TITLE: Comparison of Rademacher processes
QUESTION [5 upvotes]: Suppose that $T$ is a bounded set in $\mathbb{R}^n$ and $f,g$ are two nonnegative functions such that $0\leq f(x)\leq g(x)$ for all $x\geq 0$.
Let $\epsilon_1,\epsilon_2,\dots,$ be a Rademacher sequence. Does it hold that
$$
\mathbb{E} \sup_{t\in T} \left| \sum_i \epsilon_i f(|t_i|) \right| \leq C\cdot \mathbb{E} \sup_{t\in T} \left| \sum_i \epsilon_i g(|t_i|) \right| \qquad (*)
$$
for some absolute constant $C$?
It looks intuitively true to me but I cannot find it anywhere in the literature. To compare it with the contraction principle that
$$
\mathbb{E} \sup_{t\in T} \left| \sum_i \epsilon_i \alpha_i g(|t_i|) \right| \leq \mathbb{E} \sup_{t\in T} \left| \sum_i \epsilon_i g(|t_i|) \right|, \quad |\alpha_i|\leq 1
$$
it seems that the $\alpha_i$ in my case would depend on $t$ and the proof of the contraction principle above does not seem to go through.
Question: Does (*) hold? Is there a reference in the literature or is there a particularly simple proof?

REPLY [3 votes]: $\newcommand{\ep}{\varepsilon}$
Letting $a_{it}:=f(|t_i|)$ and $b_{it}:=g(|t_i|)$, rewrite your inequality ($\ast$) as 
\begin{equation}
 E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i a_{it}\Big|\le C\,E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i b_{it}\Big|
\end{equation}
for any natural $N$ and any set $T$, 
with the condition that $0\le a_{it}\le b_{it}$ for all $i,t$. 
Let now $T:=2^{[N]}$, the power set of the set $[N]:=\{1,\dots,N\}$. Let $b_{it}:=1$ 
and 
\begin{equation}
 a_{it}:=1_{i\in t}
\end{equation}
for all $i\in[N]$ and $t\in T=2^{[N]}$. Then 
\begin{equation}
 E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i a_{it}\Big|
 \ge E\sum_{i=1}^N\max(0,\ep_i)=N/2, 
\end{equation}
whereas
\begin{equation}
 E\sup_{t\in T}\Big|\sum_{i=1}^N\ep_i b_{it}\Big|
 =E\Big|\sum_{i=1}^N\ep_i\Big|\le\sqrt{E\Big(\sum_{i=1}^N\ep_i\Big)^2}=\sqrt N. 
\end{equation}
So, your inequality ($\ast$) will fail to hold for any $C>0$ if $N>4C^2$.