The following concentration inequality for the supremum of a Gaussian process indexed by a separable metric space appears here: <http://math.iisc.ac.in/~manju/GP/6-Concentration%20and%20comparison%20again.pdf>
(this, to the best of my knowledge, is one of several lecture notes prepared by Prof. Manjunath Krishnapur, IISC Bangalore), page-22 (or page 1 in the pdf), exercise-2. 


Let $X$ be a centered, continuous Gaussian process on a separable metric space $T$ and suppose that $X^\* := \sup\_{t \in T} X\_t$ is finite with probability $1$. Then, show that: $$\lim\_{x\rightarrow +\infty}\frac{1}{x^2} \log \mathbb{P}\left(X^\* \geq x\right) = -\frac{1}{2\sigma\_T^2}~,$$
where $\sigma\_T^2 := \sup\_{t \in T} \mathbb{E} (X\_t^2)$.


I need to use this result in one of my research works, so I need a proper reference, where this, or anything similar is proved. I basically want a reference, where an exponential concentration of the supremum of a continuous Gaussian process in a separable metric space (for me, Euclidean spaces suffice) in terms of its maximum variance, is proved. Can anyone help me? Thanks in advance.



Can anyone give a reference to the proof of this concentration inequality?
This follows immediately from the [Borell-TIS inequality](https://en.wikipedia.org/wiki/Borell%E2%80%93TIS_inequality). 


Indeed, for any $x>EX^\*$ this inequality yields 
$$\frac1{x^2}\,\ln P(X^\*\ge x)\le-\frac{(x-EX^\*)^2}{2\sigma\_T^2 x^2}\to-\frac1{2\sigma\_T^2}
$$
as $x\to\infty$. 


On the other hand, for any $t\in T$
$$\frac1{x^2}\,\ln P(X^\*\ge x)\ge\frac1{x^2}\,\ln P(X\_t\ge x)\to-\frac1{2\sigma\_t^2}
$$
as $x\to\infty$. It remains to choose $t\in T$ so that $\sigma\_t^2$ be arbitrarily close to $\sigma\_T^2$.


