Supremum of Brownian local time

Question

Exercise 1.22 in Chapter 6 of Revuz and Yor's "Continuous Martingales and Brownian Motion" states that if $L$ is the local time for a standard Brownian Motion then $$\sup_{s \le t} L_s^{B_s} = \sup_a L_t^a.$$ Because local time is non-decreasing, we can easily show $$\sup_{s \le t} L_s^{B_s} \le \sup_{s \le t} L_t^{B_s} \le \sup_a L_t^a,$$ but showing the reverse inequality seems somewhat more difficult. A hint in the book suggested using the Laplace method, but I have had minimal success attempting that. Intuitively the result makes sense because if $a^*$ maximizes $L_t^a$ then the Brownian Motion must have reached $a^*$ at some time, and once it leaves $a^*$ the local time will no longer increase. More precisely, if $\tau := \sup\{s \le t : B_s = a \}$ then we should have that $L_{\tau}^{B_\tau} = L_t^a$, but I'm not sure how to formalize that into a proof.

Is this the correct approach? And if so, how would I go about turning this plausibility argument into a proof?

Answer 1

I believe the intuition you described can be formulated as a proof, with the help of Corollary 1.9 from Chapter 6.1 of Revuz and Yor (3rd ed.). Slightly paraphrased, it says:

If $M$ is a continuous local martingale, then almost surely, $$ L_t^a(M) = \lim_{\epsilon \downarrow 0} \frac{1}{2 \epsilon} \int_0^t 1_{]a - \epsilon, a + \epsilon[}(M_s) \, d \langle M,M \rangle_s $$ for all $a$ and $t$.

With $M = B$, a standard Brownian motion, $\langle B,B \rangle_s = s$. Fix an outcome on the almost-sure event for which the above statement holds. Let $\delta > 0$ and choose $a^\ast$ such that $L_t^{a^\ast} > \sup_a L_t^a - \delta$. (I'm doing this to avoid arguing that $a \mapsto L_t^a$ attains its supremum.) If $\sup_a L_t^a = 0$, then the inequality we desire holds, so let's assume this sup is positive and that $\delta$ is chosen so that $L_t^{a^\ast} > 0$. There are now two claims we wish to verify:

1. The Brownian motion attains the value $a^\ast$ in the interval $[0,t]$.
2. If $\tau$ is the last time the Brownian motion is equal to $a^\ast$, then $L_{\tau}^{B_\tau} = L_t^{a^\ast}$.

We prove these in turn. Consider claim 1. Since $L_t^{a^\ast} > 0$, the Corollary above implies that $$ \int_0^t 1_{(a^\ast -\epsilon,a^\ast+\epsilon)}(B_s) \, ds > 0 $$ for every $\epsilon > 0$. Then for every positive integer $n$ we can find a time $t_n$ such that $B_{t_n} \in (a^\ast - 1/n, a^\ast + 1/n)$. By taking a convergent subsequence of $(t_n)_{n \geq 1}$ and using continuity of $B$, we find that $B$ attains the value $a^\ast$.

Now consider claim 2. First, continuity of $B$ tells us that $B_\tau = a^\ast$. Note that the integral appearing in the corollary is unchanged if we replace its upper limit $t$ with $\tau_\epsilon = \sup \{s \leq t | B_s \in (a^\ast-\epsilon,a^\ast+\epsilon) \}$. Notice that $\tau_\epsilon \downarrow \tau$ as $\epsilon \downarrow 0$. Fix $\eta > 0$ and suppose $\epsilon > 0$ is small enough that $\tau_\epsilon < \tau + \eta$. Taking limits of $$ \frac{1}{2\epsilon} \int_0^{\color{red} \tau} 1_{(a^\ast-\epsilon,a^\ast+\epsilon)}(B_s) \, ds \leq \frac{1}{2\epsilon} \int_0^{\color{red}{\tau_\epsilon}} 1_{(a^\ast-\epsilon,a^\ast+\epsilon)}(B_s) \, ds \leq \frac{1}{2\epsilon} \int_0^{\color{red}{\tau+\eta}} 1_{(a^\ast-\epsilon,a^\ast+\epsilon)}(B_s) \, ds$$ as $\epsilon \downarrow 0$ then yields $L_\tau^{a^\ast} \leq L_t^{a^\ast} \leq L_{\tau + \eta}^{a^\ast}$. But the local time is continuous in the time variable, so taking $\eta \downarrow 0$ and using $B_\tau = a^\ast$, we have $L_{\tau}^{B_\tau} = L_t^{a^\ast}$. This finally shows that $$ \sup_{s \leq t} L_s^{B_s} \geq \sup_{a} L_t^a .$$