Controll of all moments of a gaussian random variable given that the second is finite

Question

I need to prove the following: Let $X$ be a gaussian random variable with mean $0$. Then there exists a constant $c_p$ such that $$\mathbb{E}[|X|^p] \leq c_p (\mathbb{E}[|X|^2])^{\frac{p}{2}}.$$ I have tried that with the Hölder Inequality and got a result, but since I need $p>2$ Hölder does not apply here and i unfortunatelly have no more ideas. Please help.

Answer 1

Assuming the mean is $0$ we have a simple argument for this. Let $\sigma^{2}$ be the variance of $X$ and $Y=\frac 1 {\sigma} X$. Then $Y \sim N(0,1)$. The inequality becomes $\sigma^{p} E|Y|^{p} \leq c_p (\sigma ^{2})^{p/2}$ which obviously holds with $c_p=E|Y|^{p}$. In fact we have equality!

Note that $E|Y|^{p}=\frac 1 {\sqrt {2\pi}} \int |y|^{p} e^{-y^{2}/2} dy$ which depends only on $p$.