# On the continuity of map $\Gamma$

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.

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.

• Thanks for your answer. For the reservse inequality, I think there is a gap. Indeed, $\{\tau^{\ell}\le t_0\}\neq \cup_{i\in\mathbb N}\{\inf_{s\in [0,t_0)}X^{\ell}_s\le -1/i\}$ as the scenario $X_s>0$ for $s\in [0,t_0)$ and $X_{t_0}=0$ belongs to the first set but not the later one. Jun 27 at 19:52
• To make your arguments rigorous, I think we need to show Therefore, we need to argue that $\{\tau^{\ell}\le t_0\}$ coincides with $\cup_{i\in\mathbb N}\{\inf_{s\in [0,t_0)}X^{\ell}_s\le -1/i\}$ almost surely Jun 27 at 19:54
• Ah yes, I’ve implicitly argued that the scenario you mentioned has probability zero. Do you need me to add that as a detail to the answer? Jun 27 at 20:58
• Could you please provide the details? Many thanks! Jun 27 at 21:43
• Many thanks. I will accept your answer as soon as you complete it. Btw, there is some typo, e.g. $1/2k_0$ instead of $k_0/2$ Jun 28 at 15:18