Suppose that $\{X_n(t),t\in\mathbb{R}\}_{n=0,1,...}$ is a collection of stochastic processes, i.e., for any fixed $n$, we have and stochastic process $\{X_n(t),t\in\mathbb{R}\}$.
Assume that for any fixed $n$ we have that $X_n(t)\overset{d}{\to}N(0,t^2)$ for all $t\in \mathbb{R}$ when $n\to\infty$ ($N(0,0)$ is the constant $0$.) Now, if $(T_n)_{n\geq 0}\subset\mathbb{R}$ is a sequence of random variables such that $T_n\overset{P}{\to}0$ when $n\to\infty$, is it true that $$ X_n(T_n)\overset{P}{\to}0 $$ when $n\to\infty$? This seems true since, intuitively, $X_n(T_n)\sim N(0, T_n^2)$ for large $n$. My only "formal" approach is to calculate $$ P(|X_n(T_n)|>\epsilon)=\int_{\mathbb{R}}P(|X_n(t)|>\epsilon|T_n=t)dF_n(t), $$ where $F_n(t)=P(T_n\leq t)$. Since $F_n(t)\to \mathbf{1}_{[0,\infty)}(t)$ when $n\to\infty$ then $dF_n(t)/dt=0$ for $t\neq 0$. I don't know where to go from here.
Any ideas on how to prove it? If it is not true, is there any other assumption under it is true?
Assuming that a.s. the family $(X_n)_n$ is equicontinuous at $0$. The convergence might hold under some weaker assumption but I found this quite natural.
Let $\epsilon >0$, then there exists $\delta >0$ such that $0\leq t \leq \delta \Longrightarrow |X_n(t)-X_n(0)| \leq \epsilon$ for every $n$. Now write $$\mathbb{E}\left|e^{iX_n(T_n)} - 1\right| \leq \mathbb{E}\left|e^{iX_n(0)} - 1\right| + \mathbb{E}\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|.$$ The first term goes to $0$ since $X_n(0) \to 0$ in distribution. For the second term, we have the decomposition $$\begin{align}\mathbb{E}\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|&= \mathbb{E}\left[\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|\mathbf{1}_{T_n \leq \delta}\right] + \mathbb{E}\left[\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|\mathbf{1}_{T_n >\delta}\right]\\ &\leq \mathbb{E}\left[\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right|\mathbf{1}_{|X_n(T_n)-X_n(0)|\leq \epsilon}\right] + 2 \mathbb{P}(T_n >\delta) \\ &\leq \mathbb{E}\left[\left|X_n(T_n) - X_n(0)\right|\mathbf{1}_{|X_n(T_n)-X_n(0)|\leq \epsilon}\right] + 2 \mathbb{P}(T_n >\delta)\\ &\leq \epsilon + 2 \mathbb{P}(T_n >\delta). \end{align}$$ Since $T_n \to 0$ in probability, it follows that $$\limsup_n \mathbb{E}\left|e^{iX_n(T_n)} - e^{iX_n(0)}\right| \leq \epsilon.$$ This holds for every $\epsilon >0$, so letting $\epsilon \to 0$ gives the convergence in distribution $X_n(T_n) \to 0$.