Let $T \subset \mathbb R^n$, and assume it's a finite set if that helps. Consider the symmetric Gaussian process $(X\_t)\_{t\in T}$ defined by $X\_t = \langle G, t\rangle$, where $G$ is a standard Gaussian in $\mathbb{R}^n$ and $\langle\cdot, \cdot \rangle$ is the standard inner product. In other word, $\mathbb{E} X\_t = 0$, and $\mathbb{E} (X\_s - X\_t)^2 = \|s-t\|\_2^2$ for all $s,t\in T$



> 
> I want to know what happens to $\mathbb E \sup\_{t\in T} X\_t$ when we apply a contraction map to the index set $T$?
> 
> 
> 


I.e. take a contraction map $f:\mathbb R^n \to \mathbb R^n$, $\|f(x)-f(y)\|\_2 \leq \|x-y\|\_2$ for all $x,y\in\mathbb R^n$, and define the process $(Y\_t)\_{t\in T}$ as $Y\_t = \langle G, f(t)\rangle$. (Equivalently, $\mathbb{E} Y\_t = 0$, and $\mathbb{E} (Y\_s - Y\_t)^2 = \|f(s)-f(t)\|\_2^2$.) An application of Talagrand's majorizing measures theorem shows that for some absolute constant $L$,


$$
\mathbb E \sup\_t Y\_t \leq L\ \mathbb E \sup\_t X\_t.
$$


Here are two questions:



> 
> 1. Is the bound true with $L=1$?
> 2. Is there a direct proof that does not use Talagrand's theorem?
> 
> 
> 


I am even interested in the simple map $f(t\_1, \ldots, t\_n) = (|t\_1|, \ldots, |t\_n|)$. Is there a simple direct proof in this case? Is the inequality true with constant 1 in this case?



Does a Gaussian process shrink under a contraction map
It is true, and it follows from the following fact, sometimes referred to as Sudakov-Fernique inequality, sometimes as Slepian-Fernique lemma:


Assume that $(X\_t)$ and $(Y\_t)$ are two centered Gaussian processes. If $\|Y\_s - Y\_t\|\_2 \leqslant \|X\_s - X\_t\|\_2$ then $\mathbb{E} \sup Y\_t \leqslant \mathbb{E} \sup X\_t$.


In this inequality the index set $T$ is assumed to be countable so that we do not have any problems with measurability.


