Comparison of Rademacher processes

Question

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?

Answer 1

$\newcommand{\ep}{\varepsilon}$ 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 $N>4C^2$.