# A stochastic process $X$ with values in a separable Banach space $E$ is a martingale iff $f(X)$ is a martingale for all $f\in E^\ast$

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
• $(E,\left\|\;\cdot\;\right\|)$ be a separable Banach space and $E^*$ be the dual space of $E$
• $(X_t)_{t\ge 0}$ be an $\mathcal F$-adapted $E$-valued stochastic process on $(\Omega,\mathcal A,\operatorname P)$

I want to show that $X$ is a $\mathcal F$-martingale $\Leftrightarrow$ $$f(X)\text{ is a }\mathcal F\text{-martingale}\;\;\;\text{for all }f\in E^*\tag 1\;.$$

Using

Lemma$\;\;\;$Let $Y$ be a $E$-valued Bochner integrable random variable $\Rightarrow$ $$\operatorname E[f(Y)]=f\left(\operatorname E[Y]\right)\;\;\;\text{for all }f\in E^*\tag 2\;.$$

it's easy to prove "$\Rightarrow$":

• Let $f\in E^*$
• $f$ is continuous $\Rightarrow$ $f$ is $\mathcal B(E)$-measurable $\Rightarrow$ $f(X_t)$ is $\mathcal A$-measurable and $$\operatorname E\left[\left|f(X_t)\right|\right]\le\sup_E|f|<\infty$$ for all $t\ge 0$ since $f$ is bounded $\Rightarrow$ $f(X)$ is integrable
• Let $t\ge s\ge 0$ and $A\in\mathcal F_s$ $\Rightarrow$ $$\begin{split} \operatorname E\left[1_Af(X_t)\right]&=\operatorname E\left[f(1_AX_t)\right]\\ &\stackrel{(2)}=f\left(\operatorname E\left[1_AX_t\right]\right)\\ &=f\left(\operatorname E\left[1_AX_s\right]\right)\\ &\stackrel{(2)}=\operatorname E\left[f(1_AX_s)\right]\\ &=\operatorname E\left[1_Af(X_s)\right] \end{split}$$ since $X$ is a $\mathcal F$-martingale

How can we prove "$\Leftarrow$"?

$f(E(X_t|\mathcal F_s)) = E(f(X_t)|\mathcal F_s)\tag{1}$
Since $f(X_t)$ is martingale, we have
$f(E(X_t|\mathcal F_s)) = f(X_s)$
$E(X_t|\mathcal F_s)= X_s$