Proof of The Riccati Coefficients of the Wick-Square Problem: Existence, the Gaussian Profile and Summable Constants

Each result cited below is universally quantified over the data in its own statement.

Throughout, order and arithmetic in $\mathbb{R}$ are handled with Elementary Order Arithmetic in an Ordered Field and Elementary Arithmetic in an Ordered Field, whose claims are cited by number, together with the field identities of Zero Products and Elementary Identities in a Field; we write $4=2\cdot2$ and $8=2\cdot4$, which are positive by claims 8 and 5 of Elementary Order Arithmetic in an Ordered Field.

Preliminaries. Fix a mode $k\in\mathbb{Z}^{n}$ and write $\mu=\mu_{k}$. By The Wick-Square Problem on the Torus: Standing Notation §modes we have $1\le\mu$; since $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, claim 2 of that lemma gives $0<\mu$. By The Wick-Square Problem on the Torus: Standing Notation §parameters the numbers $\nu$, $\beta$ and $\gamma$ are positive. Put

$$b=\gamma+2\mu .$$

Since $0<2$ and $0<\mu$, claim 5 of Elementary Order Arithmetic in an Ordered Field gives $0<2\mu$, and then claim 3 of that lemma, applied to $0<2\mu$ and $0\le\gamma$, gives $0<2\mu+\gamma=b$. The equation of clause 1 reads $2q^{2}+bq=\beta$.

Clause 1 (existence). Put $D=b^{2}+8\beta$. We have $0\le b^{2}$ by claim 2 of Nonnegativity of Squares in an Ordered Field and $0<8\beta$ by claim 5 of Elementary Order Arithmetic in an Ordered Field; hence $b^{2}<b^{2}+8\beta=D$ by claim 1 of that lemma (add $b^{2}$ to $0<8\beta$), and $0\le D$ by claim 2 of Elementary Arithmetic in an Ordered Field. By Existence and Uniqueness of the Nonnegative Square Root there is a real number $r$ with $0\le r$ and $r^{2}=D$. Since $0\le b$, $0\le r$ and $b^{2}<r^{2}$, claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $b<r$, so $0<r-b$ by claim 1 of Elementary Order Arithmetic in an Ordered Field (add $-b$). Put

$$q_{k}=\frac{r-b}{4}.$$

Then $0<q_{k}$ by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field. Using the field axioms (distributivity and commutativity) and the difference of squares, claim 4 of Zero Products and Elementary Identities in a Field,

$$2q_{k}^{2}+bq_{k}=\frac{(r-b)^{2}}{8}+\frac{2b(r-b)}{8}=\frac{(r-b)\bigl((r-b)+2b\bigr)}{8}=\frac{(r-b)(r+b)}{8}=\frac{r^{2}-b^{2}}{8}=\frac{8\beta}{8}=\beta .$$

Thus $q_{k}$ is a positive solution of $2q_{k}^{2}+(\gamma+2\mu_{k})q_{k}=\beta$.

Clause 1 (uniqueness). Let $q,q'$ be positive real numbers with $2q^{2}+bq=\beta$ and $2q'^{2}+bq'=\beta$. Subtracting and using claim 4 of Zero Products and Elementary Identities in a Field together with distributivity,

$$0=2(q^{2}-q'^{2})+b(q-q')=(q-q')\bigl(2(q+q')+b\bigr).$$

Now $0<q+q'$ by claim 3 of Elementary Order Arithmetic in an Ordered Field, so $0<2(q+q')$ by claim 5 of that lemma and $0<2(q+q')+b$ by claim 3 of that lemma; in particular $2(q+q')+b\ne0$. By claim 3 of Zero Products and Elementary Identities in a Field, $q-q'=0$, that is $q=q'$. So $q_{k}$ is the only positive root.

Clause 2. Write $q=q_{k}$. The equation of clause 1 gives $\beta-2\mu q=2q^{2}+\gamma q=q(2q+\gamma)$. Dividing by $2\mu$, which is positive, hence invertible by claim 7 of Elementary Order Arithmetic in an Ordered Field,

$$\frac{\beta}{2\mu}-q=\frac{\beta-2\mu q}{2\mu}=\frac{q(\gamma+2q)}{2\mu}. \tag{$\ast$}$$

Since $0<q$, $0<2q$ and $0<\gamma$, the number $\gamma+2q$ is positive (claims 5 and 3 of Elementary Order Arithmetic in an Ordered Field), so the right-hand side of $(\ast)$ is positive by claims 5 and 7 of that lemma. This proves the lower bound $0\le\frac{\beta}{2\mu}-q$, and with it $q\le\frac{\beta}{2\mu}$ by claim 3 of Elementary Arithmetic in an Ordered Field. Multiplying the last inequality by $2$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $2q\le\frac{\beta}{\mu}$, and adding $\gamma$ gives $\gamma+2q\le\gamma+\frac{\beta}{\mu}$. Since $0\le q$ and $0\le\gamma+2q$, two applications of claim 5 of Elementary Arithmetic in an Ordered Field (and transitivity of $\le$) give

$$q(\gamma+2q)\le\frac{\beta}{2\mu}\,(\gamma+2q)\le\frac{\beta}{2\mu}\Bigl(\gamma+\frac{\beta}{\mu}\Bigr),$$

where in the second step the factor $\frac{\beta}{2\mu}$ is nonnegative by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field. Multiplying by the positive number $\frac{1}{2\mu}$ (claim 5 of Elementary Arithmetic in an Ordered Field) and using $(\ast)$,

$$\frac{\beta}{2\mu}-q\le\frac{\beta}{4\mu^{2}}\Bigl(\gamma+\frac{\beta}{\mu}\Bigr)=\frac{\beta\gamma+\beta^{2}\,\frac{1}{\mu}}{4\mu^{2}} .$$

Finally, $0<1\le\mu$ gives $\frac{1}{\mu}\le1$ by Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal, so $\beta^{2}\frac{1}{\mu}\le\beta^{2}$ by claim 5 of Elementary Arithmetic in an Ordered Field (as $0\le\beta^{2}$ by claim 2 of Nonnegativity of Squares in an Ordered Field); adding $\beta\gamma$ (claim 3 of Elementary Arithmetic in an Ordered Field) and multiplying by the positive number $\frac{1}{4\mu^{2}}$ (claim 5 of Elementary Arithmetic in an Ordered Field) yields

$$0\le\frac{\beta}{2\mu_{k}}-q_{k}\le\frac{\beta\gamma+\beta^{2}}{4\mu_{k}^{2}} .$$

Clause 3 (the bounds). By The Free-Field Variances of the Fourier Modes §variances, $c_{k}=\frac{\nu}{2\mu_{k}}$, so $\beta c_{k}=\nu\frac{\beta}{2\mu_{k}}$ and

$$\nu q_{k}-\beta c_{k}=-\nu\,\delta_{k},\qquad\delta_{k}=\frac{\beta}{2\mu_{k}}-q_{k},\qquad a_{k}=-\frac{\nu}{\gamma}\,\delta_{k}$$

by distributivity and claim 2 of Zero Products and Elementary Identities in a Field. The number $\frac{\nu}{\gamma}$ is positive by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field. Put $E_{k}=\frac{\beta\gamma+\beta^{2}}{4\mu_{k}^{2}}$. Clause 2 gives $0\le\delta_{k}\le E_{k}$, so by claim 5 of Elementary Arithmetic in an Ordered Field and claim 1 of Zero Products and Elementary Identities in a Field, $0\le\frac{\nu}{\gamma}\delta_{k}\le\frac{\nu}{\gamma}E_{k}$, and sign reversal, claim 4 of Elementary Order Arithmetic in an Ordered Field, gives

$$-\frac{\nu(\beta\gamma+\beta^{2})}{4\gamma\,\mu_{k}^{2}}=-\frac{\nu}{\gamma}E_{k}\le a_{k}\le0 .$$

Clause 3 (cube-summability). Put $K=\frac{\nu(\beta\gamma+\beta^{2})}{4\gamma}$, so that $\frac{\nu}{\gamma}E_{k}=K\,\frac{1}{\mu_{k}^{2}}$ for every mode $k$; note $0\le K\frac{1}{\mu_{k}^{2}}$ because the lower bound just proved gives $-K\frac{1}{\mu_{k}^{2}}\le0$ (claim 4 of Elementary Order Arithmetic in an Ordered Field). Since $a_{k}\le0\le K\frac{1}{\mu_{k}^{2}}$ and $-K\frac{1}{\mu_{k}^{2}}\le a_{k}$, claim 6 of Properties of the Absolute Value in an Ordered Field gives

$$|a_{k}|\le K\,\frac{1}{\mu_{k}^{2}}\qquad\text{for every }k\in\mathbb{Z}^{n}.$$

By The Wick-Square Problem on the Torus: Standing Notation §dimension, $n\le3<4=2\cdot2$ in $\mathbb{N}$, so Lattice Sums of the Fourier Weights of the Torus, and the Negative Sobolev Norms as Lattice Sums §convergent, applied with $s=2$ (its setting is that of The Flat Torus: Standing Notation with $1\le n$, and its weights $\mu_{k}$ are those of The Wick-Square Problem on the Torus: Standing Notation §modes), shows that the family $k\mapsto\frac{1}{\mu_{k}^{2}}$ is cube-summable. By Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §linear (with $\lambda=K$) the family $k\mapsto K\frac{1}{\mu_{k}^{2}}$ is cube-summable, and by Lattice Sums along Cubes: Linearity, Nonnegative Families, Absolute Summability, Comparison and Finitely Supported Families §comparison the family $k\mapsto a_{k}$ is cube-summable. This completes the proof.