Proof of The Scale Set for the Van Trees Assembly: Eventual Validity of the Copy-Side Constraints, Lower Bounds for the Chernoff Exponents, and Vanishing of the Error Majorants

Tools. We write $\mathsf{T}$ for Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities and use its claims as follows. By $\mathsf{T}$.1 and $\mathsf{T}$.2, for $N\in\mathbb{N}$ (so $N\ge1$) and real $a,b$: $N^{a}>0$, $N^{a}N^{b}=N^{a+b}$, $(N^{a})^{b}=N^{ab}$, $N^{a}\le N^{b}$ whenever $a\le b$, and $N^{a}\ge1$ for $a\ge0$; also $(N^{a})^{1/2}=N^{a/2}$ and $N^{1/2}=\sqrt{N}$. By $\mathsf{T}$.3(a) and the scalar-multiple law of Arithmetic of Limits of Real Sequences, $cN^{-a}\to0$ for all real $c$ and $a>0$; hence, by $\mathsf{T}$.3(d), for every $x>0$ there is $N_{0}$ with $cN^{-a}<x$ for all $N\ge N_{0}$ (eventual smallness of powers). By $\mathsf{T}$.3(c), a sequence $(u_{N})$ with $|u_{N}-u|\le v_{N}$ for all $N\ge N_{0}$ and $v_{N}\to0$ converges to $u$ (eventual domination); in particular a sequence with $0\le u_{N}\le cN^{-a}$ for all $N\ge N_{0}$ converges to $0$. By $\mathsf{T}$.3(b), $N^{k}\exp(-cN^{a})\to0$ for $a>0$, $c>0$, $k\ge0$. By $\mathsf{T}$.4, for $x\ge0$ the number $\lfloor x\rfloor+1$ is a natural number in $(x,x+1]$, and for every real $x$ there is $N_{0}\in\mathbb{N}$ with $N>x$ for all $N\ge N_{0}$. By $\mathsf{T}$.5, square roots are monotone and $\sqrt{a+b}\le\sqrt a+\sqrt b$, $\sqrt{ab}=\sqrt a\sqrt b$ for $a,b\ge0$; by $\mathsf{T}$.6, $\exp(x)-1\le x\exp(x)$ and $\exp(x)-1-x\le x^{2}\exp(x)$ for $x\ge0$, and $\exp$ is nondecreasing with $\exp(-x)\le1$ for $x\ge0$. Limits of sums, products and scalar multiples are those of Arithmetic of Limits of Real Sequences; comparison and squeeze are claims 1 and 2 of Order Properties of Limits of Real Sequences; constant sequences converge to their value. Finally, $\mathsf{T}$.3(e) gives $\sqrt{u_{N}}\to\sqrt{u}$ and $u_{N}^{1/4}\to u^{1/4}$ when $u_{N}\ge0$ and $u_{N}\to u$, and $\mathsf{T}$.3(f) gives $\exp(u_{N})\to\exp(u)$ when $u_{N}\to u$. Whenever finitely many natural numbers $N_{0},N_{0}',\dots$ have been produced, "for all $N$ beyond them" means for all $N$ at least as large as the largest of them. Each of the finitely many thresholds produced in the proof of claim 1 is a natural number; $N_{\mathrm{c}}$ is taken to be the largest of them.

Proof of claim 1. (i) By $\mathsf{T}$.4 with $x=N^{1/16}\ge0$, $\mathsf{m}_{N}\in\mathbb{N}$ and $N^{1/16}<\mathsf{m}_{N}\le N^{1/16}+1\le2N^{1/16}$, since $N^{1/16}\ge1$.

(ii) By $\mathsf{T}$.4 with $x=N(Bs+1)\ge0$, $R_{N}\in\mathbb{N}$ and $N(Bs+1)<R_{N}\le N(Bs+1)+1\le N(Bs+2)$, since $1\le N$. Also $R_{N}>N(Bs+1)\ge NBs$ and $R_{N}>N(Bs+1)\ge N$.

(iii) By $\mathsf{T}$.4 with $x=R_{N}N^{-3/4}>0$, $J_{N}\in\mathbb{N}$ and $R_{N}N^{-3/4}<J_{N}\le R_{N}N^{-3/4}+1$. Hence $\mu_{N}=R_{N}/J_{N}<R_{N}/(R_{N}N^{-3/4})=N^{3/4}$, using $N^{-3/4}N^{3/4}=1$. Since $R_{N}\ge N\ge N^{3/4}$ we have $R_{N}N^{-3/4}\ge1$, so $J_{N}\le2R_{N}N^{-3/4}$ and $\mu_{N}\ge N^{3/4}/2$. By eventual smallness of powers there is a threshold beyond which $N^{-3/4}\le\tfrac14$, i.e. $N^{3/4}\ge4$ and $\mu_{N}\ge2$. Next, $d_{N}=l(l-1)J_{N}\ge2J_{N}>2R_{N}N^{-3/4}\ge2NN^{-3/4}=2N^{1/4}\ge N^{1/4}$, and $J_{N}\le R_{N}N^{-3/4}+1\le(Bs+2)NN^{-3/4}+1=(Bs+2)N^{1/4}+1\le(Bs+3)N^{1/4}$, whence $d_{N}\le l(l-1)(Bs+3)N^{1/4}$.

(iv) $D_{N}=N^{3/64}\le N^{4/64}=N^{1/16}<\mathsf{m}_{N}$. Hence $\mathsf{m}_{N}+l(l-1)(D_{N}+\mathsf{m}_{N})\le\mathsf{m}_{N}(1+2l(l-1))$ and $A_{N}\le C_{A}\mathsf{m}_{N}\le2C_{A}N^{1/16}$ by (i). As $A_{N}\ge0$, $0\le\Lambda_{1}sA_{N}<\Lambda_{1}sA_{N}+1=L_{N}$. Since $L_{N}\ge1>0$, $\mathsf{T}$.4 gives $M_{N}\in\mathbb{N}$ and $L_{N}<M_{N}\le L_{N}+1$. Finally $M_{N}+2\le M_{N}+3\le L_{N}+4=\Lambda_{1}sA_{N}+5\le2\Lambda_{1}sC_{A}N^{1/16}+5N^{1/16}=C_{M}N^{1/16}$, using $N^{1/16}\ge1$.

(v) By eventual smallness of powers ($N^{-3/64}\le\tfrac14$ beyond a threshold) we have $D_{N}=N^{3/64}\ge4$ beyond it. Since $N\ge1$, $N^{-3/8}\le N^{0}=1$ and $N^{-1/8}\le1$, while all powers are positive; so $0<\eta_{N}\le1$, $0<\delta_{N}\le1$, $\zeta_{N}>0$. By $\mathsf{T}$.4 with $x=N^{1/4}$, $\theta_{N}\in\mathbb{N}$ and $N^{1/4}<\theta_{N}\le N^{1/4}+1\le2N^{1/4}$.

(vi) By (i) and (iii), $\mathsf{m}_{N}\le2N^{1/16}$ and $\mu_{N}/2\ge N^{3/4}/4$, and $2N^{1/16}<N^{3/4}/4$ holds as soon as $8N^{-11/16}<1$, since $N^{3/4}N^{-1/16}=N^{11/16}$; this holds beyond a threshold by eventual smallness of powers.

(vii) By (iv), $\varepsilon_{0,N}=\Gamma A_{N}/(N\underline{b})\le(2\Gamma C_{A}/\underline{b})N^{1/16}N^{-1}=(2\Gamma C_{A}/\underline{b})N^{-15/16}$; by (v), $2\theta_{N}\varepsilon_{0,N}\le(8\Gamma C_{A}/\underline{b})N^{1/4}N^{-15/16}=(8\Gamma C_{A}/\underline{b})N^{-11/16}$; by (iv), $\bar{E}_{N}\le(4\tilde{l}s\Gamma^{2}C_{A}^{2}/\underline{b})N^{1/8}N^{-1}=(4\tilde{l}s\Gamma^{2}C_{A}^{2}/\underline{b})N^{-7/8}$; with $C_{E}=\tilde{l}s\tilde{B}(2\Gamma C_{A}/\underline{b})^{2}$, $\mathsf{E}^{\star}_{N}\le C_{E}N\,N^{-15/8}=C_{E}N^{-7/8}$ and $\mathsf{E}^{\mathrm{ch}}_{N}\le8C_{E}\,N\,(2N^{1/4})^{2}N^{-15/8}=32C_{E}N^{-3/8}$. By (i) and (iii), $\mathsf{m}_{N}^{2}/\mu_{N}\le4N^{1/8}/(N^{3/4}/2)=8N^{-5/8}\le8$, so $\kappa_{0,N}\le1+4\exp(8)$ (as $\exp$ is nondecreasing) and $\bar{\mathsf{j}}_{N}\le8(1+4\exp(8))N^{-5/8}$; and $\kappa^{\mathrm{mv}}_{N}=\mathsf{m}_{N}^{2}N^{-1}N^{3/8}\le4N^{1/8-1+3/8}=4N^{-1/2}$. Each of these seven majorants is a constant times a negative power of $N$, so by eventual smallness of powers each of the seven inequalities of (vii) holds beyond a threshold.

(viii) By (iii) and monotonicity of the square root, $x_{N}=\mu_{N}^{1/2}N^{1/32}\le(N^{3/4})^{1/2}N^{1/32}=N^{3/8}N^{1/32}=N^{13/32}$. Hence, by (i), $\mathsf{m}_{N}(x_{N}+\mathsf{m}_{N})\le2N^{1/16}(N^{13/32}+2N^{1/16})=2N^{15/32}+4N^{1/8}\le6N^{15/32}$, since $N^{1/8}\le N^{15/32}$. On the other hand $\delta_{N}\mu_{N}/2\ge N^{-1/8}N^{3/4}/4=N^{5/8}/4$. Now $6N^{15/32}\le N^{5/8}/4$ holds as soon as $24N^{-5/32}\le1$, since $N^{5/8}N^{-15/32}=N^{5/32}$; this holds beyond a threshold. This completes the proof of claim 1, with $N_{\mathrm{c}}$ the largest of the thresholds.

Proof of claim 2. Let $k>0$ and $0\le x\le x'$. Then $x^{2}\le x'^{2}$ (Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field), so $\varpi_{k}(x)\le x^{2}/(4k)\le x'^{2}/(4k)$ and $\varpi_{k}(x)\le x/2\le x'/2$; hence $\varpi_{k}(x)\le\varpi_{k}(x')$, the smaller of the two bounds. Let $n\ge0$, $N\in\mathbb{N}$ and $y=(n+2)^{1/2}N^{1/32}$. Then $y^{2}=(n+2)N^{1/16}$, so $y^{2}/(4(n+2))=N^{1/16}/4\ge N^{1/32}/4$, and $y/2\ge\sqrt{2}\,N^{1/32}/2\ge N^{1/32}/4$ because $(n+2)^{1/2}\ge\sqrt2$ by monotonicity of the square root and $\sqrt2/2\ge1/4$ (as $\sqrt2\ge1/2$, since $2\ge1/4$); hence $\varpi_{n+2}(y)\ge N^{1/32}/4$. Now let $N\ge N_{\mathrm{c}}$. First, $x_{N}^{2}=\mu_{N}N^{1/16}$ gives $x_{N}^{2}/(4\mu_{N})=N^{1/16}/4\ge N^{1/32}/4$, and $\mu_{N}\ge2\ge1$ gives $\mu_{N}^{1/2}\ge1$ and $x_{N}/2\ge N^{1/32}/2$; so $\varpi_{\mu_{N}}(x_{N})\ge N^{1/32}/4$. Second, by claim 1(vi), $\mu_{N}-\mathsf{m}_{N}\ge\mu_{N}/2\ge0$, so by monotonicity $\varpi_{\mu_{N}}(\mu_{N}-\mathsf{m}_{N})\ge\varpi_{\mu_{N}}(\mu_{N}/2)=(\mu_{N}/16)\wedge(\mu_{N}/4)=\mu_{N}/16\ge N^{3/4}/32$. Third, by claim 1(v), $D_{N}\ge4$, so $D_{N}-2\ge D_{N}/2\ge0$, and by claim 1(iv), $M_{N}+2\le C_{M}N^{1/16}$; hence

the last step because $C_{M}\ge5$ gives $c_{\mathrm{w}}\le\tfrac1{80}$. Thus $\varpi_{M_{N}+2}(D_{N}-2)\ge c_{\mathrm{w}}N^{1/32}$.

Proof of claim 3. First, for every $N\in\mathbb{N}$, all quantities of the statement are well-defined nonnegative real numbers: they are formed from nonnegative data by sums, products, quotients by positive numbers, square roots of nonnegative numbers and exponentials, together with the differences $\exp(u)-1\ge0$ for $u\ge0$ ($\exp$ being nondecreasing with $\exp(0)=1$) and $\exp(u)-1-u\ge0$ for $u\ge0$ ($\mathsf{T}$.6), the product $\kappa_{0,N}-1=\frac{\mathsf{m}_{N}^{2}}{2\mu_{N}}\exp(\mathsf{m}_{N}^{2}/\mu_{N})\ge0$ of nonnegative numbers, and, inside $\mathsf{g}_{N}$, the arguments $\mu_{N}-\mathsf{m}_{N}$ and $D_{N}-2$ of $\varpi$, which are nonnegative whenever the formula for $\mathsf{g}_{N}$ is read; in particular $\mathsf{J}_{N}\ge0$ and $\mathcal{I}_{N}\ge0$. Throughout the rest of this proof, $N\ge N_{\mathrm{c}}$, so that all statements of claims 1 and 2 are available; every bound below is of the form "$0\le u_{N}\le$ (constant)$\cdot N^{-a}$" or "$0\le u_{N}\le$ (constant)$\cdot N^{k}\exp(-cN^{a})$", and the corresponding limit $u_{N}\to0$ then follows from eventual domination together with $cN^{-a}\to0$ or $N^{k}\exp(-cN^{a})\to0$. We also use the bounds $\mathsf{w}_{N}\le\sqrt{2}\Lambda l(l-1)(Bs+2)N\,N^{-1/16}=C_{\mathrm{w}}N^{15/16}$ with $C_{\mathrm{w}}=\sqrt{2}\Lambda l(l-1)(Bs+2)$ (from claim 1(i),(ii)), hence $\mathsf{w}_{N}^{2}/N\le C_{\mathrm{w}}^{2}N^{7/8}$, and $\alpha_{N}\le C_{\alpha}N^{1/8}$ with $C_{\alpha}=(2l(l-1)(Bs+3))^{1/2}\mathsf{c}_{0}\bar\Phi^{2}$ (from claim 1(iii) and monotonicity of the square root, $(N^{1/4})^{1/2}=N^{1/8}$).

(a) $\varepsilon_{S,N}$ and $\varepsilon_{\mathrm{ctl},N}$ are constant multiples of $N^{-1/4}$, so they converge to $0$. The bounds recorded in the proof of claim 1(vii) give $\varepsilon_{0,N}\to0$ and $\bar{E}_{N}\to0$. By $\mathsf{T}$.6 and $\bar{E}_{N}\le1$, $\mathsf{c}_{N}=\exp(\bar{E}_{N})-1\le\bar{E}_{N}\exp(\bar{E}_{N})\le\exp(1)\bar{E}_{N}$, so $\mathsf{c}_{N}\to0$. By $\mathsf{E}^{\star}_{N}\le1$ and $\mathsf{T}$.6, $\exp(9\mathsf{E}^{\star}_{N})-1\le9\mathsf{E}^{\star}_{N}\exp(9)$, so, writing $t^{3/2}=t\,t^{1/2}$ for $t\ge0$, using $(\mathsf{E}^{\star}_{N})^{2}=\mathsf{E}^{\star}_{N}(\mathsf{E}^{\star}_{N})^{1/2}(\mathsf{E}^{\star}_{N})^{1/2}\le(\mathsf{E}^{\star}_{N})^{3/2}$ (as $(\mathsf{E}^{\star}_{N})^{1/2}\le1$ by monotonicity of the square root) and $\sqrt{ab}=\sqrt a\sqrt b$,

where $(C_{E}N^{-7/8})^{3/2}=C_{E}^{3/2}N^{-7/8}N^{-7/16}$ by multiplicativity of the square root and $(N^{-7/8})^{1/2}=N^{-7/16}$, and monotonicity of $t\mapsto t^{3/2}$ on $[0,\infty)$ follows from that of $t\mapsto t$ and $t\mapsto t^{1/2}$; so $\mathsf{e}^{\star}_{N}\to0$. Next, $0\le\kappa_{0,N}-1=\frac{\mathsf{m}_{N}^{2}}{2\mu_{N}}\exp(\mathsf{m}_{N}^{2}/\mu_{N})\le4N^{-5/8}\exp(8)$ by the bound $\mathsf{m}_{N}^{2}/\mu_{N}\le8N^{-5/8}\le8$ of claim 1(vii); so $\kappa_{0,N}\to1$, and $\bar{\mathsf{j}}_{N}\le8(1+4\exp(8))N^{-5/8}$ gives $\bar{\mathsf{j}}_{N}\to0$. Since $\bar{\mathsf{j}}_{N}\le1$, $(1+\bar{\mathsf{j}}_{N})^{1/2}+1\le\sqrt2+1$, so $\mathsf{j}^{\star}_{N}\le(\sqrt2+1)\bar{\mathsf{j}}_{N}^{1/2}\le(\sqrt2+1)\bigl(8(1+4\exp(8))\bigr)^{1/2}N^{-5/16}$ by monotonicity of the square root and $(N^{-5/8})^{1/2}=N^{-5/16}$; so $\mathsf{j}^{\star}_{N}\to0$.

For $\bar\Pi_{N}$: by claim 2 and monotonicity of $\exp$, $\exp(-\varpi_{\mu_{N}}(x_{N}))\le\exp(-N^{1/32}/4)$; by claim 1(v),(vii), $\theta_{N}\delta_{N}/2\ge N^{1/4}N^{-1/8}/2=N^{1/8}/2\ge N^{1/32}/4$ and $\mathsf{E}^{\mathrm{ch}}_{N}\le1$, so $\exp(-\theta_{N}\delta_{N}/2+\mathsf{E}^{\mathrm{ch}}_{N})\le\exp(1)\exp(-N^{1/32}/4)$. Hence, with claim 1(iii),

and $\bar\Pi_{N}\to0$ by $\mathsf{T}$.3(b). For $\mathsf{g}_{N}$: by claim 2, claim 1(ii),(iii),(iv) and $N^{1/16}\le N$,

with $C_{g}=l(l-1)\bigl((Bs+3)+2(Bs+3)C_{M}\bigr)$ (using $(Bs+2)N+1\le(Bs+3)N$ and $N\,N^{1/16}\le N^{2}$); both terms converge to $0$ by $\mathsf{T}$.3(b), so $\mathsf{g}_{N}\to0$ by the sum law and squeeze.

For $\mathsf{e}_{F,N}$: its first summand is a constant multiple of $N^{-1/4}$; in the second, by claim 1(iii),(iv), $A_{N}/\mu_{N}\le2C_{A}N^{1/16}/(N^{3/4}/2)=4C_{A}N^{-11/16}\le4C_{A}$, so $3+2(\Lambda_{1}sA_{N}+\mu_{N})/\mu_{N}\le5+8\Lambda_{1}sC_{A}$ and the second summand is at most $2\Lambda l(l-1)(5+8\Lambda_{1}sC_{A})N^{-1/4}$ (as $\mu_{N}/N\le N^{-1/4}$). Hence $\mathsf{e}_{F,N}\to0$. For $\epsilon_{\psi,N}$: $\mathsf{w}_{N}/N\le C_{\mathrm{w}}N^{-1/16}$, and $D_{N}+\Lambda_{2}sA_{N}^{2}/N\le N^{3/64}+4\Lambda_{2}sC_{A}^{2}N^{1/8-1}\le(1+4\Lambda_{2}sC_{A}^{2})N^{3/64}$, so the middle summand inside the bracket is at most $\sqrt2l(l-1)C_{\mathrm{w}}(1+4\Lambda_{2}sC_{A}^{2})N^{-1/64}$, since $N^{-1/16}N^{3/64}=N^{-1/64}$; the third summand is a constant multiple of $N^{-1/4}$. Hence each of the three summands converges to $0$, and by the sum and scalar-multiple laws $\epsilon_{\psi,N}\to0$. Then $\kappa_{N}\to0$ by the sum and product laws, as $\kappa_{N}$ is a polynomial expression in $\varepsilon_{S,N}$ and $\epsilon_{\psi,N}$ (both null) with constant coefficients and no constant term. For $\mathsf{Q}_{N}$: $\zeta_{N}\to0$, so $(1+\zeta_{N})(\mathsf{Q}+\tilde{l}s\kappa_{N})\to\mathsf{Q}$ by the sum and product laws; and, using claim 1(iv) and $1+1/\zeta_{N}=1+N^{1/2}\le2N^{1/2}$,

which converges to $0$; hence $\mathsf{Q}_{N}\to\mathsf{Q}$.

For the bad term, put $\mathsf{b}_{N}=2d_{N}\mathsf{w}_{N}^{2}\mathsf{B}_{N}/N$. By claim 1(iii) and the bound on $\mathsf{w}_{N}^{2}/N$, $2d_{N}\mathsf{w}_{N}^{2}/N\le2l(l-1)(Bs+3)C_{\mathrm{w}}^{2}N^{9/8}=C_{b}N^{9/8}$. Since $\bar{\mathsf{j}}_{N}\le1$ and $\mathsf{c}_{N}\le\exp(1)\bar{E}_{N}\le\exp(1)$, we have $(1+\bar{\mathsf{j}}_{N})^{1/2}\le\sqrt2$ and, by monotonicity of the square root and $\sqrt{ab}=\sqrt a\sqrt b$, $(\mathsf{c}_{N}\bar\Pi_{N})^{1/2}\le\exp(1)^{1/2}\bar\Pi_{N}^{1/2}$, so

where $d_{N}^{3/2}=d_{N}d_{N}^{1/2}$. Write $\bar\Pi_{N}\le C_{\Pi}N^{1/4}\exp(-N^{1/32}/4)$ for the bound obtained above, with $C_{\Pi}=(1+\exp(1))l(l-1)(Bs+3)$, and note $d_{N}\le C_{d}N^{1/4}$ with $C_{d}=l(l-1)(Bs+3)$. Since $\sqrt{\exp(-u)}=\exp(-u/2)$ for real $u$ (the right-hand side is positive with square $\exp(-u)$), monotonicity and multiplicativity of the square root give $\bar\Pi_{N}^{1/2}\le C_{\Pi}^{1/2}N^{1/8}\exp(-N^{1/32}/8)$ and $\mathsf{g}_{N}^{1/2}\le C_{g}^{1/2}\bigl(N^{1/8}\exp(-N^{3/4}/64)+N\exp(-c_{\mathrm{w}}N^{1/32}/2)\bigr)$ (using $\sqrt{a+b}\le\sqrt a+\sqrt b$). Multiplying out, $\mathsf{b}_{N}\le C_{b}N^{9/8}\mathsf{B}_{N}$ is bounded by a finite sum of terms of the form (constant)$\cdot N^{k}\exp(-cN^{a})$ with $k\ge0$, $c>0$, $a>0$; explicitly, the five terms are bounded by

Each of these converges to $0$ by $\mathsf{T}$.3(b), so $\mathsf{b}_{N}\to0$ by the sum law and squeeze.

Finally, by the bound on $\mathsf{e}^{\star}_{N}$ obtained above, $\frac{\mathsf{w}_{N}^{2}}{N}\mathsf{e}^{\star}_{N}\le C_{\mathrm{w}}^{2}C_{\mathsf{e}}C_{E}^{3/2}N^{7/8-21/16}=C_{\mathrm{w}}^{2}C_{\mathsf{e}}C_{E}^{3/2}N^{-7/16}\to0$. Next $\frac{\mathsf{w}_{N}^{2}}{N}\bar{E}_{N}(N^{-1/2}+\mathsf{c}_{\star}N^{-1})\le C_{\mathrm{w}}^{2}(4\tilde{l}s\Gamma^{2}C_{A}^{2}/\underline{b})(1+\mathsf{c}_{\star})N^{7/8-7/8-1/2}\to0$, using $N^{-1}\le N^{-1/2}$; and $\frac{\mathsf{w}_{N}^{2}}{N}\mathsf{c}_{N}\mathsf{j}^{\star}_{N}\le C_{\mathrm{w}}^{2}\exp(1)(4\tilde{l}s\Gamma^{2}C_{A}^{2}/\underline{b})(\sqrt2+1)(8(1+4\exp(8)))^{1/2}N^{7/8-7/8-5/16}\to0$. Since $\delta_{N}\to0$, $\kappa_{0,N}\mathsf{P}\to\mathsf{P}$, $\mathsf{Q}_{N}\to\mathsf{Q}$ and $\mathsf{b}_{N}\to0$, the sum and product laws give $\mathsf{J}_{N}\to1\cdot(\mathsf{P}+\mathsf{Q}+0)+0=\mathsf{P}+\mathsf{Q}$.

(b) $\mathsf{e}_{2,N}$ is a sum of constant multiples of $N^{-1/2}$, $\varepsilon_{S,N}$ and $\varepsilon_{\mathrm{ctl},N}$, with no constant term, so $\mathsf{e}_{2,N}\to0$ by the sum and scalar-multiple laws. For $\mathsf{e}_{3,N}$: put $C_{\mathrm{clk}}=\Lambda_{1}s(C_{\mathrm{flw}}+1)+C_{\mathrm{Lip}}(TC_{\mathrm{ctl}})^{1/2}$, so that $w_{1,N}=C_{\mathrm{clk}}NN^{-1/4}=C_{\mathrm{clk}}N^{3/4}$, while $w_{2,N}=\mu_{N}\le N^{3/4}$; hence $w_{i,N}+3\le(C_{\mathrm{clk}}+4)N^{3/4}$ and $w_{i,N}+4\le(C_{\mathrm{clk}}+5)N^{3/4}$ for $i\in\{1,2\}$, using $1\le N^{3/4}$. By monotonicity of the square root, $(w_{i,N}+3)^{1/2}\le(C_{\mathrm{clk}}+4)^{1/2}N^{3/8}$. Next, $R_{N}\ge1$ gives $8(4096+17R_{N}^{4})\le8\cdot4113R_{N}^{4}\le14^{4}R_{N}^{4}$ (as $8\cdot4113=32904\le38416=14^{4}$), so $(8(4096+17R_{N}^{4}))^{1/4}\le14R_{N}\le14(Bs+2)N$ by monotonicity of the fourth root; and $(2(R_{N}+1)(w_{i,N}+4))^{1/4}\le(4(Bs+2)(C_{\mathrm{clk}}+5))^{1/4}(N\,N^{3/4})^{1/4}=(4(Bs+2)(C_{\mathrm{clk}}+5))^{1/4}N^{7/16}$, using $R_{N}+1\le2R_{N}$. Therefore

and multiplying by $\mathsf{c}_{0}l(l-1)(\sqrt2+H)N^{-1/2}$ gives $\mathsf{e}_{3,N}\le C_{3}\bigl(N^{-3/32}+N^{-1/2}+N^{15/16}\exp(-N^{1/32}/16)\bigr)$ with $C_{3}=\mathsf{c}_{0}l(l-1)(\sqrt2+H)\max\{2(C_{\mathrm{clk}}+4)^{1/2},4,28(Bs+2)(4(Bs+2)(C_{\mathrm{clk}}+5))^{1/4}\}$ (the largest of three reals), since $N^{3/8+1/32-1/2}=N^{-3/32}$ and $N^{1+7/16-1/2}=N^{15/16}$; each of the three terms converges to $0$, so $\mathsf{e}_{3,N}\to0$. Next, $\mathsf{e}_{4,N}=\eta_{N}^{1/2}\alpha_{N}\le N^{-3/16}C_{\alpha}N^{1/8}=C_{\alpha}N^{-1/16}\to0$. For $\mathsf{e}_{5,N}$: since $\mu_{N}\ge1$, $3\mu_{N}^{2}+\mu_{N}\le4\mu_{N}^{2}$, so $\mathsf{k}_{N}\le\alpha_{N}(4\mu_{N}^{2})^{1/4}=\sqrt2\,\alpha_{N}\mu_{N}^{1/2}\le\sqrt2C_{\alpha}N^{1/8}N^{3/8}=\sqrt2C_{\alpha}N^{1/2}$ (using $(4\mu_{N}^{2})^{1/2}=2\mu_{N}$ and $(2\mu_{N})^{1/2}=\sqrt2\mu_{N}^{1/2}$), whence $\mathsf{k}_{N}/\sqrt{N}\le\sqrt2C_{\alpha}$ and