In the proof of Corollary 5.7 in the following link:


<https://arxiv.org/pdf/1610.05200.pdf>


the author claims that $E[\lVert X\rVert^2] \lesssim E[\lVert X\rVert ]^2$ for the standard normal distribution on $\mathbb{R}^n$.


I wonder if this result is still valid for cenetered Gaussian distributions on any infinite dimensional separable Banach space $V$ .


Moreover, the covariance for the measure is a symmetric nonnegative bilinear form on $V^\*$. If the above inequality is still valid on $v$, how would the covariance affect it?


Could anyone please provide any information or reference?



For centered Gaussian measures, is $E[\lVert X\rVert^2] \lesssim E[\lVert X\rVert ]^2$ true in infinite dimensions as well?
$\newcommand{\si}{\sigma}$Yes, we have
\begin{equation\*}
 E\|X\|^2\le c(E\|X\|)^2 \tag{1}\label{1}
\end{equation\*}
for
\begin{equation\*}
 c:=1+2\pi
\end{equation\*}
and any centered Gaussian random vector $X$ in any separable Banach space $V$.


This follows from the [Borell--Tsirelson--Ibragimov--Sudakov inequality](https://en.wikipedia.org/wiki/Borell%E2%80%93TIS_inequality):  

\begin{equation\*}
 P(\|f\|\_T\ge E\|f\|\_T+u)\le\exp-\frac{u^2}{2\si\_T^2}
\end{equation\*}
for real $u\ge0$, where $(f\_t)\_{t\in T}$ is a centered Gaussian process, $\|f\|\_T:=\sup\_{t\in T}|f\_t|$, and
\begin{equation\*}
 \si\_T:=\sqrt{\sup\_{t\in T}Ef\_t^2}. 
\end{equation\*}


Indeed, let $T=B^\*$, the unit ball in the dual $V^\*$ to $V$, with $f\_t:=t(X)$ for $t\in B^\*$, so that $\|f\|\_T=\|X\|$ and
\begin{equation\*}
 \si\_T=\si:=\sqrt{\sup\_{t\in B^\*}Et(X)^2}=\sqrt{\frac\pi2\,\sup\_{t\in B^\*}(E|t(X)|)^2}
 \le\sqrt{\frac\pi2}\,E\|X\|. \tag{2}\label{2}
\end{equation\*}
So, for real $u\ge0$,
\begin{equation\*}
 P(\|X\|\ge E\|X\|+u)\le\exp-\frac{u^2}{2\si^2}. 
\end{equation\*}
(More immediately, the latter inequality is inequality (4.4) in [Ledoux's book](https://link.springer.com/content/pdf/10.1007/BFb0095676.pdf).)


So,
\begin{equation\*}
\begin{aligned}
 E\|X\|^2&=E\int\_0^\infty 2w\,dw\,1(\|X\|\ge w) \\ 
 &=\int\_0^\infty 2w\,dw\,P(\|X\|\ge w) \\ 
 &\le\int\_0^{E\|X\|} 2w\,dw+\int\_{E\|X\|}^\infty 2w\,dw\,\exp-\frac{(w-E\|X\|)^2}{2\si^2} \\ 
&=(E\|X\|)^2+2\si^2+\sqrt{2\pi}E\|X\|\si. 
\end{aligned}
\end{equation\*}


Now \eqref{1} follows from \eqref{2}. $\quad\Box$


