Let $X$ be a mean-zero random variable whose moment generating function is bounded by a symmetric exponential function, e.g: $$ \mathbb{E}[e^{\lambda X}] \leq \exp(c|\lambda|) $$ with $c > 0$.
Question: Is it true that $X$ must be bounded?
I asked one of my professors this and he told me it was the case, but I am not sure if it is true and if so how it can be proved.
For proofs like this it seems like one would taylor expand both sides and take $\lambda$ to $0$ to get some sort of bounds, but I don't think this method works here because $e^{c|\lambda|}$ decays slower than $e^{c\lambda^2}$, and the variables for which the second holds are the sub-gaussians, which of course include some unbounded variables (e.g. the usual Gaussian). I would imagine instead one should look at $\lambda$ large, but I am not sure how to carry out the argument.
For $t\in\mathbb R$, you have by Markov's inequality
$$\mathbb P(X\ge t) = \mathbb P\left(\mathrm e^{\lambda X}\ge \mathrm e^{\lambda t}\right)\le\frac{\mathbb E(\mathrm e^{\lambda X})}{\mathrm e^{\lambda t}}$$
for all $\lambda\in\mathbb R$.
Now by assumption,
$$\mathbb E(\mathrm e^{\lambda X})\le \mathrm e^{c\lvert\lambda\rvert}.$$
Therefore, letting $\lambda\to\infty$, we conclude that $\mathbb P(X>c)=0$.
The same argument works for $\mathbb P(X<-c)$ since this is $\mathbb P(-X>c)$, and $-X$ has the same bound on its moment generating function as $X$.