I recently came across this question and reckon it should be a direct application of Doob's inequality (correct me if I am wrong). But I struggle to write formal proof.
For each $n \in \mathbb{N}$, let $(X^{(n)}_{t})_{t\geq0}$ be a martingale on $(\Omega,F,P)$ with respect to F. Assume that for all $n \in \mathbb{N}$ we have that $\mathbb{E}[|X^{(n)}_1|^2]=0$ < $\infty$ and assume further that $lim_{n \rightarrow \infty} \mathbb{E}[|X^{(n)}_1|^2]=0$. Show that for any $\epsilon > 0 $, we have $$ \lim_{x\to\infty} \mathbb{P}(\sup_{t \in [0,1]}|X^{(n)}_t| > \epsilon) =0$$
I really appreciate your help :)
I guess in the limit you mean $n \to \infty$ not $x$. Moreover, your assumptions are redundants : if $\forall n \in \mathbb{N}, \hspace{0.3cm} \mathbb{E}(|X_1^{(n)}|^2) = 0$ this implies the limit is also 0.
Let us assume that it is n in the limit and just $\lim_{n\to\infty} \mathbb{E}(|X_1^{(n)}|^2) = 0 $. Using Doob’s maximal inequality :
$$\mathbb{P}( \hspace{0.1cm} \underset{t \in [0,1]}{\sup} \hspace{0.1cm} |X_t^{(n)}| > \epsilon) \le \frac{\mathbb{E}(|X_1^{(n)}|^2)}{\epsilon^2}$$
And the result is straightforward by the assumption