Schwartz functions with prescribed flatness of the Fourier transform at the origin

Reduction to even order. If $m$ is odd, replace it by $m+1$: a function whose moments vanish through order $m+1$ also has all moments vanishing through order $m$. We may therefore assume $m\ge 2$ is even, and we write $h=1\mathrm{/}(8m)$. This uses no choice.

The one-dimensional construction. Let $\theta$ be the even bump of [F1] and put $\mathrm{\Theta }(x)=\theta (x+1\mathrm{/}2)-\theta (x-1\mathrm{/}2)$. Then $\mathrm{\Theta }\in C_c^{\mathrm{\infty }}(\mathbb{R})$ is real, odd, and supported in $[-3\mathrm{/}4,-1\mathrm{/}4]\cup [1\mathrm{/}4,3\mathrm{/}4]$; moreover $\mathrm{\Theta }(x)=-\theta (x-1\mathrm{/}2)<0$ on $(1\mathrm{/}4,3\mathrm{/}4)$ and $\mathrm{\Theta }(x)=\theta (x+1\mathrm{/}2)>0$ on $(-3\mathrm{/}4,-1\mathrm{/}4)$. Define $\varphi (x)=x^{-1}{\mathrm{\Delta }}_h^m\mathrm{\Theta }(x)$. Since $\mathrm{supp}{\mathrm{\Delta }}_h^m\mathrm{\Theta }\subseteq [-3\mathrm{/}4-mh,3\mathrm{/}4+mh]=[-7\mathrm{/}8,7\mathrm{/}8]$ and is bounded away from $0$, the factor $x^{-1}$ is smooth on a neighbourhood of that support, so $\varphi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ with support in $[-7\mathrm{/}8,7\mathrm{/}8]$. The reflection operator $(Pf)(x)=f(-x)$ satisfies $P{\mathrm{\Delta }}_h=-{\mathrm{\Delta }}_{h}P$; since $\mathrm{\Theta }$ is odd and $m$ is even, ${\mathrm{\Delta }}_h^m\mathrm{\Theta }$ is odd, so $\varphi$ is even and real.

The flatness of the one-dimensional Fourier transform. For $1\le \nu \le m$, differentiating $\widehat{\varphi }(\xi )=\int e^{-2\pi i\xi x}\varphi (x)dx$ under the integral sign and using the polynomial identity ${\mathrm{\Delta }}_h^m(x^{\nu -1})=0$ (the $m$-th finite difference of a polynomial of degree $\nu -1\le m-1$ vanishes) gives ${\widehat{\varphi }}^{(\nu )}(0)=(-2\pi i{)}^{\nu }{\int }_{\mathbb{R}}{x}^{\nu -1}{\mathrm{\Delta }}_h^m\mathrm{\Theta }(x)dx=(-2\pi i{)}^{\nu }{\int }_{\mathbb{R}}\left({\mathrm{\Delta }}_h^m{x}^{\nu -1}\right)(x)\mathrm{\Theta }(x)dx=0,$ where the middle equality is the self-adjointness $\int f({\mathrm{\Delta }}_h^{m}g)=\int ({\mathrm{\Delta }}_h^{m}f)g$ of the even-order finite difference, which follows from the translation invariance of Lebesgue measure and $({\mathrm{\Delta }}_h{)}^{\ast }=-{\mathrm{\Delta }}_h$.

Nonvanishing of the mean. Using self-adjointness again, ${\int }_{\mathbb{R}}\varphi ={\int }_{\mathbb{R}}{x}^{-1}{\mathrm{\Delta }}_h^m\mathrm{\Theta }(x)dx={\int }_{\mathbb{R}}\left({\mathrm{\Delta }}_h^m{x}^{-1}\right)(x)\mathrm{\Theta }(x)dx\mathrm{.}$ On the support of $\mathrm{\Theta }$ one has $\mathrm{\mid }x\mathrm{\mid }\ge 1\mathrm{/}4$, and all points $x+s_1+\cdots +s_m$ in the iterated integral representation of [F2] stay on the same side of zero. Since $m$ is even, $(x^{-1}{)}^{(m)}=m!x^{-m-1}$ has the sign of $x$, while $\mathrm{\Theta }$ has the opposite sign on each of its two support components. Thus the integrand $({\mathrm{\Delta }}_h^m{x}^{-1})\mathrm{\Theta }$ has one constant sign and there is no cancellation. By [F2], ${\mathrm{\Delta }}_h^m(x^{-1})(x)=m!{\int }_{[-h,h{]}^m}(x+s_1+\cdots +s_m{)}^{-m-1}d{s}_1\cdots d{s}_m\mathrm{.}$ The integrand has a constant sign on this box, so its absolute value is the integral of the absolute value. The factor $m!$ comes from $(x^{-1}{)}^{(m)}$; the iterated integral contributes the box volume $(2h{)}^m$. Therefore $\mathrm{\mid }{\mathrm{\Delta }}_h^m(x^{-1})(x)\mathrm{\mid }=m!{\int }_{[-h,h{]}^m}\mathrm{\mid }x+s_1+\cdots +s_m{\mathrm{\mid }}^{-m-1}d{s}_1\cdots d{s}_m\ge (2h{)}^{m}m!(7\mathrm{/}8{)}^{-m-1}$ for every $x\in \mathrm{supp}\mathrm{\Theta }$, because $\mathrm{\mid }x+s_1+\cdots +s_m\mathrm{\mid }\le 3\mathrm{/}4+mh=7\mathrm{/}8$. Hence $\mid {\int }_{\mathbb{R}}\varphi \mid ={\int }_{\mathbb{R}}\mid {\mathrm{\Delta }}_h^m{x}^{-1}\mid \mathrm{\mid }\mathrm{\Theta }\mathrm{\mid }\ge (2h{)}^{m}m!(7\mathrm{/}8{)}^{-m-1}{\int }_{\mathbb{R}}\mathrm{\mid }\mathrm{\Theta }\mathrm{\mid }>0,$ since $\mathrm{\Theta }$ is continuous and not identically zero.

The tensor product and its moments. Put $\psi (x)=\varphi (x_1)\varphi (x_2)\cdots \varphi (x_n)$ and $\mathrm{\Psi }(x)=\psi (\lambda x)$ with $\lambda =\sqrt{n}+1>7\sqrt{n}\mathrm{/}8$; then $\mathrm{\Psi }\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ is real and even and $\mathrm{supp}\mathrm{\Psi }\subseteq [-7\mathrm{/}(8\lambda ),7\mathrm{/}(8\lambda ){]}^n\subseteq B(0,1)$, since the euclidean circumradius of that cube is $(7\mathrm{/}8)\sqrt{n}\mathrm{/}\lambda <(7\mathrm{/}8)\sqrt{n}\mathrm{/}(7\sqrt{n}\mathrm{/}8)=1$. For a multi-index $\alpha$ with $0<\mathrm{\mid }\alpha \mathrm{\mid }\le m$ the substitution $x={\lambda }^{-1}t$ gives the factorisation ${\int }_{{\mathbb{R}}^n}{x}^{\alpha }\mathrm{\Psi }(x)dx={\lambda }^{-n-\mathrm{\mid }\alpha \mathrm{\mid }}{\prod }_{j=1}^n{\int }_{\mathbb{R}}{t}^{{\alpha }_j}\varphi (t)dt\mathrm{.}$ If ${\alpha }_j=0$ the corresponding factor is $\int \varphi \mathrm{\ne }0$ by step 1.4; if ${\alpha }_j\ge 1$ then ${\alpha }_j\le \mathrm{\mid }\alpha \mathrm{\mid }\le m$, and $\int t^{{\alpha }_j}\varphi (t)dt=0$ by step 1.3 combined with [F3] applied in one dimension. Hence every factor with ${\alpha }_j\ge 1$ vanishes and the product is zero.

Normalisation and conclusion. Step 2.1 gives $\int \mathrm{\Psi }={\lambda }^{-n}(\int \varphi {)}^n\mathrm{\ne }0$, so $\phi =({\int }_{{\mathbb{R}}^n}\mathrm{\Psi }{)}^{-1}\mathrm{\Psi }$ is real, even, smooth and compactly supported in $B(0,1)$, with $\int \phi =1$ and all moments $\int x^{\alpha }\phi$, $0<\mathrm{\mid }\alpha \mathrm{\mid }\le m$, still vanishing. The Fourier form of the statement follows from the differentiation identity of [F3] and $\widehat{\mathrm{\Psi }}(0)=\int \mathrm{\Psi }\mathrm{\ne }0$, together with the linearity of the Fourier transform under the real scalar normalisation. This proves the lemma.