I am confused on [the Lemma 5.7 (Lipschitz maximal inequality) here](https://web.math.princeton.edu/%7Ervan/APC550.pdf). Let me first restate the definition and the lemma:


**Def**
$\{X\_t\}\_{t\in T}$ is called Lipschitz for metric $d$ on $T$ if there exists a random variable $C$ such that
$$|X\_t-X\_s|\leq Cd(t,s),\text{ for all }t,s\in T.$$


**Lemma**
Suppose $\{X\_t\}\_{t\in T}$ is a Lipschitz process and is $\sigma^2$-subgaussian for every $T$. Then $$E[\sup\_{t\in T}X\_t]\leq \inf\_{\epsilon>0}\{\epsilon E[C]+\sqrt{2\sigma^2\log N(T,d,\epsilon)}$$


To prove the lemma, we split $X\_t$ into two parts, i.e. $$\sup\_{t\in T}X\_t\leq \sup\_{t\in T}\{X\_t-X\_{\pi(t)}\}+\sup\_{t\in T}X\_{\pi(t)}$$ then take expectation we have $$E[\sup\_{t\in T}X\_t]\leq \epsilon E[C]+\sqrt{2\sigma^2\log|N|}$$


I am a bit confused on the applicability of this lemma on random process that its $E[\sup X\_t]$ is negative.


In the proof, we use the Lipschitz bound to control $\sup\_{t\in T}\{X\_t-X\_{\pi(t)}\}\leq Cd(t,\pi(t))$. Clearly, this bound $Cd(t,\pi(t))$ is non-negative, thus the the resulting bound $ \inf\_{\epsilon>0}\{\epsilon E[C]+\sqrt{2\sigma^2\log N(T,d,\epsilon)}$ in the lemma is also non-negative.


But, what if the expectation of sup of random process is negative? Then the bound given by the lemma is very loose.


For example, I guess this process $\sum\_{i=1}^n(-100+w\_i)\sin(\theta\_i)$ where $w\_i$ is iid standard gaussian would have negative expectation of supreme.


Am I misunderstood something on this lemma?



Lipschitz maximal inequality for random process
$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\td{\tilde d}$You wrote: "I am a bit confused on the applicability of this lemma on random process that its $E[\sup X\_t]$ is negative."


One way to use this result, whether $E\sup\_{t\in T} X\_t$ is negative or not, is as follows. We have



> 
> $$E\sup\_{t\in T}X\_t\le \inf\_{\ep>0}\big\{\ep EC+\sqrt{2\si^2\ln N(T,d,\ep)}\big\} \tag{10}\label{10} $$
> given that
> $$|X\_t-X\_s|\le Cd(t,s)\text{ for all }(t,s)\in T\times T. \tag{20}\label{20}$$
> 
> 
> 


Here $N(T,d,\ep)$ is the smallest number of balls of radius $\ep$ in metric $d$ needed to cover the set $T$, and $C>0$ is a random variable.


By appropriately truncating the $X\_t$'s and $d$ (or by going through the proof of this highlighted result), here we may allow $d$ to be a generalized metric -- such that $d(t,s)$ may take the value $\infty$ for some $(t,s)\in T\times T$.


Let a set $S$ be a copy of the set $T$ **disjoint** from $T$, so that we have a bijection $f$ from $S$ onto $T$. For all $u\in U:=T\cup S$, let
$$Y\_u:=\begin{cases}
X\_u&\text{ if }u\in T, \\
-X\_{f(u)}&\text{ if }u\in S.
\end{cases}
$$
Then
$$\sup\_{u\in U}Y\_u=\max\Big(\sup\_{u\in T}X\_u,\sup\_{u\in S}(-X\_{f(u)})\Big) \\ 
=\max\Big(\sup\_{t\in T}X\_t,\sup\_{t\in T}(-X\_t)\Big)
=\sup\_{t\in T}|X\_t|. \tag{30}\label{30}
$$
Extend the generalized metric $d$ on $T\times T$ to the generalized metric $\td$ on $U\times U$ by the formula
$$\td(u,v):=\begin{cases}
d(u,v)&\text{ if }(u,v)\in T\times T, \\
d(f(u),f(v))&\text{ if }(u,v)\in S\times S, \\
\infty&\text{ otherwise }.
\end{cases}
$$
Then $|Y\_u-Y\_v|\le C\td(u,v)\text{ for all }(u,v)\in U\times U$; cf. \eqref{20}. So, by the highlighted result,
$$E\sup\_{u\in U}Y\_u\le \inf\_{\ep>0}\big\{\ep EC+\sqrt{2\si^2\ln N(U,\td,\ep)}\big\}. \tag{40}\label{40}$$


But $N(U,\td,\ep)\le2N(T,d,\ep)$ for real $\ep>0$.
So, by \eqref{30} and \eqref{40},
$$E\sup\_{t\in T}|X\_t|\le \inf\_{\ep>0}\big\{\ep EC+\sqrt{2\si^2\ln (2N(T,d,\ep))}\big\}. $$


