Vanishing bias of a smoothed regular-grid spatial periodogram

Write ${\varphi }_n(x):=(2\pi n{)}^{-1}\mathrm{\mid }{\sum }_{a=0}^{n-1}{e}^{-iax}{\mathrm{\mid }}^2$, so ${\mathrm{\Phi }}_n(x)={\prod }_{r=1}^d{\varphi }_n(x_r)$ by [F1]. Fix $0<\delta <\pi$. If the circular distance of $x$ from $0$ is at least $\delta$, choose its representative in $[-\pi ,\pi ]$; then $\mathrm{\mid }\sin (x\mathrm{/}2)\mathrm{\mid }\ge \sin (\delta \mathrm{/}2)>0$ by [F2]. The finite geometric-series formula therefore gives ${\varphi }_n(x)=\frac{1}{2\pi n}{\mid \frac{\sin (nx\mathrm{/}2)}{\sin (x\mathrm{/}2)}\mid }^2\le \frac{1}{2\pi n{\sin }^2(\delta \mathrm{/}2)}\mathrm{.}$ By the union bound over coordinates and the unit integral of each factor from [F1], ${\int }_{\{x:{\rho }_{\mathrm{\infty }}(x,0)\ge \delta \}}{\mathrm{\Phi }}_n(x)dx\le \frac{d}{n{\sin }^2(\delta \mathrm{/}2)},$ where ${\rho }_{\mathrm{\infty }}$ is the maximum circular distance over the $d$ coordinates.

Fix $\epsilon >0$. Continuity of $f$ at ${\lambda }_0$ gives $0<\delta <\pi$ such that $\mathrm{\mid }f(\theta )-f({\lambda }_0)\mathrm{\mid }<\epsilon$ whenever ${\rho }_{\mathrm{\infty }}(\theta ,{\lambda }_0)<2\delta$. For any ${\lambda }^{\mathrm{\prime }}$ with ${\rho }_{\mathrm{\infty }}({\lambda }^{\mathrm{\prime }},{\lambda }_0)<\delta$, split the convolution in [F1] into that neighborhood and its complement. On the neighborhood its contribution to $\mathrm{\mid }{g}_n({\lambda }^{\mathrm{\prime }})-f({\lambda }_0)\mathrm{\mid }$ is at most $\epsilon$ because ${\mathrm{\Phi }}_n$ has integral one. On the complement, ${\rho }_{\mathrm{\infty }}({\lambda }^{\mathrm{\prime }}-\theta ,0)\ge \delta$; with $M:=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\mid }f({\lambda }_0)\mathrm{\mid }<\mathrm{\infty }$, step 1.1 gives $\mathrm{\mid }{g}_n({\lambda }^{\mathrm{\prime }})-f({\lambda }_0)\mathrm{\mid }\le \epsilon +\frac{Md}{n{\sin }^2(\delta \mathrm{/}2)}\mathrm{.}$ This bound holds uniformly for all such ${\lambda }^{\mathrm{\prime }}$.

Since $K$ is continuous and $K(0)>0$, there is $a>0$ such that $K(u)>0$ whenever $\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}<a$. A nearest Fourier-grid point to ${\lambda }_0$ is within $\pi \mathrm{/}n$ in each circular coordinate. Since $n{b}_n\to \mathrm{\infty }$, eventually $\pi \mathrm{/}(n{b}_n)<a$, so the periodized kernel is positive at that grid point and the denominator in [A1] is positive. Moreover, if a weight in [A1] is positive, compact support of $K$ implies its grid point is within circular distance $R{b}_n$ of ${\lambda }_0$ in every coordinate, where $[-R,R{]}^d$ contains the support of $K$. Since $b_n\to 0$, all positive-weight grid points eventually lie within the $\delta$-neighborhood from step 2.1.

For all sufficiently large $n$, [F3] and [A1] give $\mathbb{E}[{\tilde{f}}_{n,b_n}({\lambda }_0)]={\sum }_{q\in {\mathrm{\Lambda }}_n}{w}_{n,q,b_n}({\lambda }_0)g_n({\lambda }_{n,q})\mathrm{.}$ The weights are nonnegative and sum to one, and every positive weight is on a grid point covered by step 2.1. Hence $\mid \mathbb{E}[{\tilde{f}}_{n,b_n}({\lambda }_0)]-f({\lambda }_0)\mid \le \epsilon +\frac{Md}{n{\sin }^2(\delta \mathrm{/}2)}\mathrm{.}$ Taking $n\to \mathrm{\infty }$ and then arbitrary $\epsilon \downarrow 0$ proves the smoothed bias limit. Taking ${\lambda }^{\mathrm{\prime }}={\lambda }_0$ in step 2.1 proves the raw expectation limit as well.