Let $S_n:=\sum_{i=1}^n X_i$ where $\{X_i\}$ are IID with $\mathbb{E}X_i = 0$ and $\mathbb{E}X_i^2=1$ for all $i\in \mathbb{N}$. We define stopping times: $\tau_n:=\inf\{1\leq k\leq n: S_n = \max_{1\leq k\leq n}S_k\}$ and $T_n:=\sup\{1\leq k\leq n: S_n = \max_{1\leq k\leq n}S_k\}$, with respect to $\mathcal{F}_n:=\sigma(X_1,\dots,X_n)$, i.e., the first and the last times until $n$ at which maxima are attained. I want to show that $(T_n-\tau_n)/n\to 0$ in probability.
I know that, if $\psi:C([0,1])\to \mathbb{R}$, is continuous $\mathbb{P}_0$-almost surely, with respect to the supremum norm and probability measure of the standard Brownian motion $(B_t)_{t\geq 0}$, then $\psi (S(n\cdot)/\sqrt{n})\to \psi(B_\cdot)$ in distribution, where $t\mapsto S(nt)$ is the linear interpolation of $\{S_k\}_{k=1}^n$. Then, by showing $\psi(\omega):=\inf\{t\in [0,1]:\omega(t) = \sup_{s\in [0,1]}\omega(s)\}$ and $\phi(\omega):=\sup\{t\in [0,1]:\omega(t) = \sup_{s\in [0,1]}\omega(s)\}$ are continuous $\mathbb{P}_0$-a.e., we would have $\psi(S(n\cdot)/\sqrt{n}) = \tau_n$ and $\phi(S(n\cdot)/\sqrt{n}) = T_n$ converging to the same distribution. This is where I got stuck. I'm not sure how to proceed and don't know if there's a better way to do this.
I'd appreciate any help.
Firstly, as suggested by @Snoop the random variables $T_n$ and $\tau_n$ are not stopping times. But this is also not relevant for the solution you suggest.
Another small correction is needed. Since $\phi$ is $[0,1]$-valued you have $\phi(S(n\cdot)/\sqrt{n})=T_n/n$ and not $\phi(S(n\cdot)/\sqrt{n})=T_n$ (clearly $S(n\cdot)$ can only assume its maximum at the time points $0,1/n,\dotsc,1$). Similarly, $\psi(S(n\cdot)/\sqrt{n})=\tau_n/n$.
Now, the functions $\phi$ and $\psi$ are bounded and continuous so we have the convergence $$ E[T_n/n]=E[\phi(S(n\cdot)/\sqrt{n})]\to E[\phi(B)]\qquad\text{and}\qquad E[\tau_n/n]=E[\psi(S(n\cdot)/\sqrt{n})]\to E[\psi(B)]. $$ The final ingredient is the inequality $T_n\geq\tau_n$. We get $$ E[\lvert (T_n-\tau_n)/n\rvert]=E[T_n/n]-E[\tau_n/n]\to E[\phi(B)]-E[\psi(B)]=0 $$ since $\phi(B)=\psi(B)$ a.s. This proves the convergence.