First hitting time for non-homogeneous diffusion martingale This question can be seen as a continuation of Lipschitz continuity of $\mathbb P[\tau>t]$ with respect to $t$
Consider the martingale given as
$$X_t=1+\int_0^t a(s,X_s)dW_s,\quad \forall t\ge 0.$$
Denote $\tau:=\inf\{t\ge 0: X_t\le0\}$. My question is whether $t\mapsto \mathbb P[\tau>t]$ is Holder continuous, i.e. $\exists C>0, \alpha\in (0,1]$ s.t.
$$0\le \mathbb P[\tau>t]-\mathbb P[\tau>t+\Delta t]\le C\Delta t^{\alpha}.$$
Any solution, references or comments are appreciated. Here we assume that $0<\underline a \le \inf_{(t,x)} a(t,x)\le \sup_{(t,x)} a(t,x)\le \overline a$, $t\mapsto a(t,x)$ is continuous and $|a(t,x)-a(t,y)|\le L|x-y|$ for some $L>0$.
PS : My idea is as follows :
\begin{eqnarray}
\mathbb P[\tau>t]-\mathbb P[\tau>t+\Delta t] &=& \mathbb P\left[\inf_{0\le s\le t}X_s>0, X_t+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\right] \\
&=& \mathbb P\left[\inf_{0\le s\le t}X_s>0, X_t>0, X_t+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\right] \\
&\le &\mathbb P\left[X_t>0, X_t+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\right] \\
&=& \int_{(0,\infty)}\mathbb P\left[x+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\Big|X_t=x\right]\mathbb P[X_t\in dx].
\end{eqnarray}
Therefore, it suffices to estimate the conditional probability
$$\mathbb P\left[x+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\Big|X_t=x\right]$$
 A: For $h:=\Delta t>0$, you had
$$P(\tau>t)-P(\tau>t+h)=\int_{(0,\infty)}
P(M\le-x|X_t=x)\,P(X_t\in dx),$$
where
$$M:=\inf_{t\le u\le t+h}J_u,\quad J_u:=\int_t^u a(s,X_s)\,dW_s.$$
By Doob's martingale inequality,
$$P(M\le-x|X_t=x)\le x^{-2}\,E(J_{t+h}^2|X_t=x)
\le x^{-2}\,ha_2^2,$$
where $a_2:=\overline a$ and $a_1:=\underline a$.
The crucial point is that the pdf $p_t$ of $X_t$ for $t>0$ is bounded so that
$$p_t(x)\le\frac c{\sqrt t}\,e^{-bx^2/t}\le\frac c{\sqrt t}$$
for some positive real constants $c,b$ depending only on $a_1,a_2,L$ and for all real $x$. So,
$$
\begin{align}
P(\tau>t)-P(\tau>t+h)&\le
\int_{(0,\infty)}\min(1,x^{-2}\,ha_2^2)\,\frac c{\sqrt t}\,dx \\ 
&=
\sqrt h\int_{(0,\infty)}\min(1,u^{-2}\,a_2^2)\,\frac c{\sqrt t}\,du \\ 
&=\frac C{\sqrt t}\,\sqrt h,
\end{align}
$$
where $C>0$ is a real constant depending only on $a_1,a_2,L$.

To get now a uniform Hölder continuity, we can reason as follows:
$$P(\tau>t)-P(\tau>t+h)\le1-P(\tau>t+h)=P(\tau\le t+h)\le a_2^2(t+h)/1^2,$$
again by Doob's martingale inequality. So,
$$P(\tau>t)-P(\tau>t+h)\le\min\Big(1,\frac C{\sqrt t}\,\sqrt h,a_2^2(t+h)\Big)\le C_1h^{1/3},$$
where $C_1>0$ is a real constant depending only on $a_1,a_2,L$ (to verify the latter inequality, consider separately the three cases when (i) $h\ge1$, (ii) $t\le h^{1/3}<1$, or (iii) $t>h^{1/3}$).
Using here an exponential inequality (see e.g. Theorem 3.1) instead of Doob's one, one can improve the factor $h^{1/3}$ to  $h^{1/2}\ln\frac1h.$
