Let $F: \mathbb{R}^d \to \mathbb{R},$ $F(v):= (\pi T)^{-d/2}\mathrm{e}^{-\frac{|v|^2}{T}}$ and $ T > 0 $ is a real constant. So we have for the moments
$$(1) \int_{\mathbb{R}^d} F(v) \, \mathrm{d}v = 1$$ $$(2 )\int_{\mathbb{R}^d} vF(v) \, \mathrm{d}v = 0$$ $$(3) \int_{\mathbb{R}^d} |v|^2F(v) \, \mathrm{d}v = \frac{d}{2}T.$$
Let $g: \Omega \to \mathbb{R}^d$ another function with $g \in L^2(\Omega;\mathbb{R}^d)$ and $\Omega \subset \mathbb{R}^d$ a bounded, connected and sufficient regular subset.
Now consider the integral (by Fubini) $$ \int_{\Omega \times \mathbb{R}^d} |g(x) \cdot v|^2 F(v) \, \mathrm{d}(v,x) = \int_{\Omega} \int_{\mathbb{R}^d} |g(x) \cdot v|^2 F(v) \, \mathrm{d}v\mathrm{d}x. $$ I would like to write this integral like $$ \dots = \int_{\Omega} \int_{\mathbb{R}^d} |g(x)|^2|v|^2 F(v) \, \mathrm{d}v\mathrm{d}x,$$ without the dot product. Maybe with a multiplicative constant. By Cauchy-Schwarz we would easily get the estimate from above but I would need equality. (Jensen is unfortunately not helpful here.) Preferably without explicit computation by use of $F.$ An computation by explicitly using $F$ would work but is there a better way?
Defining $f$ as the version of $F$ when $d=1$, we have $F(v) = \prod_{k=1}^d f(v_k)$. Let's also write $|g(x)\cdot v|^2$ as : \begin{align*} |g(x)\cdot v|^2 &= \left(\sum_{i=1}^d g_i(x)v_i\right)^2 = \sum_{i=1}^d g_i(x)v_i \sum_{j=1}^d g_j(x)v_j \\ &= \sum_{i=1}^d {g_i(x)}^2{v_i}^2 + \sum_{\substack{i,j=1 \\ i\neq j}}^d g_i(x)g_j(x)v_iv_j \end{align*}
By Fubini, we have : \begin{align*} \int_{\Omega\times\mathbb{R}^d} |g(x)\cdot v|^2 F(v)\,\mathrm{d}(v,x) &= \sum_{i=1}^d \int_\Omega {g_i(x)}^2 \underbrace{\int_{\mathbb{R}^d} {v_i}^2 F(v)\,\mathrm{d}v}_{(i)}\,\mathrm{d}x \\ & + \sum_{\substack{i,j=1 \\ i\neq j}}^d g_i(x)g_j(x) \underbrace{\int_{\mathbb{R}^d} v_iv_j F(v)\,\mathrm{d}v}_{(ii)}\,\mathrm{d}x \end{align*}
We now use $f$ to take care of $(i)$ and $(ii)$. First, for $(i)$ : \begin{align*} \int_{\mathbb{R}^d} {v_i}^2 F(v)\,\mathrm{d}v &= \int_{\mathbb{R}^d} {v_i}^2 \prod_{k=1}^d f(v_k)\,\mathrm{d}v \\ &= \left(\vphantom{\int_R}\right. \underbrace{\int_{\mathbb{R}} {v_i}^2 f(v_i)\,\mathrm{d}v_i}_{=\frac{1}{2}T} \left.\vphantom{\int_R}\right) \left(\vphantom{\int_R}\right. \prod_{\substack{k=1\\k\neq i}}^d \underbrace{\int_{\mathbb{R}} f(v_k)\,\mathrm{d}v_k}_{=1} \left.\vphantom{\int_R}\right) \\ &=\frac{T}{2}, \end{align*} and for $(ii)$ : \begin{align*} \int_{\mathbb{R}^d} v_iv_j F(v)\,\mathrm{d}v &= \int_{\mathbb{R}^d} v_iv_j \prod_{k=1}^d f(v_k)\,\mathrm{d}v \\ &= \left(\vphantom{\int_R}\right. \underbrace{\int_{\mathbb{R}} v_i f(v_i)\,\mathrm{d}v_i}_{=0} \left.\vphantom{\int_R}\right) \left(\vphantom{\int_R}\right. \underbrace{\int_{\mathbb{R}} v_j f(v_j)\,\mathrm{d}v_j}_{=0} \left.\vphantom{\int_R}\right) \left(\vphantom{\int_R}\right. \prod_{\substack{k=1\\k\neq i,j}}^d \underbrace{\int_{\mathbb{R}} f(v_k)\,\mathrm{d}v_k}_{=1} \left.\vphantom{\int_R}\right) \\ &= 0, \end{align*} and so our whole integral becomes : \begin{align*} \int_{\Omega\times\mathbb{R}^d} |g(x)\cdot v|^2 F(v)\,\mathrm{d}(v,x) &= \frac{T}{2} \sum_{i=1}^d \int_\Omega {g_i(x)}^2 \,\mathrm{d}x \\ &= \frac{T}{2}\|g\|_{L^2(\Omega)}^2. \end{align*}