I've done the standard proof of Talagrand's contraction lemma for Gaussian processes (see Exercise 7.2.13 in Vershynin's High-Dimensional Probability) using the Sudakov-Fernique inequality as suggested. Here's my proof:
We let $T\subset \mathbb{R}^n$ be bounded and $Z$ be a standard $n$-variate normal vector. Letting $f : \mathbb{R}^n\to \mathbb{R}^n$ be a contraction mapping, we will show that $$\mathbb{E}[\sup_{t\in T} \langle Z, f(t)\rangle]\leq \mathbb{E}[\sup_{t\in T} \langle Z, t\rangle]$$ We recall the Sudakov-Fernique inequality: for mean-zero Gaussian processes $(X_t)_{t\in T}$ and $(Y_t)_{t\in T}$, if for all $s, t\in T$ we have $\mathbb{E}[(X_t-X_s)^2]\geq \mathbb{E}[(Y_t-Y_s)^2]$, then $$\mathbb{E}[\sup_{t\in T} X_t]\geq \mathbb{E}[\sup_{t\in T} Y_t]$$ We define $X_t := \langle Z, t\rangle$ and $Y_t := \langle Z, f(t)\rangle$, and we note that $$(X_t-X_s)^2 = \left[\langle Z, t-s\rangle\right]^2 = (t-s)^{\mathrm{T}}ZZ^{\mathrm{T}}(t-s)$$ Recalling that $\mathbb{E}[ZZ^{\mathrm{T}}] = I$, we have $$\mathbb{E}[(X_t-X_s)^2] = \lvert t-s\rvert^2$$ Similarly, we will have $$\mathbb{E}[(Y_t-Y_s)^2] = \lvert f(t)-f(s)\rvert^2$$ By the fact that $f$ is a contraction, we therefore have that $\mathbb{E}[(Y_t-Y_s)^2]\leq \mathbb{E}[(X_t-X_s)^2]$, which directly implies the problem inequality as the result of Sudakov-Fernique.
However, I'm wondering if there's a way to leverage the existence of a fixed point of $f$ to achieve the same result without using Sudakov-Fernique. I think this proof would go something like rewriting the problem inequality as $$\mathbb{E}[\sup_{t\in T} \langle Z, f(t)-f(s)\rangle]\leq \mathbb{E}[\sup_{t\in T} \langle Z, t-s\rangle]$$ for $f(s) = s$ and then using again the fact that $f$ is a contraction, but I'm not quite seeing how to proceed. Does anybody else have any ideas? Thank you!
