the problem I am considering is as follows:
For every probability space $(\Omega , \mathcal F , \mathbb P)$; every separable $\mathbb R$-Banach space $(V, ||\cdot||_{V})$, and every centered Gaussian distributed random variable $X: \Omega \to V$, it holds that:
$limsup_{\epsilon \to 0^{+}} sup_{r \in \mathbb R} \Big[e^{\epsilon r^{2}} \mathbb P \Big(||X||_{V} \geq r\Big)\Big]$ is finite.
What I have tried is to employ Markov's Inequality.
Suppose, $r > 0$. (This positivity enables us to employ Markov's inequality.)
By Markov's inequality, we have:$ \quad e^{\epsilon r^{2}} \mathbb P \Big(||X||_{V} \geq r\Big) \leq \mathbb E(||X||_{V}) \frac {e^{\epsilon r^{2}}}{r}$.
Consequently, first taking the supremum on both side over $r \in \mathbb R$, and then letting $\epsilon \to 0^{+}$, we can conclude that the desired limit is finite since $\quad limsup_{\epsilon \to 0^{+}} sup_{r \in \mathbb R} \mathbb E(||X||_{V}) \frac {e^{\epsilon r^{2}}}{r}$ is $0$.
Is my argument okay ??
Please let me know if I am missing something!!
Thanks in advance!!!
P.S. (EDIT):- As ntt commented, if $r=0$, then the result is false.
As mentioned in the comments, this is obviously false as written, but is true if we restrict the supremum to be over $r\ge0$. As the comments also noted, Fernique's theorem is the way to go. Fernique's theorem states that there exists $\alpha>0$ such that $\mathbb E[e^{\alpha\|X\|^2}]<\infty$. Then for $0<\epsilon<\alpha$ and $r>0$, we have $$e^{\epsilon r^2}\mathbb P(\|X\|\ge r)=e^{\epsilon r^2}\mathbb P(e^{\alpha\|X\|^2}\ge e^{\alpha r^2})\le e^{-(\alpha-\epsilon)\|r\|^2}\mathbb E[e^{\alpha\|X\|^2}]\le\mathbb E[e^{\alpha\|X\|^2}]$$ where the second to last step uses Markov's inequality. Taking suprema over $r\ge0$ and letting $\epsilon\to0$ completes the proof. Note that you could also fairly easily show that $$\lim_{\epsilon\to0}\limsup_{r\to\infty}e^{\epsilon r^2}\mathbb P(\|X\|\ge r)=0.$$