let $X_t$ be an Ito process where $X_t = \int_{0}^t v_t dB_t$ where $v_t$ is a stochastic process, $B_t$ is a Wiener Process, $\mathcal{F}_t$ be a filtration: $\sigma\{B_t, 0 \leq t \leq T\}$, and $v_t$ is adapted to $\mathcal{F}_t$
Let $M_t = X_t^2 - \int_{0}^t |v_s|^2ds$.
I have to prove that $M_t$ is a martingale if $v_t$ is bounded. If $v_t$ is bounded, the expectation exists.
If $s \leq t$ Then:
$$E[X_t|\mathcal{F_s}] = (|\int_{0}^t v_r dB_r|^2|\mathcal{F_s})$$
$$E[X_t|\mathcal{F_s}] = (|\int_{0}^s v_r dB_r + \int_{s}^t v_r dB_r|^2|\mathcal{F_s})$$
$$E[X_t|\mathcal{F_s}] = (\int_{0}^s v_r^2 dB_r^2 + H+ \int_{s}^t v_r^2 dB_r^2|\mathcal{F_s})$$
$$E[X_t|\mathcal{F_s}] = \int_{0}^s v_r^2 dB_r^2 + (H + \int_{s}^t v_r^2 dr|\mathcal{F_s})$$
Then, for $M_t$:
$$E[M_t|\mathcal{F_s}] = \int_{0}^s v_r^2 dB_r^2 + (H + \int_{s}^t v_r^2 dr - \int_{0}^t v_r^2 dr|\mathcal{F_s})$$
$$E[M_t|\mathcal{F_s}] = \int_{0}^s v_r^2 dB_r^2 + (H - \int_{0}^s v_r^2 dr|\mathcal{F_s})$$
$$E[M_t|\mathcal{F_s}] = \int_{0}^s v_r^2 dB_r^2 + H - \int_{0}^s v_r^2 dr$$
$$E[M_t|\mathcal{F_s}] = X_s^2 + H - \int_{0}^s v_r^2 dr$$
So, $M_t$ is a martingale if the magic term $H$ is zero. How do I know that $H$ is zero?
Edit:
Oh, gosh, I finally got it!
$$H = 2\int_{0}^s v_rdB_r \int_{s}^tv_rdB_r$$
$$E[H|\mathcal{F_s}] = 2E[\int_{0}^s v_rdB_r \int_{s}^tv_rdB_r|\mathcal{F_s}]$$ $$E[H|\mathcal{F_s}] = 2\int_{0}^s v_rdB_r E[ \int_{s}^tv_rdB_r|\mathcal{F_s}]$$
by the independence of the increments: $$E[H|\mathcal{F_s}] = 2(E[\int_{0}^s v_rdB_r)0$$ $$E[H|\mathcal{F_s}] = 0$$
Please, tell me if I'm wrong!