Boundedness in $L^2$ implies convergence in $L^2$ for martingales. a. Show that if $X = (X_n)_n$ is an $L^2$ martingale bounded in $L^{2}$ then there exists a positive r.v. $U \in L^{2}$ that satisfies :
$$
P\left(\forall n \geq 1,\left|X_{n}\right| \leq U\right)=1
$$
b. Deduce from a) or otherwise that $X$ converges in $L^2$.
What I could do so far : since $(X^2_n)_n$ is a submartingale then $\sup_nEX^2_n = \sum_nEX^2_n < \infty$ and thus : $EX^2_n \to 0$ so $X_n$ converges to $0$ in $L^2$.
my questions :

*

*is my work for b, correct?

*how to show a) and can it actually be used to deduce b) ?

Thanks!

Your work for $b)$ is not correct as Kavi Rama Murthy already mentioned in the comments. The statement $\sup_n EX_n^2=\sum_n EX_n^2$ is NOT true in general. Here is a solution for $a)$:
The r.v. $U$ you are looking for is given by $U:=\sup_n |X_n|$. So the $|X_n|\leq U$-part is trivial and you just have to show $U\in L^2$. This can be done by using the monotone convergence theorem for the sequence $U_n:=\sup_{m\leq n} |X_m|$,  Doob's maximal inequality and the fact that $M:=\sup_n EX_n^2<\infty$ by assumption
$$EU^2=E(\lim_nU_n)^2=\lim_nEU_n^2\leq\lim_n4 EX_n^2\leq4M<\infty.
$$
To your $b)$-part: From $a)$ you can conclude that $X^2$ is a uniformly integrable submartingale. Now use Doob's martingale convergence theorem to conclude that it converges a.s. an in $L^1$ to $X_\infty^2$. By Vitali's convergence theorem you can now conclude that $X$ converges in $L^2$. (The last part you can also be done different depending on which version of the theorems you are using.)
