= Solution
For a <kernel for density estimation> $K$ and <smoothing bandwidth> $h>0$, the <kernel density estimator> is
$$
\boxed{f_n^K(x,h)=\frac1{nh}\sum_{i=1}^nK\!\left(\frac{x-X_i}{h}\right).}
$$
We use the convention that a <kernel of order ell> has
$$
\int_{\mathbb R}K(u)\,du=1,\qquad \int_{\mathbb R}u^jK(u)\,du=0\ (1\leq j<\ell),\qquad \int_{\mathbb R}|u|^\ell|K(u)|\,du<\infty.
$$
The <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:
$$
\boxed{K_\ell(u)=\mathbf1_{[-1,1]}(u)\sum_{m=0}^{\ell-1}\phi_m(0)\phi_m(u).}
$$
Every <polynomial> $p$ of degree at most $\ell-1$ has the expansion $p=\sum_{m=0}^{\ell-1}\langle p,\phi_m\rangle\phi_m$, since the displayed <Legendre polynomials> have those successive degrees and form an <orthonormal basis> of this polynomial space. Therefore
$$
\int_{-1}^1K_\ell(u)p(u)\,du=\sum_{m=0}^{\ell-1}\phi_m(0)\langle p,\phi_m\rangle=p(0).
$$
Taking $p=1,u,\ldots,u^{\ell-1}$ 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 by
$$
\widetilde K_\ell=K_\ell+(\phi_\ell(0)+1)\phi_\ell\mathbf1_{[-1,1]}.
$$
The lower moments are unchanged by <orthogonality>. If $a_\ell=\langle u^\ell,\phi_\ell\rangle$, then $a_\ell\ne0$ because otherwise $u^\ell$ would lie in the polynomial space of degree at most $\ell-1$. Evaluating its full Legendre expansion at zero gives $\int u^\ell K_\ell=-\phi_\ell(0)a_\ell$, so $\int u^\ell\widetilde K_\ell=a_\ell\ne0$.
For instance, a simple kernel satisfying the moment conditions needed for three derivatives is
$$
K_3(u)=\frac{9-15u^2}{8}\mathbf1_{[-1,1]}(u),\qquad \int K_3=1,\quad\int uK_3=\int u^2K_3=0,\quad\int K_3^2=\frac98.
$$
Its negative wings cancel its central contribution to the second moment.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-31-moment-cancellation.png]
{title=Signed kernel and cancellation of its second moment}
Here is the <third-order pointwise kernel error bound>. Write $B=\|f\|_\infty$ and $M_3=\|f^{(3)}\|_\infty$, 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 gives
$$
\mathbb Ef_n^K(x,h)=\int K(u)f(x-hu)\,du.
$$
By the <Taylor theorem with Lagrange remainder>,
$$
f(x-hu)=f(x)-hu f'(x)+\frac{h^2u^2}{2}f''(x)+R(x,u,h),\qquad |R(x,u,h)|\leq\frac{M_3h^3|u|^3}{6}.
$$
The moment cancellations consequently give the uniform <bias of a kernel density estimator> bound
$$
|\mathbb Ef_n^K(x,h)-f(x)|\leq\frac{M_3\mu_3}{6}h^3.
$$
For the <variance>, independence and another change of variables give
$$
\operatorname{Var}(f_n^K(x,h))=\frac1n\operatorname{Var}\!\left(h^{-1}K\!\left(\frac{x-X_1}{h}\right)\right)\leq\frac1{nh}\int K(u)^2f(x-hu)du\leq\frac{B\|K\|_2^2}{nh}.
$$
The <triangle inequality> and <Cauchy-Schwarz inequality> now yield
$$
\mathbb E|f_n^K(x,h)-f(x)|\leq\frac{M_3\mu_3}{6}h^3+\frac{\sqrt B\|K\|_2}{\sqrt{nh}}.
$$
Balancing $h^3$ against $(nh)^{-1/2}$ gives
$$
\boxed{h_n=n^{-1/7},\qquad \sup_{x\in\mathbb R}\mathbb E|f_n^K(x,h_n)-f(x)|\leq\left(\frac{M_3\mu_3}{6}+\sqrt B\|K\|_2\right)n^{-3/7}.}
$$
The 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 $\widehat g(t)=\int e^{-itx}g(x)dx$, choose a real even $C^\infty$ function $\eta$ equal to one on $[-1,1]$ and zero outside $[-2,2]$, and put
$$
\boxed{K(x)=\frac1{2\pi}\int_{-2}^2e^{itx}\eta(t)\,dt.}
$$
Such 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> give
$$
\int K=\eta(0)=1,\qquad \int x^jK(x)dx=i^j\eta^{(j)}(0)=0\quad(j\geq1).
$$
Thus this kernel can be used with $h_n=n^{-1/7}$ for the preceding general bound.
For the sharper example choose the <probability density function>
$$
\boxed{f(x)=\frac{1-\cos x}{\pi x^2}\ (x\ne0),\qquad f(0)=\frac1{2\pi},\qquad h_n=1.}
$$
To check this is a density and identify its <Fourier transform>, direct integration gives
$$
\frac1{2\pi}\int_{-1}^1(1-|t|)e^{itx}dt=\frac1\pi\int_0^1(1-t)\cos(tx)dt=\frac{1-\cos x}{\pi x^2}.
$$
The 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 $\widehat f(t)=(1-|t|)_+$, and in particular $\int f=\widehat f(0)=1$. Differentiation of this compact frequency integral also shows that $f$ and $f^{(3)}$ are bounded. Since $\eta=1$ on the support of $\widehat f$, the <convolution theorem> gives
$$
\widehat{K*f}=\eta\widehat f=\widehat f,\qquad K*f=f.
$$
Both 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 gives
$$
\boxed{\sup_{x\in\mathbb R}\mathbb E|f_n^K(x,1)-f(x)|\leq\sqrt B\|K\|_2\,n^{-1/2}.}
$$
This 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.
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