I need to show that two random processes $\{X_t\}_{t\geq0}$ are martingale.
(1) $X_t=\int^t_af(u)dB_u$ ($f$ is continuous)
(2) $dX_t=b(t,X_t)dB_t$ ($b$ is continuous)
My attempt: To prove the second one, I tried
$E_s(X_t)=X_s+E[\int^t_sb(w,X_w)dB_u]=X_s$.
But I'm not sure this is correct. Also I can't even figure out where I should start for the first one. Any suggestions please?
There is known result that states: Let $\left(B_{t}\right)_{t \geq 0}$ be a Brownian motion and $f(t)$ be a deterministic function such that $\sigma^{2}(t)=\int_{0}^{t} f^{2}(s) d s<\infty$. Then $\int_{0}^{t} f(s) d B_{s} \sim$ $N\left(0, \sigma^{2}(t)\right)$. In your case: $\sigma^2(t) = \int_{0}^t f^{2} (u) du \leq \max_{u \in [0,t]} f^{2}(u) < \infty$ as $f$ is continuous and its composition with $x \mapsto x^2$ is also continuous as composition of continuous functions. Moreover, any continuous continuous function on compact domain is bounded. Thus: $$ \int_{0}^t f(u) dB_u \sim \mathcal{N}(0, \sigma^2(t))$$ Therefore (I will not prove integrability and adaptedness as they follow from what was stated above): $$ \mathbb{E} [\int_{0}^t f(u) dB_u | \mathcal{F}_s] = \int_{0}^s f(u) dB_u + \mathbb{E} [\int_{s}^t f(u) dB_u | \mathcal{F}_s] \stackrel{adapted}{=} \int_{0}^s f(u) dB_u + \underbrace{\mathbb{E} [\int_{s}^t f(u) dB_u]}_{=0} = \int_{0}^s f(u) dB_u $$ Thus it is a martingale.
Note that the process I consider in this answer is: $\int_{0}^{t} b(u,X_u) dB_u$. In other words, I assume the SDE has solution. For it to be a martingale you can use another known result, but you need the following assumptions to be fulfilled: $b(t,X_t)$ is adapted and satisfy: $\int_{0}^t b(u,X_u)^2 du < \infty$ for every $t \geq 0$. Then you can say indeed it is a martingale.