Proof of The Trilinear Estimate for Cyclic Gradients of Differences of Unitary Laws

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

We prove the estimate with the explicit trilinear constant $K=128$.

Conventions. Fix $\mu,\nu\in\mathcal{L}_{d}$ and $n\in\mathbb{N}$, and write $\theta=\theta_{d}=\frac{1}{6144\,d}$ as in The Word Gauge: Weights, the Space of Weighted Square-Summable Word Functions, Its Norm and Inner Product, and the Length-Weighted Gauge §weights. Since nothing beyond $\mu,\nu\in\mathcal{L}_{d}$ and $n\in\mathbb{N}$ is used, proving the displayed inequality with $K=128$ for these data proves the lemma. For $z\in W_{2d}$ put $\Delta(z)=\mu(z)-\nu(z)\in\mathbb{C}$; then $\Delta(\varnothing)=1-1=0$ by Laws of d-Tuples of Unitaries §normalised. By The Word Gauge Space Is a Real Hilbert Space, and the Unitary Laws Form a Compact Metric Space in It §sobolev, $x=\iota_{d}(\mu)-\iota_{d}(\nu)$ is the map $w\mapsto\Delta(w)$ on $W^{\circ}_{d}$, the map $w\mapsto c_{w}|w||x(w)|^{2}$ is summable, and $\lVert x\rVert_{d,1}$ is defined. For $k\in\mathbb{N}$ put $X_{k}=\sum_{w\in W^{\circ}_{d,k}}c_{w}|x(w)|^{2}$, which is nonnegative by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §nonnegative since $c_{w}>0$ (The Word Gauge: Weights, the Space of Weighted Square-Summable Word Functions, Its Norm and Inner Product, and the Length-Weighted Gauge §weights). The empty word is cyclically reduced (Reduced and Cyclically Reduced Words in Unitary Letters §reduced, Reduced and Cyclically Reduced Words in Unitary Letters §cyclically-reduced) and $x(\varnothing)=\Delta(\varnothing)=0$; the $k$-th block sum of $w\mapsto c_{w}|w||x(w)|^{2}$ is $kX_{k}$, because $|w|=k$ on $W^{\circ}_{d,k}$. Hence The Word Gauge: Weights, the Space of Weighted Square-Summable Word Functions, Its Norm and Inner Product, and the Length-Weighted Gauge §sums, The Word Gauge: Weights, the Space of Weighted Square-Summable Word Functions, Its Norm and Inner Product, and the Length-Weighted Gauge §norm and The Word Gauge: Weights, the Space of Weighted Square-Summable Word Functions, Its Norm and Inner Product, and the Length-Weighted Gauge §sobolev give, the square of a nonnegative square root being the number itself, $\lVert x\rVert_{d}^{2}=\sum_{k=1}^{\infty}X_{k}$ and $\lVert x\rVert_{d,1}^{2}=\sum_{k=1}^{\infty}kX_{k}$, both series converging and having nonnegative terms; so by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §dominates, for every $M\in\mathbb{N}$,

$$\sum_{k=1}^{M}X_{k}\le\lVert x\rVert_{d}^{2},\qquad\sum_{k=1}^{M}k\,X_{k}\le\lVert x\rVert_{d,1}^{2}.\tag{0.1}$$

Real arithmetic. Rearrangements of real equalities and inequalities (adding, multiplying by a nonnegative number, dividing by a positive number, taking reciprocals of positive numbers) are those of Elementary Arithmetic in an Ordered Field, Elementary Order Arithmetic in an Ordered Field and Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §reciprocal. For a nonnegative real $r$, $\sqrt{r}$ is its nonnegative square root (Existence and Uniqueness of the Nonnegative Square Root). For nonnegative reals $r,r'$ one has $r\le r'$ if and only if $r^{2}\le r'^{2}$, and $r=r'$ if and only if $r^{2}=r'^{2}$, by the weak and equality forms of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field; in particular $\sqrt{rr'}=\sqrt{r}\sqrt{r'}$ and $\sqrt{r^{2}}=r$. Put $\vartheta=\sqrt{\theta}$. For a real $r$ and $N\in\mathbb{N}$, $r^{N}$ is the natural power, and $r^{0}=1$; then $r^{N+N'}=r^{N}r^{N'}$ for $N,N'\in\{0\}\cup\mathbb{N}$, by Addition of Exponents for Natural Number Powers in a Field when both are in $\mathbb{N}$ and trivially otherwise; $(rr')^{N}=r^{N}r'^{N}$ and $1^{N}=1$, and $0\le r\le r'$ implies $0\le r^{N}\le r'^{N}$, by claims 3, 2 and 5 of Properties of Natural Number Powers in a Field (trivially for $N=0$). In particular $(\vartheta^{N})^{2}=(\vartheta\vartheta)^{N}=\theta^{N}$.

Finite sums. A sum over an empty index set is $0$. For sums of real numbers over finite sets: termwise inequalities pass to sums by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §comparison; sums of nonnegative terms are nonnegative and do not decrease when the index set is enlarged, by Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §nonnegative and Real Sums over a Finite Index Set: Comparison, Nonnegativity, Monotonicity, Term Bounds, Absolute Values, Counting and Limits §monotone; sums agree with numerical ranges, may be reindexed along bijections, and are additive and homogeneous, by claims 1, 2, 3 and 4 of Properties of a Sum over a Finite Index Set; they split over disjoint unions by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §disjoint-union; iterated sums are sums over dependent pairs by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §pairs; terms equal to $0$ may be dropped by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §vanishing; and a product of two sums is the sum over the Cartesian product, by The Product of Two Sums over Finite Index Sets is a Sum over the Cartesian Product. Finite unions of finite sets are finite by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §finite-union, subsets and images of finite sets are finite by claims 3 and 4 of Basic Properties of Finite Sets, and finite products of finite sets are finite by claim 1 of Finiteness of Cartesian Products, Tuple Sets, and Permutation Sets. For complex numbers we use the rules for conjugation, the modulus of a conjugate, multiplicativity of the modulus, the triangle inequality and the modulus of a nonnegative real, claims 1, 3, 4, 7 and 8 of Properties of Complex Conjugation and Modulus, and the modulus of a finite sum, Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §modulus. In particular $|\mathrm{i}|=1$ (as $|\mathrm{i}|^{2}=\mathrm{i}\,\overline{\mathrm{i}}=1$) and $|\varepsilon(l)|=1$ for every letter $l$.

Words. For $L\in\mathbb{N}$ let $[2d]^{L}$ be the set of words of length $L$, nonempty and finite by Counting Cyclically Reduced Words: Finiteness and the Bound (2d)^k §finite, and put $[2d]^{0}=\{\varnothing\}$. For $M\in\{0\}\cup\mathbb{N}$ let $U_{M}$ be the union of the sets $[2d]^{L}$, $L\in\{0,1,\dots,M\}$, a finite set containing every word of length at most $M$. By Words over a Finite Alphabet: the Empty Word, Concatenation and Reversal §concatenation and Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §monoid, if $u$ has length $L$ and $v$ length $L'$ then $uv$ has length $L+L'$, its first $L$ letters are those of $u$ and its last $L'$ letters those of $v$, and concatenation is associative. Consequently, for fixed $L\in\{0\}\cup\mathbb{N}$, the maps $(u,v)\mapsto uv$ on $[2d]^{L}\times W_{2d}$ and on $W_{2d}\times[2d]^{L}$ are injective, and for $L\le L'$ the map $(u,v)\mapsto uv$ is a bijection from $[2d]^{L}\times[2d]^{L'-L}$ onto $[2d]^{L'}$ (split a word after its $L$-th letter).

Step 0 (elementary inequalities).

(C1) Finite Cauchy--Schwarz. Let $F$ be a nonempty finite set and $a,b:F\to\mathbb{R}$ with $a,b\ge0$. Then

$$\sum_{y\in F}a(y)b(y)\le\sqrt{\textstyle\sum_{y\in F}a(y)^{2}}\;\sqrt{\textstyle\sum_{y\in F}b(y)^{2}},$$

and, with $b$ constantly $1$ and $N_{F}=\sum_{y\in F}1$, $\bigl(\sum_{y\in F}a(y)\bigr)^{2}\le N_{F}\sum_{y\in F}a(y)^{2}$. (This is The Cauchy-Schwarz Inequality in a Real Inner Product Space for the standard pairing on $\mathbb{R}^{F}$; we give the direct proof.) Put $A=\sqrt{\sum a^{2}}$ and $B=\sqrt{\sum b^{2}}$. If $A=0$, then $\sum a^{2}=0$, and since each nonnegative term $a(y)^{2}$ is at most the sum (monotonicity in the index set), $a(y)^{2}=0$, so $a(y)=0$ for all $y$ and the left side is $0$; likewise if $B=0$. Otherwise, for each $y$, $0\le\bigl(\frac{a(y)}{A}-\frac{b(y)}{B}\bigr)^{2}$ by Nonnegativity of Squares in an Ordered Field, that is $2\frac{a(y)b(y)}{AB}\le\frac{a(y)^{2}}{A^{2}}+\frac{b(y)^{2}}{B^{2}}$; summing gives $\frac{2}{AB}\sum ab\le1+1$. The second form follows by squaring the nonnegative sides.

(C2) Numerical facts. $0<\theta\le\frac{1}{6144}$, because $6144\le6144\,d$ and reciprocals reverse the order; hence $0\le2\theta\le1$ and $\theta\le\frac14$, so $\vartheta^{2}\le(\frac12)^{2}$ and $0\le\vartheta\le\frac12$. Moreover $2d\,\theta=\frac{1}{3072}$, as in The Embedding of Unitary Laws into the Word Gauge Space and the Gauge Distance §embedding.

(C3) For every $N\in\mathbb{N}$, $N\le2^{N}$ and hence $N\theta^{N}\le1$. Indeed $1\le2=2^{1}$, and if $N\le2^{N}$ then, as $1=1^{N}\le2^{N}$, $N+1\le2^{N}+2^{N}=2^{N+1}$ by claim 1 of Properties of Natural Number Powers in a Field; so $N\theta^{N}\le2^{N}\theta^{N}=(2\theta)^{N}\le1^{N}=1$ by (C2).

(C4) For $k\in\{0\}\cup\mathbb{N}$, $\sum_{j=0}^{k}\vartheta^{j}\le2$; and $\sum_{t=0}^{n}(2d\,\theta)^{t}\le2$. For the first, the case $k=0$ reads $1\le2$; for $k\ge1$, $\vartheta^{j}\le(\frac12)^{j}$ by (C2), and $\sum_{j=1}^{k}(\frac12)^{j}\le1$ by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric (the halving series has sum $1$) and Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §dominates. For the second, with $r=\frac{1}{3072}$, Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric gives $\sum_{t=1}^{n}r^{t}=\frac{r-r^{n+1}}{1-r}\le\frac{r}{1-r}=\frac{1}{3071}\le1$.

(C5) Let $N\in\{0\}\cup\mathbb{N}$ and $k,k',p\in\mathbb{N}$ with $k\le k'$ and $k+k'=p+2N$. Then $\theta^{N}k'\le4p$. Indeed, if $k'\le4p$, then $\theta^{N}\le1^{N}=1$ by (C2), so $\theta^{N}k'\le k'\le4p$. Otherwise $4p+1\le k'$; since $1\le k$, $2N=k+k'-p\ge k'-p$, so $8N\ge4k'-4p\ge4k'-(k'-1)=3k'+1$. Hence $N\ge1$ and $3k'<8N\le9N$, so $k'\le3N$, and by (C3) $\theta^{N}k'\le3N\theta^{N}\le3\le4p$.

Step 1 (facts about words).

(W1) For every letter $l\in[2d]$, $(l^{-1})^{-1}=l$. This is read off Words in Unitary Letters: Generators, Signs, Inverse Letters, Lengths and Adjoint Words §letters: if $l\le d$ then $l^{-1}=d+l$, whose inverse letter is the unique $j$ with $d+l=d+j$, namely $l$; if $d<l$ then $l=d+j$ with $j\in[d]$, $l^{-1}=j$ and $j^{-1}=d+j=l$.

(W2) Let $z\in W_{2d}$ have length $L\in\mathbb{N}$. Then $z^{*}$ has length $L$ and $(z^{*})_{i}=(z_{i'})^{-1}$ where $i+i'=L+1$, by Words in Unitary Letters: Generators, Signs, Inverse Letters, Lengths and Adjoint Words §adjoint and Words over a Finite Alphabet: the Empty Word, Concatenation and Reversal §reversal; for a letter $l$, $l^{*}=l^{-1}$ (the case $L=1$); $(z^{*})^{*}=z$ for every $z\in W_{2d}$, as noted in Word Polynomials in Unitary Letters: the Normed *-Algebra of Finitely Supported Word Functions §adjoint, so $z\mapsto z^{*}$ is injective; and $(uv)^{*}=v^{*}u^{*}$ for all $u,v\in W_{2d}$. For the last, Word Polynomials under a Unitary Law: Linearity, Traciality, Positivity, the Cauchy-Schwarz Inequality and the l1 Bound §algebra gives $e_{(uv)^{*}}=(e_{uv})^{*}=(e_{u}e_{v})^{*}=(e_{v})^{*}(e_{u})^{*}=e_{v^{*}}e_{u^{*}}=e_{v^{*}u^{*}}$, and evaluating at $(uv)^{*}$, where the left side takes the value $1$, gives $(uv)^{*}=v^{*}u^{*}$.

(W3) If $z$ is reduced, then so is every word formed by consecutive letters of $z$ (its consecutive letters are consecutive letters of $z$), and so is $z^{*}$. For the latter let $z$ have length $L$, $i\in[L]$ with $i<L$, and $i'$ with $i+i'=L+1$, so that $(z^{*})_{i}=(z_{i'})^{-1}$ and $(z^{*})_{i+1}=(z_{i'-1})^{-1}$ by (W2). Since $((z^{*})_{i})^{-1}=z_{i'}$ by (W1), an equality $(z^{*})_{i+1}=((z^{*})_{i})^{-1}$ would read $z_{i'}=(z_{i'-1})^{-1}$, contradicting reducedness of $z$ at position $i'-1<L$ (note $1<i'$ as $i<L$).

(W4) Let $k\in\mathbb{N}$, $w\in W^{\circ}_{d,k}$ and $m\in[k]$. Put $q=m$ if $\varepsilon(w_{m})=1$; $q=m+1$ if $\varepsilon(w_{m})=-1$ and $m<k$; and $q=1$ if $\varepsilon(w_{m})=-1$ and $m=k$. By Rotations and Cyclic Derivatives of Words in Unitary Letters §rotations, $r_{m}(w)=w$ if $q=1$, and otherwise $r_{m}(w)$ is the concatenation of the word $w_{q}w_{q+1}\cdots w_{k}$ and the word $w_{1}\cdots w_{q-1}$. Hence $r_{m}(w)$ has length $k$, its $i$-th letter is $w_{q-1+i}$ for $i\le k-q+1$ and its $(k-q+1+i)$-th letter is $w_{i}$ for $i\le q-1$; so $w$ is determined by $r_{m}(w)$ together with $m$ and $\varepsilon(w_{m})$. Moreover $r_{m}(w)$ is reduced: two consecutive letters of $r_{m}(w)$ are either consecutive letters $w_{i},w_{i+1}$ of $w$, which satisfy $w_{i+1}\ne(w_{i})^{-1}$, or (when $1<q$, so $1<k$) the letters $w_{k},w_{1}$, and $w_{1}=(w_{k})^{-1}$ would give $(w_{1})^{-1}=w_{k}$ by (W1), contradicting Reduced and Cyclically Reduced Words in Unitary Letters §cyclically-reduced.

(W5) For all $a,c,b\in W_{2d}$ and $\lambda\in\mathcal{L}_{d}$, $\lambda(a\,c\,c^{*}\,b)=\lambda(ab)$. By induction on the length of $c$: for $c=\varnothing$ there is nothing to prove, as $\varnothing^{*}=\varnothing$. If $c$ has length $j+1$, then by Basic Properties of Words: Associativity, Reversal, Finitely Many Factorisations, and Countability §last-letter $c=c'l$ with a letter $l$ and $c'$ of length $j$ (or $c'=\varnothing$ if $j=0$), so $c^{*}=l^{-1}c'^{*}$ by (W2), and $a\,c\,c^{*}\,b=(ac')\,l\,l^{-1}\,(c'^{*}b)$; by Laws of d-Tuples of Unitaries §cancellation and the induction hypothesis, $\lambda(acc^{*}b)=\lambda(ac'c'^{*}b)=\lambda(ab)$.

Step 2 (maximal cancellation and stripping). Let $v,v'$ be reduced words of lengths $k,k'\in\mathbb{N}$. Let $J$ be the set of $j\in\{0,1,\dots,\min(k,k')\}$ such that $v'_{i}=(v_{i'})^{-1}$ whenever $i\in[j]$ and $i+i'=k+1$; it contains $0$ and is finite, and we let $j(v,v')$ be its largest element. With $j=j(v,v')$, let $a$ be the word of length $k-j$ with $a_{i}=v_{i}$, $c$ the word of length $j$ with $c_{i}=v_{k-j+i}$, and $b$ the word of length $k'-j$ with $b_{i}=v'_{j+i}$ (each read as $\varnothing$ when its length is $0$). Then:

(2a) $v=ac$ and $v'=c^{*}b$. The first is the letter description of concatenation. For the second, by (W2) the $i$-th letter of $c^{*}$, $i\in[j]$, is $(c_{j+1-i})^{-1}=(v_{k+1-i})^{-1}=v'_{i}$ by the definition of $J$; the remaining letters of $v'$ are those of $b$.

(2b) $ab$ is reduced, of length $k+k'-2j$. The words $a$ and $b$ are reduced by (W3). If one of them is empty, $ab$ is the other. If both are nonempty, then $j<k$ and $j<k'$, so $j+1\le\min(k,k')$ and $j+1\notin J$ by maximality; since the defining condition of $J$ holds for all $i\in[j]$, it fails at $i=j+1$, that is $v'_{j+1}\ne(v_{k-j})^{-1}$. The consecutive letters of $ab$ are consecutive letters of $a$, consecutive letters of $b$, or the pair $a_{k-j}=v_{k-j}$, $b_{1}=v'_{j+1}$; so $ab$ is reduced.

(2c) $\Delta(vv')=\Delta(ab)$, by (2a), associativity and (W5) applied to $\mu$ and to $\nu$.

We write $\rho(v,v')=ab$. Since $|ab|=k+k'-2j\le k+k'$, $\rho(v,v')\in U_{2n}$ whenever $k,k'\le n$.

For every reduced $z\in U_{2n}$ fix, by Unitary Laws: Adjoints, the Bound One, and Reduction to Cyclically Reduced Words §strip-exists, words $q_{z}\in W_{2d}$ and $u_{z}\in W^{\circ}_{d}$ with $z=q_{z}\,u_{z}\,q_{z}^{*}$ and $|z|=2|q_{z}|+|u_{z}|$, and put $t_{z}=|q_{z}|\in\{0\}\cup\mathbb{N}$. By Unitary Laws: Adjoints, the Bound One, and Reduction to Cyclically Reduced Words §strip-invariance, applied to $\mu$ and to $\nu$,

$$\Delta(z)=\Delta(u_{z}),\tag{2.1}$$

which is $0$ if $u_{z}=\varnothing$ and equals $x(u_{z})$ otherwise (Conventions).

Weights. For $k\in\mathbb{N}$ and $w\in W^{\circ}_{d,k}$ put $\hat c_{w}=\frac{\vartheta^{k}}{(k+1)^{2}}\ge0$; then $\hat c_{w}^{2}=\frac{\theta^{k}}{(k+1)^{4}}=c_{w}$ by The Word Gauge: Weights, the Space of Weighted Square-Summable Word Functions, Its Norm and Inner Product, and the Length-Weighted Gauge §weights. For $p\in\mathbb{N}$ and $u\in W^{\circ}_{d,p}$ put $Y(u)=\sqrt{p}\,\hat c_{u}\,|x(u)|\ge0$, so that $Y(u)^{2}=p\,c_{u}|x(u)|^{2}$. Define $h:W_{2d}\to\mathbb{R}$ by $h(z)=\vartheta^{t_{z}}\,Y(u_{z})$ if $z\in U_{2n}$ is reduced and $u_{z}\ne\varnothing$, and $h(z)=0$ otherwise; $h\ge0$.

Step 3 (slots, and expansion of the left side). Let $W^{\circ}_{\le n}$ be as in The Truncated Cyclic Gradient of a Gauge Vector §gradient; it is finite and nonempty, as it contains $W^{\circ}_{d,1}$ (Counting Cyclically Reduced Words: Finiteness and the Bound (2d)^k §finite). Let $\mathcal{S}$ be the set of slots $\alpha=(w,m,s)$ with $w\in W^{\circ}_{\le n}$, $m\in[|w|]$ and $s\in\{1,-1\}$; as a set of dependent pairs $(w,(m,s))$ it is nonempty and finite. For such $\alpha$ write $w_{\alpha}=w$, $k_{\alpha}=|w|$, and put

$$v_{\alpha}=r_{m}(w)\ \text{if }s=1,\qquad v_{\alpha}=r_{m}(w)^{*}\ \text{if }s=-1;\qquad\sigma_{\alpha}=c_{w}\,|x(w)|;$$ $$\kappa_{\alpha}=\tfrac12\,c_{w}\,\overline{x(w)}\;\mathrm{i}\,\varepsilon(w_{m})\ \text{if }s=1,\qquad\kappa_{\alpha}=\overline{\tfrac12\,c_{w}\,\overline{x(w)}\;\mathrm{i}\,\varepsilon(w_{m})}\ \text{if }s=-1.$$

By (W4), (W3) and (W2), $v_{\alpha}$ is a reduced word of length $k_{\alpha}\in[n]$, so $v_{\alpha}\in U_{n}$; and $|\kappa_{\alpha}|=\frac12\sigma_{\alpha}$, since $c_{w}>0$, $|\overline{x(w)}|=|x(w)|$ and $|\mathrm{i}|=|\varepsilon(w_{m})|=1$. For $\alpha=(w,m,s)$ put $\bar\alpha=(w,m,-s)\in\mathcal{S}$; then $v_{\bar\alpha}=v_{\alpha}^{*}$ (using $(r^{*})^{*}=r$ from (W2)), $\kappa_{\bar\alpha}=\overline{\kappa_{\alpha}}$, $\sigma_{\bar\alpha}=\sigma_{\alpha}$ and $k_{\bar\alpha}=k_{\alpha}$. For $i\in[d]$ let $\mathcal{S}_{i}$ be the set of $(w,m,s)\in\mathcal{S}$ with $g(w_{m})=i$; since $g(w_{m})\in[d]$ is unique (Words in Unitary Letters: Generators, Signs, Inverse Letters, Lengths and Adjoint Words §letters), the sets $\mathcal{S}_{i}$ are pairwise disjoint with union $\mathcal{S}$, and $\alpha\in\mathcal{S}_{i}$ if and only if $\bar\alpha\in\mathcal{S}_{i}$.

(3a) Let $i\in[d]$ and $\lambda\in\mathcal{L}_{d}$, and write $\Xi^{i}=\Xi^{i}_{x,n}$, $Z^{i}=Z^{i}_{x,n}$. If $\mathcal{S}_{i}$ is empty then $\lambda(\Xi^{i}\Xi^{i})=0$; otherwise

$$\lambda\bigl(\Xi^{i}\Xi^{i}\bigr)=\sum_{(\alpha,\beta)\in\mathcal{S}_{i}\times\mathcal{S}_{i}}\kappa_{\alpha}\kappa_{\beta}\,\lambda\bigl(v_{\alpha}v_{\beta}\bigr).$$

Proof. Let $\mathcal{S}^{+}_{i}$ and $\mathcal{S}^{-}_{i}$ be the slots in $\mathcal{S}_{i}$ with $s=1$ and $s=-1$; $\alpha\mapsto\bar\alpha$ is a bijection from $\mathcal{S}^{+}_{i}$ onto $\mathcal{S}^{-}_{i}$. By The Truncated Cyclic Gradient of a Gauge Vector §gradient and Rotations and Cyclic Derivatives of Words in Unitary Letters §derivative, and since sums and complex multiples of word polynomials are pointwise (Word Polynomials in Unitary Letters: the Normed *-Algebra of Finitely Supported Word Functions §polynomials), $Z^{i}$ is the sum over $w\in W^{\circ}_{\le n}$ of $c_{w}\overline{x(w)}\,D^{i}_{w}$, and $D^{i}_{w}$ is a sum over the $m\in[|w|]$ with $g(w_{m})=i$. A word $w$ with no such $m$ has $D^{i}_{w}=0$, and its term is dropped by Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §vanishing. If every word is dropped, then $\mathcal{S}^{+}_{i}$ is empty and $Z^{i}=0$ by the same clause, which is the sum over the empty index set $\mathcal{S}^{+}_{i}$. Otherwise the remaining words have nonempty sets of such $m$, so iterated sums over dependent pairs (Sums over Finite Index Sets: Finite Unions, Disjoint Unions, Vanishing Terms, Dependent Pairs, Conjugation and the Modulus §pairs), reindexed along $(w,m)\mapsto(w,m,1)$, and homogeneity give $Z^{i}=\sum_{\alpha\in\mathcal{S}^{+}_{i}}2\kappa_{\alpha}\,e_{v_{\alpha}}$. By Word Polynomials under a Unitary Law: Linearity, Traciality, Positivity, the Cauchy-Schwarz Inequality and the l1 Bound §algebra (adjoints of sums and of complex multiples, $(e_{u})^{*}=e_{u^{*}}$), applied finitely often, and reindexing along $\alpha\mapsto\bar\alpha$, $(Z^{i})^{*}=\sum_{\alpha\in\mathcal{S}^{+}_{i}}\overline{2\kappa_{\alpha}}\,e_{v_{\alpha}^{*}}=\sum_{\alpha\in\mathcal{S}^{-}_{i}}2\kappa_{\alpha}\,e_{v_{\alpha}}$. Hence $\Xi^{i}=\frac12\bigl(Z^{i}+(Z^{i})^{*}\bigr)=\sum_{\alpha\in\mathcal{S}_{i}}\kappa_{\alpha}e_{v_{\alpha}}$ by the disjoint union $\mathcal{S}_{i}=\mathcal{S}^{+}_{i}\cup\mathcal{S}^{-}_{i}$. If $\mathcal{S}_{i}$ is empty, all these sums are the zero word polynomial, so $\Xi^{i}=0$, $\Xi^{i}\Xi^{i}=0$ by Word Polynomials in Unitary Letters: the Normed *-Algebra of Finitely Supported Word Functions §product, and $\lambda(0)=0$ by Evaluation of Word Polynomials by a Unitary Law §evaluation. Otherwise distributivity, compatibility with complex multiples and $e_{u}e_{u'}=e_{uu'}$ from Word Polynomials under a Unitary Law: Linearity, Traciality, Positivity, the Cauchy-Schwarz Inequality and the l1 Bound §algebra, applied finitely often, give $\Xi^{i}\Xi^{i}=\sum_{(\alpha,\beta)\in\mathcal{S}_{i}\times\mathcal{S}_{i}}\kappa_{\alpha}\kappa_{\beta}\,e_{v_{\alpha}v_{\beta}}$, and Word Polynomials under a Unitary Law: Linearity, Traciality, Positivity, the Cauchy-Schwarz Inequality and the l1 Bound §linear together with $\lambda(e_{z})=\lambda(z)$ (Evaluation of Word Polynomials by a Unitary Law §evaluation) gives the formula.

(3b) Put

$$S=\sum_{(\alpha,\beta)\in\mathcal{S}\times\mathcal{S}}\sigma_{\alpha}\sigma_{\beta}\,\bigl|\Delta(v_{\alpha}v_{\beta})\bigr|\ \ge0.$$

Then

$$\Bigl|\sum_{i\in[d]}\Bigl(\mu\bigl(\Xi^{i}\Xi^{i}\bigr)-\nu\bigl(\Xi^{i}\Xi^{i}\bigr)\Bigr)\Bigr|\le\frac14\,S.\tag{3.1}$$

Indeed, by (3a) and additivity, the $i$-th term is $0$ if $\mathcal{S}_{i}$ is empty and otherwise equals $\sum_{\mathcal{S}_{i}\times\mathcal{S}_{i}}\kappa_{\alpha}\kappa_{\beta}\Delta(v_{\alpha}v_{\beta})$, whose modulus is at most $\sum_{\mathcal{S}_{i}\times\mathcal{S}_{i}}\frac14\sigma_{\alpha}\sigma_{\beta}|\Delta(v_{\alpha}v_{\beta})|$. The modulus of the sum over $i\in[d]$ is at most the sum of the moduli, and the sets $\mathcal{S}_{i}\times\mathcal{S}_{i}$ with $\mathcal{S}_{i}$ nonempty are pairwise disjoint subsets of $\mathcal{S}\times\mathcal{S}$; as all terms are nonnegative, splitting over the disjoint union and enlarging the index set give (3.1).

Step 4 (the short word may be placed on the left). Let $\mathcal{P}_{1}$ be the set of $(\alpha,\beta)\in\mathcal{S}\times\mathcal{S}$ with $k_{\alpha}\le k_{\beta}$ (nonempty, as it contains the pairs $(\alpha,\alpha)$), $\mathcal{P}_{2}$ its complement in $\mathcal{S}\times\mathcal{S}$, and $S_{1}$, $S_{2}$ the sums defining $S$ restricted to $\mathcal{P}_{1}$, $\mathcal{P}_{2}$ ($S_{2}=0$ if $\mathcal{P}_{2}$ is empty), so $S=S_{1}+S_{2}$. The map $(\alpha,\beta)\mapsto(\bar\beta,\bar\alpha)$ is injective (it is its own inverse on $\mathcal{S}\times\mathcal{S}$) and maps $\mathcal{P}_{2}$ into $\mathcal{P}_{1}$, since $k_{\bar\beta}=k_{\beta}<k_{\alpha}=k_{\bar\alpha}$. By Step 3 and (W2), $v_{\bar\beta}v_{\bar\alpha}=v_{\beta}^{*}v_{\alpha}^{*}=(v_{\alpha}v_{\beta})^{*}$, so by Unitary Laws: Adjoints, the Bound One, and Reduction to Cyclically Reduced Words §adjoint for $\mu$ and $\nu$ and the rules for conjugation, $\Delta(v_{\bar\beta}v_{\bar\alpha})=\overline{\Delta(v_{\alpha}v_{\beta})}$, which has the same modulus; also $\sigma_{\bar\beta}\sigma_{\bar\alpha}=\sigma_{\alpha}\sigma_{\beta}$. Reindexing $S_{2}$ along this map and enlarging the index set to $\mathcal{P}_{1}$ (nonnegative terms) gives $S_{2}\le S_{1}$, hence

$$S\le2S_{1}.\tag{4.1}$$

Step 5 (one term). For a slot $\alpha$ with $w=w_{\alpha}$ and $k=k_{\alpha}$ put

$$\phi_{\alpha}=\frac{\hat c_{w}\,|x(w)|}{(k+1)^{2}},\qquad\psi_{\alpha}=\frac{\hat c_{w}\,|x(w)|}{\sqrt{k}},$$

both nonnegative, with $\phi_{\alpha}^{2}=\frac{c_{w}|x(w)|^{2}}{(k+1)^{4}}$ and $\psi_{\alpha}^{2}=\frac{c_{w}|x(w)|^{2}}{k}$. We show: for every $(\alpha,\beta)\in\mathcal{P}_{1}$,

$$\sigma_{\alpha}\sigma_{\beta}\,\bigl|\Delta(v_{\alpha}v_{\beta})\bigr|\le8\,\vartheta^{\,j(v_{\alpha},v_{\beta})}\,\phi_{\alpha}\,\psi_{\beta}\,h\bigl(\rho(v_{\alpha},v_{\beta})\bigr).\tag{5.1}$$

Write $w=w_{\alpha}$, $w'=w_{\beta}$, $k=k_{\alpha}\le k'=k_{\beta}$, $v=v_{\alpha}$, $v'=v_{\beta}$, $j=j(v,v')$ and $z=\rho(v,v')$, a reduced word in $U_{2n}$ of length $k+k'-2j$ by (2b). By (2c) and (2.1), $\Delta(vv')=\Delta(u_{z})$. If $u_{z}=\varnothing$, the left side of (5.1) is $0$, and so is the right side. Otherwise put $t=t_{z}$, $p=|u_{z}|\in\mathbb{N}$, $u=u_{z}\in W^{\circ}_{d,p}$ and $N=j+t$; then $|\Delta(vv')|=|x(u)|$, $h(z)=\vartheta^{t}Y(u)$, and $k+k'=|z|+2j=p+2N$. Since $c_{w}=\hat c_{w}\frac{\vartheta^{k}}{(k+1)^{2}}$, $c_{w'}=\hat c_{w'}\frac{\vartheta^{k'}}{(k'+1)^{2}}$ and $\vartheta^{k}\vartheta^{k'}=\vartheta^{k+k'}=\vartheta^{2N}\vartheta^{p}$, and since $\vartheta^{p}|x(u)|=(p+1)^{2}\hat c_{u}|x(u)|=\frac{(p+1)^{2}}{\sqrt{p}}Y(u)$,

$$\sigma_{\alpha}\sigma_{\beta}|x(u)|=\hat c_{w}|x(w)|\;\hat c_{w'}|x(w')|\;\vartheta^{2N}\;\frac{(p+1)^{2}}{(k+1)^{2}(k'+1)^{2}\sqrt{p}}\;Y(u).$$

Now $p\le k+k'\le2k'$, so $0<p+1\le2(k'+1)$ and $(p+1)^{2}\le4(k'+1)^{2}$; hence the right side is at most $4\,\phi_{\alpha}\;\hat c_{w'}|x(w')|\;\vartheta^{N}Y(u)\cdot\frac{\vartheta^{N}}{\sqrt{p}}$. By (C5) (with $N=j+t$), $(\vartheta^{N}\sqrt{k'})^{2}=\theta^{N}k'\le4p=(2\sqrt{p})^{2}$, so $\vartheta^{N}\sqrt{k'}\le2\sqrt{p}$, that is $\frac{\vartheta^{N}}{\sqrt{p}}\le\frac{2}{\sqrt{k'}}$. Therefore

$$\sigma_{\alpha}\sigma_{\beta}|\Delta(vv')|\le8\,\phi_{\alpha}\,\frac{\hat c_{w'}|x(w')|}{\sqrt{k'}}\,\vartheta^{j}\,\vartheta^{t}Y(u)=8\,\vartheta^{j}\phi_{\alpha}\psi_{\beta}\,h(z),$$

which is (5.1).

Step 6 (grouping the slots by their words). For $v\in W_{2d}$ let $\mathcal{S}(v)$ be the set of slots $\alpha$ with $v_{\alpha}=v$, and put $f(v)=\sum_{\alpha\in\mathcal{S}(v)}\phi_{\alpha}$ and $\ell(v)=\sum_{\alpha\in\mathcal{S}(v)}\psi_{\alpha}$ (both $0$ if $\mathcal{S}(v)$ is empty); $f,\ell\ge0$. Let $V=\{v_{\alpha}:\alpha\in\mathcal{S}\}\subseteq U_{n}$, nonempty and finite; $f$ and $\ell$ vanish outside $V$, and the sets $\mathcal{S}(v)$, $v\in V$, are nonempty, pairwise disjoint, with union $\mathcal{S}$. For $k\in[n]$ let $\mathcal{S}_{(k)}$ be the set of slots with $k_{\alpha}=k$; it is the set of dependent pairs $(w,(m,s))$ with $w\in W^{\circ}_{d,k}$ and $(m,s)\in[k]\times\{1,-1\}$, a set with $2k$ elements for each $w$ (by The Product of Two Sums over Finite Index Sets is a Sum over the Cartesian Product and claim 1 of Properties of a Sum over a Finite Index Set), so that for every map $\chi$ on $W^{\circ}_{d,k}$

$$\sum_{\alpha\in\mathcal{S}_{(k)}}\chi(w_{\alpha})=\sum_{w\in W^{\circ}_{d,k}}2k\,\chi(w).\tag{6.1}$$

(6a) If $v\in V$ has length $k$, then $\sum_{\alpha\in\mathcal{S}(v)}1\le4k$. Indeed, the map $\alpha=(w,m,s)\mapsto(m,s,\varepsilon(w_{m}))$ from $\mathcal{S}(v)$ to $[k]\times\{1,-1\}\times\{1,-1\}$ (note $k_{\alpha}=|v|=k$) is injective: given $(m,s,e)$, the word $r_{m}(w)$ equals $v$ if $s=1$ and $v^{*}$ if $s=-1$ (by (W2)), and by (W4) $w$ is determined by $r_{m}(w)$, $m$ and $e=\varepsilon(w_{m})$. Reindexing onto the image and enlarging to the product set, which has $4k$ elements, gives the claim.

(6b) For $k\in[n]$, $\Phi_{k}:=\sqrt{\sum_{v\in[2d]^{k}}f(v)^{2}}\le\frac{\sqrt{8}}{k}\sqrt{X_{k}}$. Indeed, for $v\in V$ of length $k$, (C1) and (6a) give $f(v)^{2}\le4k\sum_{\alpha\in\mathcal{S}(v)}\phi_{\alpha}^{2}=\frac{4k}{(k+1)^{4}}\sum_{\alpha\in\mathcal{S}(v)}c_{w_{\alpha}}|x(w_{\alpha})|^{2}$, and $f(v)=0$ for $v\in[2d]^{k}\setminus V$. The sets $\mathcal{S}(v)$, $v\in V\cap[2d]^{k}$, are pairwise disjoint with union $\mathcal{S}_{(k)}$ (as $|v_{\alpha}|=k_{\alpha}$). Summing and using (6.1),

$$\sum_{v\in[2d]^{k}}f(v)^{2}\le\frac{4k}{(k+1)^{4}}\sum_{\alpha\in\mathcal{S}_{(k)}}c_{w_{\alpha}}|x(w_{\alpha})|^{2}=\frac{8k^{2}}{(k+1)^{4}}X_{k}\le\frac{8}{k^{2}}X_{k},$$

the last step because $k^{4}\le(k+1)^{4}$. Taking square roots gives (6b).

(6c) $\Lambda:=\sqrt{\sum_{z\in U_{2n}}\ell(z)^{2}}\le\sqrt{8}\,\lVert x\rVert_{d,1}$. Indeed, for $v\in V$ of length $k$, (C1) and (6a) give $\ell(v)^{2}\le4k\sum_{\alpha\in\mathcal{S}(v)}\psi_{\alpha}^{2}=4\sum_{\alpha\in\mathcal{S}(v)}c_{w_{\alpha}}|x(w_{\alpha})|^{2}$, and $\ell$ vanishes on $U_{2n}\setminus V$. Since the $\mathcal{S}(v)$, $v\in V$, partition $\mathcal{S}$, and $\mathcal{S}$ is the disjoint union of the $\mathcal{S}_{(k)}$, $k\in[n]$, (6.1) and (0.1) give

$$\sum_{z\in U_{2n}}\ell(z)^{2}\le4\sum_{\alpha\in\mathcal{S}}c_{w_{\alpha}}|x(w_{\alpha})|^{2}=4\sum_{k=1}^{n}2k\,X_{k}\le8\,\lVert x\rVert_{d,1}^{2}.$$

(6d) Regrouping. For every map $\Psi:V\times V\to\mathbb{R}$,

$$\sum_{(\alpha,\beta)\in\mathcal{S}\times\mathcal{S}}\Psi(v_{\alpha},v_{\beta})\,\phi_{\alpha}\psi_{\beta}=\sum_{(v,v')\in V\times V}\Psi(v,v')\,f(v)\,\ell(v'),$$

because $\mathcal{S}\times\mathcal{S}$ is the disjoint union of the sets $\mathcal{S}(v)\times\mathcal{S}(v')$, $(v,v')\in V\times V$, on each of which $\Psi(v_{\alpha},v_{\beta})=\Psi(v,v')$, and $\sum_{\mathcal{S}(v)\times\mathcal{S}(v')}\phi_{\alpha}\psi_{\beta}=f(v)\ell(v')$ by The Product of Two Sums over Finite Index Sets is a Sum over the Cartesian Product.

Step 7 (the Cauchy--Schwarz argument over cancellations). Put $\Omega=\sqrt{\sum_{z\in U_{2n}}h(z)^{2}}$. For $k\in[n]$ and $j\in\{0,1,\dots,k\}$ let $Q_{k,j}$ be the set of $(v,v')\in V\times V$ with $|v|=k$ and $j(v,v')=j$. We show

$$\sum_{(v,v')\in Q_{k,j}}f(v)\,\ell(v')\,h\bigl(\rho(v,v')\bigr)\le\Phi_{k}\,\Lambda\,\Omega\tag{7.1}$$

(the left side being $0$ if $Q_{k,j}$ is empty). Put $\mathcal{A}=[2d]^{k-j}$, $\mathcal{C}=[2d]^{j}$ and $\mathcal{B}=U_{n}$. By Step 2, each $(v,v')\in Q_{k,j}$ determines words $a\in\mathcal{A}$, $c\in\mathcal{C}$, $b\in\mathcal{B}$ (indeed $|b|=|v'|-j\le n$) with $v=ac$, $v'=c^{*}b$ and $\rho(v,v')=ab$, and $(v,v')\mapsto(a,c,b)$ is injective since $(v,v')=(ac,c^{*}b)$. Reindexing onto the image and enlarging to $\mathcal{C}\times(\mathcal{A}\times\mathcal{B})$ (nonnegative terms; iterated sums as sums over pairs), the left side of (7.1) is at most

$$\sum_{c\in\mathcal{C}}\ \sum_{(a,b)\in\mathcal{A}\times\mathcal{B}}f(ac)\,\ell(c^{*}b)\,h(ab).$$

Fix $c\in\mathcal{C}$ and put $\eta(c)=\sqrt{\sum_{a\in\mathcal{A}}f(ac)^{2}}$ and $\zeta(c)=\sqrt{\sum_{b\in\mathcal{B}}\ell(c^{*}b)^{2}}$. By (C1) on $\mathcal{A}\times\mathcal{B}$ and The Product of Two Sums over Finite Index Sets is a Sum over the Cartesian Product,

$$\sum_{(a,b)}f(ac)\ell(c^{*}b)h(ab)\le\sqrt{\textstyle\sum_{(a,b)}f(ac)^{2}\ell(c^{*}b)^{2}}\;\sqrt{\textstyle\sum_{(a,b)}h(ab)^{2}}=\eta(c)\,\zeta(c)\,\sqrt{\textstyle\sum_{(a,b)}h(ab)^{2}}.$$

The map $(a,b)\mapsto ab$ is injective on $\mathcal{A}\times\mathcal{B}$ (the length of $a$ being fixed), with values in $U_{2n}$ since $|ab|\le(k-j)+n\le2n$; so $\sum_{(a,b)}h(ab)^{2}\le\Omega^{2}$ by reindexing and enlarging the index set, and the inner sum is at most $\eta(c)\zeta(c)\Omega$. By (C1) on $\mathcal{C}$, $\sum_{c}\eta(c)\zeta(c)\le\sqrt{\sum_{c}\eta(c)^{2}}\sqrt{\sum_{c}\zeta(c)^{2}}$. Here $\sum_{c}\eta(c)^{2}=\sum_{(a,c)\in\mathcal{A}\times\mathcal{C}}f(ac)^{2}=\sum_{v\in[2d]^{k}}f(v)^{2}=\Phi_{k}^{2}$, because $(a,c)\mapsto ac$ is a bijection from $\mathcal{A}\times\mathcal{C}$ onto $[2d]^{k}$ (Conventions); and $\sum_{c}\zeta(c)^{2}=\sum_{(c,b)\in\mathcal{C}\times\mathcal{B}}\ell(c^{*}b)^{2}\le\Lambda^{2}$, because $(c,b)\mapsto c^{*}b$ is injective on $\mathcal{C}\times\mathcal{B}$ (the first $j$ letters of $c^{*}b$ form $c^{*}$, which determines $c=(c^{*})^{*}$ by (W2), and the rest is $b$) with values in $U_{2n}$, as $|c^{*}b|\le j+n\le2n$. Combining the three estimates gives (7.1).

Step 8 (the norm of $h$). We show $\Omega^{2}\le2\,\lVert x\rVert_{d,1}^{2}$. Let $Z$ be the set of reduced $z\in U_{2n}$ with $u_{z}\ne\varnothing$; $h$ vanishes on $U_{2n}\setminus Z$, so $\Omega^{2}=\sum_{z\in Z}h(z)^{2}$, which is $0$ if $Z$ is empty. For $z\in Z$, $h(z)^{2}=\theta^{t_{z}}Y(u_{z})^{2}$, $u_{z}$ lies in the finite set $\mathcal{U}=\bigcup_{p\in[2n]}W^{\circ}_{d,p}$ (as $1\le|u_{z}|\le|z|\le2n$), and $t_{z}\in\{0,1,\dots,n\}$ (as $2t_{z}\le|z|\le2n$). For $t\in\{0,\dots,n\}$ and $u\in\mathcal{U}$ let $Z_{t,u}$ be the set of $z\in Z$ with $t_{z}=t$ and $u_{z}=u$; these sets are pairwise disjoint with union $Z$. On $Z_{t,u}$ the map $z\mapsto q_{z}\in[2d]^{t}$ is injective, since $z=q_{z}\,u\,q_{z}^{*}$; hence $\sum_{z\in Z_{t,u}}1\le1$ if $t=0$, and $\sum_{z\in Z_{t,u}}1\le(2d)^{t}$ if $t\ge1$ and $Z_{t,u}$ is nonempty, by Counting Cyclically Reduced Words: Finiteness and the Bound (2d)^k §count applied to the image with $a$ constantly $1$ and $M=1$. Since $\theta^{t}(2d)^{t}=(2d\,\theta)^{t}$, we get $\sum_{z\in Z_{t,u}}h(z)^{2}\le(2d\,\theta)^{t}\,Y(u)^{2}$ in all cases, and summing over $(t,u)$ with (C4),

$$\Omega^{2}\le\sum_{u\in\mathcal{U}}Y(u)^{2}\sum_{t=0}^{n}(2d\,\theta)^{t}\le2\sum_{u\in\mathcal{U}}Y(u)^{2}=2\sum_{p=1}^{2n}\ \sum_{u\in W^{\circ}_{d,p}}p\,c_{u}|x(u)|^{2}=2\sum_{p=1}^{2n}p\,X_{p}\le2\,\lVert x\rVert_{d,1}^{2},$$

using that $\mathcal{U}$ is the disjoint union of the $W^{\circ}_{d,p}$, $p\in[2n]$, and (0.1). Hence $\Omega\le\sqrt{2}\,\lVert x\rVert_{d,1}$.

Step 9 (conclusion). By (5.1), enlarging the index set from $\mathcal{P}_{1}$ to $\mathcal{S}\times\mathcal{S}$ (the right sides of (5.1) being nonnegative), and (6d) with $\Psi(v,v')=8\,\vartheta^{\,j(v,v')}h(\rho(v,v'))$,

$$S_{1}\le8\sum_{(v,v')\in V\times V}\vartheta^{\,j(v,v')}f(v)\,\ell(v')\,h\bigl(\rho(v,v')\bigr).$$

Every $(v,v')\in V\times V$ has $|v|=k\in[n]$ and $j(v,v')\in\{0,\dots,\min(k,|v'|)\}\subseteq\{0,\dots,k\}$, so $V\times V$ is the disjoint union of the sets $Q_{k,j}$, $k\in[n]$, $j\in\{0,\dots,k\}$, and $\vartheta^{\,j(v,v')}=\vartheta^{j}$ on $Q_{k,j}$. Splitting over the nonempty $Q_{k,j}$, applying (7.1), adding the nonnegative bounds for the empty ones, and using (C4),

$$S_{1}\le8\,\Lambda\,\Omega\sum_{k=1}^{n}\Phi_{k}\sum_{j=0}^{k}\vartheta^{j}\le16\,\Lambda\,\Omega\sum_{k=1}^{n}\Phi_{k}.$$

By (6b), (C1) on $[n]$, Order Reversal under Reciprocals, and Summability of the Reciprocals of the Squares §partial (which gives $\sum_{k=1}^{n}\frac{1}{k^{2}}\le2-\frac1n\le2$) and (0.1),

$$\sum_{k=1}^{n}\Phi_{k}\le\sqrt{8}\sum_{k=1}^{n}\frac1k\sqrt{X_{k}}\le\sqrt{8}\,\sqrt{\textstyle\sum_{k=1}^{n}\frac{1}{k^{2}}}\,\sqrt{\textstyle\sum_{k=1}^{n}X_{k}}\le\sqrt{8}\,\sqrt{2}\,\lVert x\rVert_{d}=4\,\lVert x\rVert_{d}.$$

With (6c) and Step 8, $\Lambda\,\Omega\le\sqrt{8}\,\sqrt{2}\,\lVert x\rVert_{d,1}^{2}=4\,\lVert x\rVert_{d,1}^{2}$, so $S_{1}\le16\cdot4\cdot4\,\lVert x\rVert_{d}\lVert x\rVert_{d,1}^{2}=256\,\lVert x\rVert_{d}\lVert x\rVert_{d,1}^{2}$. By (3.1) and (4.1),

$$\Bigl|\sum_{i\in[d]}\Bigl(\mu\bigl(\Xi^{i}_{x,n}\Xi^{i}_{x,n}\bigr)-\nu\bigl(\Xi^{i}_{x,n}\Xi^{i}_{x,n}\bigr)\Bigr)\Bigr|\le\frac14\,S\le\frac12\,S_{1}\le128\,\lVert x\rVert_{d}\,\lVert x\rVert_{d,1}^{2}.$$

As $\mu,\nu\in\mathcal{L}_{d}$ and $n\in\mathbb{N}$ were arbitrary, $K=128$ is a trilinear constant.