Periodic PIC regularization gives a smooth compatible Poisson solution

For $0<\alpha <1$, the two geometric series ${\sum }_{n\ge 0}{\alpha }^n{e}^{in\theta }$ and ${\sum }_{n\ge 1}{\alpha }^n{e}^{-in\theta }$ converge absolutely and uniformly. Adding them and simplifying the denominator gives ${\sum }_{n\in \mathbb{Z}}{\alpha }^{\mathrm{\mid }n\mathrm{\mid }}{e}^{in\theta }=1+\frac{\alpha e^{i\theta }}{1-\alpha e^{i\theta }}+\frac{\alpha e^{-i\theta }}{1-\alpha e^{-i\theta }}=\frac{1-{\alpha }^2}{1-2\alpha \cos \theta +{\alpha }^2}\mathrm{.}$ The denominator is $(1-\alpha {)}^2+2\alpha (1-\cos \theta )>0$. The constant Fourier coefficient is $1$, so integrating the absolutely convergent series over one period gives mean one. Applying this in every coordinate proves the positivity, evenness, smoothness, and unit integral of ${\eta }_{\epsilon }$.

Multiplying the coordinate Fourier series from step 1.1 gives ${\eta }_{\epsilon }(x)=L^{-d}{\sum }_{k\in {\mathbb{Z}}^d}\left({\prod }_a{\alpha }_{\epsilon }^{\mathrm{\mid }{k}_a\mathrm{\mid }}\right)e^{i{\kappa }_k\cdot x}\mathrm{.}$ The product coefficients are absolutely summable, so uniform convergence permits termwise integration against the finite measure $\mu$. The $k=0$ term is $Q\mathrm{/}{L}^d$ and is removed by neutralization. For each $k\mathrm{\ne }0$, $\mathrm{\mid }\widehat{\mu }(k)\mathrm{\mid }\le \mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$. For every integer $m\ge 0$, ${\sum }_{n\in \mathbb{Z}}(1+\mathrm{\mid }n\mathrm{\mid }{)}^m{\alpha }_{\epsilon }^{\mathrm{\mid }n\mathrm{\mid }}<\mathrm{\infty }$ by the ratio test; taking products bounds the corresponding $d$-dimensional weighted sums. Hence the source and every formally differentiated Fourier series converge absolutely and uniformly, proving that ${\rho }_{\epsilon }^{\mu }\in C^{\mathrm{\infty }}$ and has zero mean.

For $k\mathrm{\ne }0$, $-\mathrm{\Delta }{e}^{i{\kappa }_k\cdot x}=\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2{e}^{i{\kappa }_k\cdot x}$. Dividing the nonzero source coefficients in step 2.1 by this positive eigenvalue gives the displayed ${\varphi }_{\epsilon }^{\mu }$; the weighted geometric summability in step 2.1 justifies termwise differentiation and the equation. Its zero Fourier coefficient is set to zero, so it has mean zero. If two mean-zero $C^2$ potentials solve the equation, their difference $u$ satisfies $\mathrm{\Delta }u=0$. Periodic integration by parts gives ${\int }_{{\mathbb{T}}_L^d}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2=-\int u\mathrm{\Delta }u=0$, so $\mathrm{\nabla }u=0$, $u$ is constant, and its zero mean forces $u=0$. This proves uniqueness.