Bound on the MGF of the square of a random variable

Question

Suppose for a random variable $X$, one knows that $ \mathbb{E}\exp(\lambda^2 X^2) \leq \exp(C \lambda^2)$ for all $\lambda \in \mathbb{R}$ and some constant $K$. I would like to show that the random variable is bounded. It is guaranteed to be subgaussian, but one only needs the convergence of the MGF of $X^2$ around some neighborhood of $0$. Here, we have something much stronger.

Answer 1

For all $C'>C$ Markov's inequality gives

$$\mathbb P[X^2 \geq C']\leq\frac{\mathbb E\exp(\lambda^2X^2)}{\exp(\lambda^2C')}\leq \exp(\lambda^2(C-C')) \to 0$$ as $\lambda\to\infty.$ Hence $\mathbb P[X^2> C]=0.$