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Choose a real even smooth compactly supported Fourier transform $\eta$ equal to one near zero and let $K$ be its inverse Fourier transform. Repeated integration by parts makes $K$ a Schwartz function. It is real, integrable and square-integrable, with unit integral. Moment differentiation of the Fourier transform gives $\int x^{j}K(x)dx=i^j{\eta }^{(j)}(0)=0$ for every positive integer $j$. Such a kernel for density estimation therefore satisfies arbitrary finite-order moment cancellation, and necessarily takes negative values. If the transform plateau contains the frequency support of a bandlimited function $f$, the convolution theorem proves $K\ast f=f$.

For a kernel for density estimation $K$ and smoothing bandwidth $h>0$, the kernel density estimator isWe use the convention that a kernel of order ell hasThe kernel for density estimation is integrable and can have negative values; requiring nonnegativity would prevent its second moment from vanishing. Some conventions additionally require a nonzero moment of order $\ell$. The construction below covers that convention too.

The Legendre polynomial kernel construction uses the polynomial reproducing kernel at zero:Every polynomial $p$ of degree at most $\ell -1$ has the expansion $p={\sum }_{m=0}^{\ell -1}⟨p,{\varphi }_m⟩{\varphi }_m$, since the displayed Legendre polynomials have those successive degrees and form an orthonormal basis of this polynomial space. ThereforeTaking $p=1,u,\dots ,u^{\ell -1}$ proves all the required moment identities. Boundedness and compact support give every absolute moment and a finite L2 norm. If a nonzero moment of order $\ell$ is required, replace this kernel byThe lower moments are unchanged by orthogonality. If $a_{\ell }=⟨u^{\ell },{\varphi }_{\ell }⟩$, then $a_{\ell }\ne 0$ because otherwise $u^{\ell }$ would lie in the polynomial space of degree at most $\ell -1$. Evaluating its full Legendre expansion at zero gives $\int u^{\ell }{K}_{\ell }=-{\varphi }_{\ell }(0)a_{\ell }$, so $\int u^{\ell }{\tilde{K}}_{\ell }=a_{\ell }\ne 0$.

For instance, a simple kernel satisfying the moment conditions needed for three derivatives isIts negative wings cancel its central contribution to the second moment.

Here is the third-order pointwise kernel error bound. Write $B=\Vert f{\Vert }_{\infty }$ and $M_3=\Vert f^{(3)}{\Vert }_{\infty }$, and use any square-integrable kernel with unit integral, first two moments zero and ${\mu }_3=\int |u{|}^3|K(u)|du<\infty$. A change of variables givesBy the Taylor theorem with Lagrange remainder,The moment cancellations consequently give the uniform bias of a kernel density estimator boundFor the variance, independence and another change of variables giveThe triangle inequality and Cauchy-Schwarz inequality now yieldBalancing $h^3$ against $(nh{)}^{-1/2}$ givesThe constant is independent of both $n$ and $x$. The original PDF has exponent $3/7$; the converted TeX's $3/2$ is a transcription error.

To obtain the final parametric rate, we choose a flat-top kernel for density estimation. This kernel also satisfies all the preceding moment conditions, so a single kernel works for both constructions, with different bandwidth choices. With the Fourier transform convention $\hat{g}(t)=\int e^{-itx}g(x)dx$, choose a real even $C^{\infty }$ function $\eta$ equal to one on $[-1,1]$ and zero outside $[-2,2]$, and putSuch a cutoff is explicit: if $\rho (s)=e^{-1/s}$ for $s>0$ and zero otherwise, take $\eta (t)=\rho (2-|t|)/[\rho (2-|t|)+\rho (|t|-1)]$. It is constant near zero, and its endpoint transitions are smooth. Repeated integration by parts shows that $K$ and all its derivatives decay faster than every power, so $K$ lies in the Schwartz space. The Fourier inversion and moment differentiation of the Fourier transform giveThus this kernel can be used with $h_n=n^{-1/7}$ for the preceding general bound.

For the sharper example choose the probability density functionTo check this is a density and identify its Fourier transform, direct integration givesThe continuous value at zero is as displayed. This function is nonnegative and integrable, being bounded near zero and $O(x^{-2})$ at infinity. Fourier inversion gives $\hat{f}(t)=(1-|t|{)}_{+}$, and in particular $\int f=\hat{f}(0)=1$. Differentiation of this compact frequency integral also shows that $f$ and $f^{(3)}$ are bounded. Since $\eta =1$ on the support of $\hat{f}$, the convolution theorem givesBoth sides are continuous, so the equality holds at every $x$. There is therefore exactly zero bias of a kernel density estimator at bandwidth one, and the variance bound above givesThis is parametric-rate kernel estimation of a bandlimited density. A fixed bandwidth is appropriate because it reproduces this particular density exactly; a generic density would retain a nonzero smoothing bias.

A wavelet with $p$ vanishing moments satisfies $x^k\psi \in L^1$ and $\int x^k\psi (x)dx=0$ for $0\le k<p$. The moment differentiation of the Fourier transform theorem states that these weighted integrability hypotheses make $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported scaling function with a finite refinement filter. The wavelet identity givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ still gives $m(\pi +t)=o(|t{|}^{p-1})$ for the representative defined by the quotient: a Peano zero. This establishes the value at $\pi$, and for $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe Fourier coefficients of $\theta$ are absolutely summable. The Wiener algebra is closed under multiplication and the exponential series, so $a$ and $a^{-1}$ have absolutely summable Fourier coefficients. Thus $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves orthogonality and unit norm of its integer function translations; the inverse phase preserves their complete span. The same approximation spaces therefore give a multiresolution analysis.

The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ gives $a(2t)=a(t)a(t+\pi )$, so the periodic phase change of a scaling function produces $m(t)=a(t+\pi )m_0(t)$. In the high-pass formula the phases cancel:The associated wavelet is therefore exactly the original one, with unchanged vanishing moments.

At any dyadic point $t_0=2\pi k/2^j$, the difference quotient of the terms with $j\le n\le {\log }_2(1/|h|)$ is $1+O((2^{n}h{)}^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of differentiability. Dyadic points are dense, and $m_0(\pi +t)$ is nonzero for sufficiently small $t\ne 0$. Thus $m$ is not differentiable on any neighborhood of $\pi$, so an ordinary second derivative at $\pi$, understood as the derivative of a locally defined first derivative, need not exist. The Peano zero remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.

Suppose an integrable orthonormal scaling function has a low-pass filter of a multiresolution analysis that is $C^{p-1}$ near $\pi$, and its Fourier transform is $C^{p-1}$ near zero. If the associated wavelet has $p$ integrable vanishing moments, then $m^{(k)}(\pi )=0$ for $k<p$. Indeed, moment differentiation of the Fourier transform makes ${\hat{\psi }}^{(k)}(0)=0$, while $|\hat{\phi }(0)|=1$. In $\hat{\psi }(2t)=e^{-it}\overline{m(t+\pi )}\hat{\phi }(t)$, division by the nonzero smooth factor proves the conclusion. With only continuity of that factor one still obtains a Peano zero, but not automatically higher ordinary derivatives.

Use the Fourier transform convention $\hat{\psi }(\omega )=\int \psi (x)e^{-i\omega x}dx$. If $\psi$ and $x^r\psi$ for $r<q$ are integrable functions, moment differentiation of the Fourier transform yields a $C^{q-1}$ Fourier transform andThe 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 $\hat{\psi }(\omega )=o(|\omega {|}^{q-1})$; for $q=1$ it gives $\hat{\psi }(\omega )=o(1)$. 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 boundFor 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.^[ref]