Let $B_t$ be a standard Brownian motion, and let $a_t$ be progressively measurable and so that $a_t \in [0,1]$ almost surely. It's well known that $e^{\theta B_t - \frac{1}{2} \theta^2 t}$ is a martingale for constants $\alpha.$
Let $M_t = a_t B_t$. I want to show now that $e^{\theta M_t - \frac{1}{2} \theta^2 t}$ is a super-martingale. Intuitively, this should be true because $a_t \in [0,1]$, and the function $e^x$ is convex, so the large positive fluctuations of the Brownian motion are down-weighted. I tried using Ito's lemma to write $$e^{\theta M_t} - 1 = \int_0^t e^{\theta M_s} \theta dM_s + \frac{\theta^2}{2} \int_0^t e^{\theta M_s} a_s^2 ds,$$ though I feel like this is kind of circular, since I don't quite know how to show that this last term is less than $e^{\theta^2t/2}$ (in expectation, which would show that this process is a super-martingale).
Any advice would be great!
First Round
To have an easier life I will modify slightly your term in the exponential to be more of a standard form when $\alpha$ is not constant.
From the integration-by-parts formula we get $$ \theta M_t=\theta\alpha_tB_t=\theta\int_0^t\alpha_s\,dB_s+\theta\int_0^tB_s\,d\alpha_s=:X_t+A_t\,. $$ From the Ito formula we get \begin{align} e^{\theta M_t-\frac12\langle X\rangle_t}&=1+\int_0^te^{\theta M_s-\frac12\langle X\rangle_s}\,dX_s+\int_0^te^{\alpha_sM_s-\frac12\langle X\rangle_s}\,dA_s \end{align} where $$ \langle X\rangle_t=\theta^2\int_0^t\alpha_s^2\,ds\,. $$ Since $\int_0^te^{\theta M_s-\frac12\langle X\rangle_s}\,dX_s$ is a martingale it follows from the uniqueness of the Doob-Meyer decomposition that $e^{\theta M_t-\frac12\langle X\rangle_t}$ is a supermartingale if and only if $$ \int_0^te^{\theta M_s-\frac12\langle X\rangle_s}\,dA_s $$ is decreasing. But this is not the case since the integrand is $$ e^{\theta M_s-\frac12\langle X\rangle_s}\,\theta\,\color{red}{B_s}\,\alpha'_s $$ which is not guaranteed to stay negative.
Second Round
If you want instead of $\frac12\langle X\rangle_t$ the compensator $\frac12\theta^2\,t$ we get from the Ito formula
\begin{align} e^{\theta M_t-\frac12\theta^2t}&=1+\int_0^te^{\theta M_s-\frac12\theta^2s}\,dX_s\\[2mm] &\quad+\underbrace{\int_0^te^{\alpha_sM_s-\frac12\theta^2s}\,dA_s-\frac12\int_0^te^{\theta M_s-\frac12\theta^2s}\;\theta^2\,ds+\frac12\int_0^te^{\theta M_s-\frac12\theta^2s}\,d\langle X\rangle_s}\,. \end{align} The conjecture that $e^{\theta M_t-\frac12\theta^2t}$ is a supermartingale is now equivalent to the underbraced term being decreasing. This is equivalent to the sum of the integrands being negative. That is: to $$ \theta\,B_s\,\alpha'_s-\theta^2\frac{1-\alpha^2_s}{2} $$ being negative which is not quaranteed, unless $\alpha$ is constant and in $[0,1]\,.$