Let $\tau = \inf\{ t>0 : W_t = 1 \}$. The conjecture is true and the essence of the proof outlined below appears to be the following peculiar property of the hitting time $\tau$: $$
\lim_{\epsilon \searrow 0} P_{\epsilon}[\tau > \epsilon-\epsilon^{3/2}] = 1 \;. $$
(This can be computed directly using the fact that distribution of $\tau$ is inverse gamma with parameters $1/2$ and $1/2$.) In order to leverage this property, one must carefully split up $e_t:= X_t - Y_{t/\epsilon}$ as outlined below.
Theorem. It holds that: $$\lim_{\epsilon \searrow 0} E[ \sup_{0 \le t \le \epsilon} |e_t|^2 ] = 0 \;.$$
Proof.
The proof shows that for any $T \in [0, \epsilon]$ $$
E_{\epsilon} [ \sup_{0 \le t \le T} |e_t|^2 ] \le C_1(\epsilon) + C_2 (\epsilon) \int_0^T E_{\epsilon} [ \sup_{0 \le r \le s} |e_r|^2 ] ds
$$ where $C_1(\epsilon)$ and $C_2 (\epsilon) $ are non-negative and $\lim_{\epsilon \searrow 0} C_1(\epsilon) =0$ and $\lim_{\epsilon \searrow 0} C_2(\epsilon) \epsilon = O(1)$. By Grönwall's inequality, $$
E_{\epsilon} [ \sup_{0 \le t \le \epsilon} |e_t|^2 ] \le C_1(\epsilon) \exp(C_2 (\epsilon) \epsilon) \;,
$$ and then passing to the limit gives the required result. The remaining details follow.
By Itô's formula, \begin{align*}
& |e_t|^2 = \mathrm{I} + \mathrm{II} + \mathrm{III} \quad \text{where} \\
& \mathrm{I}:= \frac{2}{\epsilon} \int_0^t e_s (\sigma(X_s) - \sigma(Y_{s/\epsilon})) ds \;, \\
& \mathrm{II}:= \frac{2}{\epsilon} \int_0^t e_s \sigma(X_s) (\epsilon dW_s - ds) \;, \\
& \mathrm{III}:= \int_0^t \sigma(X_s)^2 ds \;.
\end{align*}
Estimate for $\mathrm{I}$.
This term exclusively contributes to $C_2(\epsilon)$.
Since $\sigma$ is $L$-Lipschitz for some $L>0$ $$
I \le \frac{2 L}{\epsilon} \int_0^t |e_s|^2 ds
$$ and thus $$
\sup_{0 \le t \le \epsilon} I \le \frac{2 L}{\epsilon} \int_0^{\epsilon} |e_s|^2 ds \le \frac{2 L}{\epsilon} \int_0^{\epsilon} \sup_{0 \le r \le s} |e_r|^2 ds \;.
$$
Thus, $C_2(\epsilon) = 2 L / \epsilon$.
Estimate for $\mathrm{II}$.
This term contributes to $C_1(\epsilon)$, and here is where we leverage the aforementioned peculiar property of $\tau$.
\begin{align*}
& \lim_{\epsilon \searrow 0} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \mathrm{II} \right| ] =
\lim_{\epsilon \searrow 0} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \mathrm{II} \right| \mathbf{1}_{ \{ \tau > \epsilon - \epsilon^{3/2} \} } ] \\
& \quad =
\lim_{\epsilon \searrow 0} E [ \sup_{0 \le t \le \epsilon} \left| 2 \int_0^t e_s \sigma(X_s) dW_s \right| ] = 0
\end{align*}
Here we took 3 steps that are explained in detail below.
In the first step, we used Cauchy-Schwarz to show that $$
\left( E_{\epsilon} \sup_{0 \le t \le \epsilon} |\mathrm{II}| \mathbf{1}_{\{\tau < \epsilon - \epsilon^{3/2} \} } \right)^2 \le \underbrace{E[\sup_{0 \le t \le \epsilon} |\mathrm{II}|^2 ]}_{\to O(1)} \, \underbrace{P_{\epsilon}[ \tau < \epsilon - \epsilon^{3/2}]}_{\to 0}
$$
In the second step, we used a natural splitting and the triangle inequality to write, \begin{align*}
& E_{\epsilon} \sup_{0 \le t \le \epsilon} |\mathrm{II}| \mathbf{1}_{\{\tau > \epsilon - \epsilon^{3/2} \} } \le \mathrm{II}_a + \mathrm{II}_b \quad \text{where} \\
& \mathrm{II}_a := E_{\epsilon} \sup_{0 \le t \le \epsilon} \frac{2}{\epsilon} \left| \int_0^{t \wedge \tau} e_s \sigma(X_s) (\epsilon dW_s - ds) \right| \mathbf{1}_{\{\tau > \epsilon - \epsilon^{3/2} \} }\\
& \mathrm{II}_b := E_{\epsilon} \sup_{0 \le t \le \epsilon} \frac{2}{\epsilon} \left| \int_{t \wedge \tau}^t e_s \sigma(X_s) (\epsilon dW_s - ds) \right| \mathbf{1}_{\{\tau > \epsilon - \epsilon^{3/2} \} }
\;.
\end{align*}
To estimate these terms, there are two cases to consider:
- Case 1: $t \le \tau$. Then $\mathrm{II}_b=0$ and $\mathrm{II}_a$ can be written in terms of the piece of the Brownian bridge up to time $t$; and,
- Case 2: $t > \tau$. Then $t>\epsilon - \epsilon^{3/2}$ and hence $\mathrm{II}_b=O(\epsilon^{1/2})$ and again $\mathrm{II}_a$ can be written in terms of the piece of the Brownian bridge up to time $\tau$.
In other words, conditioned on the event $(\tau < \epsilon)$, the law of $\epsilon W_s - s$ is equal to the law of a standard Brownian bridge.
In the third and last step, we used Doob's martingale inequality, Itô isometry, and (standard) a priori bounds on $X_t$ and $Y_{t/\epsilon}$ over $(0,\epsilon)$. Since the estimate of this term is almost identical to the estimate of $\mathrm{III}$ given below, the details are suppressed.
Estimate for $\mathrm{III}$.
This term also contributes to $C_1(\epsilon)$. Noting that $\sigma$ is $L$-Lipschitz, \begin{align*}
E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \mathrm{III} ] &= E_{\epsilon}[ \int_0^{\epsilon} [ \sigma(X_s)^2 ds ] \\
&\le 2 \epsilon \sigma(0)^2 + 2 L^2 E_{\epsilon}[ \int_0^{\epsilon} |X_s|^2 ds ] \\
&\le 2 \epsilon ( \sigma(0)^2 + L^2 |x_0|^2 ) + 2 L^2 \epsilon \int_0^{\epsilon} \sigma(X_s)^2 ds \\
& \quad + 2 L^2 \epsilon \int_0^{\epsilon} X_s \sigma(X_s) dW_s
\end{align*}
and as long as $2 L^2 \epsilon \le 1/2$, it follows that
$$
E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \mathrm{III} ] \le 4 \epsilon ( \sigma(0)^2 + L^2 |x_0|^2 ) + 4 L^2 \epsilon \int_0^{\epsilon} X_s \sigma(X_s) dW_s
$$ The last term in this expression can be treated in a similar way as the last step in the estimate for $\mathrm{II}$, namely Doob's martingale inequality, Cauchy-Schwarz, Itô isometry, and (standard) a priori bounds on $X_t$ over $(0,\epsilon)$. In particular, \begin{align*}
\left( E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \int_0^t X_s \sigma(X_s) d W_s \right| ] \right)^2 &\le
E \sup_{0 \le t \le \epsilon} \left| \int_0^t X_s \sigma(X_s) dW_s \right|^2\\
&\le 4 E \left| \int_0^{\epsilon} X_s \sigma(X_s) dW_s \right|^2 \\
&\le 4 E \int_0^{\epsilon} X_s^2 \sigma(X_s)^2 ds \\
&\le 4 \tilde{C}_2 (1+ x_0^4) e^{\tilde{C}_1 \epsilon} \epsilon
\end{align*}
where in turn we used Cauchy-Schwarz, Doob's martingale inequality with $p=2$, Itô's isometry, and then an a priori bound on the second/fourth moment of $X_t$ over $(0, \epsilon)$.
$\Box$
Best Answer
I think so. The martingale $X_t$ is a time-changed Brownian motion: $X_t = B_{A_t}$, where $$A_T = \int_0^T \sigma^2(t, X_t) dt$$ and $B_t$ is some Brownian motion. Now write $$ \int_0^T \mathbb 1_{[-\delta,\delta]}(X_t) dt = \int_0^{A_T} \mathbb 1_{[-\delta,\delta]}(B_s) dA^{-1}_s = \int_0^{A_T} \frac{\mathbb 1_{[-\delta,\delta]}(B_s)}{\sigma^2(A^{-1}_s, B_s)} ds \leqslant \int_0^{A_T} \frac{\mathbb 1_{[-\delta,\delta]}(B_s)}{M^2} ds . $$ If $\sigma(t, x)$ is bounded above by some number $N$, then we get $$ \int_0^T \mathbb 1_{[-\delta,\delta]}(X_t) dt \leqslant \int_0^{N^2 T} \frac{\mathbb 1_{[-\delta,\delta]}(B_s)}{M^2} ds , $$ and the desired result follows directly from the corresponding statement for the Brownian motion. The general case requires a more delicate bound on $A_T$. I do not have time now to work out the details, but if need be, I will get back to this when I am more available.