Let $X = (X\_1, \ldots, X\_d) \in \mathbb{R}^d$ be a mean-zero Gaussian random vector with identity covariance matrix. Are there upper bounds for 
$$E \left(\|X\|\_{\infty}^k \right)$$ for $k=1, \ldots, 6$ ? It is easy to see for example for $k=1$, the upper bounded is of the order $\sqrt{2\log(2d)}$. In particular for lower order moments (like when $k=6$) can one still obtain upper bounds that are logarithmic in $d$ ?.



Moments of maximum of independent Gaussian random variables
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Let $M:=\|X\|\_{\infty}$. Then for $x>0$
\begin{equation}
 \PP(M>x)\le d\,\PP(|X\_1|>x)=2d\, G(x)<2d\, f(x)/x,
\end{equation}
where $G:=1-F$, $F$ is the standard normal cdf, and $f$ is the standard normal cdf. Also, trivially $\PP(M>x)\le1$. So,
\begin{equation}
 \E M^k=\int\_0^\infty kx^{k-1} \PP(M>x)dx\le I\_1+2kdI\_2,
\end{equation}  

where $x\_d:=\sqrt{2\ln d}$, 
\begin{equation}
 I\_1:=\int\_0^{x\_d} kx^{k-1}dx=x\_d^k=(2\ln d)^{k/2},
\end{equation}
\begin{equation}
 I\_2:=\int\_{x\_d}^\infty x^{k-2}f(x)\,dx
 \sim x\_d^{k-3}f(x\_d)\ll(\ln d)^{(k-3)/2}/d=o((\ln d)^{k/2}/d), 
\end{equation}
where the asymptotic equivalence $\sim$ follows by l'Hospital's rule. So,
\begin{equation}
 \E M^k\le(2\ln d)^{k/2}(1+o(1));
\end{equation}
the convergence everywhere here is as $d\to\infty$. 




---


This asymptotic bound is in fact the best possible one. Indeed, take any $c\in(0,2)$. Then for $y=y\_{d,c}:=\sqrt{c\ln d}$ we have 
\begin{multline}
 \PP(M>y)=1-(1-G(y))^d\ge1-\exp\{-d\,G(y)\} \\ 
 =1-\exp\Big\{-d\,\exp\Big(-\frac{y^2}{2+o(1)}\Big)\Big\}
 =1-\exp\big\{-d^{1-c/2+o(1)}\big\}\to1
\end{multline}
and hence $\PP(M>x)\to1$ uniformly over all $x\in[0,y]$. So, 
\begin{multline}
 \E M^k=\int\_0^\infty kx^{k-1} \PP(M>x)dx\ge \int\_0^y kx^{k-1}\,dx\,(1-o(1)) \\ 
 =y^k(1-o(1))=(c\ln d)^{k/2}(1+o(1)). 
\end{multline}
Taking now $c$ to be arbitrarily close to $2$, we conclude that 
\begin{equation}
 \E M^k=(2\ln d)^{k/2}(1+o(1)). 
\end{equation}


