Let $Y_i$ be a sequence of independent and identically distributed non-degenerate variables, $X_n$ their n-th partial sum, $\phi(u)=\log E[\exp(uY_1)]$ and define $M_n(u)=\exp(uX_n-n\phi(u))$.
Then $M_n(u)$ for u such that $\phi(u)\in\mathbb{R}$ is a martingale, converges almost surely, and for $u\ne0$ it converges almost surely to zero.
Verifying that $M_n(u)$ is a martingale is just a matter of plugging things in and calculating. In addition, $M_n(u)$ converges almost surely by the martingale convergence theorem since it is $L^1$-bounded (their expectation is $1$).
Now I am having trouble verifying the convergence to zero. I found that, noting that with $Z_i:=uY_i-\phi(u)$ we can write $\ln M_n(u)$ as the $n$-th partial sum of the $Z_i$'s. Noting the independence of the $Z_i$'s and that $EZ_i=EuY_i-\phi(u)\lt 0$ by Jensen, we have, by the law of large numbers, almost sure convergence of $\ln M_n(u)/n$ to $EZ_1\lt 0$. Now I feel like that this should enable me to show that the $M_n(u)$ must converge to zero, but I can not see how. Any help would be very much appreciated.
Jensen is the right tack, but apply it to the exponential, not the exponent: The limit $M_\infty(u)$ is a.s. equal to $\exp(uX_1-\phi(u))\cdot M'_\infty(u)$, where $M'_\infty(u):=\lim_n \exp((X_n-X_1)-(n-1)\phi(u))$, and the two factors are independent. Take square roots and then expectations. Either $\Bbb E[\sqrt{M_\infty(u)}]=0$ (in which case you are done) or $\Bbb E[\sqrt{\exp(uX_1-\phi(u)}]=1$. This latter equality would violate the strict concavity of the square root function unless $X_1$ were degenerate, which you have assumed it not to be.