= Vanishing moments and Fourier zeros
{title2=$\widehat\psi^{(r)}(0)=(-i)^r\int x^r\psi(x)\,dx$}
Use the <Fourier transform> convention $\widehat\psi(\omega)=\int\psi(x)e^{-i\omega x}\,dx$. If $ψ$ and $x^r\psi$ for $r<q$ are <integrable functions>, <moment differentiation of the Fourier transform> yields a $C^{q-1}$ <Fourier transform> and
$$
\widehat\psi^{(r)}(0)=(-i)^r\int_{\mathbb R}x^r\psi(x)\,dx,\qquad 0\leq r<q.
$$
The <dominated convergence theorem> justifies these <derivatives>. Under these assumptions, $q$ <vanishing moments> are equivalent to vanishing <derivatives> through order $q-1$ at zero. For $q>1$ this gives a <Peano zero> $\widehat\psi(\omega)=o(|\omega|^{q-1})$; for $q=1$ it gives $\widehat\psi(\omega)=o(1)$. It does not by itself justify a $q$th <derivative> or an $O(|\omega|^q)$ estimate.
If also $\int|x|^q|\psi(x)|\,dx<\infty$, subtract the degree-$(q-1)$ <Taylor polynomial> of $e^{-i\omega x}$ inside the <integral>. Its <Taylor remainder> gives the stronger bound
$$
|\widehat\psi(\omega)|\leq\frac{|\omega|^q}{q!}\int_{\mathbb R}|x|^q|\psi(x)|\,dx.
$$
For an <integrable function> with <compact support>, every weighted <integral> exists and its <Fourier transform> extends to an <entire function>. The language “zero of order at least $q$” is then literal <analytic function> terminology. If the next <integral> is nonzero, the order is exactly $q$. The <smooth-mask vanishing-moment criterion> relates these cancellations to a refinement mask under its additional hypotheses.<A Wavelet Tour of Signal Processing by Stéphane Mallat/Wavelet Bases>[Wavelet Bases, §7.2.1, printed p. 206, Theorem 7.4 and proof of (i) ⇔ (ii): decay assumptions and differentiation of the Fourier transform]{ref}
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