= Solution
In <kernel density estimation>, take a <kernel for density estimation> $K$ that is nonnegative, symmetric, and has integral one, finite $\mu_2(K)=\int u^2K(u)\,du>0$, and finite $R(K)=\int K(u)^2\,du$. Its <smoothing bandwidth> $h>0$ gives $K_h(u)=h^{-1}K(u/h)$ and the <kernel density estimator>
$$
\widehat f_{h,K}(x)=\frac1n\sum_{i=1}^nK_h(x-X_i)
$$
from an <independent and identically distributed> sample with <probability density function> $f$. Nonnegativity and unit mass make the estimate itself a <probability density function>. We compare <kernel for density estimation> shapes by their optimized <mean integrated squared error>, rather than by their performance at an arbitrarily common numerical <smoothing bandwidth>.
For the usual second-order theory, assume, for example, that $f$ belongs to the <Sobolev space> $H^2(\mathbb R)$ and $0<R(f'')<\infty$, and let $h\to0$, $nh\to\infty$. The <expectation> is the <convolution> $K_h*f$. The vanishing first moment of the symmetric <kernel for density estimation> and the integral <Taylor remainder> give
$$
K_h*f-f=h^2\int_{\mathbb R}u^2K(u)\int_0^1(1-t)f''(\,\cdot-thu)\,dt\,du.
$$
Continuity of translations in the <L2 norm>, with domination by $u^2K(u)\|f''\|_2$, shows that $h^{-2}(K_h*f-f)\to\mu_2(K)f''/2$ in the <L2 norm>. Hence the integrated squared <bias of a kernel density estimator> is
$$
\|K_h*f-f\|_2^2=\frac14h^4\mu_2(K)^2R(f'')+o(h^4).
$$
By <independence>, the exact <integrated variance of a kernel density estimator> is
$$
\int\operatorname{Var}\{\widehat f_{h,K}(x)\}\,dx=\frac1n\left\{\frac{R(K)}h-R(K_h*f)\right\}.
$$
Indeed the integrated second moment of a single summand is $R(K)/h$, and its integrated squared <mean> is $R(K_h*f)$. The latter quantity is bounded by $R(f)$, by <Young's convolution inequality>, so its contribution is $O(n^{-1})$. The <bias-variance decomposition of mean squared error> therefore yields the leading criterion
$$
\operatorname{AMISE}_K(h)=\frac{R(K)}{nh}+\frac14h^4\mu_2(K)^2R(f'').
$$
This <asymptotic mean integrated squared error> displays the <bias-variance tradeoff>: a small <smoothing bandwidth> increases integrated <variance>, while a large one increases squared <bias of an estimator>.
Differentiating the <asymptotic mean integrated squared error> with respect to $h$ gives $-R(K)/(nh^2)+h^3\mu_2(K)^2R(f'')$. It changes sign once, from negative to positive, so its unique minimum is
$$
\boxed{h_{*,K}=\left\{\frac{R(K)}{n\mu_2(K)^2R(f'')}\right\}^{1/5}.}
$$
At this minimum the integrated <variance> term is four times the squared <bias of an estimator> term. Substitution gives
$$
\boxed{\inf_{h>0}\operatorname{AMISE}_K(h)=\frac54C(K)R(f'')^{1/5}n^{-4/5},\qquad C(K)=R(K)^{4/5}\mu_2(K)^{2/5}.}
$$
Thus the leading rate is $n^{-4/5}$, and the shape comparison reduces to minimizing $C(K)$.
There is an essential scale ambiguity: if $K_c(u)=c^{-1}K(u/c)$, then $\widehat f_{h,K_c}=\widehat f_{ch,K}$. Changing the scale of the <kernel for density estimation> merely changes the numerical convention for the <smoothing bandwidth>. Moreover
$$
R(K_c)=c^{-1}R(K),\qquad \mu_2(K_c)=c^2\mu_2(K),\qquad C(K_c)=C(K).
$$
A comparison at the same raw $h$ is therefore not a fair comparison of shape. One can normalize the second moment to one, but that does not make the optimal <smoothing bandwidth> independent of shape.
The <canonical kernel for density estimation> instead chooses the unique scale
$$
\boxed{c_K=\left\{\frac{R(K)}{\mu_2(K)^2}\right\}^{1/5},\qquad K^{\mathrm{can}}(u)=c_K^{-1}K(u/c_K).}
$$
It satisfies $R(K^{\mathrm{can}})=\mu_2(K^{\mathrm{can}})^2=C(K)$, and so
$$
\operatorname{AMISE}_{K^{\mathrm{can}}}(h)=C(K)\left\{\frac1{nh}+\frac14h^4R(f'')\right\}.
$$
Consequently \b[all canonical kernels share the same leading optimal bandwidth $h_*=[nR(f'')]^{-1/5}$]. Only the multiplicative constant $C(K)$ depends on shape. This is precisely the separation of shape choice and scale choice that makes the <canonical kernel> normalization useful. For an original unscaled <kernel for density estimation>, its corresponding optimal <smoothing bandwidth> is $c_Kh_*$.
To prove optimal shape, scale a candidate <kernel for density estimation> to have second moment one. Let $B=\sqrt5$ and
$$
E(u)=\frac{3}{4B^3}(B^2-u^2)_+=\frac3{4\sqrt5}\left(1-\frac{u^2}5\right)_+.
$$
Direct integration gives $\int E=1$, $\int u^2E(u)\,du=B^2/5=1$, and $R(E)=3/(5B)$. Thus $E$ is a rescaled <Epanechnikov kernel>. For any admissible $K$ with the same mass and second moment,
$$
\int_{\mathbb R}(B^2-u^2)(K-E)\,du=0.
$$
Outside $[-B,B]$, $E=0$ and $(B^2-u^2)K\leq0$, because $K\geq0$. The integral inside $[-B,B]$ is therefore nonnegative, and multiplying by $3/(4B^3)$ gives $\int E(K-E)\geq0$. Expanding the square now proves
$$
R(K)-R(E)=\int(K-E)^2+2\int E(K-E)\geq0.
$$
Equality forces $K=E$ almost everywhere. At fixed second moment, minimizing $R(K)$ is equivalent to minimizing $C(K)$. Since $C(K)$ is scale invariant, this proves the <optimality of the Epanechnikov kernel> over all nonnegative symmetric second-order <kernels for density estimation>, including the equality case up to rescaling.
In its usual scale, the optimal <Epanechnikov kernel> is
$$
\boxed{K_E(u)=\frac34(1-u^2)\mathbf1_{\{|u|\leq1\}},\qquad \mu_2(K_E)=\frac15,\qquad R(K_E)=\frac35.}
$$
Its canonical scale is $c_E=15^{1/5}$, giving
$$
K_E^{\mathrm{can}}(u)=\frac3{4\,15^{1/5}}\left(1-\frac{u^2}{15^{2/5}}\right)_+.
$$
The optimal <Epanechnikov kernel> shape is therefore a truncated parabola, rather than a particular support radius. The <canonical kernel> and unit-second-moment version represent the same shape but different scale conventions.
The numerical gains over other reasonable <kernels for density estimation> are modest. Define <asymptotic relative efficiency> against the optimal <Epanechnikov kernel> by the sample-size fraction giving the same optimized leading <mean integrated squared error>:
$$
\operatorname{Eff}(K)=\left\{\frac{C(K_E)}{C(K)}\right\}^{5/4}=\frac{R(K_E)\sqrt{\mu_2(K_E)}}{R(K)\sqrt{\mu_2(K)}}\leq1.
$$
For the <Gaussian density kernel> $K_N(u)=(2\pi)^{-1/2}e^{-u^2/2}$, $R(K_N)=1/(2\sqrt\pi)$ and $\mu_2(K_N)=1$, so its efficiency is $6\sqrt\pi/(5\sqrt5)\simeq0.9512$. For the <uniform smoothing kernel> $K_U(u)=\tfrac12\mathbf1_{\{|u|\leq1\}}$, $R(K_U)=1/2$ and $\mu_2(K_U)=1/3$, giving efficiency $6\sqrt3/(5\sqrt5)\simeq0.9295$. The <Gaussian density kernel> is infinitely smooth, whereas the optimal <Epanechnikov kernel> has compact support and allows each observation to affect only a finite window. Such computational and smoothness considerations may outweigh a small leading-efficiency difference.
In practice $R(f'')$ is unknown. A <plug-in estimator> estimates this roughness functional from a pilot estimate of the <second derivative> and substitutes it into the optimal <smoothing bandwidth> formula. Alternatively, <Leave-one-out cross-validation> minimizes
$$
\operatorname{CV}(h)=\int\widehat f_{h,K}(x)^2\,dx-\frac2n\sum_{i=1}^n\widehat f_{-i,h,K}(X_i),
$$
where $\widehat f_{-i,h,K}$ uses the other $n-1$ observations. By <independence>, the <expectation> of the second term is $2\int(K_h*f)f$, so $\mathbb E\operatorname{CV}(h)=\operatorname{MISE}(h)-R(f)$. Thus it targets integrated error without knowing $f$. The choice of <smoothing bandwidth> remains important: even for the optimal <Epanechnikov kernel>, setting $h=t h_{*,K}$ multiplies the leading minimum error by $(4t^{-1}+t^4)/5$.
The preceding optimum concerns nonnegative second-order <kernels for density estimation> and twice-smooth densities on the real line. Signed <kernels of order ell> can cancel higher moments for smoother densities and yield different rates, but may produce negative density estimates. A support boundary or a density jump also requires separate <bias of a kernel density estimator> analysis. These settings do not contradict the proved second-order <optimality of the Epanechnikov kernel>.
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