Positive Coulomb bilinear form, Cauchy–Schwarz, and screening bound

For $r>0$, substitute $u=sr$ in the Gaussian integral [F3] and use symmetry to get $\frac{1}{r}=\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-s^2{r}^2}ds\mathrm{.}$ The diagonal $x=y$ is measurable and product-measure null: every section is a singleton, so Tonelli [F5] gives measure zero. By [F2], the absolute Coulomb integrand is integrable; Tonelli [F5] on its nonnegative product with the $s$-integral therefore permits insertion of the Laplace representation into $\mathcal{C}(f,f)$.

At $s=0$ the double integral equals $\mathrm{\mid }\int f{\mathrm{\mid }}^2\ge 0$; for $s>0$, factor $e^{-s^2\mathrm{\parallel }x-y{\mathrm{\parallel }}^2}=e^{-s^2\mathrm{\parallel }x{\mathrm{\parallel }}^2}{e}^{-s^2\mathrm{\parallel }y{\mathrm{\parallel }}^2}{\prod }_{j=1}^3{e}^{2s^2{x}_j{y}_j}\mathrm{.}$ Expanding the three exponentials gives terms $(2s^2{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{x}^{\alpha }{y}^{\alpha }\mathrm{/}\alpha !$. The pointwise absolute sum after multiplying the Gaussian factors is at most one, because ${\sum }_j\mathrm{\mid }{x}_j{y}_j\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ by [F6] and $\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2-2\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }=(\mathrm{\parallel }x\mathrm{\parallel }-\mathrm{\parallel }y\mathrm{\parallel }{)}^2\ge 0$. Its product with $\mathrm{\mid }f(x)\mathrm{\mid }\mathrm{\mid }f(y)\mathrm{\mid }$ is integrable, with double integral at most $\mathrm{\parallel }f{\mathrm{\parallel }}_1^2$. Tonelli and Fubini [F5] therefore permit termwise integration and factor each term, yielding $\int \int \overline{f(x)}f(y)e^{-s^2\mathrm{\parallel }x-y{\mathrm{\parallel }}^2}dydx={\sum }_{\alpha \in {\mathbb{N}}_0^3}\frac{(2s^2{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}}{\alpha !}{\mid \int f(x)x^{\alpha }{e}^{-s^2\mathrm{\parallel }x{\mathrm{\parallel }}^2}dx\mid }^2\ge 0.$

Integrating the nonnegative quantity in step 1.2 over $s$ with the factor $2\mathrm{/}\sqrt{\pi }$ from step 1.1 gives $\mathcal{C}(f,f)\ge 0$. Absolute convergence [F2] also makes $\mathcal{C}$ Hermitian and sesquilinear.

For $\mathcal{C}(g,g)>0$, expand $0\le \mathcal{C}(f+zg,f+zg)$ for arbitrary $z\in \mathbb{C}$ and choose $z=-\mathcal{C}(g,f)\mathrm{/}\mathcal{C}(g,g)$; the resulting quadratic inequality is $\mathrm{\mid }\mathcal{C}(f,g){\mathrm{\mid }}^2\le \mathcal{C}(f,f)\mathcal{C}(g,g)$. If $\mathcal{C}(g,g)=0$ and $a=\mathcal{C}(f,g)\mathrm{\ne }0$, choosing $z=-t\bar{a}$ for arbitrarily large real $t>0$ would give $0\le \mathcal{C}(f,f)-2t\mathrm{\mid }a{\mathrm{\mid }}^2$, a contradiction. Hence $\mathcal{C}(f,g)=0$ and the inequality still holds.

By [F1], each two-electron entry is $\mathcal{C}(f_{ab},f_{cd})$. Applying step 3.1 gives its diagonal Cauchy–Schwarz bound and nonnegative diagonals. Substitute the four $C$ factors from [F7] into the finite two-body energy and reorder its finite sums; the coefficient of $G_{ab,cd}$ is $W_{ab}{W}_{cd}$. The absolute omitted contribution is at most the sum of the absolute values of its weighted terms, giving $B_S$. The sum is retained in the report, so no omitted entry is called exactly zero.