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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 kernel for density estimation is an integrable function $K:\mathbb{R}\to \mathbb{R}$ with $\int K=1$. For bandwidth $h>0$, the kernel density estimator isA kernel of order ell has

Let ${\mathcal{H}}_{n,\delta }$ and $\Gamma$ be as in the question, put $h_{\max }=\max {\mathcal{H}}_{n,\delta }$, and define the positive pilot boundFor every $h\in {\mathcal{H}}_{n,\delta }$, formThe Lepski bandwidth selection method takesSmall bandwidths have little bias but wide intervals; the rule increases the bandwidth until the estimates cease to be mutually compatible.

The grid is geometric with ratio two. Its lower endpoint has order $\log (\log n/\delta )/n$, while its upper endpoint is of orderFor fixed $\beta \le \ell$, the oracle bandwidtheventually lies between these endpoints. Its largest grid predecessor ${\tilde{h}}_{\mathrm{opt}}$ therefore exists, and the dyadic spacing gives

On ${\Omega }_0$, the true value $f(x)$ belongs to every interval with $h\le {\tilde{h}}_{\mathrm{opt}}$. Their intersection is therefore nonempty, soThe intervals at ${\hat{h}}_{\delta }$ and ${\tilde{h}}_{\mathrm{opt}}$ have a common point. Since ${\hat{\sigma }}_{n,h,\delta }$ decreases with $h$,Using ${\hat{f}}_{\infty }\le A$ and ${\tilde{h}}_{\mathrm{opt}}\ge h_{\mathrm{opt}}/2$ givesBecause $|{\mathcal{H}}_{n,\delta }|=O(\log n)$, enlarging the constant gives the required

For the final claim take ${\delta }_n=n^{-3}$ for sufficiently large $n$ and use $\hat{h}={\hat{h}}_{{\delta }_n}$. The squared error on ${\Omega }_0$ is at most a constant timesOff ${\Omega }_0$, boundedness of $K$ and the lower endpoint of the bandwidth grid give a deterministic polynomial bound on $|{\hat{f}}_{n,\hat{h},K}(x)|$; multiplying its square by $\mathbb{P}({\Omega }_0^c)\le n^{-3}$ is $O(n^{-1})$. Since $f(x)\le A$ and $N$ is independent of ${\delta }_n$, this term is no larger than the target rate after changing the constant. Enlarging it once more covers the finitely many $n<N$, proving the claimed adaptive mean squared error bound.

A kernel for density estimation is an integrable function $K$ with $\int K=1$. With $K_h(u)=h^{-1}K(u/h)$, the kernel density estimator isA kernel of order ell satisfies

Split the integrated $L^p$ risk into stochastic and bias terms:For fixed $x$, put $A_i(x)=K_h(x-X_i)$ and $W_i(x)=A_i(x)-\mathbb{E}{A}_i(x)$. The supplied Rosenthal inequality givesNowAlso $\mathbb{E}{W}_1(x{)}^2\le (K_h^2\ast f)(x)$. The Young convolution inequality with exponent $p/2$ givesAfter integration and multiplication by the outer factor $2^{p-1}$, the stochastic contribution is at most

It remains to control the bias. Here $\ell =\lceil \beta \rceil =m+1$. The Taylor formula with integral remainder, the vanishing kernel moments, and Minkowski integral inequality giveThe defining Nikol'skii smoothness bound is $L|hu{|}^{\beta -m}$, soRaising to the $p$th power and applying the outer factor $2^{p-1}$ proves