Let $\Omega$ be a compact subset of $\mathbb R^p$ and let $f\_1,\ldots,f\_k$ be zero mean identically distrubuted Gaussian processes on $\Omega$ such that $f\_1(x),\ldots,f\_k(x)$ are independent $x \in \Omega$. Thus $f:=(f\_1,\ldots,f\_k)$ can be seen as a vector-valued Gaussian process with iid components at each point in space. Let $\lambda:= \mathbb E[f\_1(0)^2]$. Now, given $x \in \Omega$, define $\nu(x) := \|f(x)\| := (\sum\_{j=1}^k f\_j(x)^2)^{1/2}$. I'm interested in concentration inequalities for $\|\nu\|\_\infty := \sup\_{x \in \Omega}\nu(x)$.


In the special case where $k=1$, the [Borell-TIS inequality](http://Kernel%20methods%20for%20deep%20learning) kicks-in to give


$$
\begin{split}
&\forall u\ge 0,\; \mathbb P(\|\nu\|\_\infty > \mathbb E[\|\nu\|\_\infty] + u) \le e^{-u^2/(2\sigma^2)},\\
&\text{ where }\sigma^2 := \sup\_{x \in \Omega}\mathbb E[\nu(x)^2] = \mathbb E[\nu(0)^2] = \lambda
\end{split}
$$



> 
> **Question.** How to get concentration ienqualities for $\|\nu\|\_\infty$ when $k \ge 2$ ?
> 
> 
> 


Some notes on the case $k \to \infty$
-------------------------------------


Define a random field $Z\_k$ on $\Omega$ by setting $Z\_k(x):= \dfrac{1}{\sqrt{2\lambda k}}(\|f(x)\|^2-\lambda k) = \dfrac{1}{\sqrt{2\lambda k}}(\sum\_{j=1}^kf\_j(x)^2-\lambda k) $, for all $x \in \Omega$.



> 
> I don't know if there is an appropriate notion of CLT which could kick-in here, to give the limiting distribution of the random field $Z\_k$.
> 
> 
> 



Concentration inequality for the supremum of $L_2$ norm of a vector-valued Gaussian process with iid components
Consider the real-valued centered Gaussian process $(X\_{t,a}\colon(t,a)\in T\times B\_k)$, where
$$X\_{t,a}:=\sum\_{j\in[k]}a\_j f\_j(t),$$
$T:=\Omega$, $B\_k$ is the unit ball in $\mathbb R^k$, and $[k]:=\{1,\dots,k\}$. Then
$$\|\nu\|\_\infty=\|X\|\_\infty:=\sup\{|X\_{t,a}|\colon t\in T, a\in B\_k\}$$
and
$$EX\_{t,a}^2=\sum\_{j\in[k]}a\_j^2 Ef\_j(t)^2\le\sup\_{t\in T,j\in[k]}Ef\_j(t)^2=:\sigma^2.$$
So, by the Borell--TIS inequality for the process $(X\_{t,a})$, for all real $u\ge0$
$$
P(\|\nu\|\_\infty>E\|\nu\|\_\infty+u)=P(\|X\|\_\infty>E\|X\|\_\infty+u)\le e^{-u^2/(2\sigma^2)}.
$$


