Weak Borell-TIS inequality for a subgaussian process

Question

It is a known fact (Borell-TIS inequality) that, given an almost surely bounded Gaussian centered process $X(t), t \in T$, where $T$ is a topological space, $$\mathbb{P}\{\sup_t X(t)-\mathbb{E} \sup_t X(t)>u\}\leq \exp\left(-\frac{u^2}{2\sigma_T^2}\right)\},$$ where $\sigma_T^2=\sup_{t \in T} \mathbb{E} X(t)^2$ for $u>0$. In particular, $\sigma_T$ and $\mathbb{E} \sup_t X(t)$ are finite.

Is it true that if instead $X(t), t \in T$ is not necessarily Gaussian and not necessarily centered, but almost surely bounded from above (that is,$\sup_{t \in T} X(t)<\infty$ almost surely) and subgaussian in the sense that $$\limsup_{u \to \infty} \sup_{t \in T} \frac{\log\mathbb{P}\{X(t)>u\}}{u^2}\leq -\frac{1}{2\sigma^2}$$ for some $\sigma>0$, then also $$\limsup_{u \to \infty} \frac{\log\mathbb{P}\{\sup_{t \in T} X(t)>u\}}{u^2}\leq -\frac{1}{2\sigma^2}?$$

Answer 1

Of course not. E.g., let $T$ be the set of all natural numbers, let $U$ be any random variable (r.v.), let the $Y(t)$'s be iid standard normal, and let $X(t):=\min(U,Y(t))$ for all $t$.

Then the condition in your penultimate display holds with $\sigma=1$, whereas for any real $u$ $$P(\sup_{t\in T}Y(t)\le u)\le P(\sup_{t=1}^n Y(t)\le u)=P(Y(1)\le u)^n\to0$$ as $n\to\infty$, so that $P(\sup_{t\in T}Y(t)\le u)=0$ and hence $$P(\sup_{t\in T}X(t)\le u)\le P(U\le u)+P(\sup_{t\in T} Y(t)\le u)=P(U\le u),$$ so that $P(\sup_{t\in T}X(t)>u)\ge P(U>u)$. Now let $U$ be a r.v. such that (say) $P(U>u)=1/u$ for real $u>1$. $\quad\Box$