If I have a $L$-Lipschitz function $\psi \colon \mathbb{R} \longrightarrow \mathbb{R}$ with $\psi(0) = 0$, and a random variable $X \sim \mathcal{N}(0, \sigma^2)$, is it true that
$$ \|\psi(X)\|^2_{\phi_2} \leq L^2 \sigma^2 \|Y\|^2_{\phi_2},$$
where $Y \sim \mathcal{N}(0,1)$ and $\| \cdot \|_{\phi_2}$ denotes the subgaussian norm?
I found a similar result which states $$ | \mathbb{E}[\psi(X)^2] - \mathbb{E}[\psi(Z)^2] | \leq L^2 |\sigma^2 - \rho^2| $$ for another random variable $Z \sim \mathcal{N}(0, \rho^2)$, but I don't see how to use this or if it is even helpful. I don't think I can set $Z$ to zero can I? And even if I could, I would be missing the term with the norm of the standard normal in the desired result... any help would be appreciated.
This seems to follow directly from the definition of the Orlicz norm. By the Lipshitz condition $|\psi(X)| \le L |X|$. Hence, since $\phi_2$ is increasing,
$$ \| \psi(X) \|_{\phi_2 } = \inf \{ u> 0 : E \left[\phi_2\left( \frac{|\psi(X)|}{u} \right)\right] \le 1\} \le \inf \{ u> 0 : E \left[\phi_2\left( \frac{L| X|}{u} \right)\right] \le 1\} = L\|X \|_{\phi_2 }. $$ Replacing $X=\sigma Y$, where $Y$ is standard normal, we see $L\|X \|_{\phi_2 }=L\sigma \| Y \|_{\phi_2 }.$