= Solution
A <kernel for density estimation> is an integrable function K of unit integral, ordinarily nonnegative and centered. Square integrability is needed for integrated <variance>; a symmetric second-order <kernel for density estimation> also has $\int uK(u)du=0$ and finite nonzero second moment $\mu_2(K)$. Its scaled form is $K_h(u)=h^{-1}K(u/h)$, and the <kernel density estimator> is
$$
\widehat f_h(x)=\frac1n\sum_{i=1}^nK_h(x-X_i).
$$
Higher-order signed kernels are possible, but need not give a nonnegative <probability density function> estimate.
The <mean integrated squared error> is $\mathbb E\int(\widehat f_h(x)-f(x))^2dx$. Put $g_h=K_h*f$. <Independence> gives the pointwise <mean> $g_h$ and <variance> $n^{-1}[(K_h^2*f)-g_h^2]$. Integrating, use $\int f=1$ and Fubini to obtain $\int(K_h^2*f)=R(K_h)$. The bias-variance decomposition proves the exact formula
$$
\boxed{\operatorname{MISE}(\widehat f_h)=R(K_h*f-f)+\frac1n\{R(K_h)-R(K_h*f)\}.}
$$
Equivalently this is $R(f)-2\int f(K_h*f)+(1-n^{-1})R(K_h*f)+R(K)/(nh)$. These identities assume the displayed integrals are finite; they are exact, rather than a small-<smoothing bandwidth> expansion.
For a symmetric second-order <kernel for density estimation>, <Taylor expansion> gives leading <bias> $h^2\mu_2(K)f''/2$, and the leading integrated <variance> is $R(K)/(nh)$. Thus the <asymptotic mean integrated squared error> balances inverse <smoothing bandwidth> against fourth-power <bias>, with optimal <smoothing bandwidth> $[R(K)/(n\mu_2(K)^2R(f''))]^{1/5}$. Its appearance of the <integrated squared density curvature> motivates the scale comparison.
For $f_a(x)=af(ax)$, $a>0$, two differentiations give $f_a''(x)=a^3f''(ax)$. A change of variables therefore gives $R(f_a'')=a^5R(f'')$, which tends to zero as $a\downarrow0$ if the original curvature integral is finite. The distribution with <probability density function> $f_a$ is that of X divided by a, so $\sigma(f_a)=\sigma(f)/a$. Consequently
$$
\boxed{D(f_a)=\sigma(f_a)^5R(f_a'')=\sigma(f)^5R(f'')=D(f).}
$$
This <scale-invariant density curvature> measures shape without dependence on the units used to measure X.
For the final minimization, write $c=35/32$ and use the <triweight kernel> $f_0=c(1-x^2)^3$ on $(-1,1)$, zero outside. Integration gives total mass one, <mean> zero and <variance> $1/9$. Its value and its first two derivatives vanish at both endpoints, so its extension is twice continuously differentiable. On the interior,
$$
f_0''=c(-6+36x^2-30x^4),\qquad f_0'''=c(72x-120x^3),\qquad f_0^{(4)}=c(72-360x^2).
$$
In particular $f_0'''(1)=-48c$ and $f_0'''(-1)=48c$.
Put $e=h-f_0$. The mass and moment constraints give $\int e=\int x^2e=0$, since both means vanish and their variances match. If $R(h'')=\infty$ the claim is immediate; otherwise expand the squared norm and integrate its cross term twice by parts over $[-1,1]$. The first endpoint term vanishes because $f_0''(\pm1)=0$, but the next one must be retained:
$$
\begin{aligned}
\int e''f_0''&=-[ef_0''']_{-1}^1+c\int_{-1}^1(72-360x^2)e(x)dx\\
&=48c\{h(1)+h(-1)\}+c\int_{|x|>1}(360x^2-72)h(x)dx\geq0.
\end{aligned}
$$
For the second equality use the zero mass and second moment of e and $e=h\geq0$ outside the support. All terms in the last expression are nonnegative, including the endpoint values of a <probability density function>. It follows that
$$
\boxed{R(h'')=R(f_0'')+R(e'')+2\int e''f_0''\geq R(f_0'')=35.}
$$
This proves that the <triweight density minimizes integrated squared curvature> under the specified constraints. Equality forces $e''=0$; a globally affine integrable difference of <probability density functions> is zero, so equality occurs only at $h=f_0$.
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