Is there an example of a Lipschitz function of a Gaussian vector for which $f(Z)-\mathbb{E}[f(Z)]$ is not sub-Gaussian Definitions:
A random variable $X$ is called sub-Gaussian with parameter $\sigma^2$ if there exists $\sigma \in \mathbb{R}$ such that
$$\forall \lambda \in \mathbb{R} \quad \mathbb{E}[e^{\lambda X}]\le e^{\frac{\sigma^2 \lambda^2}{2}}$$
A function $f:\mathbb{R}^n\to\mathbb{R}$ is called $L$-Lipschitz if $|f(X)-f(Y)|\le L \|x-y\|$.
Note:Usually, We take the $L_2$-norm for the RHS.

Theorem:
If $Z_1,Z_2,\dots,Z_n$ are standard normal variables ($Z_i\sim \mathcal{N}(0,1)$), and $f:\mathbb{R}^n\to\mathbb{R}$ is a $L$-Lipschitz function, then $f(Z)-\mathbb{E}[f(Z)]$ is sub-Gaussian. (Taken from Concentration of Measure 3 by Jarosław Błasiok, Page 21)

Question:
Assume that we apply the same $L$-Lipschitz function $f:\mathbb{R}^n\to\mathbb{R}$ on $Z_1,Z_2,\dots,Z_n$ such that $Z=(Z_1,Z_2,\dots,Z_n)\sim\mathcal{N}(0,\Sigma)$, meaning that the random variables are normal (but not standard) and we may have $\mathbb{E}[Z_i Z_j] \neq 0$ for some $i,j$.
Can we define $f$ such that $f(Z)-\mathbb{E}[f(Z)]$ is no more sub-Gaussian?
To clarify: We can define $f$ and $\Sigma$ (the covariance matrix of the Gaussian random vector) as we would like to. The question says: "Can we select $f$ and $\Sigma$ such that $f(Z)-\mathbb{E}[f(Z)]$ is not sub-Gaussian?"

My work:
I've found out that the theorem may not hold if $f$ is Lipschitz but not convex (in this link, page 212). However, I believe in my case, reaching a counterexample should be much easier. I couldn't find any theorem which states that if the variables $Z_i$ have correlation, then $f(Z)-\mathbb{E}[f(Z)]$ is sub-Gaussian. So, I thought maybe if we define $Z_i$'s somehow that they are correlated, then the counterexample could be made. The difficulty for me is that for each Lipschitz function that I define, I should prove that
$$
\forall \sigma\in\mathbb{R} \quad \exists \lambda \in \mathbb{R} \quad \mathbb{E}[e^{\lambda (f(Z)-\mathbb{E}(f(Z))}] \gt e^{\frac{\sigma^2 \lambda^2}{2}}
$$
And proving this is where I can't proceed further. I'm not sure but maybe my approach is wrong.
Note: There is a related question here, and also another one here.

If $Z_i\sim N(,\Sigma)$ then $Z_i=\Sigma^{1/2}Y_i$ are $N(0,1)$, where $\Sigma^{1/2}$ is any square root of $\Sigma$.  So $f(Z_1,\dots,Z_n)=(f\circ g)(Y_1,\dots,Y_n)$, where $g$ is multiplication by $\Sigma^{1/2}$.
Clearly $f\circ g$ is still Lipschitz. If $\lambda$ is the largest eigenvalue of $\Sigma$, the Lipschitz constant is bounded above by $L\sqrt{\lambda}$.
So, no. But you do need to take the extra factor of $\sqrt{\lambda}$ into account, especially if you're going to look at a collection of $f$s and $Z$s that might have different $\Sigma$s.
