \sigma^2|\xi|^{p-2}\right]\right|\\
&\hspace{0.1cm}\leq |x_0|^p + pG \E_b\left[ |\xi^{p-1} b(\xi)|\right] + \frac{p(p-1)}{2} G\E_b\left[ \sigma^2 |\xi|^{p-2}\right].
\end{align*}

Using the $L^p(\Pr_b)$-bound of $\xi$ from Lemma $17$ in [1], there exists a constant $c_{mo}\geq 1$ independent of $b$ such that

\[ \E_b\left[\sigma^2 |\xi|^{p-2}\right] \leq  c_{mo}^{p-2}\sigma^2 (p-2)^{p-2}.\]

By the at most linear growth condition and the same lemma, we may further conclude

\begin{align*}
\E_b\left[ |\xi^{p-1} b(\xi)|\right] &\leq \E_b\left[ |\xi|^{p-1} C(1+|\xi|)\right] \\
&= C\E_b\left[ |\xi|^{p-1} + |\xi|^p\right] \\
&\leq C \left( c_{mo}^{p-1} (p-1)^{p-1} + c_{mo}^p p^p\right) \\
&\leq 2C c_{mo}^p p^p.
\end{align*}

In consequence,

\begin{align*}
\E_b\left[ |X_t^{x_0}|^p\right] &\leq |x_0|^p + p2Cc_{mo}^p p^p G + G\frac{p(p-1)}{2} \sigma^2 c_{mo}^{p-2} (p-2)^{p-2} \\
&\leq |x_0|^p + 2CG c_{mo}^p p^{p+1} + \frac12 G\sigma^2 c_{mo}^{p-2} p^p \\
&\leq |x_0|^p + p^{p+1}  G c_{mo}^p \left( 2C + \sigma^2 2^{-1}c_{mo}^{-2}\right).
\end{align*}

Now using $(a^p+b^p)^\frac1p \leq a+b$ for $a,b\geq 0$, $|x_0|\leq A$ and $p^\frac1p \leq 2$, we arrive at

\begin{align*}
\E_b\left[ |X_t^{x_0}|^p\right]^\frac1p &\leq \left(  |x_0|^p + p^{p+1}  G c_{mo}^p \left( 2C + \sigma^2 2^{-1}c_{mo}^{-2}\right) \right)^\frac1p \\
&\leq |x_0| + p^{1+1/p} c_{mo} \left( 2GC + G\sigma^2 2^{-1}c_{mo}^{-2}\right)^\frac1p \\\
&\leq |x_0| + 2p c_{mo}\max\left\{ \left( 2GC + G\sigma^2 2^{-1}c_{mo}^{-2}\right),1\right\} \\
&\leq p\left( A + 2c_{mo}\max\left\{ \left( 2GC + G\sigma^2 2^{-1}c_{mo}^{-2}\right),1\right\} \right).
\end{align*}

Thus, we have the desired inequality $\E_b\left[|X_t^{x_0}|^p\right]^\frac1p\leq \C_{mo}' p$ for all $p\geq 2$ and $\C_{mo}' = A + 2c_{mo}\max\left\{ \left( 2GC + G\sigma^2 2^{-1}c_{mo}^{-2}\right),1\right\} $. For $1\leq p<2$,

\[ \E_b\left[ |X_t^{x_0}|^p\right] \leq \E_b\left[ |X_t^{x_0}|^{2p}\right]^\frac12 \leq \sqrt{ (\C_{mo}')^{2p} (2p)^{2p}} \leq (2\C_{mo}')^p p^p\]

and consequently $\E_b\left[ |X_t^{x_0}|^p\right]^\frac1p \leq 2\C_{mo}' p$. Combining both cases for the value of $p$, we arrive at the claim by setting $\C_{mo} = 2\C_{mo}'$.

We now turn to the necessary modifications in the proofs of [1] to adapt them to the case of the diffusion $X^{x_0}$ that is not started in the invariant law but at a fixed point $x_0$. The following numbers of sections and results refer to those in [1]. We first remark that the results of Section $2$ and $3$ in this paper are formulated in a general setting of continuous semimartingales, which means nothing has to be adapted to our setting here.

(1) First, we check Assumption $1$. As our diffusion coefficient is constant, the function $\phi_2(t) := \sigma\sqrt{t}$ can be used for the same reason as in the stationary case considered in the paper. With the at most linear growth condition on $b$ and Lemma $3.4.8$ in [40],

\begin{align*}
&\|x_0\|_{L^p(\Pr_b)} + \left\| \int_0^t |b(X_s^{x_0})|ds\right\|_{L^p(\Pr_b)}\\
&\hspace{1.5cm}\leq |x_0| + C\E_b\left[  t^p\left( \frac{1}{t} \int_0^t 1+|X_s^{x_0}| ds \right)^p\right]^\frac1p \\
&\hspace{1.5cm}\leq |x_0| + Ct \E_b\left[ \frac1t \int_0^t (1+|X_s^{x_0}|)^p ds \right]^\frac1p\\
&\hspace{1.5cm}\leq |x_0| + C t^{1-1/p} \left( (t+G) \E_b\left[ (1+|\xi|)^p\right]\right)^\frac1p\\
&\hspace{1.5cm}\leq |x_0| + 2C t (1+G)\left(1+c_{mo}p\right)
\end{align*}

for the constant $c_{mo}$ from Lemma $17$ in [1] that is independent of $b$. Hence, we may choose $\phi_1$ as

\[ \phi_1(t) := \max\left\{ |x_0| + 2C(1+G)(1+c_{mo}), \sigma^2\right\} t,\]

where the maximum ensures that $\phi_2(t)\leq \sqrt{\phi_1(t)}$.

(2) Corollary $6$ directly transfers to $X^{x_0}$ as it only uses results from Section $2$ and the validity of Assumption $1$ that was checked in (1).

(3) Next, we check Proposition $7$. The first step where stationarity is needed is from (D.3) to (D.4). By Jensen's inequality  and Lemma $3.4.8$ in [40] we get for $p\geq 2$,

\begin{align*}
t^{-p/2} \E_b\left[ \left(\int_0^t (1+X_s^{x_0})^{2\eta}) ds\right)^\frac{p}{2}\right] &\leq \E_b\left[ \frac1t \int_0^t (1+|X_s^{x_0}|)^{p\eta} ds\right] \\
&\leq (1+G) \E_b\left[ (1+|\xi|)^{p\eta}\right] \\
&\leq (1+G) 2^{p\eta}\left(1+(c_{mo} p\eta)^{p\eta}\right) \\
&\leq p^{p\eta} (1+G) 2^{p\eta}\left(1+(c_{mo} \eta)^{p\eta}\right),
\end{align*}

where we used in the third step that $\E_b[|\xi|^p]\leq (c_{mo}p)^p$ from Lemma $17$ in [1].

The inequality for $1\leq p<2$ can be modified in the same way. Then the moment bound of $|\mathbb{M}_t^f|$ follows with an additional factor $2^{\eta+1/2}(1+G)$, but the same exponent of $p$.

Bounding the remainder term $\R_t^f$ follows the same lines as in the original proof and leads to the upper bound

\[ \mathcal{S} \overline{\Lambda}_{\textrm{prox}}\left( x_0 +\frac{1}{\eta+1} x_0^{\eta+1} + \|X_t^{x_0}\|_{L^p(\Pr_b)} + \frac{1}{\eta+1}\|X_t^{x_0}\|_{L^p(\Pr_b)}\right). \]

Now, we use the $L^p(\Pr_b)$-bound of Lemma <ref> instead of the moment bound $c_{mo}p$ of Lemma $17$ in [1] and have the same upper bound as in the paper concerning the order of $p$ and $\mathcal{S}$, but with modified constant.

The estimates for $\mathbb{M}_t^{b_0}$ and $\R_t^{b_0}$ in the second part of the proof where $f=b_0$ can be modified in the same way.

(4) The proof of Theorem $9$ can be completely transfered, one just has to change the constants as Proposition $7$ is used. Note however, that the important exponents of $p$ and $\mathcal{S}$ did not have changed there.

(5) The proof of Theorem $10$ carries over (with modified constants) as it only uses Corollary $6$ and Proposition $7$ which are both available as seen in (2) and (3) with modified constants. The important dependence on $p$ remains the same as this was the case in both other results.

(6) Corollary $12$ is a consequence of Theorem $9$ and thus still valid with modified constants, see (4).

(7) Most of the proof of Theorem $13$ does not take into account the distribution of $X_0$. Just in the final steps where Proposition $7$ and Theorem $10$ are applied, we have to take their (modified) constants. Furthermore, as $X_0$ is not distributed under the invariant measure, we have to substitute it by $\xi$ in the deviation expression

\[ \sup_{b\in\Sigma} \Pr_b\left( \left\| \sqrt{t} \left( \frac1t \int_0^t f(X_s) dX_s - \E_b[f(\xi) b(\xi)]\right) \right\|_{\mathcal{F}_{K,h}} \geq \phi(u)\right) \leq e^{-u}.\]

(8) The proof of Corollary $14$, which is our desired moment estimate of the deviation of the local time from the invariant density (Proposition <ref>), carries over from Corollary $12$ and Theorem $13$.

§ PROOF OF THEOREM <REF> AND GENERAL MULTISCALE THEORY

This part of the appendix is organized in the following way: In Subsection <ref> we present the multiscale theory that is used in our context, where the key result is Theorem <ref>. In Subsection <ref> we then provide the proof of Theorem <ref> as an application of this Theorem <ref>. In addition, we use it to prove boundedness of the random variable $S_b$ given in Theorem <ref>.

§.§ Auxiliary results from empirical process theory

The key ingredient for the proof of Theorem <ref> will by a general result about the supremum of a stochastic process $X=(X_s)_{s\in\mathcal{S}}$ that is defined on a totally bounded metric space $(\mathcal{S},\rho)$. Totally bounded means that for arbitrary $u>0$ the capacity number

\[ D(u,\mathcal{S},\rho) := \max\left\{ \#\mathcal{S}_0: \mathcal{S}_0\subset\mathcal{S}, \rho(s,t)>u\textrm{ for different }s,t\in\mathcal{S}_0\right\}\]

is finite, where $\#S$ denotes the cardinality of a finite set $S$. Additionally, we consider a function $\sigma:\mathcal{S}\rightarrow (0,1]$, where $\sigma(s)$ measures the spread of $X(s)$. We assume that

\begin{align}\label{eqp: B2}
|\sigma(t) - \sigma(s)| \leq \rho(s,t)\quad \textrm{ for all }\ s,t\in\mathcal{S}
\end{align}

\begin{align}\label{eqp: B12}
\left\{ t\in\mathcal{S}:\ \sigma(t)\geq \delta\right\}
\end{align}

is compact for any $\delta\in (0,1]$. A result of this kind was first established by Dümbgen and Spokoiny in [20] to derive a multiscale test in a regression setting and substantially refined by Dümbgen and Walther in [19]. A setup with random metrics and time-dependent metric spaces $\mathcal{S}_T$ was treated by Rohde in [53]. In our scenario, the metric $\rho$ and the spread measure $\sigma$ are random and depend on the time horizon $T$, but are defined on the same fixed metric space $\mathcal{S}$ for all $T$. We follow the proofs given in technical report [19] to weaken the assumption on the tails of $X$, by allowing for an additional $\log$-factor compared to Theorem $8$ and Corollary $9$ in this reference [19].

Let $L>0$ be some constant, and for $\delta>0$ let $G(\cdot, \delta)$ be a nondecreasing function on $[0,\infty)$ such that for all $\lambda\geq 0$ and $s,t\in\mathcal{S}$ with $\rho(s,t)\geq\delta$ we have

\[ \Pr\left( \frac{|X(s)-X(t)|}{\rho(s,t)}> G(\lambda,\delta)\right) \leq L e^{-\lambda}.\]

Then for arbitrary $\delta >0$ and $a\geq 1$

\[ \Pr\left( |X(s)-X(t)| \geq 12 J(\rho(s,t),a)\textrm{ for some } s,t\in\mathcal{S}^*\textrm{ with }\rho(s,t)\leq \delta\right)\leq \frac{L\delta}{2a},\]

where $\mathcal{S}^*$ is a dense subset of $\mathcal{S}$, and

\[ J(\epsilon,a) := \int_0^\epsilon G\left(\log\left(\frac{aD(u,\mathcal{S},\rho)^2}{u}\right),u\right) du.\]

The following theorem is a special case of Theorem $8$ in the technical report [19], besides the somewhat relaxed assumption in $(i)$. The proof of our version is given below for the reader's convenience, where the key idea follows the lines in [19]. See the subsequent Remarks <ref> and <ref> for a discussion, why this result is interesting on its own.

Suppose that the following two conditions are satisfied:

(a) There exists a constant $L>0$ such that for arbitrary $\lambda \geq 1$ we have

\[ \Pr\left( |X(s)\geq \sigma(s) \lambda \right)\leq L\exp\left( -\frac{\lambda^2}{2} + \log(\lambda)\right).\]

(b) There exist positive constants $A_0,B,V>0$ such that

\[ D\left( u\delta, \{ t\in\mathcal{S}: \sigma(s)\leq\delta\}, \rho\right) \leq A_0 u^{-B}\delta^{-V}\quad \textrm{ for all }u,\delta\in (0,1].\]

For a constant $Q>0$ define the events

\[ \mathcal{A}(Q,\delta) :=\left\{  \sup_{s,t\in\mathcal{S}: \rho(s,t)\leq\delta}\frac{|X(s)-X(t)|}{\rho(s,t) \log(e/\rho(s,t))} \leq Q\right\}.\]

Then there exists a constant $C_0=C_0( A_0,B,V,Q)>0$ and $0<\delta_0\leq 1$ such that for $0<\delta\leq \delta_0$ the probability of the event

\begin{align*}
&\left\{ |X|\leq \sigma \sqrt{ 2V\log(1/\sigma) + 2C_0\log\log(e/\sigma)} + C_0\sigma\log(e/\sigma)^{-1} \right.\\
&\hspace{7cm}\left. \textcolor{white}{\sqrt{V}}\textrm{ on } \{ s\in\mathcal{S}: \sigma(s)\leq\delta\}\right\}
\end{align*}

is at least $\Pr(\mathcal{A}(Q,2\delta)) - C_0\log(e/(2\delta))^{-\frac12}$.

The preceeding theorem can be stated and proven in greater generality and basically in the setting of Section $7$ in the technical report [19]. In this setup, we consider a function $G:[0,\infty)\times (0,1]\rightarrow [0,\infty)$ with

\[G_0 := \sup_{\lambda\geq 0, 0<\delta\leq 1} G(\lambda,\delta)<\infty\]

and replace the first condition in Theorem <ref> by

\[ \Pr\left( |X(s)| \geq \sigma(s) G(\lambda,\delta)\right) \leq L\exp\left( -\lambda +\frac12\log(2\lambda)\right)\]

for all $s\in\mathcal{S}$ with $\sigma(s)\geq \delta$ and $\delta \in (0,1]$. Then the probability of

\begin{align*}
&\left\{ |X|\leq \sigma G(V\log(1/\sigma) + 2C_0\log\log(e/\sigma), \sigma) + C_0\sigma\log(e/\sigma)^{-1}\right. \\
&\hspace{8cm}\left. \textrm{ on } \{ s\in\mathcal{S}: \sigma(s)\leq\delta\}\right\}
\end{align*}

is at least $\Pr(\mathcal{A}(Q,2\delta) - C_0\log(e/(2\delta))^{-\frac12}$.
Our formulation in Theorem <ref> is the special case $G(\lambda,\delta) = \sqrt{2\lambda}$.

The proof will show that an additional term $+c\log(\lambda)$ in condition (a) of Theorem <ref> can only be added to the exponent $-\lambda^2/2$ for $c <2$. In the general setting of Remark <ref>, an additional term $+c'\log(2\lambda)$ can only be added for a factor $c'<1$.

We define a set $\mathcal{S}^*$ inductively as follows: Let $s_1$ be any point in $\mathcal{S}$ maximizing $\sigma$. Next, let $u$ be some continuous, non-decreasing function from $(0,1]$ into iteself that will be specified later. Suppose we already picked $s_1,\dots, s_m$ and the set

\[\mathcal{S}':= \left\{ s\in\mathcal{S}:\ \min_{i=1,\dots, m} \rho(s,s_i)\geq u(\sigma(s)) \sigma(s)\right\}\]

is non-empty. Then we define $s_{m+1}$ to be an element of it with maximal value $\sigma(s)$, which is well-defined as the displayed set $\mathcal{S}'$ is closed and $\{ \sigma\geq\delta\}$ is compact for any $\delta>0$ by (<ref>). Following this procedure we end up with a finite or countable set $\mathcal{S}^*=\{ s_1,s_2,\dots\}$ and its construction entails that

\[ \sigma(s_1)\geq \sigma(s_2)\geq \dots\]

For $0<\delta\leq 1$ define

\[ \mathcal{S}^*(\delta) := \left\{ s\in\mathcal{S}^*:\ \frac{\delta}{2}<\sigma(s)\leq\delta\right\}, \]

which is contained in $\{ s\in\mathcal{S}: \sigma(s)\leq\delta\}$ with $\rho(s,t)\geq u(\delta/2)\frac{\delta}{2}$ for different $s,t\in\mathcal{S}^*(\delta)$. Consequently, we have from assumption (b) that

\begin{align}\label{eqp: B1}
\#\mathcal{S}^*(\delta) \leq A_02^B u(\delta/2)^{-B} \delta^{-V}.
\end{align}

In particular, we have $\lim_{n\to\infty}\sigma(s_n)=0$ in the case $\mathcal{S}^*$ is infinite. An important property of this set $\mathcal{S}^*$ is that for any $s\in\mathcal{S}$ there exists a point $t\in\mathcal{S}^*$ such that

\begin{align}\label{eqp: B18}
\sigma(s)\leq \sigma(t)\quad \textrm{ and }\quad \rho(s,t) < u(\sigma(s))\sigma(s).
\end{align}

This can be seen in the following way: for $s\in\mathcal{S}$ let $m$ be a maximal index such that $\sigma(s_m)\geq \sigma(s)$. If $\rho(s,s_i)\geq u(\sigma(s))\sigma(s)$ for every $i\leq m$, then $s$ would belong to the set $\mathcal{S}'$, whence $\sigma(s_{m+1})\geq \sigma(s)$ contradicting the definition of $m$.

In order to bound $\frac{|X(t)|}{\sigma(t)}$ for all $t\in\mathcal{S}^*$ we define

\[ H(t) := \sqrt{ 2V\log(1/\sigma(t)) + 2B\log(1/u(\sigma(t))) +4\log\log(e/\sigma(t)) }\]

\[ u(\delta) := \log(e/\delta)^{-2}.\]

With this choice of $u$ it is possible to choose $\delta_0'$ small enough such that $H(t)\geq 1$ for $\sigma(t)\leq \delta_0'$.
Then with assumption (a) we have for $0<\delta\leq \delta_0'$,

\begin{align*}
&\Pr\left( \sup_{t\in\mathcal{S}^*: \sigma(t)\leq\delta}\left(\frac{|X_t|}{\sigma(t)} - H(t)\right) >0\right)\\
&\hspace{0.2cm}\leq \sum_{t\in\mathcal{S}^*: \sigma(t)\leq\delta} \Pr\left( \frac{|X(t)|}{\sigma(t)}\geq H(t)\right)\\
&\hspace{0.2cm}\leq \sum_{t\in\mathcal{S}^*: \sigma(t)\leq\delta} L\exp\left( -V\log(1/\sigma(t))  -B\log(1/u(\sigma(t))) - 2\log\log(e/\sigma(t))\right)\\
&\hspace{1.1cm} \cdot\exp\left(\frac12\log\left(2V\log(1/\sigma(t)) + 2B\log(1/u(\sigma(t))) + 4\log\log(e/\sigma(t))\right)\right)\\
& \hspace{0.2cm}= L\sum_{k=0}^\infty \sum_{t\in\mathcal{S}^*(2^{-k}\delta)} \sigma(t)^V u(\sigma(t))^B\log(e/\sigma(t))^{-2}\\
&\hspace{2.5cm}\cdot\sqrt{2V\log(1/\sigma(t)) + 2B\log(1/u(\sigma(t))) + 4\log\log(e/\sigma(t))}.
\end{align*}

The choice of $u$ implies $2B\log(1/u(\sigma(t))) = 4B\log\log(e/\sigma(t))$ and since we have additionally $\log\log(e/\sigma(t))\leq \log(e/\sigma(t))$, the square root factor is bounded by $\sqrt{2(V+2B+2)\log(e/\sigma(t))}$. We proceed by

\begin{align*}
&\Pr\left( \sup_{t\in\mathcal{S}^*: \sigma(t)\leq\delta}\left(\frac{|X_t|}{\sigma(t)} - H(t)\right) >0\right)\\
&\hspace{0.2cm}\leq L\sum_{k=0}^\infty \sum_{t\in\mathcal{S}^*(2^{-k}\delta)} \sigma(t)^V u(\sigma(t))^B\log(e/\sigma(t))^{-2} \sqrt{2(V+2B+2)\log(e/\sigma(t))}\\
&\hspace{0.2cm}\leq C_1\sum_{k=0}^\infty \sum_{t\in\mathcal{S}^*(2^{-k}\delta)} (2^{-k}\delta)^V u(2^{-k}\delta)^B\log(e/(2^{-k}\delta))^{-\frac32}\\
&\hspace{0.2cm}\leq  C_1\sum_{k=0}^\infty \left(\#\mathcal{S}^*(2^{-k}\delta)\right)(2^{-k}\delta)^V u(2^{-k}\delta)^B \left( \log(e/\delta)+ \log(2)k\right)^{-\frac32},
\end{align*}

where $C_1 := L\sqrt{2(V+2B+2)}$. Now, by (<ref>) we have

\begin{align*}
&\left(\#\mathcal{S}^*(2^{-k}\delta)\right)(2^{-k}\delta)^V u(2^{-k}\delta)^B\\
&\hspace{2cm} \leq A_02^{B}\left(\sup_{0<x\leq 1}\frac{u(x)}{u(x/2)} \right)^B\leq A_02^{B}\log(2e)^{2B} =:C_2.
\end{align*}

Here we used that

\[ \frac{u(x)}{u(x/2)} = \left(\frac{\log(2e/x)}{\log(e/x)} \right)^2= \left(\frac{\log(2)}{\log(e/x)} +1\right)^2,\]

which is increasing for growing $x$ and equals $\log(2e)$ for $x=1$. By the integral test for convergence,

\begin{align*}
&\sum_{k=0}^\infty (\log(e/\delta) + \log(2)k)^{-\frac32}\\
&\hspace{2cm}\leq \log(e/\delta)^{-\frac23} + \int_0^\infty (\log(e/\delta) + \log(2)x)^{-\frac32} dx\\
&\hspace{2cm} =\log(e/\delta)^{-\frac32} + \left[ -\frac{2}{\log(2)}(\log(e/\delta) + \log(2)x)^{-\frac12}\right]_0^\infty\\
&\hspace{2cm}\leq \left(1+\frac{2}{\log(2)}\right) \log(e/\delta)^{-\frac12}.
\end{align*}

Finally, for $C_3=C_1C_2(1+2/\log(2))$ and $\delta\leq \delta_0'$ we have shown so far

\begin{align}\label{eqp: B3}
\Pr\left( \sup_{t\in\mathcal{S}^*: \sigma(t)\leq\delta}\left(\frac{|X_t|}{\sigma(t)} - H(t)\right) >0\right) \leq C_3 \log(e/\delta)^{-\frac12}.
\end{align}

To proceed, let $s\in\mathcal{S}$ be arbitrary and choose $t\in\mathcal{S}^*$ to satisfy $\sigma(s)\leq\sigma(t)$ and $\rho(s,t)<\log(e/\sigma(s))^{-2}\sigma(s)$, which is possible by (<ref>) and our specific choice of $u$. Then

\begin{align}\label{eqp: B17}
\frac{\sigma(t)}{\sigma(s)}-1 \leq \frac{\rho(s,t)}{\sigma(s)} < \log(e/\sigma(s))^{-2} = u(\sigma(s)),
\end{align}

where we used (<ref>) for the first inequality. Now we have on the set $\mathcal{A}(Q,2\sigma(s))$,

\begin{align*}
\frac{|X(s)|}{\sigma(s)} - \frac{|X(t)|}{\sigma(t)} &\leq \frac{|X(s)-X(t)|}{\sigma(s)} + \frac{|X(t)|}{\sigma(t)}\left(\frac{\sigma(t)}{\sigma(s)}-1\right)\\
&\leq \frac{Q\rho(s,t)\log(e/\rho(s,t))}{\sigma(s)} + \frac{|X(t)|}{\sigma(t)}\frac{\rho(s,t)}{\sigma(s)}\\
&\leq Q u(\sigma(s))\log\left(e/(u(\sigma(s))\sigma(s))\right) + \frac{|X(t)|}{\sigma(t)} u(\sigma(s))\\
&\leq C_4\log(e/\sigma(s))^{-1} + \frac{|X(t)|}{\sigma(t)}\log(e/\sigma(s))^{-2}
\end{align*}

for $C_4=3Q$. In the third step we used that $x\log(e/x)$ is increasing on $(0,1]$ and in the forth one that

\begin{align*}
&\hspace{2cm}= \log(e/\sigma(s))^{-2} \log(e/((\log(e/\sigma(s))^{-2}\sigma(s)))\\
&\hspace{2cm}= \log(e/\sigma(s))^{-2} \left( \log(e/\sigma(s)) + 2\log\log(e/\sigma(s))\right) \\
&\hspace{2cm}\leq 3\log(e/\sigma(s))^{-1}.
\end{align*}

Consequently, if in addition $\frac{|X(t)|}{\sigma(t)}\leq H(t)$, then

\begin{align*}
\frac{|X(s)|}{\sigma(s)}&\leq H(t) + C_4\log(e/\sigma(s))^{-1} + H(t)\log(e/\sigma(s))^{-2}\\
&\leq H(s) + C_4\log(e/\sigma(s))^{-1} + H(s)\log(e/\sigma(s))^{-2}\\
&\leq H(s) + C_4\log(e/\sigma(s))^{-1}\\
&\hspace{0.7cm} + \left(1+2V\log(1/\sigma(s)) + (2B +4)\log\log(e/\sigma(s))\right)\log(e/\sigma(s))^{-2}\\
&\leq H(s) + C_5\log(e/\sigma(s))^{-1}
\end{align*}

for some constant $C_5=C_5(B,V,Q)$, where we used $H(t)\leq H(s)$ because $\sigma(s)\leq \sigma(t)$ and $\sqrt{x}\leq 1+x$ in the third step. Finally, note that $\sigma(s)\leq\delta$ implies $\sigma(t)\leq 2\delta$ by (<ref>). Consequently, on $\{\sigma(s)\leq\delta\}$ the ratio $\frac{|X(s)|}{\sigma(s)}$ is not greater than

\[ \sqrt{ 2V\log(1/\sigma(s)) + 2(B+2)\log(e/\sigma(s))} + C_5\log(e/\sigma(s))^{-1}\]

with probability at least $\Pr(\mathcal{A}(Q,2\delta) )- C_3\log(e/(2\delta))^{-\frac12}$ by (<ref>) for every $\delta\leq \delta_0'/2$. This yields the statement of the theorem if we take a suitable $C_0=C_0(A_0,B,V,Q)$ and set $\delta_0 := \delta_0'/2$.

Suppose $X_1,\dots, X_n$ are random variables such that for all $1\leq i\leq n$,

\[ \Pr \left( X_i >\lambda\right) \leq \exp\left(-\frac{\lambda^2}{2} + \log(\lambda)\right).\]

\[ \Pr\left( \max_{i=1,\dots, n} X_i >\lambda\right) \leq \exp\left( \log(n) - \frac{\lambda^2}{2} + \log(\lambda)\right) \]

\[ \Pr\left( \max_{i=1,\dots, n} |X_i| >\lambda\right) \leq \exp\left(\log(2n)-\frac{\lambda^2}{2} + \log(\lambda)\right).\]

The first statement follows by the union bound

\begin{align*}
\Pr\left( \max_{i=1,\dots, n} X_i >\lambda\right) &=\Pr\left( \bigcup_{i=1}^n \{ X_i>\lambda\}\right) \\
&\leq \sum_{i=1}^n \Pr(X_i>\lambda)\\
& = n\exp\left( -\frac{\lambda^2}{2} + \log(\lambda)\right).
\end{align*}

For the second one, note that

\[ \max_{i=1,\dots, n} |X_i| = \max_{i=1,\dots, 2n} X_i,\]

where we set $X_{n+i} := -X_i$, $i=1,\dots, n$. Then apply the first part again.

Suppose there exists a family of random sets $(\mathcal{C}_T)_{T\geq 0}$ with $\lim_{T\to\infty}\Pr(\mathcal{C}_T)=1$ and random $\sigma_T$, $\rho_T$ such that the following three conditions are satisfied:

(a) For arbitrary $s\in\mathcal{S}$ and $\lambda\geq 1$ there exists a constant $L>0$ such that

\[ \Pr\left(\{ |Z_T(s)| \geq \sigma_T(s)\lambda\}\cap\mathcal{C}_T\right) \leq L\exp\left(-\frac{\lambda^2}{2} + \log(\lambda)\right).\]

(b) There exists a constant $L'\geq 1$ such that for arbitrary $s,t\in\mathcal{S}$ and $\lambda\geq 0$,

\[ \Pr\left(\{ |Z_T(s) - Z_T(t)| \geq \rho_T(s,t)\lambda\}\cap \mathcal{C}_T\right) \leq L'e^{-\lambda}. \]

(c) For some constants $A_0,B,V>0$ we have the following bound for the capacity numbers on $\mathcal{C}_T$:

\[ D\left( u\delta, \{ s\in\mathcal{S}: \sigma_T(s)\leq \delta\}, \rho_T\right)\leq A_0 u^{-B}\delta^{-V}\quad \textrm{ for all } u,\delta\in (0,1].\]

Then for every $\epsilon>0$ there exists $T_\epsilon$ and $\lambda_\epsilon>0$ such that for $T\geq T_\epsilon$,

\[ \Pr\left( \sup_{s\in\mathcal{S}^*} \frac{|Z_T(s)|/\sigma_T(s) - \sqrt{2V\log(1/\sigma_T(s))}}{D(\sigma_T(s))} \geq\lambda_\epsilon\right)\leq \epsilon, \]

where $D(\delta):= \log(e/\delta)^{-\frac12}\log\log(e^e/\delta)$ and $\mathcal{S}^*\subset\mathcal{S}$ is a countable dense subset. Moreover, the sequence of random variables

\[ \left( \sup_{s\in\mathcal{S}^*} \frac{|Z_T(s)|/\sigma_T(s) - \sqrt{2V\log(1/\sigma_T(s))}}{D(\sigma_T(s))} \right)_{T\geq T_0} \]

is tight for some $T_0>0$, provided that for some $T_0'>0$,

\begin{align}\label{eqp: B16}
\inf_{T\geq T_0'} \sup_{s\in\mathcal{S}}\left( \sigma_T(s)|_{\mathcal{C}_T} \right)>0.
\end{align}

First of all, we note the following: For $\tilde{q},q>0$ and $G(\lambda, \delta) = \tilde{q}\lambda^q$, Remark $3$ in the technical report [19] entails that under assumption (c) for $\delta=1$,

\[ J(\epsilon,1) \leq C'\epsilon \log(e/\epsilon)^q \quad \textrm{ for all } 0<\epsilon \leq 1,\]

for $C'=\tilde{q}\max(1+2B,\log(A_0^2))^q \int_0^1\log(e/z)^q dz$ and $J(\cdot,\cdot)$ given in Proposition <ref>.

Then, with condition (b) and $\tilde{q}=q=1$, Proposition <ref> implies for $Q=1+12C'$ and

\[ \mathcal{A}_T(Q,\delta) := \left\{ \sup_{s,t\in\mathcal{S}: \rho_T(s,t)\leq\delta}\frac{|Z_T(t) - Z_T(s)|}{\rho_T(s,t)\log(e/\rho_T(s,t))} \leq Q\right\} \]

we have

\begin{align}\label{eqp: B4}
\Pr\left(\mathcal{A}_T(Q,\delta)^c \cap\mathcal{C}_T\right) \leq \frac{L'\delta}{2}.
\end{align}

For the rest of the proof define

\[ \Psi_T(s) := \frac{|Z_T(s)|/\sigma_T(s) - \sqrt{2V\log(1/\sigma_T(s))}}{D(\sigma_T(s))}. \]

Next, we bound from above,

\begin{align*}
&\Pr\left( \sup_{s\in\mathcal{S}^*}\Psi_T(s)\geq\lambda\right) \\
&\hspace{1cm}= \Pr\left(\left\{ \sup_{s\in\mathcal{S}^*}\Psi_T(s)\geq\lambda\right\}\cap\mathcal{C}_T\right) + \Pr\left(\left\{ \sup_{s\in\mathcal{S}^*}\Psi_T(s)\geq\lambda\right\}\cap\mathcal{C}_T^c\right)\\
&\hspace{1cm}\leq \Pr\left(\left\{ \sup_{s\in\mathcal{S}^*}\Psi_T(s)\geq\lambda\right\}\cap\mathcal{C}_T\right) + \Pr\left(\mathcal{C}_T^c\right)\\
&\hspace{1cm}\leq \Pr\left(\left\{\sup_{s\in\mathcal{S}^*: \sigma_T(s)<\delta} \Psi_T(s)\geq \lambda\right\}\cap\mathcal{C}_T\right)  +\Pr\left(\mathcal{C}_T^c\right) \\
&\hspace{3cm} +  \Pr\left(\left\{\sup_{s\in\mathcal{S}^*: \sigma_T(s)\geq\delta} \Psi_T(s)\geq \lambda\right\}\cap\mathcal{C}_T\right)
\end{align*}

for some $0<\delta<1$. Let $\epsilon>0$. We are now going to show that for appropriate $\delta=\delta_\epsilon>0$ there exists $T_\epsilon$ and $\lambda_\epsilon$ such that each summand is $<\frac{\epsilon}{3}$ for $T>T_\epsilon$ and $\lambda=\lambda_\epsilon$.

(i) It is clear by $\lim_{T\to\infty}\Pr(\mathcal{C}_T)=1$ that there exists $T_0$ such that for all $T\geq T_0$ we have $\Pr(\mathcal{C}_T^c)<\frac{\epsilon}{3}$ .

(ii) The remark in the technical report of [19] after Theorem $8$ entails that we have for $\delta \searrow 0$,

\begin{align*}
&\sqrt{2V\log(1/\delta) + C_0\log\log(e/\delta)} + C_0\log(e/\delta)^{-1}\\
&\hspace{1.5cm} = \sqrt{2V\log(1/\delta)} + \mathcal{O}\left(\log\log(e/\delta) \log(e/\delta)^{-\frac{1}{2}}\right).
\end{align*}

Conditions (a) and (c) are clearly the same as those of Theorem <ref> and we may apply it to get for $\lambda$ large enough,

\begin{align*}
&\Pr\left(\left\{\sup_{s\in\mathcal{S}^*: \sigma_T(s)<\delta} \Psi_T(s)\geq \lambda\right\}\cap\mathcal{C}_T\right)\\
&\hspace{1cm} \leq 1-\Pr(\mathcal{A}(Q,2\delta)\cap\mathcal{C}_T) + C_0\log(e/\delta)^{-\frac12}\\
&\hspace{1cm} \leq \Pr(\mathcal{A}(Q,2\delta)^c\cap\mathcal{C}_T) +\Pr\left( \mathcal{C}_T^c\right)+ C_0\log(e/\delta)^{-\frac12}\\
&\hspace{1cm} \leq L'\delta +\Pr\left(\mathcal{C}_T^c\right)+ C_0\log(e/\delta)^{-\frac12},
\end{align*}

where we used

\begin{align*}
1-\Pr(A\cap B) &= \Pr((A\cap B)^c) = \Pr( A^c\cup B^c)\\
&= \Pr( (A^c\cap B) \cup B^c) \leq \Pr(A^c\cap B) + \Pr(B^c)
\end{align*}

in the second step and (<ref>) in the last one. Now choose $\delta_\epsilon>0$ fullfilling $L'\delta_\epsilon + C_0\log(e/\delta_\epsilon)^{-\frac12}<\frac{\epsilon}{6}$ and $T_1$ large enough such that $\Pr(\mathcal{C}_T^c)<\frac{\epsilon}{6}$ for $T\geq T_1$.

(iii) Lastly, we need to bound the remaining probability by $\frac{\epsilon}{3}$ where we have to use $\delta=\delta_\epsilon$ from step (ii). On $\{\sigma_T(s)\geq \delta\}$ we have the bound

\begin{align}\label{eqp: B6}
\Psi_T(s) \leq \frac{C''}{\delta} \sup_{s\in\mathcal{S}^*} |Z_T(s)|,
\end{align}

where the constant $C''$ emerges from minimizing $D(\cdot)$ on $[\delta, 1]$.
Now fix $T$ and choose $\kappa>0$ together with a maximal subset $\{s_1, \dots, s_N\}\subset\mathcal{S}$ that fullfills $\rho_T(s_i, s_j)>\kappa$ for all $i\neq j$. From assumption (c) we know that $N\leq A_0\kappa^{-B}$. Now, on the set $\mathcal{A}_T(Q,\kappa)$,

\[ \sup_{s\in\mathcal{S}^*}|Z_T(s)| \leq Q\kappa \log(e/\kappa) + \max_{i=1,\dots, N} |Z_T(s_i)|.\]

Hence, by Lemma <ref> and assumption (a) we have for $\lambda':= \lambda \delta/C''$ with $\lambda$ large enough and $\kappa$ small enough to ensure $\lambda'-Q\kappa \log(e/\kappa)\geq 1$,

\begin{align*}
&\Pr\left(\left\{ \sup_{s\in\mathcal{S}^*} |Z_T(s)| >\lambda'\right\}\cap\mathcal{C}_T\cap\mathcal{A}_T(Q,\kappa)\right)\\
&\hspace{0.7cm}\leq \Pr\left(\left\{ Q\kappa\log(e/\kappa) + \max_{i=1,\dots, N} |Z_T(s_i)| >\lambda' \right\}\cap\mathcal{C}_T\right)\\
&\hspace{0.7cm}\leq \Pr\left(\left\{ \max_{i=1,\dots, N} |Z_T(s_i)| >\lambda' - Q\kappa\log(e/\kappa)\right\}\cap\mathcal{C}_T\right)\\
&\hspace{0.7cm}\leq \exp\left( \log(N) - \frac{(\lambda'-Q\kappa\log(e/\kappa))^2}{2} + \log(\lambda' - Q\kappa\log(e/\kappa))\right)\\
&\hspace{0.7cm} =  \exp\left( \log(A_0) -B\log(\kappa) - \frac{(\lambda'-Q\kappa\log(e/\kappa))^2}{2} \right.\\[-2\jot]
&\hspace{6cm} \left. \textcolor{white}{\frac12}+ \log(\lambda' - Q\kappa\log(e/\kappa))\right).
\end{align*}

Thus, with (<ref>) and (<ref>), we have

\begin{align*}
&\Pr\left(\left\{ \sup_{s\in\mathcal{S}^*: \sigma_T(s)\geq\delta}\Psi_T(s)>\lambda \right\}\cap\mathcal{C}_T\right)\\
&\hspace{0.5cm} \leq \Pr\left(\left\{ \sup_{s\in\mathcal{S}^*} |Z_T(s)| >\lambda'\right\}\cap\mathcal{A}_T(Q,\kappa)\cap\mathcal{C}_T\right) + \Pr\left(\mathcal{A}_T(Q,\kappa)^c \cap \mathcal{C}_T\right)\\
&\hspace{0.5cm}\leq \exp\left( \log(A_0) -B\log(\kappa) - \frac12 (\lambda'-Q\kappa\log(e/\kappa))^2\right.\\[-3\jot]
&\hspace{5.5cm} \left.\textcolor{white}{\frac12}+ \log(\lambda' - Q\kappa\log(e/\kappa))\right) + \frac{L'\kappa}{2}.
\end{align*}

Note that we accomplished to bound the probability independent of $T$. Now, choose $\kappa_\epsilon$ small enough such that $\frac{L'\kappa_\epsilon}{2} <\frac{\epsilon}{6}$ and after that $\lambda_\epsilon$ large enough in order to ensure

\begin{align*}
&\exp\left( \log(A_0) -B\log(\kappa_\epsilon) - \frac{(\lambda_\epsilon \delta/C -Q\kappa_\epsilon\log(e/\kappa_\epsilon))^2}{2} \right. \\
&\hspace{5cm} \left.\textcolor{white}{\frac12} + \log(\lambda_\epsilon \delta/C - Q\kappa_\epsilon\log(e/\kappa_\epsilon))\right) <\frac{\epsilon}{6}.
\end{align*}

For those choices we then have

\[ \Pr\left(\left\{ \sup_{s\in\mathcal{S}^*: \sigma_T(s)\geq\delta}\Psi_T(s)>\lambda\right\}\cap\mathcal{C}_T\right) \leq \frac{\epsilon}{3}.\]

In conclusion, for all $T>\max\{T_0,T_1\}$,

\[ \Pr\left(\sup_{s\in\mathcal{S}^*} \Psi_T(s) >\lambda_\epsilon\right)<\epsilon,\]

which gives the first claim of the theorem. For the tightness statement it remains to show that for any $\epsilon>0$ there exists $\lambda_\epsilon'>0$ and $T_\epsilon'$ such that for $T\geq T_\epsilon'$

\begin{align}\label{eqp: B9}
\Pr\left( \sup_{s\in\mathcal{S}^*} \Psi_T(s) < - \lambda_\epsilon'\right) \leq \epsilon.
\end{align}

By non-negativity of $|Z_T(s)|/\sigma_T(s)$ and $D(\cdot)$, it is enough to show that

\begin{align*}
&\Pr\left( - \inf_{s\in\mathcal{S}^*}\sqrt{2V\log(1/\sigma_T(s))} < - \lambda_\epsilon'\right) \\
& \hspace{1cm} \leq \Pr\left( -\inf_{s\in\mathcal{S}^*}\sqrt{2V\log(1/\sigma_T(s))} < - \lambda_\epsilon', \mathcal{C}_T\right) + \Pr\left( \mathcal{C}_T^c\right)
\end{align*}

is bounded from above by $\epsilon$ for $T\geq T_\epsilon'$. The second term is bounded by $\frac{\epsilon}{2}$ for $T\geq T_1'$ large enough by assumption on $\mathcal{C}_T$. For the first one, this bound follows for $T\geq T_0'$ directly from the assumption (<ref>) and $\lambda_\epsilon'$ large enough. Now (<ref>) follows for $T\geq T_\epsilon' := T_1'\vee T_0'$.

§.§ Proof of Theorem <ref>

The proof of Theorem <ref> will be an application of Theorem <ref>. One major part is establishing the exponential inequalities on appropriate sets - the other bounding the covering numbers. Our proof of the first part relies substantially on the following result.

Let $(M_t)_{t\geq 0}$ be a continuous martingale with $M_0=0$. Then for every $T>0$, $\theta>0$, $S\geq 1$ and $\lambda\geq 1$ we have

\[ \Pr\left( \left| M_T\right| > \lambda\sqrt{\langle M\rangle_T}, \theta\leq \sqrt{\langle M\rangle_T}\leq \theta S\right) \leq 4\lambda\sqrt{e}(1+\log(S))\exp\left(-\frac{\lambda^2}{2}\right).\]

The proof is an application of Theorem <ref> for the process

\[ Z_T(y,h) := \frac{1}{\sqrt{T}\hat\sigma_{T,\max}} \int_0^T K_{y,h}(X_s) dW_s\]

and the set

\begin{align}\label{eqp: B13}
\overline{\mathcal{T}} := \left\{ (y,h)\in\R^2\mid 0<h\leq A\textrm{ and } -A+h \leq y\leq A-h\right\}
\end{align}

with dense subset $\mathcal{T}=\overline{\mathcal{T}}\cap\Q^2$. The spread measure of $Z_T$ is given by $\overline\sigma_T(y,h) := \frac{\hat\sigma_T(y,h)}{\hat\sigma_{T,\max}}\leq 1$ and the corresponding semimetric by

\[ \overline\rho_T((y,h), (y',h'))^2 := \frac{1}{T\hat\sigma_{T,\max}^2} \int_0^T \left( K_{y,h}(X_s) - K_{y',h'}(X_s)\right)^2 ds.\]

The random map $\hat\sigma_T^2:\overline{\mathcal{T}}\rightarrow (0,1]$ is continuous, as

\[ \left| \hat\sigma_T(y_n,h_n)^2-\hat\sigma_T(y,h)^2\right| \leq \left\|K_{y_n,h_n}^2-K_{y,h}^2\right\|_{[-A,A]} \]

and the right-hand side converges to zero for $(y_n,h_n)\to (y,h)$. In particular, it follows that $\overline\sigma_T$ is continuous in $(y,h)$ and hence the level sets in (<ref>) are compact because they are bounded and closed as the preimage of the closed set $[\delta,1]$. To see that $|\overline\sigma_T(y,h) -\overline\sigma_T(y',h')|\leq \overline\rho_T((y,h),(y',h'))$, we compare the squares on both sides. This gives on the one hand,

\begin{align*}
&|\overline\sigma_T(y,h) -\overline\sigma_T(y',h')|^2\\
&\hspace{1cm} = \frac{1}{T \hat\sigma_{T,\max}^2} \left( \sqrt{\int_0^T K_{y,h}(X_s)^2 ds} -\sqrt{\int_0^T K_{y',h'}(X_s) ds} \right)^2\\
&\hspace{1cm}=\frac{1}{T \hat\sigma_{T,\max}^2}\left( \int_0^T K_{y,h}(X_s)^2 + K_{y',h'}(X_s)^2 ds \right. \\
&\hspace{5cm} \left. - 2\sqrt{\int_0^T K_{y,h}(X_s)^2 ds \int_0^T K_{y',h'}(X_s)^2 ds}\right)
\end{align*}

and on the other hand $\overline\rho_T((y,h), (y',h'))^2$ equals

\[ \frac{1}{T\hat\sigma_{T,\max}^2} \left( \int_0^T K_{y,h}(X_s)^2 + K_{y',h'}(X_s)^2 ds - 2\int_0^T K_{y,h}(X_s) K_{y',h'}(X_s) ds\right).\]

Now we have the desired inequality (<ref>), since by Cauchy-Schwarz' inequality,

\[ \int_0^T K_{y,h}(X_s) K_{y',h'}(X_s) ds\leq \sqrt{\int_0^T K_{y,h}(X_s)^2 ds \int_0^T K_{y',h'}(X_s)^2 ds}.\]

Moreover, for the application of Theorem <ref>, we define the set

\begin{align*}
\mathcal{C}_{T,b_0} := \left\{ \sup_{z\in [-A,A]} \left| \frac{1}{\sigma^2 T}L_T^z(X) - q_{b_0}(z)\right| <\frac12 L_*\right\}.
\end{align*}

By Corollary <ref> we have $\lim_{T\to\infty}\Pr(\mathcal{C}_{T,b_0})= 1$ and on $\mathcal{C}_{T,b_0}$,

\begin{align}\label{eqp: B10}
\frac12 L_*\leq \frac12 q_b(z) \leq \frac{1}{\sigma^2 T} L_T^z(X)\leq \frac32 q_b(z) \leq \frac32 L^*.
\end{align}

Now we check each of the conditions (a)-(c) of Theorem <ref>.

(a) Using the occupation times formula <ref>,

\[ \hat\sigma_T(y,h)^2 = \frac1T \int_0^T K_{y,h}^2(X_s) ds= \int_\R K_{y,h}(z)^2 \frac{1}{\sigma^2 T}L_T^z(X) dz.\]

Hence, on $\mathcal{C}_{T,b_0}$ we have by (<ref>),

\[ \sqrt{\frac12 L_*} \|K_{y,h}\|_{L^2} \leq \hat\sigma_T(y,h) \leq \sqrt{\frac32 L^*} \|K_{y,h}\|_{L^2}. \]

Note furthermore that $Z'_t(y,h)=\hat\sigma_{t,\max} \sqrt{t}Z_t(y,h)$ is a martingale (as a process indexed in $t$) with quadratic variation $\langle Z'(y,h)\rangle_t = t\hat\sigma_t(y,h)^2$. Setting $\theta=\sqrt{TL_*/2}\|K_{y,h}\|_{L^2}$ and $S=\sqrt{3L^*/L_*}\geq 1$ in Proposition <ref> gives for $\lambda\geq 1$,

\begin{align*}
&\Pr\left( |Z_T(y,h)|\geq \lambda\overline\sigma_T(y,h), \mathcal{C}_{T,b_0}\right)\\
&\hspace{2cm} = \Pr\left( | Z'_T(y,h)|\geq \lambda\sqrt{T}\hat\sigma_T(y,h), \mathcal{C}_{T,b_0}\right)\\
&\hspace{2cm} \leq 4\sqrt{e}\left(1+\log\big(\sqrt{3L^*/L_*}\big)\right) \exp\left(-\frac{\lambda^2}{2} + \log(\lambda)\right).
\end{align*}

(b) We proceed similarly as for the verification of (a) and set

\begin{align*}
\tilde{Z}_T((y,h),(y',h')) &:=  Z_T(y,h)- Z_T(y',h') \\
&= \frac{1}{\sqrt{T}\hat\sigma_{T,\max}}\int_0^T \left( K_{y,h}(X_s) - K_{y',h'}(X_s)\right) dW_s
\end{align*}

and note that

\[\tilde Z'_t((y,h),(y',h')) =\hat\sigma_{t,\max}\sqrt{t} \tilde{Z}_t((y,h),(y',h'))\]

is a martingale with quadratic variation process

\[ \langle\tilde{Z}'((y,h),(y',h'))\rangle_t = t\hat\sigma_{t,\max}^2\overline\rho_t((y,h), (y',h'))^2, \]

where both sides are given by the occupation times formula <ref> as

\begin{align*}
&\int_0^t \left( K_{y,h}(X_s) - K_{y',h'}(X_s)\right)^2 ds \\
&\hspace{2cm}= t\int_\R \left( K_{y,h}(z) - K_{y',h'}(z)\right)^2 \frac{1}{\sigma^2 t}L_t^z(X) dz.
\end{align*}

Again, on $\mathcal{C}_{T,b_0}$ we have for all $(y,h),(y',h')\in\overline{\mathcal{T}}$,

\begin{align*}
&\sqrt{\frac{TL_*}{2}}\|K_{y,h}(z) - K_{y',h'}(z)\|_{L^2} \\
&\hspace{0.7cm}\leq \sqrt{\langle \tilde{Z}'((y,h),(y',h'))\rangle_T}\leq \sqrt{\frac{3TL^*}{2}}\|K_{y,h}(z) - K_{y',h'}(z)\|_{L^2}.
\end{align*}

We conclude with Proposition <ref> for $\lambda\geq 1$,

\begin{align*}
&\Pr\left( |\tilde{Z}_T| >\lambda\overline\rho_T((y,h), (y',h')), \mathcal{C}_{T,b_0}\right) \\
&\hspace{2cm} = \Pr\left( |\tilde{Z}'_T| >\lambda \sqrt{\langle \tilde Z'_T\rangle}, \mathcal{C}_{T,b_0}\right) \\
&\hspace{2cm}\leq 4\sqrt{e}\left(1+ \log\big(\sqrt{3L^*/L_*}\big)\right)\exp\left(-\frac{\lambda^2}{2} + \log(\lambda)\right),
\end{align*}

by setting $\theta = \sqrt{TL_*/2}\|K_{y,h}(z) - K_{y',h'}(z)\|_{L^2}$ and $S=\sqrt{3L^*/L_*}\geq 1$ as in the verification of (a). The sub-exponential tail follows then for $L'>0$ large enough because $\exp\left(-\lambda^2/2 +\log(\lambda)+\lambda\right)$ is bounded from above for $\lambda\geq 1$.

(c) We split the interval $[-A,A]$ into a partition $(M_k)_{1\leq k\leq N}$ such that

\[ \lambda(M_k)\leq\frac{\hat\sigma_{T,\max}^2}{6L^*\|K\|_{TV}}(u\delta)^2\]

for $1\leq k\leq N$, where equality holds for $k\neq N$. Here and subsequently, $\|K\|_{TV}$ denotes the total variation of $K$ and $\lambda(\cdot)$ the Lebesgue measure on $\R$.

Then we have

\[ 2A=\lambda([-A,A]) = \sum_{k=1}^N \lambda(M_k)\geq (N-1)\frac{\hat\sigma_{T,\max}^2}{6L^*\|K\|_{TV}}(u\delta)^2, \]

or equivalently

\begin{align}\label{eqp: B15}
N\leq 1+ \frac{12AL^*\|K\|_{TV}}{\hat\sigma_{T,\max}^2}(u\delta)^{-2}.
\end{align}

Take $(y,h)\in\overline{\mathcal{T}}$ such that $\overline\sigma_T(y,h)\leq\delta$, i.e.

\[ \frac{1}{T\hat\sigma_{T,\max}^2}\int_0^T K_{y,h}(X_s)^2 ds = \frac{1}{\hat\sigma_{T,\max}^2}\int_\R K_{y,h}(z)^2 \frac{1}{\sigma^2 T} L_T^z(X) dz \leq \delta^2\]

by the occupation times formula <ref>. Now using the lower bound (<ref>) for the averaged local time on $\mathcal{C}_{T,b_0}$ we get

\[ \delta^2\geq \frac{L_*}{2\hat\sigma_{T,\max}^2}\int_\R K_{y,h}(z)^2 dz  = \frac{hL_*\| K\|_{L^2}^2}{2\hat\sigma_{T,\max}^2},\]

which implies

\[ h\leq 2\hat\sigma_{T,\max}^2\delta^2 (L_*\|K\|_{L^2}^2)^{-1}. \]

Now suppose $y-h\in M_i$ and $y+h\in M_j$. As we know the length of the intervals $M_k$, we conclude

\[ (i-j-1)\frac{\hat\sigma_{T,\max}^2(u\delta)^2}{6L^*\|K\|_{TV}} \leq \frac{4\delta^2\hat\sigma_{T,\max}^2}{L_*\|K\|_{L^2}^2}\quad\Leftrightarrow\quad i-j\leq 1+\frac{24 L^*\|K\|_{TV}}{L_*\|K\|_{L^2}^2} u^{-2}.\]

From this and (<ref>), we conclude that there are at most

\begin{align}\label{eqp: B8}
\begin{split}
&\left( 1+\frac{24 L^*\|K\|_{TV}}{L_*\|K\|_{L^2}^2} u^{-2}\right)\left( 1+\frac{12AL^*\|K\|_{TV}}{\hat\sigma_{T,\max}^2}(u\delta)^{-2}\right)\\
&\hspace{0.5cm}\leq  \left( 1+ 12L^* \|K\|_{TV} \left( \frac{A}{\hat\sigma_{T,\max}^2}+\frac{2}{L_* \|K\|_{L^2}^2 } \right.\right. \\
&\hspace{5.5cm} \left.\left.+\frac{24AL^* \|K\|_{TV}}{\hat\sigma_{T,\max}^2 L_*\|K\|_{L^2}^2}\right)\right) u^{-4}\delta^{-2}\\
&\hspace{0.5cm} \leq \left( 1+ 12L^* \|K\|_{TV} \left( \frac{1}{L_*}+\frac{2}{ L_* \|K\|_{L^2}^2}+\frac{24 L^* \|K\|_{TV}}{(L_*)^2\|K\|_{L^2}^2}\right)\right) u^{-4}\delta^{-2}
\end{split}
\end{align}

such pairs $(i,j)$, where we used the lower bound $\hat\sigma_{T,\max}^2\geq AL_*$ on $\mathcal{C}_{T,b_0}$. Let $\mathcal{T}'\subset\overline{\mathcal{T}}$ be a maximal subset of $\{ (y,h)\in\overline{\mathcal{T}}: \overline\sigma_T(y,h)\leq \delta\}$ with $\overline\rho_T((y,h),(y',h'))>u\delta$ for arbitrary $(y,h),(y',h')\in\mathcal{T}'$.
Then, the proof is finished by (<ref>) if we could show that for all $(i,j)$ there is at most one such point $(y_{ij},h_{ij})$ in $\mathcal{T}'$ with $y_{ij}-h_{ij}\in M_i$ and $y_{ij}+h_{ij}\in M_j$. To this aim, we pick $(y,h), (y',h')\in\overline{\mathcal{T}}$ such that $y-h, y'-h'\in M_i$ and $y+h, y'+h'\in M_j$ and are done if we can show that $\overline\rho_T((y,h),(y',h'))^2\leq (u\delta)^2$.

As the kernel $K$ is of bounded variation, there exists a probability measure $\mu$ on $[-1,1]$ and a measurable function $g$ with $|g|\leq \|K\|_{TV}$ such that for almost all $x\in [-1,1]$ we have

\[ K(x) = \int_{-1}^x g d\mu.\]

Since $K$ is bounded by one, we have on $\mathcal{C}_{T,b_0}$,

\begin{align*}
&\frac1T\int_0^T \left( K_{y,h}(X_s)-K_{y',h'}(X_s)\right)^2 ds\\
&\hspace{1.5cm} \leq \frac2T\int_0^T \left| K_{y,h}(X_s)-K_{y',h'}(X_s)\right| ds\\
&\hspace{1.5cm} = 2\int_\R\left| K\left(\frac{z-y}{h}\right) - K\left(\frac{z-y'}{h'}\right)\right| \frac{1}{\sigma^2 T} L_T^z(X) dz\\
&\hspace{1.5cm} \leq 3L^* \int_\R \left| \int_{-1}^{\frac{z-y}{h}} g(u) d\mu(u)- \int_{-1}^{\frac{z-y'}{h'}} g(u) d\mu(u)\right| dz\\
&\hspace{1.5cm} \leq 3L^* \int_\R \left|\int_{\min\left\{\frac{z-y}{h},\frac{z-y'}{h'}\right\}}^{\max\left\{\frac{z-y}{h}, \frac{z-y'}{h'}\right\}} g(u) d\mu(u)\right| dz\\
&\hspace{1.5cm} \leq 3L^* \int_\R \int_{\min\left\{\frac{z-y}{h},\frac{z-y'}{h'}\right\}}^{\max\left\{\frac{z-y}{h}, \frac{z-y'}{h'}\right\}} |g(u)| d\mu(u) dz\\
&\hspace{1.5cm} \leq 3L^* \|K\|_{TV} \int_\R\int_{-1}^1 \1_{B_z((y,h),(y',h'))} (u) d\mu(u) dz\\
&\hspace{1.5cm} = 3L^* \|K\|_{TV} \int_{-1}^1 \int_\R \1_{B_z((y,h),(y',h'))} (u) dz d\mu(u),
\end{align*}

\[B_z((y,h),(y',h')) = \left[\min\left\{\frac{z-y}{h},\frac{z-y'}{h'}\right\}, \max\left\{\frac{z-y}{h},\frac{z-y'}{h'}\right\}\right]\]

and an application of Fubini's theorem in the last step. To proceed, note that $u\in B_z((y,h),(y',h'))$ if and only if

\[ \frac{z-y}{h}\leq u\leq \frac{z-y'}{h'}\quad \textrm{ or }\quad \frac{z-y'}{h'}\leq u\leq \frac{z-y}{h}.\]

The first condition is equivalent to $uh'+y'\leq z\leq uh+y$, the second one to $uh+y\leq z\leq uh'+y'$. Thus,

\[ \int_\R \1_{B_z((y,h),(y',h'))} (u) dz = |y' + uh' - y - uh| \]

and we get

\begin{align*}
&\frac1T\int_0^T \left( K_{y,h}(X_s)-K_{y',h'}(X_s)\right)^2 ds\\
&\hspace{1cm}\leq 3L^* \|K\|_{TV}\int_{-1}^1 |y'+uh'-y-uh| d\mu(u)\\
&\hspace{1cm}\leq 3L^*  \|K\|_{TV}\max_{u=\pm 1} |y-y' +u( h-h')|\\
&\hspace{1cm}\leq 3L^* \|K\|_{TV} \lambda([y-h,y+h]\bigtriangleup [y'-h',y'+h']),
\end{align*}

where the second inequality follows from the fact, that $|y-y' + u(h-h')|$ is maximized for $u=\textrm{sign}(y-y')\textrm{sign}(h-h')$. Here, $\bigtriangleup$ denotes the symmetric difference of two set.

As we chose $(y,h),(y',h')\in\mathcal{T}'$ to fullfill $y-h,y'-h'\in M_i$ and $y+h,y'+h'\in M_j$, we know that

\[\lambda([y-h,y+h]\bigtriangleup [y'-h',y'+h']) \leq 2\frac{(u\delta)^2\hat\sigma_{T,\max}^2}{6L^* \|K\|_{TV}}.\]

\begin{align*}
\overline\rho_T((y,h),(y',h')) &=\frac{1}{T\hat\sigma_{T,\max}^2}\int_0^T \left( K_{y,h}(X_s)-K_{y',h'}(X_s)\right)^2 ds\\
&\leq \frac{3L^*\|K\|_{TV}}{\hat\sigma_{T,\max}^2}\frac{2(u\delta)^2\hat\sigma_{T,\max}^2}{6L^* \|K\|_{TV}} = (u\delta)^2
\end{align*}

and the bound of the covering numbers follows with $B=4$ and $V=2$.

To conclude the tightness, it suffices by Theorem <ref> to verify

\[ \inf_{T\geq T_0} \sup_{(y,h)\in\overline{\mathcal{T}}} \overline\sigma_T(y,h) >0 \]

on $\mathcal{C}_{T,b_0}$. But this follows directly by the occupations times formula <ref>, as (<ref>) holds on $\mathcal{C}_{T,b_0}$, and hence

\begin{align*}
\overline\sigma_T(y,h)^2 = \frac{\int_0^T K_{y,h}(X_s)^2 ds}{\int_0^T \1_{[-A,A]} (X_s) ds} = \frac{\int_\R K_{y,h}(z) \frac{1}{\sigma^2 T}L_T^z(X) dz}{\int_\R \1_{[-A,A]}(z) \frac{1}{\sigma^2 T}L_T^z(X) dz} \geq \frac{\|K_{y,h}\|_{L^2}^2 L_*}{6AL^*}.
\end{align*}

The assertion of Theorem <ref> now follows from the observation that $D(\cdot)$ is bounded and stricly positive on $(0,1]$ with $\lim_{r\searrow 0}D(r)=0$.

To conclude this section, we establish the following result. It proves boundedness of the random variable $S_b$ given in Theorem <ref>.

Let $K$ be a compactly supported kernel function of bounded variation with $\|K\|_\infty\leq 1$. For $b\in\Sigma(C,A,\gamma,\sigma)$ define

\[ \sigma_b(y,h)^2 := \int_\R K_{y,h}(z)^2 q_b(z) dz\]

\[ \sigma_{b,\max}^2 := \int_\R \1_{[-A,A]}(z) q_b(z) dz.\]

Then for $\mathcal{T}$ given in (<ref>), $\Pr_b$-almost surely,

\[ \sup_{(y,h)\in\mathcal{T}} \left( \frac{\left| \int_\R K_{y,h}(z) \sqrt{\rho_b(z)} dW_z\right|}{\|K_{y,h}\sqrt{\rho_b}\|_{L^2}} - C\left(\sigma_b(y,h)^2/\sigma_{b,\max}^2\right) \right)<\infty.\]

We apply Theorem <ref> to the process

\[ Z_b(y,h) := \sigma_{b,\max}^{-1}\int_\R K_{y,h}(z)\sqrt{\rho_b} dW_z\]

and $\overline{\mathcal{T}}$ given in (<ref>). Define

\[ \rho_b((y,h), (y',h'))^2 = \sigma_{b,\max}^{-2}\int_\R \left( K_{y,h}(z) - K_{y',h'}(z)\right)^2\rho_b(z) dz.\]

Condition (<ref>) and (<ref>) are verified as in the proof of Theorem <ref>. Conditions (a) and (b) of Theorem <ref> are immediate as the random variable $Z_b(y,h)$ is Gaussian with variance

\[ \overline\sigma_b^2:=\frac{\sigma_b(y,h)^2}{\sigma_{b,\max}^2}\]

and $Z_b(y,h)-Z_b(y',h')$ is Gaussian with variance $\rho_b((y,h),(y',h'))^2$, respectively. It remains to show that

\begin{align}\label{eqp: B5}
D\left( u\delta, \{ (y,h)\in\overline{\mathcal{T}}: \overline\sigma_b(y,h)\leq \delta\}, \rho_b\right) \leq A_0u^{-4}\delta^{-2}
\end{align}

for all $u,\delta\in (0,1]$ and some constant $A_0>0$. But this estimate follows the lines of verifying (c) in the proof of Theorem <ref>. We use the fact that $L_*\leq \rho_b\leq L^*$ and split $[-A,A]$ into a partition $(M_k)_{1\leq k\leq N}$ such that

\[ \lambda(M_k) \leq \frac{\sigma_{b,\max}^2}{4L^*\|K\|_{TV}}(u\delta)^2 \]
for $1\leq k\leq N$, where equality holds for $k\neq N$. Here and subsequently, $\|K\|_{TV}$ denotes the total variation of $K$ and $\lambda(\cdot)$ Lebesgue measure on $\R$. Then we have

\begin{align}\label{eqp: B14}
N\leq 1+ \frac{8AL^*\|K\|_{TV}}{\sigma_{b,\max}^2}(u\delta)^{-2}.
\end{align}

Take $(y,h)\in\overline{\mathcal{T}}$ such that $\overline\sigma_b(y,h)\leq \delta$ which is equivalent to

\[ \delta^2\geq \frac{1}{\sigma_{b,\max}^2}\int_\R K_{y,h}(z)^2\rho_b(z) dz \geq \frac{L_*}{\sigma_{b,\max}^2}\int_\R K_{y,h}(z)^2 dz = \frac{L_* \|K\|_{L^2}^2}{\sigma_{b,\max}^2} h\]

and yields $h\leq \delta^2 \sigma_{b,\max}^2 L_*^{-1}\|K\|_{L^2}^{-2}$. Now suppose that $y-h\in M_i$ and $y+h\in M_j$, which implies

\[ i-j \leq 1+ \frac{8 L^*\|K\|_{TV}}{L_* \|K\|_{L^2}^2} u^{-2}.\]

By this bound and (<ref>), there are at most

\begin{align}\label{eqp: B7}
\begin{split}
&\left( 1+ \frac{8 L^*\|K\|_{TV}}{L_* \|K\|_{L^2}^2} u^{-2}\right) \left( 1+ \frac{8AL^* \|K\|_{TV}}{\sigma_{b,\max}^2} (u\delta)^{-2}\right)\\
&\hspace{0.3cm} \leq \left( 1+ 8L^* \|K\|_{TV}\left( \frac{A}{\sigma_{b,\max}^2}+ \frac{1}{L_* \|K\|_{L^2}^2} + \frac{8A L^* \|K\|_{TV}}{\sigma_{b,\max}^2 L_*\|K\|_{L^2}^2}\right)\right) u^{-4}\delta^{-2}\\
&\hspace{0.3cm} \leq \left( 1+ 8L^* \|K\|_{TV}\left( \frac{1}{2L_*}+ \frac{1}{L_* \|K\|_{L^2}^2} + \frac{4 L^* \|K\|_{TV}}{ (L_*)^2\|K\|_{L^2}^2}\right)\right) u^{-4}\delta^{-2}
\end{split}
\end{align}

of such pairs $(i,j)$, where we use in the last step that $\sigma_{b,\max}^2\geq 2AL_*$. Let $\mathcal{T}'\subset\overline{\mathcal{T}}$ be a maximal subset of $\{(y,h)\in\overline{\mathcal{T}} : \overline\sigma_b(y,h)\leq \delta\}$ with the constraint that $\rho_b((y,h), (y',h'))>u\delta$ for arbitrary $(y,h), (y',h')\in\mathcal{T}'$. The proof is finished by (<ref>) if we show that for each $(i,j)$ there is at most one point $(y_{ij},h_{ij})\in\mathcal{T}'$ with $y_{ij}-h_{ij}\in M_i$ and $y_{ij}+h_{ij}\in M_j$.

Assume that $y-h, y'-h'\in M_i$ and $y+h, y'+h'\in M_j$ for $(y,h),(y',h')\in\overline{\mathcal{T}}$. Following the steps for verifying (c) in the proof of Theorem <ref> we get that

\begin{align}\label{eqp: B11}
\begin{split}
&\int_\R\left( K_{y,h}(z) - K_{y',h'}(z)\right)^2\rho_b(z) dz\\
&\hspace{2cm} \leq 2L^* \|K\|_{TV}\lambda([y-h, y+h]\bigtriangleup [y'-h', y'+h']).
\end{split}
\end{align}

As we know the length of the intervals $M_k$, we see that

\[\lambda([y-h, y+h]\bigtriangleup [y'-h', y'+h'])\leq 2\frac{\sigma_{b,\max}^2}{4L^*\|K\|_{TV}}(u\delta)^2\]

and hence

\[ \rho_b((y,h),(y',h'))^2 \leq \frac{2L^*\|K\|_{TV}}{\sigma_{b,\max}^2} \frac{2\sigma_{b,\max}^2(u\delta)^2}{4L^*\|K\|_{TV}} = (u\delta)^2.\]

This completes the bound (<ref>) on the capacity numbers which makes Theorem <ref> applicable. Then the claim follows as $D(\cdot)$ is bounded and strictly positive on $(0,1]$ with $\lim_{r\searrow 0} D(r)=0$.

§ PROOF OF THEOREM <REF> AND UNIFORM WEAK CONVERGENCE

This section presents the proof of the uniform weak convergence result in Theorem <ref>. In Subsection <ref> we introduce the notion of uniform weak convergence using the dual bounded Lipschitz metric. Furthermore, it comprises many auxiliar results on uniform weak convergence that are of interest on their own, in particular the important Proposition <ref>. The proof of Theorem <ref>, which is a consequence of Lemma <ref> and this Proposition <ref>, is then given in Subsection <ref>.

§.§ Preliminaries on uniform weak convergence

In this part we deal with general results about uniform weak convergence. Here and subsequently, let $(E,d)$ denote some separable metric space. To define weak convergence that is uniform over some class of parameters, we use the fact that weak convergence for probability measures (and for random variables via their corresponding measure) is metrisized by the dual bounded Lipschitz metric.  To introduce this metric, we define the bounded Lipschitz norm for a Lipschitz function $f:E\rightarrow\R$ as

\[ \|f\|_{BL} := \left( \|f\|_\infty  + \sup_{x\neq y} \frac{|f(x) - f(y)|}{d(x,y)}\right)\]

and set

\begin{align}\label{eq: F_BL}
\mathcal{F}_{BL}(E) := \{ f:E\rightarrow \R \mid \|f\|_{BL}\leq 1\}.
\end{align}

A sequence of $E$-valued random variables $(X_n)_{n\in\N}$ converges weakly to $X$ if and only if

\[ d_{BL}(X_n, X) := \sup_{f\in\mathcal{F}_{BL}(E)} \left| \E[f(X_n)] - \E[f(X)]\right| \stackrel{n\to\infty}{\longrightarrow} 0,\]

see for example [65], p. 73. This can be used to define uniform weak convergence in the following way.

Let $\Theta$ be a set of parameters and $(X_n)_{n\in\N}$ a sequence of $E$-valued random variables, where $(E,d)$ is a separable metric space. We say that $(X_n)_{n\in\N}$ converges uniformly (over $\Theta$) to $X$ if and only if

\[ \sup_{\theta\in\Theta} d_{BL}^\theta(X_n,X) :=\sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(E)} \left| \E_\theta[f(X_n)] - \E_\theta[f(X)]\right| \stackrel{n\to\infty}{\longrightarrow} 0.\]

The following result is immediate.

Let $(E,d)$ be a separable metric space, $f:E\rightarrow\R$ be a bounded Lipschitz function and $\sup_{\theta\in\Theta} d_{BL}^\theta (X_n, X)\rightarrow 0$. Then we have

\[ \sup_{\theta\in\Theta} \E_\theta[ | f(X_n) -f(X)|] \stackrel{n\to\infty}{\longrightarrow} 0.\]

Let $B,L\geq 1$ be chosen such that $|f(x)-f(y)|\leq Ld(x,y)$ and $|f(x)|\leq B$ for all $x,y\in E$, which is possible as $f$ is a bounded Lipschitz function. Set

\[ \tilde{f} := \left(2BL\right)^{-1} f,\]

and note that $\tilde{f}\in\mathcal{F}_{BL}(E)$ as

\[ \|\tilde{f}\|_{BL} \leq \|\tilde f\|_\infty + \sup_{x\neq y}\frac{|f(x)-f(y)|}{2BL d(x,y)}  \leq \frac{B}{2BL} + \frac{L}{2BL} \leq 1.\]

\begin{align*}
\sup_{\theta\in\Theta} \E_\theta[ | f(X_n) -f(X)|] &=  2BL\cdot \sup_{\theta\in\Theta} \E_\theta[ | \tilde{f}(X_n) -\tilde{f}(X)|]  \\
&\leq 2BL\cdot\sup_{\theta\in\Theta} d_{BL}^\theta (X_n, X) \rightarrow 0.
\end{align*}

Let $(A_n(s))_{n\in\N}$ and $(B_n)_{n\in\N}$ be two real-valued stochastic processes, where $A_n(s)$, $n\in\N$, depends on some parameter $s\in\mathcal{S}$. Suppose that for all $\epsilon>0$,

\[ \sup_{\theta\in\Theta} \Pr_\theta\left( \sup_{s\in\mathcal{S}}\left| A_n(s) - A(s)\right|>\epsilon\right)\rightarrow 0 \]

\[ \sup_{\theta\in\Theta} \Pr_\theta\left( \left| B_n - B\right|>\epsilon\right) \rightarrow 0,\]

where $A(s)$ and $B$ are real-valued random variables satisfying $1\geq B\geq \beta >0$ and $\sup_{s\in\mathcal{S}} |A(s)|\leq 1$. Then we have for all $\epsilon>0$,

\[ \sup_{\theta\in\Theta} \Pr_\theta\left( \sup_{s\in\mathcal{S}}\left| \frac{A_n(s)}{B_n} - \frac{A(s)}{B}\right|>\epsilon\right)\rightarrow 0. \]

By adding zero, we have

\[ \sup_{s\in\mathcal{S}}\left| \frac{A_n(s)}{B_n} - \frac{A(s)}{B}\right| = \sup_{s\in\mathcal{S}}\left| \frac{B(A_n(s)-A(s)) + A(s)(B-B_n)}{BB_n}\right|.\]

With this identity, we can write

\begin{align*}
&\sup_{\theta\in\Theta} \Pr_\theta\left( \sup_{s\in\mathcal{S}}\left| \frac{A_n(s)}{B_n} - \frac{A(s)}{B}\right|>\epsilon\right) \\
&\hspace{0.5cm} = \sup_{\theta\in\Theta} \Pr_\theta\left(\sup_{s\in\mathcal{S}}\left| \frac{B(A_n(s)-A(s)) + A(s)(B-B_n)}{BB_n}\right|>\epsilon\right) \\
&\hspace{0.5cm} = \sup_{\theta\in\Theta} \Pr_\theta\left(\sup_{s\in\mathcal{S}}\left| B(A_n(s)-A(s)) + A(s)(B-B_n)\right|>\epsilon B |B_n|\right) \\
&\hspace{0.5cm}\leq \sup_{\theta\in\Theta} \Pr_\theta\left(\sup_{s\in\mathcal{S}}\left| B(A_n(s)-A(s)) + A(s)(B-B_n)\right|>\epsilon B |B_n|, |B_n|\geq\frac{\beta}{2}\right)\\
&\hspace{3cm} + \sup_{\theta\in\Theta}\Pr_\theta\left( |B_n|<\frac{\beta}{2}\right).
\end{align*}

The second summand is easy to treat because as $B\geq\beta$,

\[\sup_{\theta\in\Theta}\Pr_\theta\left( |B_n|<\frac{\beta}{2}\right) \leq \sup_{\theta\in\Theta}\Pr_\theta\left( |B_n-B|>\frac{\beta}{2}\right),\]

and this term tends to zero by the assumption $\beta>0$. The first summand  is bounded from above by

\begin{align*}
&\sup_{\theta\in\Theta} \Pr_\theta\left(\sup_{s\in\mathcal{S}}\left| B(A_n(s)-A(s)) + A(s)(B-B_n)\right|>\epsilon \beta^2/2\right)\\
&\hspace{0.5cm} \leq \sup_{\theta\in\Theta} \Pr_\theta\left(\sup_{s\in\mathcal{S}}\left| B(A_n(s)-A(s))\right|+ \sup_{s\in\mathcal{S}}\left|A(s)(B-B_n)\right|>\epsilon \beta^2/2\right)\\
&\hspace{0.5cm} = \sup_{\theta\in\Theta} \Pr_\theta\left(B \sup_{s\in\mathcal{S}}\left| (A_n(s)-A(s))\right|+ \left(\sup_{s\in\mathcal{S}}|A(s)|\right)|B-B_n|>\epsilon \beta^2/2\right)\\
&\hspace{0.5cm} \leq \sup_{\theta\in\Theta} \Pr_\theta\left( \sup_{s\in\mathcal{S}}\left| A_n(s)-A(s)\right|+ |B-B_n|>\epsilon \beta^2/2\right)\\
&\hspace{0.5cm} \leq \sup_{\theta\in\Theta} \Pr_\theta\left( \sup_{s\in\mathcal{S}}\left| A_n(s)-A(s)\right|>\epsilon \beta^2/4\right) + \sup_{\theta\in\Theta} \Pr_\theta\left( \left| B_n - B\right|>\epsilon \beta^2/4\right),
\end{align*}

where we used in the third step that $0<B\leq 1$ and $\sup_{s\in\mathcal{S}} |A(s)|\leq 1$. Now, these remaining probabilities both converge to zero by the assumption on the uniform stochastic convergence for $(A_n(s))_{n\in\N}$ and $(B_n)_{n\in\N}$ and the proof is complete.

Next, we establish that a uniform continuous mapping theorem holds true for Lipschitz functions and give an example for a Lipschitz function on $l_\infty(\mathcal{S})$ that is used later. It is a reformulation of Theorem $1$ in [34] for metric spaces and a short proof is given for the reader's convenience.

Let $(E,d)$ be a separable metric space. Suppose that the $E$-valued sequence $(X_n)_{n\in\N}$ converges uniformly (over $\Theta$) in distribution to $X$. If $h:E\rightarrow \R$ is Lipschitz continuous, then $h(X_n)$ converges uniformly (over $\Theta$) in distribution to $h(X)$.

Choose $L\geq 1$ such that $|h(x)-h(y)|\leq Ld(x,y)$. We first note that $f\in\mathcal{F}_{BL}(\R)$ implies $\tilde{f}:= \frac1L f\circ h \in\mathcal{F}_{BL}(E)$ as

\begin{align*}
\left| \tilde{f}(x) - \tilde{f}(y)\right| &= L^{-1} \left| f(h(x)) - f(h(y))\right| \leq L^{-1}  |h(x)- h(y))| \leq L^{-1} L d(x,y)
\end{align*}

and $|\tilde{f}(x)|\leq 1$. Consequently,

\begin{align*}
&\sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(\R)} \left| \E_\theta[f(h(X_n))] - \E_\theta[ f(h(X))]\right| \\
&\hspace{1cm} \leq L \cdot \sup_{\tilde{f}\in\mathcal{F}_{BL}(E)} \left| \E_\theta[\tilde{f}(X_n)] - \E_\theta[ \tilde{f}(h(X))]\right|
\end{align*}

and this last term tends to zero by assumption on $(X_n)_{n\in\N}$.

Let $a,b\in l_\infty(\mathcal{S})$. Then for any $\delta>0$, the map $l_\infty(\mathcal{S}) \rightarrow \R$ with

\[ (x(s))_{s\in\mathcal{S}}\ \mapsto\ \sup_{s\in\mathcal{S}: a(s)>\delta}\left| \frac{x(s)}{a(s)} - b(s)\right| \]

is Lipschitz continuous with respect to the supremum metric $\|\cdot\|_\mathcal{S}$ on $l_\infty(\mathcal{S})$.

Take $x, \tilde{x}\in l_\infty(\mathcal{S})$. Then, by the reverse triangle inequality,

\begin{align*}
&\left| \sup_{s\in\mathcal{S}: a(s)>\delta}\left| \frac{x(s)}{a(s)} - b(s)\right| - \sup_{s\in\mathcal{S}: a(s)>\delta}\left| \frac{\tilde{x}(s)}{a(s)} - b(s)\right| \right|\\
&\hspace{1cm} \leq \sup_{s\in\mathcal{S}: a(s)>\delta}\left| \left| \frac{x(s)}{a(s)} - b(s)\right| - \left| \frac{\tilde{x}(s)}{a(s)} - b(s)\right| \right| \\
&\hspace{1cm} \leq \sup_{s\in\mathcal{S}: a(s)>\delta}\left|\frac{x(s) - \tilde{x}(s)}{a(s)} \right| \\
&\hspace{1cm}\leq \frac{1}{\delta} \|x - \tilde{x}\|_\mathcal{S}
\end{align*}

and the claim follows.

Let $(X_n)_{n\in\N}$ be a sequence of $\R^d$-valued random variables whose distribution depends on a parameter $\theta\in\Theta$ and the uniform convergence

\[ \sup_{\theta\in\Theta} \sup_{f\in \mathcal{F}_{BL}(\R^d)} \E_\theta\left[ | f(X_n) - f(X)|\right] \stackrel{n\to\infty}{\longrightarrow} 0 \]

holds true. Furthermore, assume the family $(\Pr_\theta^X)_{\theta\in\Theta}$ to be tight. Then for every $\epsilon>0$ there exists a compact set $K_\epsilon\subset \R^d$ and $n_0\in\N$ such that for every $n\geq n_0$,

\[ \inf_{\theta\in\Theta} \Pr_\theta\left( X_n\in K_\epsilon\right) \geq 1-\epsilon. \]

Let $\epsilon>0$. By assumption there exists a compact set $K\subset\R^d$ such that

\[ \inf_{\theta\in\Theta} \Pr_\theta (X\in K) \geq 1-\epsilon.\]

We assume $K$ to be a closed ball of radius $r$ centered at the origin, i.e. $K=\overline{B_r(0)}$. This can always be done, as any compact set in $\R^d$ is subset of such a ball by the Bolzano Weierstrass theorem. Now pick $\delta>0$ and define $K^\delta := \overline{B_{r+\delta}(0)}$, together with

\[ f_{K,\delta}(x) := \begin{cases}
1& \textrm{ if } x\in K,\\
1-\frac{d(x,K)}{\delta} & \textrm{ if } x\in K^\delta\setminus K, \\
0 &\textrm{ else},
\end{cases} \]

where $d(x,K) := \inf_{z\in K} \|x-z\|_2$. The function $f_{K,\delta}$ is Lipschitz continuous and we have $\1_K\leq f_{K,\delta}\leq \1_{K^\delta}$. This gives

\begin{align*}
&\inf_{\theta\in\Theta} \Pr_\theta \left(X_n\in K^\delta\right) = \inf_{\theta\in\Theta}\E_\theta \left[ \1_{K^\delta}(X_n)\right] \geq \inf_{\theta\in\Theta} \E_\theta\left[ f_{K,\delta}(X_n)\right]\\
&\hspace{1cm} \stackrel{n\to\infty}{\longrightarrow} \inf_{\theta\in\Theta}\E_\theta\left[ f_{K,\delta}(X)\right] \geq \inf_{\theta\in\Theta} \E_\theta\left[ \1_K(X)\right] = \inf_{\theta\in\Theta}\Pr_\theta(X\in K).
\end{align*}

The last term is greater than $1-\epsilon$ and we find an integer $n_0$ such that for $n\geq n_0$, we have $\inf_{\theta\in\Theta} \Pr_\theta \left(X_n\in K^\delta\right)\geq 1-2\epsilon$.

The following result is a modification of Theorem $1.12.1$ in [65] for uniform weak convergence and its proof follows the same ideas. In our later application of this result, the bounded and equicontinuous class $\mathcal{F}$ used within this result will be chosen as $\mathcal{F}_{BL}(\R^d)$.

Let $\Theta$ be a parameter space and $X, X_1, X_2\dots$ be $\R^d$-valued random variables such that the family $(\Pr_\theta^{X})_{\theta\in\Theta}$ is tight, i.e. for every $\epsilon>0$ there exists a compact set $K=K_\epsilon\subset\R^d$  such that

\[ \inf_{\theta\in\Theta} \Pr_\theta ( X\in K) \geq 1-\epsilon.\]

Furthermore, we have for any bounded and continuous $f:\R^d\rightarrow\R$,

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

Let $\mathcal{F}$ be a bounded and equicontinuous class of functions. Then the last convergence even holds true uniformly in $\mathcal{F}$, i.e.

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

Let $\epsilon>0$. Then choose $K'$ to suffice $\sup_{\theta\in\Theta} \Pr_\theta(X\in K')\geq 1-\epsilon$ according to the assumption. Using Lemma <ref>, there exists a compact set $K$ with $K'\subset K\subset\R^d$ and $n_0\in\N$ such that for $n\geq n_0$ we have

\[ \inf_{\theta\in\Theta} \Pr_\theta( X_n\in K)\geq 1-\epsilon.\]

Denote by $\mathcal{F}_K$ the set of all functions of $\mathcal{F}$ restricted to the compact set $K$. Then by the Arzelà-Ascoli theorem, $\mathcal{F}_K$ is totally bounded in $\mathcal{C}_b(K)$ and there exist finitely many balls of radius $\epsilon$ that cover $\mathcal{F}_K$. Denote their centers by $f_1, \dots, f_N$. By Tietze's theorem, we can extend them to elements of $\mathcal{C}_b(\R^d)$ which are again denoted by $f_1,\dots, f_N$. Then, by assumption,

\begin{align}\label{eqp: C1}
\max_{i=1,\dots, N}\sup_{\theta\in\Theta} \left|\E_\theta\left[ f(X_n) \right] - \E_\theta\left[f(X)\right] \right|\stackrel{n\to\infty}{\longrightarrow} 0.
\end{align}

We split

\begin{align}\label{eqp: C13}
\begin{split}
\E_\theta\left[ f(X_n) \right] - \E_\theta\left[f(X)\right] &= \E_\theta\left[ f(X_n) \right] - \E_\theta\left[f_i(X_n) \right] \\
&\hspace{1cm}+ \E_\theta\left[ f_i(X_n) \right] - \E_\theta\left[f_i(X)\right]\\
&\hspace{1cm}+\E_\theta\left[ f_i(X) \right] - \E_\theta\left[f(X)\right].
\end{split}
\end{align}

The first difference can be rewritten as

\begin{align*}
&\E_\theta\left[ f(X_n) \right] - \E_\theta\left[f_i(X_n) \right] \\
&\hspace{1cm}= \E_\theta\left[(f(X_n) - f_i(X_n))\1_K\right] + \E_\theta\left[ (f(X_n) - f_i(X_n))\1_{K^c}\right].
\end{align*}

For the right choice of $i$ the first term is bounded by $\epsilon$, as we can choose $f_i$ to be the center of the $\epsilon$-ball (in supremum norm on $K$) that contains $f$. For the latter summand we have

\begin{align*}
\sup_{\theta\in\Theta}\sup_{f\in\mathcal{F}}\E_\theta\left[ (f(X_n) - f_i(X_n))\1_{K^c}\right] \leq 2C' \sup_{\theta\in\Theta} \Pr_\theta\left( X_n\in K^c\right) \leq 2C'\epsilon,
\end{align*}

for $n\geq n_0$ and $C'$ being the uniform bound on $\mathcal{F}$. Additionally, we used

\[ \sup_{\theta\in\Theta} \Pr_\theta\left(X_n\in K^c\right) = \sup_{\theta\in\Theta} \left( 1- \Pr(X_n\in K)\right) = 1-\inf_{\theta\in\Theta} \Pr_\theta(X_n\in K) \leq \epsilon. \]

The third difference in (<ref>) can be treated in the same way and the second one is bounded by $\epsilon$ for sufficiently large $n$ by (<ref>). Thus,

\[ \sup_{\theta\in\Theta}\sup_{f\in\mathcal{F}}\left|\E_\theta\left[ f(X_n) \right] - \E_\theta\left[f(X)\right] \right| \leq 3\epsilon + 4C'\epsilon, \]

and letting $\epsilon\searrow 0$ concludes the proof.

Let $(E,d)$ be a separable metric space and $(X_n)_{n\in\N}$ and $(Y_n)_{n\in\N}$ be two sequences of $E$-valued random variables such that for all $\epsilon>0$,

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

and $\sup_{\theta\in\Theta}d_{BL}^\theta(X_n, X)\rightarrow 0$ for some random variable $X$. Then,

\[\sup_{\theta\in\Theta}d_{BL}^\theta(Y_n, X) \stackrel{n\to\infty}{\longrightarrow}\ 0.\]

We clearly have

\begin{align*}
&\sup_{\theta\in\Theta} \sup_{f\in \mathcal{F}_{BL}(E)} \left| \E_\theta[ f(Y_n)] - \E_\theta[f(X)]\right|\\
&\hspace{2cm}\leq \sup_{\theta\in\Theta} \sup_{f\in \mathcal{F}_{BL}(E)} \left| \E_\theta[ f(Y_n)] - \E_\theta[f(X_n)]\right|\\
&\hspace{3cm}+\sup_{\theta\in\Theta} \sup_{f\in \mathcal{F}_{BL}(E)} \left| \E_\theta[ f(X_n)] - \E_\theta[f(X)]\right|.
\end{align*}

The second term tends to zero for $n\to\infty$ by assumption. For the first one,

\begin{align*}
&\sup_{\theta\in\Theta} \sup_{f\in \mathcal{F}_{BL}(E)} \left| \E_\theta[ f(Y_n)] - \E_\theta[f(X_n)]\right| \\
&\hspace{1cm} \leq \sup_{\theta\in\Theta} \sup_{f\in \mathcal{F}_{BL}(E)} \E_\theta[ |f(Y_n)-f(X_n)|]\\
&\hspace{1cm} = \sup_{\theta\in\Theta} \sup_{f\in \mathcal{F}_{BL}(E)} \E_\theta[ |f(Y_n)-f(X_n)| \1_{\{d(X_n,Y_n)\leq \epsilon\}}]\\
&\hspace{2cm} + \sup_{\theta\in\Theta} \sup_{f\in \mathcal{F}_{BL}(E)} \E_\theta[ |f(Y_n)-f(X_n)| \1_{\{d(X_n,Y_n)>\epsilon\}}]\\
&\hspace{1cm} \leq \epsilon + 2\sup_{\theta\in\Theta}\Pr_\theta\left( d( X_n ,Y_n)>\epsilon\right).
\end{align*}

The probability in this term vanishes for $n\to\infty$ by the assumption of uniform convergence in probability and the statement of the lemma follows by letting $\epsilon\searrow 0$.

Let sequences $(X_n(s))_{n\in\N}$, $(\alpha_n(s))_{n\in\N}$ and $(\beta_n(s))_{n\in\N}$ of real-valued stochastic processes  together with real-valued random variables $\alpha(s),\beta(s),X(s)$ depending on a parameter $s\in\mathcal{S}$ be given. Assume the following conditions to be true for a parameter space $\Theta$:

(a) We have the uniform weak convergence

\[ \sup_{\theta\in\Theta} d_{BL}^\theta\left(\sup_{s\in\mathcal{S}}\left(\frac{X_n(s)}{\alpha(s)} -\beta(s)\right),\sup_{s\in\mathcal{S}}\left(\frac{X(s)}{\alpha(s)} -\beta(s)\right)\right) \stackrel{n\to\infty}{\longrightarrow}\ 0.\]

(b) We have $\alpha(s)\in (0,\infty)$ and for every $\epsilon>0$ there exist compact sets $K(\epsilon), K'(\epsilon)\subset\R$ with

\[ \inf_{\theta\in\Theta} \Pr_\theta\left(\sup_{s\in\mathcal{S}} \beta(s)\in K(\epsilon)\right) \geq 1-\epsilon \]

\[ \inf_{\theta\in\Theta} \Pr_\theta\left(\sup_{s\in\mathcal{S}}\left(\frac{X(s)}{\alpha(s)} -\beta(s)\right)\in K'(\epsilon)\right)\geq 1-\epsilon.\]

(c) For all $n\in\N$, $\alpha_n(s)\in (0,\infty)$ and for every $\epsilon>0$,

\[ \sup_{\theta\in\Theta} \Pr_\theta\left(\sup_{s\in\mathcal{S}} \left|\frac{\alpha(s)}{\alpha_n(s)} - 1\right|>\epsilon\right)\stackrel{n\to\infty}{\longrightarrow} 0.\]

(d) For every $\epsilon>0$,

\[ \sup_{\theta\in\Theta} \Pr_\theta\left( \sup_{s\in\mathcal{S}}|\beta_n(s) - \beta(s)|>\epsilon\right) \stackrel{n\to\infty}{\longrightarrow} 0 .\]

Then we have

\[ \sup_{\theta\in\Theta} d_{BL}^\theta\left(\sup_{s\in\mathcal{S}}\left(\frac{X_n(s)}{\alpha_n(s)} -\beta_n(s)\right),\sup_{s\in\mathcal{S}}\left(\frac{X(s)}{\alpha(s)} -\beta(s)\right)\right) \stackrel{n\to\infty}{\longrightarrow}\ 0.\]

We want to apply Lemma <ref> and therefore need to show in addition to assumption (a) that for all $\epsilon>0$,

\[ \sup_{\theta\in\Theta}\Pr_\theta\left( \left|\sup_{s\in\mathcal{S}}\left( \frac{X_n(s)}{\alpha_n(s)} - \beta_n(s)\right) - \sup_{s\in\mathcal{S}}\left( \frac{X_n(s)}{\alpha(s)}- \beta(s)\right) \right|>\epsilon\right) \stackrel{n\to\infty}{\longrightarrow} 0.\]

By upper bounding the random variable within the probability, we find

\begin{align*}
&\sup_{\theta\in\Theta}\Pr_\theta\left( \left|\sup_{s\in\mathcal{S}}\left( \frac{X_n(s)}{\alpha_n(s)} - \beta_n(s)\right) - \sup_{s\in\mathcal{S}}\left( \frac{X_n(s)}{\alpha(s)}- \beta(s)\right) \right|>\epsilon\right) \\
&\hspace{0.1cm} \leq \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}\left|\left( \frac{X_n(s)}{\alpha_n(s)} - \beta_n(s)  -\frac{X_n(s)}{\alpha(s)}+ \beta(s)\right) \right|>\epsilon\right) \\
&\hspace{0.1cm} = \sup_{\theta\in\Theta}\Pr_\theta\left(\sup_{s\in\mathcal{S}} \left| \left(\frac{X_n(s)}{\alpha(s)} - \beta(s) \right)\left( \frac{\alpha(s)}{\alpha_n(s)} -1\right) +\frac{\alpha(s)}{\alpha_n(s)}\beta(s) - \beta_n(s)  \right|>\epsilon\right) \\
&\hspace{0.1cm} \leq \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}} \left|\frac{X_n(s)}{\alpha(s)} - \beta(s) \right|\sup_{s\in\mathcal{S}}\left| \frac{\alpha(s)}{\alpha_n(s)} -1\right| \right. \\
&\hspace{6cm} \left. + \sup_{s\in\mathcal{S}}\left|\frac{\alpha(s)}{\alpha_n(s)}\beta(s) - \beta_n(s)  \right|>\epsilon\right) \\
&\hspace{0.1cm} \leq \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}} \left| \frac{X_n(s)}{\alpha(s)} - \beta(s) \right|\sup_{s\in\mathcal{S}}\left| \frac{\alpha(s)}{\alpha_n(s)} -1\right| >\frac{\epsilon}{2}\right) \\
&\hspace{4cm} + \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}\left|\frac{\alpha(s)}{\alpha_n(s)}\beta(s) - \beta_n(s)\right|>\frac{\epsilon}{2}\right) .
\end{align*}

Those last two summands will be treated separately. Let $\delta>0$ and choose a corresponding $K=K(\delta)>0$ such that

\[ \sup_{\theta\in\Theta} \Pr_\theta \left( \sup_{s\in\mathcal{S}} \left|\frac{X(s)}{\alpha(s)}- \beta(s)\right| >K\right) < \delta,\]

which is possible by assumption (b). Using $\sup_\theta\Pr_\theta(A)= 1-\inf_\theta\Pr_\theta(A^c)$
and Lemma <ref>, for some $n\geq n_0$ and $K'\geq K>0$,

\[ \sup_{\theta\in\Theta} \Pr_\theta \left( \sup_{s\in\mathcal{S}} \left|\frac{X_n(s)}{\alpha(s)}- \beta(s)\right| >K'\right) < \delta. \]

Proceeding with this,

\begin{align*}
&\sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}} \left| \frac{X_n(s)}{\alpha(s)} - \beta(s) \right|\sup_{s\in\mathcal{S}}\left| \frac{\alpha(s)}{\alpha_n(s)} -1\right| >\frac{\epsilon}{2}\right) \\
&\hspace{0.2cm} \leq \sup_{\theta\in\Theta}\Pr_\theta\left( K'\sup_{s\in\mathcal{S}}\left| \frac{\alpha(s)}{\alpha_n(s)} -1\right| >\frac{\epsilon}{2}\right)  + \sup_{\theta\in\Theta} \Pr_\theta \left( \sup_{s\in\mathcal{S}} \left|\frac{X_n(s)}{\alpha(s)}- \beta(s)\right| >K'\right)\\
&\hspace{0.2cm} \leq \sup_{\theta\in\Theta}\Pr_\theta\left(\sup_{s\in\mathcal{S}}\left| \frac{\alpha(s)}{\alpha_n(s)} -1\right| >\frac{\epsilon}{2K'}\right)  + \delta
\end{align*}

and the probability converges to zero for $n\to\infty$ by condition (c). For the other summand at the end of our initial estimate, we proceed in a similar manner. We choose $L$ large enough such that

\[ \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}|\beta(s)|>L\right) <\delta,\]

which is possible with regard to our assumption (b). Then,

\begin{align*}
& \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}\left|\frac{\alpha(s)}{\alpha_n(s)}\beta(s) - \beta_n(s)\right|>\frac{\epsilon}{2}\right) \\
&\hspace{1cm} =  \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}\left|\left(\frac{\alpha(s)}{\alpha_n(s)} -1\right)\beta(s)  + \beta(s)- \beta_n(s)\right|>\frac{\epsilon}{2}\right) \\
&\hspace{1cm} \leq \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}\left|\left(\frac{\alpha(s)}{\alpha_n(s)} -1\right)\beta(s) \right| + \sup_{s\in\mathcal{S}}| \beta(s)- \beta_n(s)|>\frac{\epsilon}{2}\right)\\[10\jot]
&\hspace{1cm} \leq \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}\left|\left(\frac{\alpha(s)}{\alpha_n(s)} -1\right)\beta(s) \right| >\frac{\epsilon}{4}\right) \\
&\hspace{3.5cm} + \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}| \beta(s)- \beta_n(s)| >\frac{\epsilon}{4}\right) \\
&\hspace{1cm} \leq \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}\left|\frac{\alpha(s)}{\alpha_n(s)} -1\right| >\frac{\epsilon}{4L}\right) + \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}|\beta(s)|>L\right)  \\
&\hspace{3.5cm} + \sup_{\theta\in\Theta}\Pr_\theta\left( \sup_{s\in\mathcal{S}}| \beta(s)- \beta_n(s)| >\frac{\epsilon}{4}\right).
\end{align*}

The middle summand here is bounded by $\delta$ and the others both converge to zero by condition (c) and (d). The assertion of the lemma follows by letting $\delta\searrow 0$.

For a pseudometric space $(\mathcal{S},\rho)$ and $u>0$, the covering numbers $N(u,\mathcal{S},\rho)$ are defined by

\begin{align}\label{eq: covering_number}
N(u,\mathcal{S},\rho) := \min\left\{ \#\mathcal{S}_0: \mathcal{S}_0\subset\mathcal{S} \textrm{ and } \inf_{s_0\in\mathcal{S}_0} \rho(s,s_0)\leq u \textrm{ for all } s\in\mathcal{S} \right\},
\end{align}

where $\#S$ denotes the cardinality of a finite set $S$.
Subsequently, a proof of Proposition <ref> is given. This result is an extension of Theorem $8$ in [51] to uniform weak convergence.

For every natural number $k\in\N$ let $\mathcal{S}_\theta^k$ be some maximal subset of $\mathcal{S}$ such that $\rho_\theta(s,s')\geq \frac1k$ for any $t,t'\in\mathcal{S}_\theta^k$, and $\mathcal{S}_\theta^1\subset\mathcal{S}_\theta^2\subset\cdots$. Now define

\[ \lambda_\theta^k(s,u) := \frac{\left( 1- k\rho_\theta(s,u)\right)_+}{\sum_{v\in\mathcal{S}_\theta^k} \left( 1- k\rho_\theta(s,v)\right)_+}\]

for all $s\in\mathcal{S}$ and $u\in\mathcal{S}_\theta^k$. It satisfies

\[ \lambda_\theta^k(s,u) =0  \quad \textrm{ if }\quad \rho_\theta(s,u))\geq\frac1k \]

\[ \sum_{u\in\mathcal{S}_\theta^k} \lambda_\theta^k(\cdot, u) =1.\]

Moreover, $0\leq \lambda_\theta^k(\cdot, u) \in\mathcal{C}_u(\mathcal{S},\rho_\theta)$, where

\[ \mathcal{C}_u(\mathcal{S}), \rho_\theta) = \left\{ f\in l_\infty(\mathcal{S}):\ \lim_{\delta\searrow 0} \sup_{s,t\in\mathcal{S}: \rho_\theta(s,t)<\delta} |f(t)- f(s)| =0\right\}. \]

Define $\pi_\theta^k: l_\infty(\mathcal{S}) \rightarrow\mathcal{C}_u(\mathcal{S}, \rho_\theta)$ by

\[ \pi_\theta^k f := \sum_{u\in\mathcal{S}_\theta^k} f(u) \lambda_\theta^k(\cdot, u).\]

Then for all $f\in l_\infty(\mathcal{S})$,

\begin{align}\label{eqp: C16}
\|\pi_\theta^k f\|_\mathcal{S} \leq \|f\|_{\mathcal{S}_\theta^k}\quad \textrm{ and }\quad \| f- \pi_\theta^k f\|_{\mathcal{S}} \leq \sup_{\rho_\theta(s,s')\leq \frac1k} |f(s) - f(s')|,
\end{align}

and $\pi_\theta^k$ is a linear map with Lipschitz constant one because for $f,g\in l_\infty(\mathcal{S})$,

\begin{align*}
\left| \pi_\theta^k(s) - \pi_\theta^k(s)\right| &= \left| \sum_{u\in\mathcal{S}_\theta^k} (f(u)-g(u))\lambda_\theta^k(s,u)\right| \\
&\leq \sup_{u\in\mathcal{S}_\theta^k}\left| f(u)-g(u)\right| \sum_{u\in\mathcal{S}_\theta^k} \lambda_\theta^k(s,u)\\
& \leq \|f-g\|_{\mathcal{S}}.
\end{align*}

In particular, for any $f: l_\infty(\mathcal{S}) \rightarrow [0,1]$ with Lipschitz constant $L$, the composition $f\circ\pi_\theta^k$ again takes values in $[0,1]$ and has Lipschitz constant $L$. Now we split

\begin{align*}
&\sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))} \left| \E_\theta\left[ f(X_n)\right] - \E_\theta\left[f(Y_n)\right]\right| \\
&\hspace{2cm}\leq  \sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))}  \E_\theta\left[ \left|f(X_n)-f(\pi_\theta^k X_n)\right|\right]\\
&\hspace{3cm} + \sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))} \E_\theta\left[ \left| f(\pi_\theta^k X_n)-f(\pi_\theta^k Y_n)\right|\right]\\
&\hspace{3cm} + \sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))}  \E_\theta\left[\left| f(\pi_\theta^k Y_n)-f(Y_n) \right|\right].
\end{align*}

We have $\{f\circ\pi_\theta^k\mid f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))\} \subset \mathcal{F}_{BL}(l_\infty(\mathcal{S}_\theta^k))$ and assumption (c) on the covering numbers yields $\sup_\theta|\mathcal{S}_\theta^k|<\infty$. By assumption (a),

\begin{align*}
&\sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))} \E_\theta\left[ \left| f(\pi_\theta^k X_n)-f(\pi_\theta^k Y_n)\right|\right]\stackrel{n\to\infty}{\longrightarrow}\ 0.
\end{align*}

Let $\epsilon>0$. By assumption (b) there exists a natural number $l=l(\epsilon)$ such that for $Z_n= X_n, Y_n$ we have

\[  \limsup_{n\to\infty} \sup_{\theta\in\Theta} \Pr_\theta\left( \sup_{\rho_\theta(s,s')\leq\frac{1}{l}} |Z_n(s) - Z_n(s')| >\epsilon \right) <\epsilon. \]

By the bounded Lipschitz property of $f$ and (<ref>) we get for this number $l$,

\begin{align*}
&\limsup_{n\to\infty}\sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))}  \E_\theta\left[ \left|f(Z_n)-f(\pi_\theta^{l} Z_n)\right|\right]\\
&\hspace{0.5cm}\leq \limsup_{n\to\infty}\sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))}  \E_\theta\left[ \left|f(Z_n)-f(\pi_\theta^{l} Z_n)\right| \1_{\{\|Z_n-\pi_\theta^{l} Z_n\|_\mathcal{S}\leq\epsilon\}}\right]\\
&\hspace{1cm} +\limsup_{n\to\infty}\sup_{\theta\in\Theta} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{S}))}  \E_\theta\left[ \left|f(Z_n)-f(\pi_\theta^{l} Z_n)\right| \1_{\{\|Z_n-\pi_\theta^{l} Z_n\|_\mathcal{S}>\epsilon\}}\right]\\
&\hspace{0.5cm} \leq  \epsilon +2\limsup_{n\to\infty}\sup_{\theta\in\Theta}\Pr_\theta\left( \|Z_n-\pi_\theta^l Z_n\|_\mathcal{S}>\epsilon\right) \\
&\hspace{0.5cm}\leq \epsilon + 2 \limsup_{n\to\infty} \sup_{\theta\in\Theta} \Pr_\theta\left( \sup_{\rho_\theta(s,s')\leq\frac1l} |Z_n(s) - Z_n(s')| >\epsilon \right) <3\epsilon.
\end{align*}

Letting $\epsilon\searrow 0$ completes the proof.

§.§ Proof of Theorem <ref>

Throughout the proof we will abbreviate $\Sigma:= \Sigma(C,A,\gamma,\sigma)$. Moreover, we define

\[Z_T(y,h) := \frac{1}{\sqrt{T}} \int_0^T K_{y,h}(X_s) dW_s \]

\[ Z_b(y,h) := \int_\R K_{y,h}(z) \sqrt{q_b(z)} dW_z.\]

We remember the notation

\[ \hat\sigma_T(y,h) ^2= \frac1T\int_0^T K_{y,h}(X_s)^2 ds\quad \textrm{ and } \quad \hat\sigma_{T,\max}^2= \frac1T\int_0^T \1_{[-A,A]}(X_s) ds \]

from Appendix <ref>, together with $\overline\sigma_T(y,h) = \frac{\hat\sigma_T(y,h)}{\hat\sigma_{T,\max}}$ and the semimetric

\[ \overline\rho_T((y,h), (y',h'))^2 = \frac{1}{T\hat\sigma_{T,\max}^2} \int_0^T \left( K_{y,h}(X_s) - K_{y',h'}(X_s)\right)^2 ds.\]

We define the deterministic counterparts

\begin{align}\label{eqp: C5}
\sigma_b(y,h) ^2 := \int_\R K_{y,h}(z)^2 q_b(z) dz\ \ \textrm{ and } \ \ \sigma_{b,\max}^2 := \int_\R \1_{[-A,A]}(z)q_b(z) dz
\end{align}

on the set $\mathcal{T}$ given in (<ref>), and the corresponding semimetric on $\mathcal{T}\times\mathcal{T}$ by

\[ \rho_b((y,h), (y',h'))^2 := \int_\R \left( K_{y,h}(z) - K_{y',h'}(z)\right)^2 q_b(z) dz. \]

Last but not least, we set

\begin{align}\label{eqp: C6}
\overline\sigma_b(y,h) := \frac{\sigma_b(y,h)}{\sigma_{b,\max}}\ \ \textrm{ and }\ \ \overline\rho_b((y,h), (y',h')) := \frac{\rho_b((y,h), (y',h'))}{\sigma_{b,\max}}.
\end{align}

For the norms and seminorms defined above, we have for every $\epsilon>0$ that

\[ \lim_{T\to\infty}\sup_{b\in\Sigma} \Pr_b\left( \| \overline\sigma_T - \overline\sigma_b\|_\mathcal{T} >\epsilon\right) =0 \]

\[ \lim_{T\to\infty}\sup_{b\in\Sigma} \Pr_b\left( \| \overline\rho_T - \overline\rho_b\|_{\mathcal{T}\times\mathcal{T}} >\epsilon\right) =0. \]

Clearly, we have  $\sup_{(y,h)\in\mathcal{T}}\hat\sigma_b(y,h)^2 \leq 1$ and

\[1\geq \sigma_{b,\max}^2 = \int_\R \1_{[-A,A]}(z) q_b(z) dz \geq L_* \int_\R \1_{[-A,A]}(z) dz = 2A L_*>0.\]

Additionally, we have

\begin{align*}
&\lim_{T\to\infty} \sup_{b\in\Sigma}\Pr_b\left( \| \hat\sigma_T^2 - \sigma_b^2\|_{\mathcal{T}} > \epsilon \right) =0, \\
&\lim_{T\to\infty} \sup_{b\in\Sigma}\Pr_b\left( \| \hat\rho_T^2 - \rho_b^2\|_{\mathcal{T}\times\mathcal{T}} > \epsilon \right) =0, \textrm{ and} \\
&\lim_{T\to\infty} \sup_{b\in\Sigma}\Pr_b\left( | \hat\sigma_{T,\max}^2 - \sigma_{b,\max}^2| > \epsilon \right) =0
\end{align*}

by Proposition <ref> applied to the respective class

\begin{align*}
&\mathcal{F}_1 := \left\{ K_{y,h}^2 \mid (y,h)\in\mathcal{T}\right\},\\
&\mathcal{F}_2 := \left\{ \left( K_{y,h}- K_{y',h'}\right)^2 \mid (y,h), (y',h')\in\mathcal{T}\right\}, \textrm{ or} \\
& \mathcal{F}_3 := \left\{ \1_{[-A,A]}\right\}.
\end{align*}

The claim now follows by an application of Lemma <ref>.

Let $0<c\leq A$. Then the set

\begin{align}\label{eqp: T_c_overline}
\overline{\mathcal{T}}_c := \left\{(y,h)\in\R^2 \mid c\leq h\leq A \textrm{ and } -A+h \leq y \leq A-h\right\}
\end{align}

is compact.

First, $\overline{\mathcal{T}}_c$ is bounded and by the theorem of Bolzano-Weierstrass it is compact if it is closed. To see closedness, define the two functions $f_1,f_2: [-A,A]\times [c,A] \rightarrow \R$ by $f_1(y,h) := y+h$ and $f_2(y,h):= y-h$. Clearly, both functions are continuous and consequently, the sets

\begin{align*}
f_1^{-1}([-A,A]) &= \left\{ (y,h)\in [-A,A]\times [c,A] \mid -A\leq y+h\leq A\right\} \\
&= \left\{ (y,h)\in [-A,A]\times [c,A] \mid -A-h\leq y\leq A-h\right\}
\end{align*}

\begin{align*}
f_2^{-1}([-A,A]) &= \left\{ (y,h)\in [-A,A]\times [c,A] \mid -A\leq y-h\leq A\right\} \\
&= \left\{ (y,h)\in [-A,A]\times [c,A] \mid -A+h\leq y\leq A+h\right\}
\end{align*}

are closed. Then their intersection $f_1^{-1}([0,A])\cap f_2^{-1}([-A,0])$ is closed as well and it is given by

\begin{align*}
&\left\{ (y,h)\in [-A,A]\times [c,A] \mid (-A-h)\vee (-A+h)\leq y\leq (A+h)\wedge (A-h) \right\} \\
&\hspace{0.5cm} = \left\{ (y,h)\in [-A,A]\times [c,A] \mid -A+h \leq y\leq A-h\right\} =\overline{\mathcal{T}}_c.
\end{align*}

The proof follows the steps (i)–(iii) given in the sketch of proof in Subsection <ref> which are given here again with some additional details:

(i) First, we establish for $0<c\leq A$ and

\begin{align}\label{eqp: C9}
\mathcal{T}_{c}=\{(y,h)\in\mathcal{T}\mid h\geq c\}
\end{align}

the uniform weak convergence of $(Z_T(y,h))_{(y,h)\in\mathcal{T}_{c}}$ to $(Z_b(y,h))_{(y,h)\in\mathcal{T}_{c}}$ with Proposition <ref>.

(ii) Secondly, we define

\begin{align}\label{eqp: T_T}
& T_T(\delta, \delta') := \sup_{\substack{(y,h)\in\mathcal{T}:\\ \delta < \overline\sigma_b(y,h)\leq \delta'}} \left(\frac{\left|Z_T(y,h)\right|}{\hat\sigma_T(y,h)} - C\bigg( \frac{\hat\sigma_T(y,h)^2}{\hat\sigma_{T,\max}^2}\bigg) \right)
\end{align}

as well as

\begin{align*}
&S_b(\delta, \delta') := \sup_{\substack{(y,h)\in\mathcal{T}:\\ \delta < \overline\sigma_b(y,h)\leq \delta'}} \left(\frac{\left|Z_b(y,h)\right|}{\sigma_b(y,h) } - C\bigg( \frac{\sigma_b(y,h)^2}{\sigma_{b,\max}^2}\bigg) \right)
\end{align*}

and prove uniform weak convergence of $T_T(\delta,1)$ to $S_b(\delta,1)$ for every $c'>\delta>0$ and some constant $c'>0$. This is based on the uniform continuous mapping result in Lemma <ref> and Lemma <ref>.

(iii) Finally, we improve the result from (ii) to establish the desired weak convergence of $T_T(0,1)$ to $S_b(0,1)$.

In the following, we will work through step (i), (ii) and (iii).

Step (i). This result is established by using Proposition <ref>. To prove prerequisite (a) of it, we first pick $(y_1,h_1),\dots, (y_N, h_N)\in\overline{\mathcal{T}}_c$ where $\overline{\mathcal{T}}_c$ is given in (<ref>). For all $\epsilon>0$ and $1\leq i,j\leq N$ we have by Proposition <ref>,

\begin{align*}
&\lim_{T\to\infty}\sup_{b\in\Sigma} \Pr\left( \left| \frac1T\int_0^T K_{y_i,h_i}(X_s) K_{y_j, h_j}(X_s)  ds \right.\right. \\
&\hspace{3.5cm} \left.\left. - \int_\R K_{y_i, h_i}(z) K_{y_j, h_j}(z) q_{b}(z) dz \right| >\epsilon\right) =0.
\end{align*}

In consequence, by Proposition $1.21$ in [39],

\[ \lim_{T\to\infty}\sup_{b\in\Sigma} \left| \E_b\left[ f((Z_T(y_k,h_k))_{1\leq k\leq N}) \right] - \E_b\left[ f((Z_b(y_k,h_k))_{1\leq k\leq N})\right]\right| =0\]

for any bounded and continuous function $f:\R^N\rightarrow\R$.
In order to apply Lemma <ref> and conclude

\begin{align}\label{eqp: C10}
\begin{split}
&\lim_{T\to\infty}\sup_{b\in\Sigma}\sup_{f\in \mathcal{F}_{BL}(\R^N)}  \left| \E_b\left[ f((Z_T(y_k,h_k))_{1\leq k\leq N}) \right] \right. \\
&\hspace{4cm} \left. - \E_b\left[ f((Z_b(y_k,h_k))_{1\leq k\leq N})\right]\right| =0,
\end{split}
\end{align}

we have to show uniform tightness of $(Z_b(y_k,h_k))_{1\leq k\leq N}$. Using Markov's inequality and Itô's isometry, we find

\begin{align*}
\sup_{b\in\Sigma} \Pr_b\left( \| Z_b(y_k,h_k)\|_2^2 \leq r\right) &= \sup_{b\in\Sigma} \Pr_b\left( \sum_{k=1}^N Z_b(y_k,h_k)^2 \leq r\right)\\
&\leq \frac1r \sup_{b\in\Sigma}  \E_b\left[ \sum_{k=1}^N Z_b(y_k,h_k)^2\right] \\
&= \frac1r \sup_{b\in\Sigma} \sum_{k=1}^N \int_\R K_{y_k,h_k}(z)^2 q_b(z) dz\\
&\leq \frac{NA\|K\|_{L^2}^2 L^*}{r},
\end{align*}

where we used $h_k\leq A$ and Lemma <ref>. For $r\to\infty$, this expression tends to zero and therefore (<ref>) holds true by Lemma <ref>. Next, we have to consider an additional $\sup_{B\subset\mathcal{T}_c: \#B\leq N}$ instead of fixed $N$, but for the statement of the theorem, it suffices to consider $B\subset\mathcal{T}_c=\overline{\mathcal{T}}_c\cap\Q^2$. Suppose that for some $n_0$,

\begin{align*}
&\sup_{B\subset\mathcal{T}_c: \#B\leq n_0} \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R^{\#B})} \left| \E_b\left[ f((Z_T(y,h))_{(y,h)\in B}) \right]  \right.\\
&\hspace{5.5cm} \left. - \E_b\left[ f((Z_b(y,h))_{(y,h)\in B})\right]\right|
\end{align*}

does not tend to zero for $T\to\infty$. Then there exists a sequence $(T_n)_{n\in\N}$ with $T_n\stackrel{n\to\infty}{\longrightarrow}\infty$ together with $(B_n)_{n\in\N}\subset \mathcal{T}_c$, $\#B_n\leq n_0$, and $\epsilon>0$ such that for all $n\in\N$,

\begin{align}\label{eqp: C11}
\begin{split}
&\sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R^{\#B_n})} \left| \E_b\left[ f((Z_{T_n}(y,h))_{(y,h)\in B_n}) \right] \right.\\
&\hspace{4cm} \left. - \E_b\left[ f((Z_b(y,h))_{(y,h)\in B_n})\right]\right| >\epsilon.
\end{split}
\end{align}

We will show that this is not possible. First of all, there exists an integer $\tilde n_0\leq n_0$ and a subsequence $(n_k)_{k\in\N}$ such that $\#B_{n_k}= \tilde n_0$, as there are only finitely many integers $\leq n_0$. Furthermore, as $\mathcal{T}_c=\overline{\mathcal{T}}_c\cap\Q^2$ and $\overline{\mathcal{T}}_c^{\tilde n_0}$ is compact by Lemma <ref>, there again exists a subsubsequence $(n_{k_l})_{l\in\N}$ such that the vector of locations $((y,h))_{(y,h)\in B_{n_{k_l}}}$ converges to a vector

\[ \left( (y_0^1, h_0^1) , \dots, (y_0^{\tilde n_0}, h_0^{\tilde n_0})\right) =: ((y,h))_{(y,h)\in B}\]

with $B\subset\overline{\mathcal{T}}_c$. We reindex this subsubsequence just by $n$ for ease of notation and have

\begin{align*}
& \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R^{\tilde n_0})} \left| \E_b\left[ f\left((Z_{T_n}(y,h))_{(y,h)\in B_n}\right) \right] - \E_b\left[ f((Z_b(y,h))_{(y,h)\in B_n})\right]\right| \\
&\hspace{0.1cm}\leq \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R^{\tilde n_0})} \E_b\left[ \left| f\left((Z_{T_n}(y,h))_{(y,h)\in B_n}\right) - f\left((Z_{T_n}(y,h))_{(y,h)\in B}\right)\right|\right]\\
&\hspace{0.5cm} + \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R^{\tilde n_0})} \E_b\left[ \left| f\left((Z_{T_n}(y,h))_{(y,h)\in B}\right) - f\left((Z_b(y,h))_{(y,h)\in B}\right)\right|\right]\\
&\hspace{0.5cm} + \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R^{\tilde n_0})} \E_b\left[ \left| f\left((Z_b(y,h))_{(y,h)\in B}\right) - f\left((Z_b(y,h))_{(y,h)\in B_n}\right)\right|\right].
\end{align*}

The second summand tends to zero due to our previous work on uniform weak convergence of a finite-dimensional marginal, see (<ref>). For the first term we use the fact that $f\in\mathcal{F}_{BL}(\R^{\tilde n_0})$ is Lipschitz continuous with constant one and that all norms on $\R^{\tilde n_0}$ are equivalent. Then we have

\begin{align*}
&\sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R^{\tilde n_0})} \E_b\left[ \left| f\left((Z_{T_n}(y,h))_{(y,h)\in B_n}\right) - f\left((Z_{T_n}(y,h))_{(y,h)\in B}\right)\right|\right] \\
&\hspace{0.5cm}\leq C' \sup_{b\in\Sigma} \E_b\left[ \left\| \left((Z_{T_n}(y,h))_{(y,h)\in B_n}\right) - \left((Z_{T_n}(y,h))_{(y,h)\in B}\right)\right\|_2\right]\\
&\hspace{0.5cm}\leq C' \sup_{b\in\Sigma} \sqrt{\E_b\left[ \left\| \left((Z_{T_n}(y,h))_{(y,h)\in B_n}\right) - \left((Z_{T_n}(y,h))_{(y,h)\in B}\right)\right\|_2^2\right] }
\end{align*}

for some constant $C'>0$. Denote $B_n = \left\{ (y_n^1, h_n^1),\dots, (y_n^{\tilde n_0}, h_n^{\tilde n_0})\right\}$. Using Itô's isometry, we have

\begin{align*}
&\E_b\left[ \left\| \left((Z_{T_n}(y,h))_{(y,h)\in B_n}\right) - \left((Z_{T_n}(y,h))_{(y,h)\in B}\right)\right\|_2^2\right]\\
&\hspace{1cm} = \E_b\left[ \sum_{k=1}^{\tilde n_0} \left(\frac{1}{\sqrt{T_n}} \int_0^{T_n} \left(K_{y_n^k, h_n^k}(X_s) - K_{y_0^k, h_0^k}(X_s)\right) dW_s\right)^2\right]\\
&\hspace{1cm} = \sum_{k=1}^{\tilde n_0} \E_b \left[ \frac{1}{T_n} \int_0^{T_n} \left(K_{y_n^k, h_n^k}(X_s) - K_{y_0^k, h_0^k}(X_s)\right)^2 ds\right]\\
&\hspace{1cm} \leq \sum_{k=1}^{\tilde n_0} \left\| K_{y_n^k, h_n^k} - K_{y_0^k, h_0^k}\right\|_{[-A,A]}^2.
\end{align*}

This last term tends to zero for $n\to\infty$ as the location parameters of the kernel function converge to each other. The third term can be treated in the same way, where we have

\begin{align*}
&\E_b\left[ \left\| \left((Z_b(y,h))_{(y,h)\in B_n}\right) - \left((Z_b(y,h))_{(y,h)\in B}\right)\right\|_2^2\right]\\
&\hspace{1cm} = \E_b\left[ \sum_{k=1}^{\tilde n_0} \left( \int_\R \left(K_{y_n^k, h_n^k}(z) - K_{y_0^k, h_0^k}(z)\right) \sqrt{q_b(z)} dW_z\right)^2\right]\\
&\hspace{1cm} = \sum_{k=1}^{\tilde n_0} \int_\R \left(K_{y_n^k, h_n^k}(z) - K_{y_0^k, h_0^k}(z)\right)^2 q_b(z) dz\\
&\hspace{1cm}\leq L^* \sum_{k=1}^{\tilde n_0} \int_\R \left(K_{y_n^k, h_n^k}(z) - K_{y_0^k, h_0^k}(z)\right)^2 dz.
\end{align*}

Here, $L^*$ is the uniform upper bound on the invariant density $q_b$ from Lemma <ref> and again this expression goes to zero due to convergence of the location parameters. Altogether we have convergence to zero of the term

\[ \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R^{\tilde n_0})} \left| \E_b\left[ f\left((Z_{T_n}(y,h))_{(y,h)\in B_n}\right) \right] - \E_b\left[ f((Z_b(y,h))_{(y,h)\in B})\right]\right|, \]

contradicting our choice of $T_n$ and $B_n$ that makes this term be bounded from below by $\epsilon$, see (<ref>). Hence, condition (a) of Proposition <ref> is fullfilled.

We turn to condition (b). We start proving it for $Z_b$ and the metric $\overline\rho_b$. Here we have (with suppressed  dependence of $\overline\rho_b$ on its parameters for shorter notation)

\begin{align*}
&\Pr_{b}\left( \sup_{\overline\rho_b((y,h),(y',h'))\leq \delta}\left| Z_b(y,h) - Z_b(y',h')\right|>\epsilon\right)\\
&\hspace{0.5cm} =\Pr_{b}\left( \sup_{\overline\rho_b((y,h),(y',h'))\leq \delta}\frac{\left| Z_b(y,h) - Z_b(y',h')\right|}{\overline\rho_b \log(e/\overline\rho_b)} \overline\rho_b \log(e/\overline\rho_b)>\epsilon \right) \\
&\hspace{0.5cm}\leq \Pr_{b}\left( \sup_{\overline\rho_b((y,h),(y',h'))\leq \delta}\frac{\left| Z_b(y,h) - Z_b(y',h')\right|}{\overline\rho_b \log(e/\overline\rho_b)} \right. \\
&\hspace{5cm} \left. \cdot \sup_{\overline\rho_b((y,h),(y',h'))\leq \delta}\overline\rho_b \log(e/\overline\rho_b)>\epsilon \right)\\
&\hspace{0.5cm} \leq  \Pr_b\left( Q\cdot \sup_{\overline\rho_b((y,h),(y',h'))\leq \delta}\overline\rho_b \log(e/\overline\rho_b)>\epsilon \right)\\
&\hspace{2.5cm} + \Pr_b\left( \sup_{\overline\rho_b((y,h),(y',h'))\leq \delta}\frac{\left| Z_b(y,h) - Z_b(y',h')\right|}{\overline\rho_b \log(e/\overline\rho_b)}>Q\right)
\end{align*}

for any $Q>0$. The first term tends to zero uniformly in $b$ as $\delta\searrow 0$ because $x\log(e/x)$ does for $x\searrow 0$ monotoneously on $(0,1)$. For a proper choice of the constant $Q$, we have seen in the proof of Theorem <ref> in equation (<ref>) that the second term is bounded by $\frac{L'\delta}{2}$ for some constant $L'$ not depending on $\delta, b$. Note that Theorem <ref> is applicable in this setting as seen in the proof of Theorem <ref>. In consequence,

\[ \lim_{\delta\searrow}\sup_{b\in\Sigma}\Pr_{b}\left( \sup_{\overline\rho_b((y,h),(y',h'))\leq \delta}\left| Z_b(y,h) - Z_b(y',h')\right|>\epsilon\right) =0.\]

To show condition (b) for the process $Z_T$, we first switch to the semi-metric $\overline\rho_T$ instead of $\overline\rho_b$ in the following way (where again the dependence of $\overline{\rho}_b, \overline\rho_T$ on its parameters is suppressed in the notation):

\begin{align*}
&\Pr_{b}\left( \sup_{\overline\rho_b\leq \delta}\left| Z_T(y,h) - Z_T(y',h')\right|>\epsilon\right)\\
& \hspace{0.3cm} \leq \Pr_{b}\left(\sup_{\overline\rho_b^2\leq \delta^2}\left| Z_T(y,h) - Z_T(y',h')\right|>\epsilon, \| \overline\rho_T^2 - \overline\rho_b^2\|_{\mathcal{T}_c\times \mathcal{T}_c} \leq \delta^2 \right)\\
&\hspace{1.3cm} + \Pr_{b}\left( \sup_{\overline\rho_b^2\leq \delta^2}\left| Z_T(y,h) - Z_T(y',h')\right|>\epsilon, \| \overline\rho_T^2 - \overline\rho_b^2\|_{\mathcal{T}_c\times \mathcal{T}_c} > \delta^2 \right)\\
&\hspace{0.3cm} \leq \Pr_{b}\left( \sup_{\overline\rho_T\leq \sqrt{2}\delta}\left| Z_T(y,h) - Z_T(y',h')\right|>\epsilon \right) + \Pr_b\left( \| \overline\rho_T^2 - \overline\rho_b^2\|_{\mathcal{T}_c\times \mathcal{T}_c} > \delta^2 \right).
\end{align*}

By Lemma <ref>,

\[ \lim_{T\to\infty}\sup_{b\in\Sigma} \Pr_{b}\left( \| \overline\rho_T^2 - \overline\rho_b^2\|_{\mathcal{T}_c\times \mathcal{T}_c} > \delta^2 \right) =0.\]

The first summand can be treated in the same way as above where we established the uniform asymptotic stochastic equicontinuity for $Z_b$ and is seen to be bounded from above by

\begin{align*}
&\Pr_b\left( Q\cdot \sup_{\overline\rho_T((y,h),(y',h'))\leq \sqrt{2}\delta}\overline\rho_T \log(e/\overline\rho_T)>\epsilon \right)\\
&\hspace{2.5cm} + \Pr_b\left( \sup_{\overline\rho_T((y,h),(y',h'))\leq \sqrt{2}\delta}\frac{\left| Z_T(y,h) - Z_T(y',h')\right|}{\overline\rho_T \log(e/\overline\rho_T)}>Q\right).
\end{align*}

The first term again vanishes uniformly in $b$ as $\delta\searrow 0$ because $x\log(e/x)\to 0$ for $x\searrow 0$ monotoneously on $(0,1)$. For a proper choice of the constant $Q$, we have seen in the proof of Theorem <ref> in equation (<ref>) that the second summand is bounded by $\frac{L'\delta}{2} + \Pr_b (\mathcal{C}_{T,b}^c)$ for some constant $L'$ not depending on $\delta,T, b$ and the set $\mathcal{C}_{T,b}$ is given by

\begin{align}\label{eqp: C2}
\mathcal{C}_{T,b} := \left\{ \left\| \frac{1}{\sigma^2 T}L_T^\cdot(X) - q_b\right\|_{[-A,A]} \leq \frac12 L_*\right\}.
\end{align}

Applicability of Theorem <ref> for $Z_T$ with this choice of $\mathcal{C}_T$ was checked in the proof of Theorem <ref>. Since

\[ \lim_{T\to\infty} \sup_{b\in\Sigma}\Pr_b\left( \mathcal{C}_{T,b}^c\right) =0 \]

by Corollary <ref>, uniform stochastic equicontinuity of $Z_T$ follows, i.e.

\[ \lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma}\Pr_{b}\left( \sup_{\overline\rho_b\leq \delta}\left| Z_T(y,h) - Z_T(y',h')\right|>\epsilon\right) =0 \]

and condition (b) of Proposition <ref> is fullfilled.

For condition (c) we first note that $N(u,\mathcal{T}_c,\overline\rho_b)\leq D(u,\mathcal{T}_c,\overline\rho_b)$. Then the uniform bound on the covering numbers follows from that for the capacity numbers in the proof of Theorem <ref> for $\mathcal{T}\supset\mathcal{T}_c$. The additional uniformity follows, since the bound on the capacity number derived within this proof did not depend on $b$ as the last bound in (<ref>) does not. Hence, condition (c) of Proposition <ref> is fullfilled.

Consequently, by Proposition <ref>,

\[ \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(l_\infty(\mathcal{T}_c))} \left| \E\left[ f((Z_T(y,h)_{(y,h)\in\mathcal{T}_c})\right] -  \E\left[ f((Z_b(y,h)_{(y,h)\in\mathcal{T}_c})\right]\right|\]

tends to zero for $T\to\infty$ and step (i) is complete.

Step (ii). We set $c = 2AL_*( L^*\|K\|_{L^2}^2)^{-1}\delta^2 $, which is smaller than $A$ for every $\delta<c':= \|K\|_{L^2}\sqrt{L^*/(2L_*)}$. Using the upper and lower bound of the invariant density from Lemma <ref> and <ref>,

\[ \overline\sigma_b(y,h)^2 := \frac{\int_\R K_{y,h}(z)^2 q_b(z) dz}{\int_\R \1_{[-A,A]}(z) q_b(z)dz}  \leq \frac{ L^* h\int_\R K(z)^2 dz}{2AL_*}.\]

Consequently, $\overline\sigma_b(y,h) \geq \delta$ implies $h\geq c$ and

\begin{align*}
&\sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\{(y,h)\in\mathcal{T}: \overline\sigma_b(y,h)>\delta\})} \left| \E\left[ f((Z_T(y,h))_{\{(y,h): \overline\sigma_b(y,h)> \delta\}}) \right.\right. \\
&\hspace{6.5cm} \left.\left.- f((Z_b(y,h))_{\{(y,h): \overline\sigma_b(y,h)> \delta\}})\right]\right|
\end{align*}

converges to zero for $T\to\infty$ by step (i). By Lemma <ref> and <ref> it follows that

\begin{align}\label{eqp: C15}
\sup_{\substack{(y,h)\in\mathcal{T}: \\ \delta<\overline\sigma_b(y,h)\leq 1}} \left(\frac{\left| Z_T(y,h)\right|}{\sigma_b(y,h)} - C\bigg( \frac{\sigma_b(y,h)^2}{\sigma_{b,\max}^2}\bigg)\right)
\end{align}

converges uniformly to $S_b(\delta,1)$. To establish the weak convergence of $T_T(\delta, 1)$ to $ S_b(\delta, 1)$ for any $0<\delta<c'$ from this finding, we apply Lemma <ref> to $Z_T$ and $Z_b$ with the specifications

\[ \alpha_T(y,h) = \hat\sigma_T(y,h) \quad \textrm{ and }\quad \alpha(y,h) =\sigma_b(y,h), \]

\[ \beta_T(y,h) = C\left(\hat\sigma_T(y,h)^2/\hat\sigma_{T,\max}^2\right) \ \ \textrm{ and }\ \ \beta(y,h) =C\left(\sigma_b(y,h)^2/\sigma_{b,\max}^2\right). \]

Its first condition (a) is true by the convergence of the term in (<ref>) to $S_b(\delta,1)$. Uniform tightness of $S_b(\delta,1)$ follows from Theorem <ref> by noting that the last bound of the capacity number in (<ref>) does not depend on $b$. Moreover, tightness of $\sup_{(y,h)\in\mathcal{T}: \overline\sigma_b(y,h)>\delta}C(\overline\sigma_b(y,h)^2)$ is clear, as $C(\overline\sigma_b(y,h)^2)$ is bounded by $C(\delta^2)$ for $\overline\sigma_b(y,h)>\delta$, because $C(\cdot)$ is decreasing on $[0,1]$. In particular, assumption (b) of Lemma <ref> holds and it remains to establish assumption (c) and (d) of it, i.e. to prove that for any $\epsilon>0$,

\begin{align}\label{eqp: C7}
\sup_{b\in\Sigma} \Pr_b\left( \sup_{\overline\sigma_b(y,h)>\delta} \left| \frac{\sigma_b(y,h)}{\hat\sigma_T(y,h)} - 1\right|>\epsilon\right) \stackrel{T\to\infty}{\longrightarrow}\ 0
\end{align}

\begin{align}\label{eqp: C8}
\sup_{b\in\Sigma}\Pr_b\left( \sup_{\overline\sigma_b(y,h)\geq\delta}\left| C\left( \frac{\hat\sigma_T(y,h)^2}{\hat\sigma_{T,\max}^2}\right) - C\left( \frac{\sigma_b(y,h)^2}{ \sigma_{b,\max}^2}\right) \right|>\epsilon\right) \stackrel{T\to\infty}{\longrightarrow}\ 0.
\end{align}

On the set $\mathcal{C}_{T,b}$ defined in (<ref>) we have $\frac{1}{\sigma^2 T}L_T^\cdot\geq \frac12 q_b$ and consequently

\begin{align*}
\sqrt{\frac1T \int_0^T K_{y,h}(X_s)^2 ds } = \sqrt{\int_\R K_{y,h}(z)^2 \frac{1}{\sigma^2 T}L_T^z(X) dz} \geq 2^{-\frac12}\|K_{y,h}\sqrt{q_b}\|_{L^2},
\end{align*}

or equivalently $\hat\sigma_T(y,h) \geq 2^{-\frac12} \sigma_b(y,h)$. From $\overline\sigma_b(y,h)>\delta$, it follows that $\sigma_b(y,h)^2>\sigma_{b,\max}^2\delta^2 \geq 2AL_*\delta^2$. Thus, we get on $\mathcal{C}_{T,b}$, applying a first order Taylor expansion of $\sqrt{x}$,

\begin{align*}
&\sup_{\overline\sigma_b(y,h)>\delta} \left| \frac{\|K_{y,h}\sqrt{\rho_{b_0}}\|_{L^2}}{\sqrt{\frac1T\int_0^T K_{y,h}(X_s)^2ds}} - 1\right|\\
&\hspace{3cm}= \sup_{\overline\sigma_b(y,h)>\delta}\frac{1}{\hat\sigma_T(y,h)}\left| \sigma_b(y,h)- \hat\sigma_T(y,h)\right| \\
&\hspace{3cm} \leq \frac{1}{2AL_*\delta^2}\cdot \sup_{\overline\sigma_b(y,h)>\delta}\left| \hat\sigma_T(y,h)^2 - \sigma_b(y,h)^2\right|,
\end{align*}

where we used $1\geq \int_{-A}^A q_b(z) dz \geq 2AL_*$. This term converges to zero uniformly in probability by Proposition <ref>. Together with the fact that $\sup_{b\in\Sigma}\Pr_b(\mathcal{C}_{T,b}^c)\rightarrow 0$ by Corollary <ref>, we have for any $\epsilon>0$ that

\begin{align*}
&\sup_{b\in\Sigma} \Pr_b\left( \sup_{\overline\sigma_b(y,h)>\delta} \left| \frac{\sigma_b(y,h)}{\hat\sigma_T(y,h)} - 1\right|>\epsilon\right)\\
&\hspace{0.3cm}\leq \sup_{b\in\Sigma} \Pr_b\left( \left\{\sup_{\overline\sigma_b(y,h)>\delta} \left| \frac{\sigma_b(y,h)}{\hat\sigma_T(y,h)} - 1\right|>\epsilon\right\} \cap\mathcal{C}_{T,b}\right) +  \sup_{b\in\Sigma}\Pr_b\left(\mathcal{C}_{T,b}^c\right)
\end{align*}

converges to zero for $T\to\infty$ and (<ref>) holds.

To verify (<ref>), we use that for $\epsilon>0$ and $\delta'>\delta>0$,

\begin{align}\label{eqp: C3}
\begin{split}
&\Pr_b\left( \sup_{\overline\sigma_b(y,h)\geq\delta}\left| C\left( \overline\sigma_T(y,h)^2\right)- C\left( \overline\sigma_b(y,h)^2\right) \right|>\epsilon\right)\\
&\hspace{1cm} \leq \Pr_b\left( \sup_{\delta<\overline\sigma_b(y,h)\leq\delta'}\left| C\left( \overline\sigma_T(y,h)^2\right) - C\left( \overline\sigma_b(y,h)^2\right)\right| >\epsilon\right) \\
&\hspace{2cm}+ \Pr_b\left( \sup_{\overline\sigma_b(y,h)\geq\delta'}\left| C\left( \overline\sigma_T(y,h)^2\right) - C\left( \overline\sigma_b(y,h)^2\right)\right| >\epsilon\right).
\end{split}
\end{align}

Next, we apply a Taylor expansion for $C(r)=\sqrt{2\log(1/r)}$ with derivative
\[ C'(r) = -\frac{1}{r \sqrt{2\log(1/r)}} =- \frac{1}{rC(r)}. \]

The function $rC(r)$ has its zeros in the endpoints of the interval $[0,1]$, but the absolute value of the derivative is bounded on the compact interval $[\delta, \delta']$ by some constant $c(\delta, \delta')$. Hence,

\begin{align*}
&\sup_{\delta<\overline\sigma_b(y,h)\leq\delta'}\left| C\left( \overline\sigma_T(y,h)^2\right)- C\left( \overline\sigma_b(y,h)^2\right) \right| \\
&\hspace{1.5cm}\leq c(\delta, \delta') \cdot \sup_{\delta<\sigma_b(y,h)\leq\delta'} \left| \overline\sigma_T(y,h)^2 - \overline\sigma_b(y,h)^2\right|,
\end{align*}

which converges to zero in probability, uniformly in $\Sigma$, by Lemma <ref>. Consequently, the first summand of the right-hand side of (<ref>) converges to zero, uniformly in $b$, for $T\to\infty$ and we now treat the second one. As $C(\cdot)$ is monotonously decreasing on $[0,1]$, for any $\epsilon'>0$,

\begin{align*}
&\Pr_b\left(\sup_{\overline\sigma_b(y,h)\geq \delta'}\left| C\left( \overline\sigma_T(y,h)^2\right)- C\left( \overline\sigma_b(y, h)^2\right) \right| >\epsilon\right) \\
&\hspace{0.3cm} \leq \Pr_b\left( \sup_{\overline\sigma_b(y,h)\geq \delta'} C\left( \overline\sigma_T(y,h)^2\right)+ C\left( (\delta')^2 \right) >\epsilon\right)\\
&\hspace{0.3cm} \leq \Pr_b\left( \sup_{\overline\sigma_b(y,h)\geq \delta'} C\left( \overline\sigma_b(y,h)^2 + (\overline\sigma_T(y,h)^2 - \overline\sigma_b(y,h)^2)\right)\right. \\
&\hspace{2cm} \left. \textcolor{white}{\frac12} + C\left( (\delta')^2 \right) >\epsilon, \|\overline\sigma_T^2 -\overline\sigma_b^2\|_\mathcal{T}<\epsilon'\right)  +\Pr_b\left( \|\overline\sigma_T^2 -\overline\sigma_b^2\|_\mathcal{T}\geq \epsilon'\right)\\
&\hspace{0.3cm} \leq \Pr_b\left( C\left( (\delta')^2 - \epsilon'\right) + C\left( (\delta')^2 \right) >\epsilon\right) + \Pr_b\left( \|\overline\sigma_T^2 -\overline\sigma_b^2\|_\mathcal{T}\geq \epsilon'\right).
\end{align*}

The latter summand vanishes  uniformly over $\Sigma$ for $T\to\infty$ by Lemma <ref>, the first one is neither depending on $T$ nor probabilistic. Letting $\epsilon'\searrow 0$ and $\delta'\nearrow 1$, it converges to zero because $C(r)\stackrel{r\to 1}{\longrightarrow} 0$, and we have shown (<ref>). In summary,  by Lemma <ref> we have proven so far that

\[ \sup_{b\in\Sigma(C,A,\gamma,\sigma)} d_{BL}^b\left( T_T(\delta,1), S_b(\delta,1)\right) \stackrel{T\to\infty}{\longrightarrow} 0.\]

Step (iii). To improve the result from (ii) to uniform weak convergence of $T_T(0,1)$ to $S_b(0,1)$, we split

\begin{align}\label{eqp: C12}
\begin{split}
&\sup_{b\in\Sigma(C,A,\gamma,\sigma)} d_{BL}^b\left( T_T(0,1), S_b(0,1)\right)\\
&\hspace{1.5cm} = \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R)} \left|\E_\theta\left[ f(T_T(0,1)) - f(S_b(0,1))\right]\right|\\
&\hspace{1.5cm} \leq \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R)} \E_\theta\left[\left| f(T_T(0,1)) - f(T_T(\delta,1))\right|\right]\\
&\hspace{2.5cm} + \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R)} \left|\E_\theta\left[ f(T_T(\delta,1)) - f(S_b(\delta,1))\right]\right| \\
&\hspace{2.5cm} + \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R)} \E_\theta\left[\left| f(S_b(\delta,1)) - f(S_b(0,1))\right|\right].
\end{split}
\end{align}

The second summand equals $ \sup_{b\in\Sigma(C,A,\gamma,\sigma)} d_{BL}^b\left( T_T(\delta,1), S_b(\delta,1)\right)$ and we have seen in step (ii) that it vanishes for $T\to\infty$. The first summand is bounded from above by

\begin{align*}
&\sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R)} \E_\theta\left[\left| f(T_T(0,1)) - f(T_T(\delta,1))\right| \1_{\{ T_T(0,1) - T_T(\delta,1) >\epsilon\}}\right] \\
&\hspace{1cm} + \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}(\R)} \E_\theta\left[\left| f(T_T(0,1)) - f(T_T(\delta,1))\right| \1_{\{ T_T(0,1) - T_T(\delta,1) \leq \epsilon\}}\right] \\
&\hspace{0.5cm}\leq \epsilon + 2\sup_{b\in\Sigma} \Pr_b\left( T_T(0,1) - T_T(\delta,1) >\epsilon\right).
\end{align*}

Writing $T_T(\delta,\delta') = \sup_{\delta<\overline\sigma(y,h)\leq \delta'} A_T(y,h)$, we can proceed with

\begin{align*}
&\sup_{b\in\Sigma} \Pr_b\left( T_T(0,1) - T_T(\delta,1) >\epsilon\right)\\
&\hspace{0.2cm} = \sup_{b\in\Sigma} \Pr_b\left( \max\left\{ \sup_{0< \overline\sigma(y,h)\leq \delta} A_T(y,h), \sup_{\delta<\overline\sigma(y,h)\leq 1} A_T(y,h)\right\} \right. \\
&\hspace{6.5cm} \left. - \sup_{\delta<\overline\sigma(y,h)\leq 1} A_T(y,h)>\epsilon\right)\\
&\hspace{0.2cm} = \sup_{b\in\Sigma} \Pr_b\left( \max\left\{ \sup_{0< \overline\sigma(y,h)\leq \delta} A_T(y,h)- \sup_{\delta<\overline\sigma(y,h)\leq 1} A_T(y,h), 0\right\} >\epsilon\right)\\
&\hspace{0.2cm} \leq \sup_{b\in\Sigma} \Pr_b\left( \max\{\dots\} >\epsilon, \sup_{\delta<\overline\sigma(y,h)\leq 1} A_T(y,h) \geq -\frac{\epsilon}{2}\right) \\
&\hspace{2cm} + \sup_{b\in\Sigma} \Pr_b\left( \max\{\dots\} >\epsilon, \sup_{\delta<\overline\sigma(y,h)\leq 1} A_T(y,h) < -\frac{\epsilon}{2}\right)\\
&\hspace{0.2cm} \leq  \sup_{b\in\Sigma} \Pr_b\left( \sup_{0<\overline\sigma(y,h)\leq \delta} A_T(y,h) >\frac{\epsilon}{2}\right) + \sup_{b\in\Sigma} \Pr_b\left( \sup_{\delta<\overline\sigma(y,h)\leq 1} A_T(y,h) <-\frac{\epsilon}{2}\right).
\end{align*}

We start by treating the first summand and show that for all $\epsilon>0$,

\begin{align}\label{eqp: C4}
\lim_{\delta\searrow 0} \limsup_{T\to\infty}\sup_{b\in\Sigma} \Pr_b\left( T_T(0,\delta) \geq \epsilon\right) =0.
\end{align}

This is based on Theorem <ref>. First of all,

\begin{align*}
\Pr_{b}\left( T_T(0,\delta) \geq \epsilon\right) &= \Pr_{b}\left( T_T(0,\delta) \geq \epsilon, \|\overline\sigma_T^2 - \overline\sigma_b^2 \|_{\mathcal{T}}\leq\delta \right) \\
&\hspace{1.5cm} +\Pr_{b}\left( T_T(0,\delta) \geq \epsilon , \|\overline\sigma_T^2 - \overline\sigma_b^2 \|_{\mathcal{T}}>\delta \right),
\end{align*}

where the second summand is bounded by $\Pr_{b}(\|\overline\sigma_T^2 - \overline\sigma_b^2 \|_{\mathcal{T}}>\delta)$ and vanishes uniformly in $b$ for $T\to\infty$ as shown in Lemma <ref>. For the other one, we have

\begin{align*}
&\Pr_{b}\left( T_T(0,\delta) \geq \epsilon, \|\overline\sigma_T^2 - \overline\sigma_b^2 \|_{\mathcal{T}}\leq\delta^2 \right) \\
&\hspace{1cm}= \Pr_{b}\left( \sup_{\overline\sigma_b(y,h)^2 \leq \delta^2} A_T(y,h) >\epsilon, \|\overline\sigma_T^2 - \overline\sigma_b^2 \|_{\mathcal{T}}\leq\delta^2 \right)\\
&\hspace{1cm}\leq  \Pr_{b}\left( \sup_{\overline\sigma_T(y,h)^2 \leq 2\delta^2} \frac{A_T(y,h)}{D(\overline\sigma_T(y,h))}\cdot D(\overline\sigma_T(y,h)) >\epsilon\right)\\
&\hspace{1cm} \leq \Pr_{b}\left( \sup_{\overline\sigma_T(y,h) \leq \sqrt{2}\delta} \frac{A_T(y,h)}{D(\overline\sigma_T(y,h))} \cdot \sup_{\overline\sigma_T(y,h) \leq \sqrt{2}\delta} D(\overline\sigma_T(y,h)) >\epsilon\right)\\
&\hspace{1cm} \leq \Pr_{b}\left( L \cdot \sup_{\overline\sigma_T(y,h) \leq \sqrt{2}\delta} D(\overline\sigma_T(y,h)) >\epsilon\right) \\
&\hspace{3cm}+ \Pr_{b}\left(\sup_{\overline\sigma_T(y,h) \leq \sqrt{2}\delta} \frac{A_T(y,h)}{D(\overline\sigma_T(y,h))} >L\right)
\end{align*}

for any constant $L>0$. Now, the first summand here tends to zero for $\delta\searrow 0$ as $D(r)\searrow 0$ monotoneously for $r\searrow 0$. The random variable in the second probability is tight by Theorem <ref> and thus tends to zero for $L\to\infty$. Note that the applicability of Theorem <ref> was checked in the proof of Theorem <ref>. For the uniformity mentioned before, note that all bounds used in the proof of Theorem <ref>, in particular the last one of (<ref>), are independent of $b$.

It remains to deal for all $\epsilon>0$ with the term

\[ \lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma} \Pr_b\left( T_T(\delta,1) <-\frac{\epsilon}{2}\right). \]

Define the Lipschitz function

\[ g_{\epsilon}(y) := \begin{cases}
1 &\textrm{ if } x\leq -\frac{\epsilon}{2},\\
1-4\frac{y+\frac{\epsilon}{2}}{\epsilon} &\textrm{ if } -\frac{\epsilon}{2}<y<-\frac{\epsilon}{4},\\
0 &\textrm{ if } x\geq -\frac{\epsilon}{4}.
\end{cases}\]

Then we have

\[  \1_{\left\{x < -\frac{\epsilon}{2}\right\}} \leq g_\epsilon(x) \leq \1_{\left\{x < -\frac{\epsilon}{4}\right\}} \]

and consequently,

\begin{align*}
&\lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma} \Pr_b\left( T_T(\delta,1) <-\frac{\epsilon}{2}\right) \\
&\hspace{0.3cm} \leq \lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma} \E_b\left[ g_\epsilon(T_T(\delta, 1))\right] \\
&\hspace{0.3cm} = \lim_{\delta\searrow 0}\sup_{b\in\Sigma} \E_b\left[ g_\epsilon(S_b(\delta, 1))\right] \\
&\hspace{1cm} + \lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma} \E_b\left[ g_\epsilon(T_T(\delta, 1))\right] -\lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma} \E_b\left[ g_\epsilon(S_b(\delta, 1))\right].
\end{align*}

By the already established uniform weak convergence of $T_T(\delta, 1)$ to $S_b(\delta, 1)$ and Lemma <ref> we get for the difference

\begin{align*}
&\lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma} \E_b\left[ g_\epsilon(T_T(\delta, 1))\right] -\lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma} \E_b\left[ g_\epsilon(S_b(\delta, 1))\right] \\
& \hspace{1cm}\leq \lim_{\delta\searrow 0} \limsup_{T\to\infty} \sup_{b\in\Sigma} \left| \E_b\left[ g_\epsilon(T_T(\delta, 1))\right]  - \E_b\left[ g_\epsilon(S_b(\delta, 1))\right] \right| = 0
\end{align*}

and we have

\begin{align*}
\lim_{\delta\searrow 0}\limsup_{T\to\infty}\sup_{b\in\Sigma} \Pr_b\left( T_T(\delta,1) <-\frac{\epsilon}{2}\right) &\leq \lim_{\delta\searrow 0}\sup_{b\in\Sigma} \E_b\left[ g_\epsilon(S_b(\delta, 1))\right]  \\
&\leq \lim_{\delta\searrow 0}\sup_{b\in\Sigma} \Pr_b\left( S_b(\delta,1) <-\frac{\epsilon}{4}\right).
\end{align*}

Set $\tilde{\delta} := 4\delta^2 (2AL^*) (L_*\|K\|_{L^2}^2)^{-1}$
and define for $-\frac{A}{\tilde\delta}\leq k\leq \frac{A}{\tilde\delta}$,

\[Z_k(b) := \|K_{k\tilde\delta, \frac{\tilde\delta}{2}}\sqrt{\rho_b}\|_{L^2}^{-1} \int_\R K_{k\tilde\delta, \frac{\tilde\delta}{2}} (z)\sqrt{\rho_b(z)} dW_z.\]

Those random variables are independent, as the support of the functions $K_{k\tilde\delta, \tilde\delta/2}$ and $K_{k'\tilde\delta, \tilde\delta/2}$ is disjoint for $k\neq k'$. Moreover, they are standard Gaussians, in particular their distribution does not depend on $b$, and hence their maximum is $(2\log (2A\tilde{\delta}^{-1}))^\frac12 + o_{\Pr}(1)$ for $\delta\searrow 0$ (see [41], Theorem $1.5.3$). Using the uniform bound on the invariant density from Lemma <ref> and <ref>,

\[ \frac{\|K_{k\tilde\delta, \frac{\tilde\delta}{2}}\sqrt{\rho_b}\|_{L^2}^2}{\| \1_{[-A,A]}\sqrt{q_b}\|_{L^2}^2} \geq \frac{L_*\|K\|_{L^2}^2}{2AL^*}\frac{\tilde\delta}{2} =2\delta^2. \]

As $C(\cdot)$ is monotoneously decreasing on $[0,1]$,

\begin{align*}
S_b(\delta,1)&\geq \max_{-A\tilde\delta^{-1}\leq k\leq A\tilde\delta^{-1}}\left( Z_k(b) - C\left( \frac{\|K_{k\tilde\delta, \frac{\tilde\delta}{2}}\sqrt{\rho_b}\|_{L^2}^2}{\| \1_{[-A,A]}\sqrt{q_b}\|_{L^2}^2}\right)\right)\\
&\geq \max_{-A\tilde\delta^{-1}\leq k\leq A\tilde\delta^{-1}} Z_k(b) - C(2\delta^2)\\
& = \sqrt{2\log( 2A\tilde\delta^{-1})} -\sqrt{2\log ((2\delta^2)^{-1})} + o_\Pr(1)\\
&= \frac{2\log(2A\tilde\delta^{-1}) - 2\log((2\delta^2)^{-1})}{\sqrt{2\log( 2A\tilde\delta^{-1})} +\sqrt{2\log (2\delta^{2})^{-1})}} +o_\Pr(1)\\
& = \frac{2\log(\tilde{c} \delta^{-2}) - 2\log(2^{-1}\delta^{-2})}{\sqrt{2\log(\tilde{c} \delta^{-2})} +\sqrt{2\log(2^{-1}\delta^{-2})}} + o_\Pr(1) \\
% & = \frac{2\log(\tilde{c}) -4\log(\delta) - 2\log(2^{-1}) + 4\log(\delta)}{\sqrt{2\log(\tilde{c} \delta^{-2})} +\sqrt{2\log(2^{-1}\delta^{-2})}} + o_\Pr(1) \\
& = \frac{2\log(\tilde{c}) - 2\log(2^{-1}) }{\sqrt{2\log(\tilde{c} \delta^{-2})} +\sqrt{2\log(2^{-1}\delta^{-2})}} + o_\Pr(1) \\
&= o_\delta(1) + o_\Pr(1)\quad \textrm{ for } \delta\searrow 0,
\end{align*}

where we used in the fifth step the constant

\[ \tilde{c}:= 2A\frac14 (2AL^*)^{-1}L_*\|K\|_{L^2}^2 = L_*\|K\|_{L^2}^2 (4L^*)^{-1}\]

that does not depend on $\delta$. Thus, we finally get

\begin{align}\label{eqp: C14}
\begin{split}
&\lim_{\delta\searrow 0}\sup_{b\in\Sigma} \Pr_b\left( S_b(\delta,1) <-\frac{\epsilon}{4}\right)\\
& \hspace{0.3cm} \leq \lim_{\delta\searrow 0}\sup_{b\in\Sigma} \Pr_b\left( \max_{-A\tilde\delta^{-1}\leq k\leq A\tilde\delta^{-1}}\left( Z_k(b) - C\left( \frac{\|K_{k\tilde\delta, \frac{\tilde\delta}{2}}\sqrt{\rho_b}\|_{L^2}^2}{\| \1_{[-A,A]}\sqrt{q_b}\|_{L^2}^2}\right)\right) <-\frac{\epsilon}{4}\right)\\
&\hspace{0.3cm}\leq \lim_{\delta\searrow 0} \sup_{b\in\Sigma}\Pr\left( \max_{-A\tilde\delta^{-1}\leq k\leq A\tilde\delta^{-1}} Z_k(b) - C(2\delta^2) <-\frac{\epsilon}{4}\right) =0
\end{split}
\end{align}

and the convergence of the first summand on the right-hand side in (<ref>) to zero for $\lim_{\delta\searrow 0}\limsup_{T\to\infty}$ is shown. It remains to study the third summand in (<ref>), i.e.

\[ \sup_{b\in\Sigma} \sup_{f\in\mathcal{F}_{BL}} \E_\theta\left[\left| f(S_b(\delta,1)) - f(S_b(0,1))\right|\right].\]

This is done in the same way as for the first summand and is only easier, as the $\limsup_{T\to\infty}$ is omitted and the argument reduces to additionally showing that for any $\epsilon>0$,

\[ \lim_{\delta\searrow 0} \sup_{b\in\Sigma} \Pr_b\left( S_b(0,\delta) \geq \epsilon\right) =0.\]

This follows along the lines of the previous part as Theorem <ref> is also applicable to $S_b$, as seen in Theorem <ref>.

Let $c>0$ and $\mathcal{T}_c$ be the set given in (<ref>). For any subset $\mathcal{T}'\subset \mathcal{T}_c$ define

\begin{align*}
& T_T(\mathcal{T}') := \sup_{(y,h)\in\mathcal{T}'} \left(\frac{\left|Z_T(y,h)\right|}{\hat\sigma_T(y,h)} - C\bigg( \frac{\hat\sigma_T(y,h)^2}{\hat\sigma_{T,\max}^2}\bigg) \right)
\end{align*}

as well as

\begin{align*}
&S_b(\mathcal{T}') := \sup_{(y,h)\in\mathcal{T}'} \left(\frac{\left|Z_b(y,h)\right|}{\sigma_b(y,h) } - C\bigg( \frac{\sigma_b(y,h)^2}{\sigma_{b,\max}^2}\bigg) \right).
\end{align*}

Then our proof of Theorem <ref> even shows that

\[ \sup_{b\in\Sigma(C,A,\gamma,\sigma)} d_{BL}^b \left( T_T(\mathcal{T}'), S_b(\mathcal{T}')\right)\stackrel{T\to\infty}{\longrightarrow} 0.\]

This can be seen as follows: Step (i) directly transfers and in step (ii) we have to show (<ref>) and (<ref>) for the supremum over $\mathcal{T}'$, which is straightforward by the same lines. Step (iii) can be omitted since $\mathcal{T}'\subset\mathcal{T}_c$.

§ PROOFS OF SECTION <REF>

In Subsection <ref> the proof of Theorem <ref> is given. This proof is sketched in Section <ref> and strongly relies on the uniform weak convergence result for the test statistic $T_T^{b}$ given in  Theorem <ref> that was proven in Subsection <ref>. In the subsequent Subsection <ref> we present a proof of Remark <ref>, i.e. we prove that the distribution of the dominating random variable $U_1\vee U_2 + 4\sqrt{A\eta/\sigma^2}$ from Theorem <ref> has no point mass.

§.§ Proof of Theorem <ref>

Let $b\in H_0(b_0,\eta)$. For ease of notation we abbreviate $H_0(b_0,\eta)$ by $H_0$ in what follows. As already indicated in Section <ref>, the key idea of the proof is to split the set $\mathcal{T}$ in three parts (two of them depending on $b$). For a rigorous description of those, we choose $\epsilon,\epsilon'>0$ and set

\begin{align*}
\mathcal{T}_1 &:= \left\{ (y,h)\in\mathcal{T} \mid h <\epsilon'\right\}, \\
\mathcal{T}_2 &:= \left\{ (y,h)\in\mathcal{T} \mid h\geq\epsilon'\textrm{ and } \min_{x\in [y-h,y+h]} q_b(x)(|b(x)-b_0(x)|-\eta) < - \epsilon\right\}, \\
\mathcal{T}_3 &:= \left\{ (y,h)\in\mathcal{T} \mid h\geq\epsilon', q_b(x)(|b(x)-b_0(x)|-\eta)\geq -\epsilon\textrm{ on } [y-h, y+h]\right\}.
\end{align*}

Note the dependences $\mathcal{T}_1=\mathcal{T}_1(\epsilon')$, $\mathcal{T}_2=\mathcal{T}_2(\epsilon,\epsilon',b)$ and $\mathcal{T}_3=\mathcal{T}_2(\epsilon,\epsilon',b)$, which most times will be suppressed in the notation to allow for shorter displays. With these definitions and the union bound,

\begin{align}\label{eqp: D2}
\begin{split}
&\Pr_b(T_T^\eta(X) \geq r)\\
&\hspace{0.2cm}= \Pr_b\left( \sup_{(y,h)\in\mathcal{T}}  \left(  |\Psi_{T,y,h}^{b_0}(X)| -\Lambda_{T,y,h}^\eta(X) - C\bigg(\frac{\hat\sigma_T(y,h)^2}{\hat\sigma_{T,\max}^2}\bigg)  \right) \geq r \right) \\
&\hspace{0.2cm}\leq \sum_{i=1}^3\Pr_b\left( \sup_{(y,h)\in\mathcal{T}_i}  \left(   |\Psi_{T,y,h}^{b_0}(X)|-\Lambda_{T,y,h}^\eta(X)  - C\bigg(\frac{\hat\sigma_T(y,h)^2}{\hat\sigma_{T,\max}^2}\bigg)  \right) \geq r \right).
\end{split}
\end{align}

Of course, the inequality remains valid if we put

\[ \limsup_{\epsilon'\searrow 0}\limsup_{\epsilon\searrow 0}\limsup_{T\to\infty}\sup_{b\in H_0(b_0,\eta)}\]

before both sides. The left-hand side of (<ref>) does not depend on $\epsilon,\epsilon'$ and $\limsup_T \sup_b \Pr_b(T_T^\eta(X)\geq r)$ equals the left-hand side of the statement of the theorem. Note that $|\Psi_{T,y,h}^{b_0}(X)| - \Lambda_{T,y,h}^\eta(X) $ is bounded from above by

\begin{align}\label{eqp: D1}
\begin{split}
&\left|\frac{\frac{1}{\sqrt{T}} \int_0^T K_{y,h}(X_s) dW_s}{\sqrt{\frac1T \int_0^T K_{y,h}(X_s)^2 ds}} \right| \\
&\hspace{1cm} + \frac{\left|\frac{1}{\sqrt{T}}\int_0^T K_{y,h}(X_s) ( b(X_s) - b_0(X_s)) ds\right|-\frac{\eta}{\sqrt{T}} \int_0^T K_{y,h}(X_s) ds}{\sigma \sqrt{\frac1T \int_0^T K_{y,h}(X_s)^2 ds}}.
\end{split}
\end{align}

Using the triangle inequality, the numerator of the second summand is smaller than

\[ \frac{1}{\sqrt{T}} \int_0^T K_{y,h}(X_s) \left( |b(X_s) -b_0(X_s)| -\eta\right) ds\]

and this term is smaller than or equal to zero for each $b\in H_0(b_0,\eta)$. Consequently, the whole second summand in (<ref>) is non-positive. Here, we used the assumption $K\geq 0$. With this preliminary observation, we give the heuristic about the decomposition of $\mathcal{T}$ into $\mathcal{T}_i$, $i=1,2,3$, and the route of the rest of the proof:

* The probability in (<ref>) with $\mathcal{T}_1$ is neglectable in the limit $\lim_{\epsilon'\searrow 0}$ which can be seen using the same techniques from empirical process theory as for the proof of Theorem <ref>.

* For $(y,h)\in\mathcal{T}_2$, $|b-b_0|-\eta$ is sufficiently bounded away from zero on a subinterval of $[y-h,y+h]$ such that the second summand in (<ref>) tends to $-\infty$, whereas the first one remains stochastically bounded, uniformly in $b$. Hence, the corresponding probability in (<ref>) tends to zero.

* The probability with $\mathcal{T}_3$ in (<ref>) does not vanish in the limit and is estimated against $\Pr( U_1\vee U_2+ 4\sqrt{A\eta/\sigma^2}\geq r)$. In particular, it is the origin of the correction term $4\sqrt{A\eta/\sigma^2}$.

From (<ref>) in step (iii) of the proof of Theorem <ref> it can be seen that both random variables $U_1$ and $U_2$ are $\geq 0$ almost surely and consequently $U_1\vee U_2 + 4\sqrt{A\eta/\sigma^2}> 0$ for $\eta>0$. Therefore, it suffices to consider $r >0$ in the following, where we will estimate each summand on the right-hand side of (<ref>) separately.

Step (i). By neglecting the non-positive second summand in (<ref>), for $b\in H_0$ we may estimate

\begin{align*}
&\Pr_b\left( \sup_{(y,h)\in\mathcal{T}_1} \left( |\Psi_{T,y,h}^{b_0}(X)| -\Lambda_{T,y,h}^\eta(X) - C\bigg(\frac{\hat\sigma_T(y,h)^2}{\hat\sigma_{T,\max}^2}\right)  \bigg) \geq r\right)\\
&\hspace{1cm}\leq \Pr_b\left( \sup_{(y,h)\in\mathcal{T}_1} \left( \left|\frac{\frac{1}{\sqrt{T}} \int_0^T K_{y,h}(X_s) dW_s}{\sqrt{\frac1T \int_0^T K_{y,h}(X_s)^2 ds}} \right| - C\bigg(\frac{\hat\sigma_T(y,h)^2}{\hat\sigma_{T,\max}^2}\right)\bigg) \geq r\right).
\end{align*}

Note that $h<\epsilon'$ implies that

\[ \frac{\|K_{y,h}\sqrt{q_b}\|_{L^2}^2}{\|\1_{[-A,A]}\sqrt{q_b}\|_{L^2}^2} = \frac{\int_\R K_{y,h}(z)^2 q_b(z)dz}{\int_{-A}^A q_b(z) dz} \leq \frac{L^* h\|K\|_{L^2}^2}{2AL_*} < \frac{L^* \|K\|_{L^2}^2 \epsilon'}{2AL_*} =:c\epsilon' \]

by using the uniform upper and lower bound of the invariant density $q_b$ given in Lemma <ref> and <ref>.
Then, with $T_T(\delta,\delta')$ given in (<ref>) within the proof of Theorem <ref>,

\begin{align*}
&\limsup_{\epsilon'\searrow 0}\limsup_{\epsilon\searrow 0}\limsup_{T\to\infty}\sup_{b\in H_0} \Pr_b\left( \sup_{(y,h)\in\mathcal{T}_1} \left( \Psi_{T,y,h}^{b_0}(X) - \Lambda_{T,y,h}^\eta(X)\right.\right.\\
&\hspace{7cm}\left.\left. - C\bigg(\frac{\hat\sigma_T(y,h)^2}{\hat\sigma_{T,\max}^2}\bigg)  \right) \geq r\right) \\
&\hspace{1cm}\leq \limsup_{\epsilon'\searrow 0} \limsup_{T\to\infty}\sup_{b\in H_0}\Pr\left( T_T\left(0,\sqrt{c\epsilon'}\right) \geq r\right) =0,
\end{align*}