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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?

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

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  • $\begingroup$ Thanks! I'm aware of the contraction principle and was curious whether something like $(*)$ would hold without the contraction assumption. I constructed an example that $(*)$ fails when $C=1$ (take $T=\{(t,\sqrt{1-t^2}): t\in[0,1]\}$, $f(x)=x$ and $g(x)=1$ on $[0,1]$) and I am more or less convinced that I shouldn't expect $(*)$ to hold. $\endgroup$ – user58955 Apr 21 '19 at 3:49
  • $\begingroup$ Nice example, thanks! $\endgroup$ – user58955 Apr 21 '19 at 4:29
  • $\begingroup$ Glad to have been able to help. $\endgroup$ – Iosif Pinelis Apr 21 '19 at 4:30

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