Let $E$ be a $\mathbb R$-Banach space, $A:E'\to E$ be nonnegative ($\langle A\varphi,\varphi\rangle\ge0$ for all $\varphi\in E'$) and symmetric ($\langle A\varphi_1,\varphi_2\rangle=\langle A\varphi_2,\varphi_1\rangle$ for all $\varphi_1,\varphi_2\in E'$) and $\varrho$ be a probability measure on $\mathcal B(E)$ with characteristic function $$\hat\varrho(\varphi)=e^{-\frac{\langle A\varphi,\:\varphi\rangle}2}\;\;\;\text{for all }\varphi\in E'\tag1.$$
How can we show that $$\int\varrho({\rm d}x)\langle x,\varphi\rangle x=A\varphi\tag2$$ for all $\varphi\in E'$?
It is given that $a\phi+b\psi \sim N(0, \langle A(a\phi+b\psi), a\phi+b\psi) \rangle$ for all $\phi$ and $\psi$, for all $a,b \in \mathbb R$.
Hence $E(a\phi+b\psi )^{2}=\langle A(a\phi+b\psi), a\phi+b\psi) \rangle$. Exapand the square on LHS. Consider the special cases $a=0, b=1$ and $a=1, b=0$ and conclude that $E \phi \psi =\langle A\phi, \psi \rangle$. Since $\psi$ is arbitrary this gives $E\phi =A\phi$, as required.