$\newcommand{\R}{\mathbb R}\renewcommand{\th}{\theta}\newcommand{\ep}{\varepsilon} $This indeed follows immediately from Theorem 15 of Talagrand, Regularity of Gaussian processes, Acta Math. 159, No. 1-2, 99-149 (1987).

Details: The mentioned theorem by Talagrand states the following:

Let $(X_t)_{t\in T}$ be a Gaussian process, and $(Y_t)_{t\in T}$ be any other centered process indexed by the same set. Assume that for each $\th\in\R$, we have \begin{equation} E \exp \th(Y_u-Y_v)\le E \exp \th(X_u-X_v) = \exp(\th^2 d(u, v)^2/2). \quad (40) \end{equation} Then we have $E\sup_{t\in T}Y_t\le K E\sup_{t\in T}X_t$ where $K$ is a universal constant.

Let now $T$ be a set of vectors $t=(t_1,\dots,t_m)\in\R^m$ such that $\|x\|_C=\sup_{t\in T}t\cdot x$ for all $x\in\R^m$, where $\cdot$ denotes the dot product. For $t=(t_1,\dots,t_m)\in\R^m$, let $Y_t:=t\cdot Y$ and $X_t:=t\cdot X$, where $X$ is a standard normal random vector in $\R^m$. Then (40) will hold with $d(u,v)^2=(u-v)\cdot(u-v)$. So, we will have $E\|Y\|\le K E\|X\|$. So, one may take $S(\ep):=KR(\ep)$. $\quad\Box$

$\newcommand{\R}{\mathbb R}\renewcommand{\th}{\theta}\newcommand{\ep}{\varepsilon} $This indeed follows immediately from Theorem 15 of Talagrand, Regularity of Gaussian processes, Acta Math. 159, No. 1-2, 99-149 (1987).

Details: The mentioned theorem by Talagrand states the following:

Let $(X_t)_{t\in T}$ be a Gaussian process, and $(Y_t)_{t\in T}$ be any other centered process indexed by the same set. Assume that for each $\th\in\R$, we have \begin{equation} E \exp \th(Y_u-Y_v)\le E \exp \th(X_u-X_v) = \exp(\th^2 d(u, v)^2/2). \quad (40) \end{equation} Then we have $E\sup_{t\in T}Y_t\le K E\sup_{t\in T}X_t$ where $K$ is a universal constant.

Let now $T$ be a set of vectors $t=(t_1,\dots,t_m)\in\R^m$ such that $\|x\|_C=\sup_{t\in T}t\cdot x$ for all $x\in\R^m$, where $\cdot$ denotes the dot product. For $t=(t_1,\dots,t_m)\in\R^m$, let $Y_t:=t\cdot Y$ and $X_t:=t\cdot X$, where $X$ is a standard normal random vector in $\R^m$. Then (40) will hold with $d(u,v)^2=(u-v)\cdot(u-v)$. So, we will have $E\|Y\|_C\le K E\|X\|_C$. So, one may take $S(\ep):=KR(\ep)$. $\quad\Box$

This indeed follows immediately from Theorem 15 of Talagrand, Regularity of Gaussian processes, Acta Math. 159, No. 1-2, 99-149 (1987).

$\newcommand{\R}{\mathbb R}\renewcommand{\th}{\theta}\newcommand{\ep}{\varepsilon} $This indeed follows immediately from Theorem 15 of Talagrand, Regularity of Gaussian processes, Acta Math. 159, No. 1-2, 99-149 (1987).

Details: The mentioned theorem by Talagrand states the following:

Let $(X_t)_{t\in T}$ be a Gaussian process, and $(Y_t)_{t\in T}$ be any other centered process indexed by the same set. Assume that for each $\th\in\R$, we have \begin{equation} E \exp \th(Y_u-Y_v)\le E \exp \th(X_u-X_v) = \exp(\th^2 d(u, v)^2/2). \quad (40) \end{equation} Then we have $E\sup_{t\in T}Y_t\le K E\sup_{t\in T}X_t$ where $K$ is a universal constant.

Let now $T$ be a set of vectors $t=(t_1,\dots,t_m)\in\R^m$ such that $\|x\|_C=\sup_{t\in T}t\cdot x$ for all $x\in\R^m$, where $\cdot$ denotes the dot product. For $t=(t_1,\dots,t_m)\in\R^m$, let $Y_t:=t\cdot Y$ and $X_t:=t\cdot X$, where $X$ is a standard normal random vector in $\R^m$. Then (40) will hold with $d(u,v)^2=(u-v)\cdot(u-v)$. So, we will have $E\|Y\|\le K E\|X\|$. So, one may take $S(\ep):=KR(\ep)$. $\quad\Box$