In the following proof
These notes assume that bounded in $L^p$ means $\sup_n E[|X|^p]< \infty$
I see clearly the if part:
If $\sum_{k=1}^{\infty} E[M_k-M_{k-1}]^2 < \infty$, then $E[M_n^2]<\infty \implies \sup_n E[M_n^2]< \infty$ from the Pythagorean theorem in the second equality of the first line of the proof But how do I prove the converse?
If $\sup_n E[M_n^2]<\infty$, the best I can do is use Holder's inequality:
$E[M_k-M_{k-1}]^2\le E[M_k]^2+E[M_{k-1}]^2+E[M_kM_{k-1}] \le E[M_k]^2+E[M_{k-1}]^2+E[M_k^2]E[M_{k-1}^2] < \infty$
but this doesn't guarantee that the series is finite. How should I proceed?
EDIT:
Following the suggestion of geetha290krm:
$\sum_{k=1}^{\infty}E(M_k-M_{k-1})^2=\lim_n \sum_{k=1}^nE(M_k-M_{k-1})^2 = \lim_n E[(M_n-M_0)^2]$ $\le \lim_n ( E[M_{n}^2]+ E[M_{0}^2] + 2E[|M_{n}M_{0}|])$ $\le \lim_n ( E[M_{n}^2]+ E[M_{0}^2] + 2E[M_{n}^2][M_{0}^2])$ $\le \lim_n (\sup_r E[M_{r}^2]+\sup_r E[M_{r}^2] + 2\sup_r E[M_{r}^2]\sup_r E[M_{r}^2])= 2K+2K^2 < \infty$, with $K=\sup_r E[M_{r}^2]$