Let
- $E$ be a $\mathbb R$-Banach space
- $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge 0},\operatorname P)$ be a filtered probability space
- $X^n$ be an $E$-valued right-continuous martingale on $(\Omega,\mathcal A,\mathcal F,\operatorname P)$ with $X^n=0$ almost surely and $$\sup_{t\ge 0}\sqrt{\operatorname E\left[\left\|X_t^n\right\|_E^2\right]}<\infty\;,$$ for $n\in\mathbb N$
Suppose $(X^n)_{n\in\mathbb N}$ is Cauchy with respect to $$\left\|X\right\|:=\sqrt{\operatorname E\left[\sup_{t\ge 0}\left\|X_t\right\|_E^2\right]}\;.$$ Then there is a subsequence $(X^{n_k})_{k\in\mathbb N}$ with $$\left\|X^{n_k}-X^{n_{k+1}}\right\|<\frac1{2^k}\;\;\;\text{for all }k\in\mathbb N\;.\tag 1$$ Now, let $$Y_k:=\sum_{i=1}^k\sup_{t\ge 0}\left\|X_t^{n_i}-X_t^{n_{i+1}}\right\|_E\;\;\;\text{for }k\in\mathbb N\;.$$
How can we show that $$\operatorname E[Y_k^2]\le 2\sum_{i=1}^k\left\|X^{n_i}-X^{n_{i+1}}\right\|\tag 2$$ for all $k\in\mathbb N$?
I've found $(2)$ in a book and don't understand why this equation holds. First I thought it would be an easy application of Young's inequality. Indeed, $$(a+b)^2\le 2(a^2+b^2)\;\;\;\text{for }a,b\ge 0\;.\tag 3$$ But a recursive application only yields $$(a+b+c)^2\le 2(a+b)^2+2c^2\le 4(a^2+b^2)+2c^2\;\;\;\text{for }a,b,c\ge 0\;.\tag 4$$
So, how can we show $(2)$?
Note that $$\mathbb E\left[Y_k^2\right] =\left\lVert \sum_{i=1}^k\sup_{t\geqslant 0}\left\|X_t^{n_i}-X_t^{n_{i+1}}\right\|_E\right\rVert_2^2\leqslant \left(\sum_{i=1}^k\left\lVert\sup_{t\geqslant 0}\left\|X_t^{n_i}-X_t^{n_{i+1}}\right\|_E\right\rVert_2\right)^2= \left(\sum_{i=1}^k\left\lVert X^{n_i}-X^{n_{i+1}}\right\rVert\right)^2 .$$ Now, it suffices to prove that if $0\leqslant c_i\leqslant 2^{-i}$, then $$2\sum_{1\leqslant i\lt j\leqslant n} c_ic_j\leqslant \sum_{i=1}^nc_i. $$ This can be done in the following way: $$\sum_{1\leqslant i\lt j\leqslant n} c_ic_j\leqslant \sum_{i=1}^{n-1}c_i\sum_{j=i+1}^n2^{-j}\leqslant \sum_{i=1}^{n-1}c_i\underbrace{2 ^{-i}}_{\leqslant 1/2} . $$