Let $M$ be the space of right continuous functions $\ell: \mathbb R\_+\to [0,1]$ that are non increasing s.t. $\ell(0)=0$. Define the map $\Gamma : M\to M$ by $\Gamma[\ell](t):=\mathbb P[\tau^{\ell}>t]$ for all $\ell \in M$ and $t\ge 0$, where $\tau^{\ell}:=\inf\{t\ge 0: X^{\ell}\_t\le 0\}$ and


$$X^{\ell}\_t:=1+t+\int\_0^t\frac{1}{1+\ell(s)}dW\_s,\quad \forall t\ge 0.$$


Here $(W\_t)\_{t\ge 0}$ denotes a Brownian motion. Let $M$ be endowed with the topology as follow: $\ell^n$ converges to $\ell$ in $M$ iff $\lim\_{n\to\infty}\ell^n(t)=\ell(t)$ for all points of continuity of $\ell$. Can we prove the continuity of $\Gamma$ with respect to this topology?


Remark : To prove $\{\inf\_{0\le s\le t}X^{\ell}\_s\le 0\}=\{\inf\_{0\le s\le t}X^{\ell}\_s< 0\}$, it suffices to use Lévy's characterization. More precisely, we can write $X^{\ell}\_t=1+t+B\_{\langle X^{\ell}\rangle\_t}\equiv 1+t+B\_{L(t)}$, where $B$ denotes a Brownian motion and $L(t):=\int\_0^t ds/(1+\ell(s))^2$. Therefore


$$\inf\_{0\le s\le t}X^{\ell}\_t = \inf\_{0\le u\le L(t)}\{1+L^{-1}(u)+B\_t\},$$


which implies the desired result as $\inf\_{0\le u\le L(t)}\{1+L^{-1}(u)+B\_t\}$ admits a density.



On the continuity of map $\Gamma$
I believe we can. Let $\ell\_n \to \ell$ in your topology, and fix $t\_0 \in \mathbb R\_+$. We show that $\Gamma(\ell\_n)(t\_0) \to \Gamma(\ell)(t\_0)$. We work over the interval $[0, T]$ with $T > t\_0$.


**Step 1:** We first note that $\ell\_n$ converges to $\ell$ in measure.


Indeed, let $\varepsilon > 0$ be arbitrary. As $\ell$ is nonincreasing, $\ell$ contains only jump discontinuities and for any $n > 0$, there exist only finitely many jumps with magnitude greater than $\frac{\varepsilon}{2^j}$. Cover these jumps with finitely many closed intervals $C\_i^j$ of total length less than $\frac{\varepsilon}{2^j}$. Divide the complement into intervals $D\_i^j$ on which $\ell$ varies by no more than $\frac{\varepsilon}{2^j}$, and consider the partition $\mathcal P\_j ;= C\_i^j \cup D\_i^j$. Applying the pointwise convergence of $\ell\_n$ to $\ell$ near the endpoints of the $C\_i^n$ and the monotonicity of $\ell\_n$ and $\ell$ now allows us to conclude.


**Step 2:** We show that $X^{\ell\_n} \to X^{\ell}$ uniformly in probability.


Namely, that $$P(\sup\_{s \in [0, T]} | X\_t^{\ell\_n} - X\_t^{\ell}|
 \geq \frac{1}{2^k}) \to 0.$$


To see this, note that we can write


$$X^{\ell}\_t:=1+t+\int\_0^t\frac{1}{1+\ell(s)} + g\_n(s) \ dW\_s \, \quad \forall t\ge 0.$$


with $g\_n := \frac{1}{1 + \ell\_n(s)} - \frac{1}{1 + \ell(s)} \to 0$ in $L^2$. Whereby the aforementioned convergence follows from the convergence in measure of $\ell\_n$ to $\ell$.


Note that $X\_s^{\ell\_n} - X\_s^{\ell}$ is a martingale, and that $X\_T^{\ell\_n} - X\_T^{\ell}$ is Gaussian with mean $0$, and variance $||g\_n||\_{L^2}$.


And so by Doob’s (sub)martingale inequality and symmetry of the Gaussian distribution, we have


$$P(\sup\_{s \in [0, T]} | X\_s^{\ell\_n} - X\_s^{\ell}|
 \geq \frac{1}{2^k}) \leq 2^{k+1} E[(X\_T^{\ell\_n} - X\_T^{\ell})^+] \to 0,$$


as $n \to \infty$.


**Step 3:** Conclusion.


Let $\varepsilon > 0$ be arbitrary. We note that by continuity of $X\_n^{\ell}$ we can write the event $\{\tau^{\ell} > t\}$ as $$\bigcup\_{k \in \mathbb N} A\_k := \bigcup\_{k \in \mathbb N} \{X\_s \geq \frac{1}{k}, \ \forall s \in [0,t\_0]\}.$$


By continuity from below, we have that for some $k\_0 \in \mathbb N$ that $P(A\_{k\_0}) > P(\{\tau^{\ell} > t\}) - \frac{\varepsilon}{2}$.


Since $X^{\ell\_n} \to X^{\ell}$ uniformly in probability, for all large enough $n$, we have


$$P(\sup\_{s \in [0, T]} | X\_s^{\ell\_n} - X\_s^{\ell}|
 \geq \frac{1}{2k\_0}) < \frac{\varepsilon}{2},$$


so that for all $n > N$ we have


$$\Gamma(\ell\_n)(t\_0) = P(\{\tau^{\ell\_n} > t\_0\}) > P(A\_{k\_0} \cup \{\sup\_{s \in [0, T]} | X\_s^{\ell\_n} - X\_s^{\ell}| > \frac{1}{2k\_0} \}^c) > P(\{\tau^{\ell} > t\_0\}) - \varepsilon$$


For the reverse inequality, we argue similarly - we note that, up to $P$-null sets, we can write the event $\{\tau^{\ell} \leq t\_0\}$ as


$$\bigcup\_{i \in \mathbb N} \{\inf\_{s \in [0, t\_0)}\ X\_s^{\ell} \leq -\frac{1}{i}\},$$


In order for the above representation to hold, we need to show that $\{\inf\_{0\le s\le t}X^{\ell}\_s\le 0\}=\{\inf\_{0\le s\le t}X^{\ell}\_s< 0\}$, up to a $P$-null set.


For this, it suffices to use Lévy's characterization. More precisely, we can write $X^{\ell}\_t=1+t+B\_{\langle X^{\ell}\rangle\_t}\equiv 1+t+B\_{L(t)}$, where $B$ denotes a Brownian motion and $L(t):=\int\_0^t ds/(1+\ell(s))^2$. Therefore


$$\inf\_{0\le s\le t}X^{\ell}\_t = \inf\_{0\le u\le L(t)}\{1+L^{-1}(u)+B\_t\},$$


which implies the desired result as $\inf\_{0\le u\le L(t)}\{1+L^{-1}(u)+B\_t\}$ admits a density.


Thus with a similar calculation, we obtain $$\Gamma(\ell\_n)(t) < P(\{\tau^{\ell} < t\_0\}) + \varepsilon$$


for all large enough $n$.


Since $\varepsilon$ was arbitrary, we conclude that $\Gamma(\ell\_n)(t\_0) \to \Gamma(\ell)(t\_0)$, as was to be shown.


