Let $(\Omega,\mathcal{F},P,\{\mathcal{F}_t\}_{t \in [0,\infty)}$ be a standaard filtered probability space.
Let $\mathcal{M}^c_b$ be the set of $\{\mathcal{F}_t\}_{t \in [0,\infty)}$-martingales $M=\{M_t\}_{t \geq 0}$ satisfying the following two conditions (i) M is a continuous process and $M_0=0$; (ii) There is a $K \in (0,\infty)$ such that $|M(t,\omega)| \leq K, t \in [0,\infty), \omega \in \Omega$.
proposition 1: If $M=\{M_n\}_{n=0}^{\infty}$ is a square integrable martingale, $E[(M_n-M_m)^2|\mathcal{F}_m]=E[M_n^2|\mathcal{F}_m]-M_m^2, n \geq m \geq 1$. In particular, $E[M_n^2]=E[M_0^2]+\sum_{k=1}^nE[(M_k-M_{k-1})^2], n \geq 0$.
Proposition 2: Let $M=\mathcal{M}_b^c$. Then $$(i) E[\sup_{s,t \in [0,\infty),|s-t| \leq \delta} |M_t-M_s|] \rightarrow 0, \delta \rightarrow 0$$ $$(ii) \sup_{n \geq 1} E[\sum_{k=1}^{\infty}(M_{k/2^n}-M_{(k-1)/2^n})^2)^2]<\infty$$.
Proof of proposition 2(i):By the assumption, there is a $K>0$ such that $\sup_{t \geq 0} |M_t| \leq K$. Then for any $\delta \in (0,1]$ and $m \geq 1$, we have $$E[\sup_{s,t \in [0,\infty),|s-t| \leq \delta} |M_t-M_s|] \leq E[\sup_{s,t \in [0,m+1],|s-t| \leq \delta} |M_t-M_s|]+2E[\sup_{t \geq m} |M_t-M_m|]$$.
Since $M_t, t \in [0,m+1]$ is uniformly continuous, for any $\omega \in \Omega$, we have $$\sup_{s,t \in [0,m+1],|s-t| \leq \delta} |M_t-M_s| \rightarrow 0, \delta \rightarrow 0$$.
Since $\sup_{t,s \in [0,m+1]} |M_t-M_s| \leq 2K$, we see $$E[\sup_{s,t \in [0,m+1],|s-t| \leq \delta} |M_t-M_s|] \rightarrow 0,\delta \rightarrow 0$$.
By Doob's ineuqaity, we see $$E[\sup_{t \geq m} |M_t-M_m|] \leq 2\sup_{ t \geq m} E[|M_t-M_m|^2]^{1/2}=2(\sup_{t \geq m} E[M_t^2]-E[M_m]^2)^{1/2}$$.
So we get $$E[\sup_{t \geq m} |M_t-M_m|] \rightarrow 0, m \rightarrow \infty$$ and this is the proof of (i).
Proof of (ii).Let $a_n=\sum_{k=1}^{\infty} E[(M_{k/2^n}-M_{(k-1)/2^n})^4],n \geq 1.$ we see $$a_n \leq 4K^2 \sum_{k=1}^{\infty} E[(M_{k/2^n}-M_{(k-1)/2^n})^2] \leq 4K^2 \sup_{k \geq 1} E[M_{k/2^n}^2] \leq 4K^4 \text{here is follows from propsition 1}$$
Also, we see $$E[(\sum_{k=1}^m(M_{k/2^n}-M_{(k-1)/2^n})^2)^2] \tag{1}$$ $$=2\sum_{k=1}^m E[E[\sum_{l=k+1}^m (M_{l/2^n}-M_{(l-1)/2^n})^2 | \mathcal{F}_{k/2^n}](M_{k/2^n}-M_{(k-1)/2^n})^2]+\sum_{k=1}^m E[(M_{k/2^n}-M_{(k-1)/2^n})^4]$$ $$\tag{2}$$ $$\leq 2 \sum_{k=1}^m E[E[M_{m/2^n}^2-M_{k/2^n}^2 | \mathcal{F}_{k/2^n}](M_{k/2^n}-M_{(k-1)/2^n})^2]+a_n \tag{3}$$ $$\leq 8K^2 \sum_{k=1}^m E[(M_{k/2^n}-M_{(k-1})/2^n)^2]+4K^4 \leq 12K^4 \text{and done}$$.
The proposition is from Stochastic Analysis by Shigeo Kusuoka. My question is I don't follow the algebraic transformations from tag (1) to tag (2) to tag (3). It would be nice if someone can explain the processes. Thanks!