Let
- $E$ be a separable$^1$ Banach space
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\ge 0}$ be a filtration of $(\Omega,\mathcal A)$
- $I_{t_0}:=[0,t_0]$ for $t_0\ge 0$ and $I_\infty:=[0,\infty)$
Let $$\mathcal M^2_{t_0}:=\left\{X\subseteq\mathcal L^0(\operatorname P,E):X\text{ is an }\mathcal L^2(\operatorname P,E)\text{-bounded}^2\text{ right-continuous }(\mathcal F_t)_{t\in I_{t_0}}\text{-martingale with }\operatorname P[X_0=0]=1\right\}$$ be equipped with $$\left\|X\right\|_{\mathcal M^2_{t_0}}:=\left(\sup_{t\in I_{t_0}}\left\|X_t\right\|_{\mathcal L^2(\operatorname P,\:E)}\right)^{\frac12}\;\;\;\text{for }X\in \mathcal M^2(\mathcal F,\operatorname P,E)\tag 1$$ for $t_0\in[0,\infty]$.
Note that if $X\subseteq\mathcal L^2(\operatorname P,E)$ is a right-continuous $\mathcal F$-martingale with $\operatorname P[X_0=0]=1$, then $$(X_t)_{t\in [0,\:t_0]}\in\mathcal M^2_{t_0}\;\;\;\text{for all }t_0\ge 0\tag 2$$ by Doob’s inequality.
Now, there are many textbooks which prove that $M^2_{t_0}$ is a complete seminormed space, for all $t_0\ge 0$. Can we use this fact to prove that even $\mathcal M^2_\infty$ is a complete seminormed space?
My idea is the following:
- Let $(X^n)_{n\in\mathbb N}\subseteq\mathcal M_\infty^2$ be Cauchy
- $\mathcal M_{t_0}^2$ is complete $\Rightarrow$ $\exists Y^{t_0}\in\mathcal M_{t_0}^2$ with $$\left\|(X_t^n)_{t\in[0,\:t_0]}-Y^{t_0}\right\|_{\mathcal M_{t_0}^2}\xrightarrow{n\to\infty}0\tag 3$$ for all $t_0\ge 0$
- It's easy to see that $Y^{s_0}$ and $(Y_t^{t_0})_{t\in[0,\:s_0]}$ are indistinguishable, i.e. $$\left\|Y^{s_0}-(Y_t^{t_0})_{t\in[0,\:s_0]}\right\|_{\mathcal M_{s_0}^2}=0\;,\tag 4$$ for all $t_0\ge s_0\ge 0$
- Now, let $$X_t:=Y_t^t\;\;\;\text{for }t\ge 0$$
- Then, $$\left\|(X_t^n)_{t\in[0,\:t_0]}-(X_t)_{t\in[0,\:t_0]}\right\|_{\mathcal M^2_{t_0}}\xrightarrow{n\to\infty}0\;\;\;\text{for all }t_0\ge 0\tag 5$$
However, we need to conclude that $$\left\|X^n-X\right\|_{\mathcal M^2_\infty}\xrightarrow{n\to\infty}0\;.\tag 6$$ Note that $(6)$ would yield $X\in\mathcal M_\infty^2$ by the triangle inequality. So, how can we show $(6)$?
$^1$ I don't see where we need the separability of $E$. Maybe someone can write the reason for this assumption into the comment section.
$^2$ i.e. $$\sup_{t\ge 0}\left\|X_t\right\|_{\mathcal L^2(\operatorname P,\:E)}<\infty\;.$$