Consider the drifted Brownian motion $X_t=1+\lambda(t)+W_t$, where $\lambda: \mathbb R\to [0,\infty)$ with $1\le \lambda'(t)\le 2$ and $(W_t)_{\ge 0}$ denotes a Brownian motion. Define the hitting time
$$\tau:=\inf\{t\ge 0: X_t\le 0\}$$
and further the function $p(t):=\mathbb P[\tau>t]$ for all $t\ge 0$. Can we show that $t\mapsto p(t)$ is Lipschitz?
I claim this is not a complete answer. I wonder whether it can be improved to obtain the Lipschitz continuity.
Define
$$\left(\frac{d\mathbb Q}{d\mathbb P}\right)_t := \exp\left(-\int_0^t \lambda'(s)dW_s-\frac{1}{2}\int_0^t(\lambda'(s))^2ds\right).$$
Then $B_t:=W_t+\lambda(t)$ is a Brownian motion under $\mathbb Q$. Thus,
$$p(t) = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t}{\bf 1}_{\{\inf_{0\le r\le t}(1+\lambda(r)+W_{r})>0\}}\right] = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t}{\bf 1}_{\{\inf_{0\le r\le t}(1+B_{r})>0\}}\right].$$
For every $0\le t<t+\Delta t$, it holds
$$0\le p(t)-p(t+\Delta t) = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t+\Delta t}\left({\bf 1}_{\{\inf_{0\le r\le t}(1+B_{r})>0\}}-{\bf 1}_{\{\inf_{0\le r\le t+\Delta t}(1+B_{r})>0\}}\right)\right].$$
Using Holder inequality for $a, b>1$ with $1/a+1/b=1$, one has
$$p(t)-p(t+\Delta t)\le \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t+\Delta t}^a\right]^{1/a} \left(\mathbb Q\left[\inf_{0\le r\le t}B_{r}>-1\right]-\mathbb Q\left[\inf_{0\le r\le t+\Delta t}B_{r}>-1\right]\right)^{1/b}.$$
Let $\Phi:\mathbb R\to (0,1)$ be the CDF defined by
$$\Phi(x):=\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-z^2/2}dz.$$
Then we finally obtain
$$p(t)-p(t+\Delta t)\le \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t+\Delta t}^a\right]^{1/a}2^{1/b} \left( \Phi\left(\frac{1}{\sqrt{t}}\right)-\Phi\left(\frac{1}{\sqrt{t+\Delta t}}\right)\right)^{1/b}.$$
It follows from some tedious computation, we have
$$ \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t+\Delta t}^a\right]^{1/a} \le \exp(2(a-1)(t+\Delta t))$$
and
$$2^{1/b} \left( \Phi\left(\frac{1}{\sqrt{t}}\right)-\Phi\left(\frac{1}{\sqrt{t+\Delta t}}\right)\right)^{1/b} \le \left(\frac{2}{\pi t^3}\right)^{1/2b} \exp\left(-\frac{1}{2bt}\right) \Delta t^{1/b}.$$
We conclude thus the (local) $1/b-$Holder continuity. Is there any chance to have the Lipschitz continuity?
