The mean-zero Fourier distribution solves the neutral periodic Poisson equation

Define $\widehat{u}$ by the displayed coefficients. For $k\mathrm{\ne }0$, $\mathrm{\mid }\widehat{u}(k)\mathrm{\mid }\le \frac{4\pi Q}{V\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2},Q={\sum }_i\mathrm{\mid }{q}_i\mathrm{\mid },$ so the sequence is polynomially bounded. Under Countable Choice [A1], [F2] therefore defines a continuous periodic distribution $u$ with exactly these coefficients; its zero Fourier coefficient is zero by construction.

For a smooth periodic $\phi$, use Countable Choice [A1], [F3], and the coefficient pairing of [F2] to compute $⟨u,-\mathrm{\Delta }\phi ⟩=V{\sum }_{k\mathrm{\ne }0}\widehat{u}(k)\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2\widehat{\phi }(-k)=4\pi {\sum }_i{q}_i{\sum }_{k\mathrm{\ne }0}{e}^{-i{\kappa }_k\cdot x_i}\widehat{\phi }(-k)\mathrm{.}$ The sums converge absolutely by [F2]. Reindex the full Fourier series by $j=-k$; [F2] gives ${\sum }_{k\in {\mathbb{Z}}^3}{e}^{-i{\kappa }_k\cdot x_i}\widehat{\phi }(-k)=\phi (x_i)$. The omitted zero term contributes $\widehat{\phi }(0){\sum }_i{q}_i=0$ by neutrality. Hence $⟨u,-\mathrm{\Delta }\phi ⟩=4\pi {\sum }_i{q}_i\phi (x_i)=4\pi ⟨\rho ,\phi ⟩$, proving $-\mathrm{\Delta }u=4\pi \rho$.

If $u_1,u_2$ are two mean-zero periodic distribution solutions, let $h=u_1-u_2$. Testing $-\mathrm{\Delta }h=0$ against each character gives $\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2\widehat{h}(k)=0$ for every $k\mathrm{\ne }0$. Thus all nonzero Fourier coefficients of $h$ vanish, and its zero coefficient vanishes because both solutions have mean zero. Under [A1], the smooth Fourier partial sums converge in every smooth seminorm by [F2]; continuity of $h$ as a distribution then makes $h$ vanish on every smooth test function. Therefore $u_1=u_2$.