Taylor series for radial function

Question

Denote by $\hat{J}$ the Fourier transform of $J\in C(\mathbb{R}^n,\mathbb{R})$, a nonnegative, radial function with \begin{equation} \displaystyle\int\limits_{\mathbb{R}^n}J(x)~\mathrm{d}x=1. \end{equation} This means that $J$ is a radial density probability, obviously implying that $|J(\xi)| ≤ 1$ with $\hat{J}(0) = 1$: furthermore we shall assume that $\hat{J}$ has an expansion of the form \begin{equation} \hat{J}(\xi) = 1 -A|\xi|^\alpha+ \mathscr{O}(|\xi|^\alpha),\label{1}\tag{1} \end{equation} for $\xi\rightarrow 0$ and $A>0$.
Equation \eqref{1} express the Taylor series of $\hat{J}(\xi)$ and is equivalent to \begin{equation} \hat{J}(\xi) = 1 + i\left<a,\xi\right>-\dfrac{1}{2}\left<\xi,B\xi\right>+ \mathscr{O}(|\xi|^2),\label{2}\tag{2} \end{equation} Why equation \eqref{2} above contains the inner product and moreover why $$ \begin{align} a_i &= \int x_iJ(x)~\mathrm{d}x<\infty,\\ \quad B_{ij} &=\int x_ix_jJ(x)~\mathrm{d}x<\infty. \end{align}\:? $$ Thank you very much

Answer 1

The Fourier transform is $\hat J(\xi)=\int J(x)f(x,\xi)dx$ with $f(x,\xi)=e^{-i\langle\xi,x\rangle}$. Consider the Taylor series of $f$ with respect to $\xi$ at $\xi=0$, this gives $$f(\xi)=f(0)+\langle\nabla_\xi f(0),\xi\rangle+\frac12\langle\xi,\text{Hess}_f(0)\xi\rangle+\cdots.$$ For the first term, $$f(0)=e^0=1.$$ For the gradient, we have $$[\nabla_\xi f(0)]_i=\frac\partial{\partial\xi_i}\left(e^{-i\langle\xi,x\rangle}\right)\Big|_{\xi=0}=-ix_ie^{-i\langle\xi,x\rangle}\Big|_{\xi=0}=-ix_i.$$ For the Hessian matrix, we have $$[\text{Hess}_f(0)]_{ij}=\frac{\partial^2}{\partial\xi_i\partial\xi_j}\left(e^{-i\langle\xi,x\rangle}\right)\Big|_{\xi=0}=-x_ix_je^{-i\langle\xi,x\rangle}\Big|_{\xi=0}=-x_ix_j.$$ Taking these calculations into the Taylor series, it follows $$f(\xi)=1-i\langle x,\xi\rangle-\frac12\langle\xi,H\xi\rangle+\cdots$$ with $H_{ij}=x_ix_j$. Taking this into $\hat J$ we obtain $(2)$ in the question. (The only difference is we have $-i$ instead of $i$ for the linear term, which might be a typo)