will vanish. Consequently, $$\begin{aligned}
\Vert \xi (x) \Vert _{\alpha , 2} ^{2} &=&\sum_{m\geq 0} (1+m) ^{\alpha } m!\frac{1}{m+1}\frac{1}{m+2}  \sum_{k,l=1; k\not= l}^{m+2} \int_{([0, 1] ^{2} ) ^{m+2} } dx_{1}dy_{1}..dx_{m+2} dy_{m+2} \\
&&\times (a_{m}^{x} (x_{k}, y_{l}))^{2}  1_{[0,x_{k}]}^{\otimes m+1} (x_{1}, .., \overline{x_{k}},.., x_{m+2} )1_{[0,y_{l}] }^{\otimes m+1} (y_{1},.., \overline{y_{l}},.., y_{m+2} )\\
&=&\sum_{m\geq 0} (1+m) ^{\alpha } m!\frac{1}{m+1}\frac{1}{m+2}  \sum_{k,l=1; k\not= l} ^{m+2}\int_{0}^{1} dx_{k} \int_{0}^{1}dy_{l}(a_{m}^{x} (x_{k}, y_{l}))^{2}  \\
&&\times \int_{[0,x_{k}]^{m+1}} dx_{1}...\overline{dx_{k}} ..dx_{m+2} \int_{[0,y_{l}]^{m+1}} dy_{1}...\overline{dy_{l}} ..dy_{m+2}\\
&=&\sum_{m\geq 0} (1+m) ^{\alpha } m!\frac{1}{m+1}\frac{1}{m+2}  \sum_{k,l=1; k\not= l}^{m+2} \int_{0}^{1} dx_{k} \int_{0}^{1}dy_{l}(a_{m}^{x} (x_{k}, y_{l}))^{2}x_{k}^{m+1}y_{l} ^{m+1}.\end{aligned}$$ Using the expression (\[amx\]) of the coefficient $a_{m}^{x}$ we have $$(a_{m}^{x} (x_{k}, y_{l}))^{2}x_{k}^{m+1}y_{l} ^{m+1} = \left( p_{x_{k}y_{l}}(x) H_{m} \left( \frac{x}{\sqrt{x_{k}y_{l}}} \right)\right) ^{2} x_{k}y_{l}.$$ Therefore $$\begin{aligned}
\Vert \xi (x) \Vert _{\alpha , 2} ^{2}=\sum_{m\geq 0} (1+m) ^{\alpha } m!\int_{0}^{1} \int_{0}^{1} dudv \left( p_{uv}(x) H_{m} \left( \frac{x}{\sqrt{uv} } \right) \right)^{2}uv.\end{aligned}$$ We use the identity, for every $y\in \mathbb{R}$, $$\label{id} H_{n}(y)e^{-\frac{y^{2}}{2}} = (-1)^{[n/2]}
2^{\frac{n}{2}} \frac{2}{n!\pi } \int _{0}^{\infty } u^{n}
e^{-u^{2}} g(uy\sqrt{2}) du$$ where $g(r)=\cos (r)$ if $n$ is even and $g(r)=\sin (r)$ if $n$ is odd. Since $\vert g(r)\vert \leq 1$, we have the bound $$\label{cn}
\left|H_{n}(y)e^{-\frac{y^{2}}{2}} \right| \leq
2^{\frac{n}{2}}\frac{2}{n!\pi } \Gamma (\frac{n+1}{2}):=c_{n}.$$ This implies that $$\begin{aligned}
\Vert \xi (x) \Vert _{\alpha , 2} ^{2}&\leq &\frac{1}{2\pi}\sum_{m\geq 0} (1+m) 