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?


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.