Convergence using Laplace transform Consider the sequence $(Y_n)_{n\ge 1}$ of iid and integrable random variables, which are not almost surely constant.
Let $\delta>0$ and for all $\lambda\in(-\delta,\delta)$ it holds $$L(\lambda):=E[e^{\lambda Y_1}]<\infty$$
Define $$\psi:(-\delta,\delta)\rightarrow\mathbb{R},\quad\psi(\lambda):=\log(L(\lambda))$$
For $n\ge 0$ define
$$X_n^\lambda:=e^{\lambda\sum_{i=1}^nY_i-n\psi(\lambda)}$$
I was able to show, that $\psi$ is strictly convex on $(-\delta,\delta)$ by using the inequality of Cauchy-Schwarz and that $e^{\frac{\lambda_1}{2}Y_1}$ and $e^{\frac{\lambda_2}{2}Y_1}$ are linear independent for $\lambda_1\ne\lambda_2$. snarksi contributed a nice answer how to show that $E[\sqrt{X_n^\lambda}]\rightarrow 0$ for $n\rightarrow \infty$.
At last I want to show that $X_n^\lambda\rightarrow 0$ almost surely for $n\rightarrow \infty$ and $\lambda\ne 0$.
Guess: This looks like I need to use the strong law of large numbers, but I am not sure about this yet.
Thanks!
 A: You made a mistake in bringing in the factor of 2. Indeed,
\begin{align*}
E[ (X_n^\lambda)^{1/2} ] &= E[e^{\frac{1}{2}(\lambda\sum_{i=1}^n Y_i - n \psi(\lambda))}] \\
&= E[e^{\frac{\lambda}{2}Y_1}]^n e^{-\frac{n}{2} \psi(\lambda)} \\
&=e^{n \psi(\lambda/2) - n\psi(\lambda)/2}.
\end{align*}
So it suffices to show that $\psi(\lambda/2) - \psi(\lambda)/2 < 0$, as then the argument of the exponential is negative and you can send $n \rightarrow \infty$. 
Since the functions are not almost surely constant, the variance is positive and $\psi''(0) > 0$. The function $\psi$ is also smooth on $(-\delta, \delta)$, and $\psi(0) = 0$. We have, for sufficiently small $\lambda > 0$,
\begin{align*}
\psi(\lambda/2) - \psi(\lambda)/2 &= \psi(0) + \frac{\lambda}{2} \psi'(0) + \frac{1}{2}\cdot \frac{\lambda^2}{4} \psi''(0) \\
&-\frac{1}{2}(\psi(0) + \lambda \psi'(0) + \frac{1}{2}\lambda^2 \psi''(0))+O(\lambda^3) \\
&= -\frac{\lambda^2}{8}\psi''(0)+O(\lambda^3).
\end{align*}
Choosing $\lambda > 0$ sufficiently small will guarantee that the expression is negative. 
Edit: previous argument was not correct, strict convexity does not guarantee $\psi''(0) > 0$.
