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?



Comparison of Rademacher processes
$\newcommand{\ep}{\varepsilon}$
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 $N>4C^2$.


