 Solution

For the unit-width box kernel, the convolution and kernel density estimator areandIn particular $\mathbb{E}{\hat{f}}_{n,h}^K=g_h$. The scaling gives $\Vert K_h{\Vert }_2^2=h^{-1}\Vert K{\Vert }_2^2=h^{-1}$. Even if $f$ is not square-integrable, Young's convolution inequality gives $g_h\in L^2$ because $f\in L^1$ and $K_h\in L^2$.

Using independence to eliminate the cross terms in the centred estimator, and Tonelli theorem to integrate the nonnegative variance, gives the exact integrated variance of a kernel density estimator:The Cauchy-Schwarz inequality now yieldsThis constant comes from the unit $L^2$ norm of the unscaled unit-width box kernel.

For the piecewise constant probability density function, integrability forces the constants on both unbounded outer intervals to be zero. Thus $f$ is bounded, has compact support, and has finitely many jumps. Let ${\Delta }_r=f(x_r+)-f(x_r-)$ denote the jump at $x_r$. For $h$ smaller than the least gap between consecutive breakpoints, the smoothing regions $[x_r-h/2,x_r+h/2]$ do not overlap. Outside these regions the convolution equals $f$.

Within a region, put $v=x-x_r$. For $-h/2<v<0$, the bias is ${\Delta }_r(1/2+v/h)$; for $0<v<h/2$, it is $-{\Delta }_r(1/2-v/h)$. Endpoint values do not affect an $L^2$ norm. Integrating these two triangular errors givesThis is the box-kernel bias of a piecewise constant density. By the triangle inequality, with $C_f=({\sum }_r{\Delta }_r^2/12{)}^{1/2}$,Balancing this bias-variance tradeoff with $h_n=n^{-1/2}$ gives, for all sufficiently large $n$,