Let $B = (B_t)_{t \in [0,T]}$ be a Brownian motion and $\alpha = (\alpha_t)_{t \in [0,T]} $ be progressively measurable. Let $$ X = \int_0^\cdot \alpha_t dt + \int_0^\cdot \alpha_tdB_t. $$ If $\alpha$ satisfies $$ E\left[ \int_0^T |\alpha_t|^2 dt \right] < \infty, $$ then $X$ is bounded in $L^2.$ Is the reverse implication also true?
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Yeah, it’s true – ofey73 Dec 29 '18 at 22:17
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Note that you need some integrability condition on $\alpha$ to ensure that the integrals $\int_0^T \alpha_t , dt$ and $\int_0^T \alpha_t , dB_t$ exist; this means that already the well-definedness of $X$ gives information on the integrability of $\alpha$. – saz Dec 30 '18 at 08:22