I want to show the following is a martingale process.($Z(t)$ continuous) $$dZ(t)=\theta_{t}Z(t)dB_{t}$$ where $\theta_{t}$ is deterministic function $$\theta_{t}=\sigma t$$
My trial:$$E[\int^{T}_{0}(Z(t)\theta_{t})^{2}dt]\leq E[C^{2}(\int^{T}_{0} \theta_{t}^{2}dt)]\leq+\infty $$ leads to the $$\theta_{t}Z(t) \in\mathcal{H}^2$$
The stochastic exponential process $z_t = 1 \cdot \exp \left[ \int_{0}^{t} \sigma_s d w_s - \frac{1}{2} \int_{0}^{t} \sigma_s^2 ds \right]$ satisfies the sde $d z_t = \sigma_t z_t dw_t$. For $0 \leq s \leq t < \infty$ we have that
$$ \begin{align*} \mathbb{E}\left[ z_t \vert \mathcal{F}_s \right]& = \mathbb{E}\left[ \frac{z_t}{z_s} z_s \vert \mathcal{F}_s \right] \\ & = z_s \mathbb{E} \left[ \exp \left[ \int_{s}^{t} \sigma_u d w_u - \frac{1}{2} \int_{s}^{t} \sigma_u^2 du \right] \vert \mathcal{F}_s \right] \\ & = z_s \exp \left[- \frac{1}{2} \int_{s}^{t} \sigma_u^2 du \right] \mathbb{E} \left[ \exp \left[ \int_{s}^{t} \sigma_u d w_u \right] \vert \mathcal{F}_s \right] \\ & =z_s \exp \left[- \frac{1}{2} \int_{s}^{t} \sigma_u^2 du \right] \mathbb{E} \left[ \exp \left[ \int_{s}^{t} \sigma_u d w_u \right] \right] \\ & =z_s \exp \left[- \frac{1}{2} \int_{s}^{t} \sigma_u^2 du \right] \exp \left[ \frac{1}{2} \int_{s}^{t} \sigma_u^2 du \right] \\ &= z_s. \end{align*} $$
It is not recommended to use inequalities to prove a process is a (strict) martingale. Instead, use them to show that a process is a (strict) sub or super martingale. Equality corresponds to a strict martingale, inequality corresponds to a sub or super martingale!