Proof. If \(s = 1\), the only divisor is \(t = 1\) and \(\mu(1) = 1\) by Definition 1, so the sum equals \(1\). Suppose \(s > 1\) and let \(p_1, \dotsc, p_r\) (\(r \ge 1\)) be the distinct primes dividing \(s\). By Definition 1, \(\mu(t) = 0\) unless \(t\) is squarefree, so only the \(2^r\) divisors of \(p_1 \dotsm p_r\) contribute, and such a divisor \(t = \prod_{i \in I} p_i\) contributes \((-1)^{|I|}\). Grouping by \(|I|\) gives \[\sum_{t \mid s} \mu(t) \;=\; \sum_{j = 0}^{r} \binom{r}{j} (-1)^{j} \;=\; (1 - 1)^{r} \;=\; 0\] by the binomial theorem, since \(r \ge 1\). ◻

Proof. By Definition 1, \(\tau_r\) is the \(r\)-fold Dirichlet convolution of the constant function \(\mathbf{1}\) with itself. Convolution preserves multiplicativity, whence \(\tau_r(mn) = \tau_r(m)\tau_r(n)\) whenever \((m,n) = 1\); this gives the asserted equality.

For the inequality in general, it suffices to produce an injection from the ordered \(r\)-fold factorisations of \(mn\) into the pairs consisting of an ordered \(r\)-fold factorisation of \(m\) and one of \(n\). Fix a factorisation \(mn = d_1 \dotsm d_r\). For each prime \(p\) distribute the exponent \(v_p(d_i)\) between \(m\) and \(n\) by the greedy rule: assign to the \(m\)-part the first \(\min(v_p(d_i), \text{remaining exponent of } p \text{ in } m)\) of the copies of \(p\) occurring in \(d_i\), processing \(i = 1, \dotsc, r\) in order, and assign the rest to the \(n\)-part. This produces factorisations \(m = a_1 \dotsm a_r\) and \(n = b_1 \dotsm b_r\) with \(d_i = a_i b_i\), and the original factorisation is recovered from the pair, so the assignment is injective. Hence \(\tau_r(mn) \le \tau_r(m)\tau_r(n)\). ◻

Proof. Both sides are multiplicative in \(n\) by Lemma 1, so for squarefree \(n = p_1 \dotsm p_s\) with distinct primes \(p_i\) it suffices to verify the identity at a single prime argument. An ordered \(r\)-fold factorisation of a prime \(p\) has exactly one factor equal to \(p\) and all others equal to \(1\), so \(\tau_r(p) = r\); likewise \(\tau_{r^2}(p) = r^2\). Hence \(\tau_r(p)^2 = r^2 = \tau_{r^2}(p)\), and multiplying over \(p_1, \dotsc, p_s\) gives \(\tau_r(n)^2 = \tau_{r^2}(n)\). (The hypothesis that \(n\) be squarefree is essential: for \(n = p^2\) one has \(\tau_r(p^2)^2 = \binom{r+1}{2}^2 < \binom{r^2+1}{2} = \tau_{r^2}(p^2)\) in general.) ◻

Proof. Both sides are multiplicative functions of \(n\): this is standard for \(\varphi\), and the divisor sum of the multiplicative function \(g\) (Definition 1) is again multiplicative. It therefore suffices to compare the two sides at a prime power \(n = p^{a}\) with \(a \ge 1\).

For \(a = 1\), \[\sum_{u \mid p} g(u) \;=\; g(1) + g(p) \;=\; 1 + (p - 2) \;=\; p - 1 \;=\; \varphi(p).\] For \(a \ge 2\), using \(g(p^{j}) = p^{j-2}(p-1)^2\) for \(2 \le j \le a\), \[\sum_{u \mid p^{a}} g(u) \;=\; 1 + (p-2) + (p-1)^2 \sum_{j=2}^{a} p^{j-2} \;=\; (p-1) + (p-1)^2 \cdot \frac{p^{a-1} - 1}{p - 1} \;=\; (p-1)\,p^{a-1},\] which is \(\varphi(p^{a})\). Multiplicativity now gives the identity for every \(n\). ◻

Proof. The function \(n \mapsto \sum_{d \mid n} \mu(d)/d\) is the Dirichlet convolution of the multiplicative function \(d \mapsto \mu(d)/d\) with the constant function \(\mathbf{1}\), hence multiplicative. Evaluating at a prime power \(p^{a}\) with \(a \ge 1\) and using \(\mu(p^{j}) = 0\) for \(j \ge 2\) (Definition 1), \[\sum_{d \mid p^{a}} \frac{\mu(d)}{d} \;=\; \frac{\mu(1)}{1} + \frac{\mu(p)}{p} \;=\; 1 - \frac{1}{p}.\] Hence, writing \(n = \prod_{p \mid n} p^{a_p}\), the divisor sum equals \(\prod_{p \mid n}(1 - 1/p)\). On the other hand \(\varphi\) is multiplicative with \(\varphi(p^{a}) = p^{a}(1 - 1/p)\) (Definition 1), so \(\varphi(n)/n = \prod_{p \mid n}(1 - 1/p)\) as well. The two displayed equalities follow. ◻

Proof. Both sides are multiplicative in \(n\); for the right-hand side this is because \(\mu^2\) and \(1/\varphi\) are multiplicative and the divisor sum of a multiplicative function is multiplicative. It therefore suffices to compare at a prime power \(n = p^{a}\) with \(a \ge 1\). Since \(\mu(p^{j})^2 = 0\) for \(j \ge 2\), \[\sum_{e \mid p^{a}} \frac{\mu(e)^2}{\varphi(e)} \;=\; \frac{1}{\varphi(1)} + \frac{1}{\varphi(p)} \;=\; 1 + \frac{1}{p-1} \;=\; \frac{p}{p-1} \;=\; \frac{p^{a}}{\varphi(p^{a})},\] the last equality because \(\varphi(p^{a}) = p^{a-1}(p-1)\). Multiplicativity gives the identity in general. ◻

Proof. Write \(n = \prod_{i=1}^{r} p_i^{a_i}\) with distinct primes \(p_i\), so \(r = \omega(n)\) and \(\tau(n) = \prod_i (a_i + 1) \ge 2^{r}\). For every prime \(p \ge 2\) we have \(1 - 1/p \ge 1/2\), hence by Lemma 1 \[\frac{\varphi(n)}{n} \;=\; \prod_{i=1}^{r}\Bigl(1 - \frac{1}{p_i}\Bigr) \;\ge\; 2^{-r} \;\ge\; \frac{1}{\tau(n)}.\] Multiplying through by \(n\,\tau(n)\) gives \(n \le \tau(n)\varphi(n)\). ◻

Proof. If \(y \notin \bigcup_x A_x\) both sides are \(0\). If \(y \in \bigcup_x A_x\), then \(y \in A_{x_0}\) for at least one \(x_0 \in X\), so the left-hand side equals \(1\) while the right-hand side is a sum of non-negative terms one of which equals \(1\); the inequality follows. ◻

Proof. Fix a prime \(p\) and set \(S_p := \{h \bmod p : h \in \mathcal{H}\} \subseteq \mathbb{Z}/p\mathbb{Z}\). There exists \(a_p\) with \(a_p \not\equiv h_i \pmod p\) for every \(i\) if and only if \(S_p \ne \mathbb{Z}/p\mathbb{Z}\): given such an \(a_p\), its residue class is missing from \(S_p\); conversely, if \(S_p \subsetneq \mathbb{Z}/p\mathbb{Z}\), any representative of a class not in \(S_p\) will do. Since \(\mathbb{Z}/p\mathbb{Z}\) is finite of cardinality \(p\), the condition \(S_p \ne \mathbb{Z}/p\mathbb{Z}\) is equivalent to \(\#S_p < p\). Applying this for every prime \(p\) gives the claim. ◻

Proof. Admissibility trivially implies the stated condition, since the latter only restricts the primes \(p \le \#\mathcal{H}\) and, as in the proof of Lemma 1, one may always replace a missed residue by its representative in \(\{0, 1, \dotsc, p-1\}\).

Conversely, assume the condition and let \(p\) be an arbitrary prime. If \(p \le \#\mathcal{H}\) there is nothing to prove. If \(p > \#\mathcal{H}\), then the image of \(\mathcal{H}\) in \(\mathbb{Z}/p\mathbb{Z}\) has at most \(\#\mathcal{H} < p\) elements, so it is a proper subset of \(\mathbb{Z}/p\mathbb{Z}\) and some residue class is missed; by Lemma 1 this is exactly what admissibility requires at \(p\). Hence \(\mathcal{H}\) is admissible. ◻

Proof. The pair \((i, j) = (k, 1)\) is admissible in the maximum of Definition 1 and contributes \(h_k - h_1\), so \(\operatorname{diam}(\mathcal{H}) \ge h_k - h_1\). Conversely, for any \(i, j\) we have \(h_i \le h_k\) and \(h_j \ge h_1\), whence \(h_i - h_j \le h_k - h_1\); taking the maximum over \(i, j\) gives \(\operatorname{diam}(\mathcal{H}) \le h_k - h_1\). ◻

Proof. By Notation 1, both \(\pi_S\) and \(\pi_S(\cdot\,; q, a)\) count only elements of \(S\) that are prime, so \[\pi_{S \cap \mathbb{P}}(x) = \pi_S(x) \qquad\text{and}\qquad \pi_{S \cap \mathbb{P}}(x; q, a) = \pi_S(x; q, a)\] for all \(x, q, a\). The two instances of the inequality in Definition 1 are therefore literally the same, so each implies the other. ◻

Proof. This is the classical theorem of Bombieri and Vinogradov, taken as an external input; see Bombieri (1965), Vinogradov (1965), and the textbook treatment in Davenport’s Multiplicative Number Theory. No proof is given here. ◻

Proof. Fix \(\theta < 1/2\). By Theorem 1 the set \(\mathbb{N}\) has level of distribution \(\theta\) relative to \(1\). Since \(\mathbb{P}= \mathbb{N} \cap \mathbb{P}\), Lemma 1 applied with \(S = \mathbb{N}\) transfers the property from \(\mathbb{N}\) to \(\mathbb{P}\). ◻

Proof. This is the prime number theorem in the form \(\pi(x) = x/\log x + O\bigl(x/(\log x)^2\bigr)\), equivalently \(\psi(x) = x + O\bigl(x/\log x\bigr)\) after partial summation and the passage from \(\psi\) to \(\theta\) to \(\pi\). It is taken as an external input; no proof is given here. ◻

Proof. By Notation 1, \(\pi(x) = \sum_{n \le x} \chi_{\mathbb{P}}(n)\), the primes \(\le x\) being counted once each by the indicator \(\chi_{\mathbb{P}}\). The set of primes \(\le N\) is contained in the set of primes \(\le 2N\), so the two counts differ by exactly the primes in \((N, 2N]\): \[\pi(2N) - \pi(N) \;=\; \sum_{n \le 2N} \chi_{\mathbb{P}}(n) \;-\; \sum_{n \le N} \chi_{\mathbb{P}}(n) \;=\; \sum_{N < n \le 2N} \chi_{\mathbb{P}}(n) \;=\; X_N,\] the last equality being Notation 1. ◻

Proof. By Lemma 1, \(X_N = \pi(2N) - \pi(N)\). Let \(C_0\) and \(N_0'\) be as furnished by Theorem 2, and take \(N \ge \max(N_0', 2)\). Applying Theorem 2 at \(2N\) and at \(N\), and using \(\log(2N) \ge \log N\) to weaken the error term at \(2N\), \[\Bigl| \pi(2N) - \frac{2N}{\log(2N)} \Bigr| \le \frac{2C_0 N}{(\log N)^2}, \qquad \Bigl| \pi(N) - \frac{N}{\log N} \Bigr| \le \frac{C_0 N}{(\log N)^2}.\] It remains to compare the main terms. Since \(\log (2N) = \log 2 + \log N\), \[\frac{2N}{\log(2N)} - \frac{2N}{\log N} \;=\; -\,\frac{2N \log 2}{(\log 2 + \log N)\log N},\] whose absolute value is at most \(2N \log 2/(\log N)^2\) because \((\log 2 + \log N)\log N \ge (\log N)^2\). Combining, and writing \[X_N - \frac{N}{\log N} = \Bigl(\pi(2N) - \frac{2N}{\log 2N}\Bigr) - \Bigl(\pi(N) - \frac{N}{\log N}\Bigr) + \Bigl(\frac{2N}{\log 2N} - \frac{2N}{\log N}\Bigr),\] the triangle inequality gives the claim with \(C = 3C_0 + 2\log 2\). ◻

Proof. Write \(Q := \{q : 1 \le q \le N,\ q < N^{1/2-\varepsilon}\}\) for the range of moduli, and put \(\theta := 1/2 - \varepsilon\), so that \(0 < \theta < 1/2\) and \(Q \subseteq \{q : 1 \le q \le N^{\theta}\}\).

Step 1: bound \(E(N,q)\) by two cumulative discrepancies. Fix \(q\) and a residue \(a\) coprime to \(q\). The window count \(\sum_{N < n \le 2N,\, n \equiv a} \chi_{\mathbb{P}}(n)\) equals \(\pi(2N; q, a) - \pi(N; q, a)\), and \(\sum_{N < n \le 2N}\chi_{\mathbb{P}}(n) = X_N = \pi(2N) - \pi(N)\) by Lemma 1. Hence the quantity inside the supremum in Notation 1 equals \[\Bigl(\pi(2N; q, a) - \frac{\pi(2N)}{\varphi(q)}\Bigr) - \Bigl(\pi(N; q, a) - \frac{\pi(N)}{\varphi(q)}\Bigr),\] so by the triangle inequality \[E(N, q) \;\le\; 1 \;+\; D(2N, q) \;+\; D(N, q), \qquad D(x, q) := \max_{(a,q)=1}\Bigl|\pi(x; q, a) - \frac{\pi(x)}{\varphi(q)}\Bigr|.\]

Step 2: sum over \(q \in Q\). Since \(\mu(q)^2 \le 1\), summing Step 1 over \(q \in Q\) gives \[\sum_{q \in Q} \mu(q)^2 E(N,q) \;\le\; \# Q \;+\; \sum_{q \in Q} D(2N, q) \;+\; \sum_{q \in Q} D(N, q),\] and \(\#Q \le N^{1/2 - \varepsilon}\) by the definition of \(Q\).

Step 3: apply the level-of-distribution bound. Set \(A' := \max(A, 1)\). Since \(\theta < 1/2\), Lemma 1 says the primes have level of distribution \(\theta\) relative to \(\Xi = 1\), so Definition 1 supplies \(c = c(A') > 0\) with \(\sum_{q \le x^{\theta}} D(x, q) \le c\,x (\log x)^{-A'}\) for all \(x \ge 3\). All summands \(D(x,q)\) are non-negative, so restricting to \(q \in Q \subseteq \{q \le x^\theta\}\) (valid at \(x = N\) by the choice of \(\theta\), and at \(x = 2N\) since \(N^{\theta} \le (2N)^{\theta}\) for \(\theta \ge 0\)) yields, for \(N \ge 3\), \[\sum_{q \in Q} D(N, q) \le \frac{c\,N}{(\log N)^{A'}} \qquad\text{and}\qquad \sum_{q \in Q} D(2N, q) \le \frac{2c\,N}{(\log 2N)^{A'}} \le \frac{2c\,N}{(\log N)^{A'}}.\] Moreover \(\log N \ge 1\) for \(N \ge 3\), so \((\log N)^{-A'} \le (\log N)^{-A}\).

Step 4: absorb the trivial term and the small \(N\). The term \(\#Q \le N^{1/2 - \varepsilon}\) satisfies \(N^{1/2-\varepsilon} \le K N (\log N)^{-A}\) for a constant \(K = K(A, \varepsilon)\) and all \(N \ge 2\): indeed \((\log N)^{A} \ll_{A,\varepsilon} N^{1/2 + \varepsilon}\), and \(N^{1/2 - \varepsilon} \cdot N^{1/2+\varepsilon} = N\). Combining Steps 2–4 gives the asserted bound with a suitable constant for all \(N\) beyond an explicit threshold; enlarging the constant to accommodate the finitely many remaining \(N \ge 1\) (for which the left-hand side is a finite sum, and is empty when \(N = 1\)) gives the lemma. ◻

Proof. By Definition 1, \(W\) is a product of distinct primes, so a prime \(p\) divides \(W\) if and only if \(p\) is one of the factors, that is, if and only if \(p \le \lfloor D_0 \rfloor\). Since \(p\) is an integer, \(p \le \lfloor D_0 \rfloor\) is equivalent to \(p \le D_0\). ◻

Proof. By Definition 1, \(W\) is a product of distinct primes, each occurring to the first power. Hence no prime square divides \(W\), i.e. \(W\) is squarefree. ◻

Proof. We use the elementary Chebyshev-type bound \(\prod_{p \le n} p \le 4^{n}\), valid for every integer \(n \ge 0\); it is proved by induction on \(n\) using the divisibility of \(\binom{2m+1}{m}\) by all primes in \((m+1, 2m+1]\) together with \(\binom{2m+1}{m} \le 4^{m}\). No appeal to the prime number theorem is required.

By Definition 1 and this bound, \[W \;=\; \prod_{p \le \lfloor D_0\rfloor} p \;\le\; 4^{\lfloor D_0 \rfloor} \;\le\; 4^{D_0}\] whenever \(D_0 \ge 0\). Now \(D_0 = \log\log\log N\) by Definition 1, so \(e^{D_0} = \log\log N\) and hence \[4^{D_0} \;=\; e^{D_0 \log 4} \;=\; \bigl(e^{D_0}\bigr)^{\log 4} \;=\; (\log\log N)^{\log 4}.\] Finally \(\log 4 \le 2\), so once \(N\) is large enough that \(\log\log N \ge 1\) — equivalently \(N \ge e^{e}\), which also forces \(D_0 \ge 0\) — we may raise the exponent from \(\log 4\) to \(2\) and conclude \(W \le (\log\log N)^{2}\). ◻

Proof. By Definition 1, \(W\) is a product of distinct primes \(p \le \lfloor D_0 \rfloor\). For each such prime \(p\), admissibility of \(\mathcal{H}\) (Definition 1) supplies a residue \(a_p\) with \(a_p \not\equiv h_i \pmod p\) for every \(i\); equivalently, setting \(b_p := -a_p\), we have \(b_p + h_i \not\equiv 0 \pmod p\) for every \(i\).

The primes dividing \(W\) are pairwise coprime, so by the Chinese remainder theorem there is an integer \(v\) with \(v \equiv b_p \pmod p\) for every prime \(p \mid W\). Put \(v_0 := v \bmod W\), so \(0 \le v_0 < W\) and \(v_0 \equiv b_p \pmod p\) for each such \(p\).

It remains to check compatibility. Fix \(i\). Since \(W\) is squarefree (Lemma 1), \((v_0 + h_i, W) = 1\) holds if and only if no prime divisor of \(W\) divides \(v_0 + h_i\); and by Lemma 1 the prime divisors of \(W\) are exactly the primes \(p \le D_0\), which are the primes handled above. For such a \(p\) we have \(v_0 + h_i \equiv b_p + h_i \not\equiv 0 \pmod p\) by the choice of \(b_p\). Hence \((v_0 + h_i, W) = 1\) for every \(i\), i.e. \(v_0\) is compatible with \(\mathcal{H}\) modulo \(W\). ◻

Proof. Set \(\gamma := \tfrac12 - \eta - (\vartheta - 2\delta)\), which is positive by hypothesis. By Definition 1, \[R^{2} \;=\; N^{\vartheta - 2\delta},\] so it suffices to show that \(W \le N^{\gamma}\) for all large \(N\), since then \(R^{2}W \le N^{\vartheta - 2\delta + \gamma} = N^{1/2 - \eta}\).

By Lemma 1, \(W \le (\log\log N)^{2}\) for all large \(N\), and \(\log\log N \le \log N\) for \(N \ge e\). Hence \(W \le (\log N)^{2}\) for all large \(N\). Since \((\log N)^{2} = o(N^{\gamma})\) for every \(\gamma > 0\) — powers of the logarithm are dominated by every positive power of \(N\) — we obtain \(W \le N^{\gamma}\) for all large \(N\), as required. ◻

Proof. Since \(d_i \mid r_i\) for every \(i\), we have \(\prod_i d_i \mid \prod_i r_i\). Because \(\prod_i r_i\) is squarefree it is in particular non-zero, so \(\prod_i d_i \le \prod_i r_i \le R\), giving (i). A divisor of an integer coprime to \(W\) is itself coprime to \(W\), giving (ii), and a divisor of a squarefree integer is squarefree, giving (iii). ◻

Proof. Axiom (D1)(iv) of Definition 1 gives \(\gamma(p) < p\), i.e. \(p - \gamma(p) > 0\); hence the quotient defining \(g_*(p)\) makes sense. (Axiom (D2) sharpens this to \(p - \gamma(p) \ge A_1 p\), a bound uniform in \(p\), which is what the analytic applications use.) Since \(\gamma(p) \ge 0\) by (D1)(i) and the denominator is positive, \(g_*(p) = \gamma(p)/(p - \gamma(p)) \ge 0\). ◻

Proof. Since \((d, V) > 1\), the integer \((d, V)\) has a prime divisor \(p\); then \(p \mid d\) and \(p \mid V\). By axiom (D3) of Definition 1, \(\gamma(p) = 0\); by Lemma 1 the denominator \(p - \gamma(p)\) is strictly positive, so Definition 1 gives \(g_*(p) = 0/(p - \gamma(p)) = 0\). Write \(d = p\,c\) with \(c \ge 1\). Total multiplicativity of \(g_*\) gives \(g_*(d) = g_*(p)\, g_*(c) = 0\), whence \(h(d) = \mu(d)^2\, g_*(d) = 0\). ◻

Proof. Write \(\gamma := \gamma_g^{(V)}\). Axiom (D5) is the hypothesis, and the squarefreeness of \(V\) together with \(\gamma(p) = 0\) for \(p \mid V\) (immediate from Definition 1) is axiom (D3). We check the remaining axioms.

(D1) Density. Every prime value \(\gamma(p)\) is non-negative — it is either \(0\) or \(p/(p-1) > 0\) — so the product over the prime factorisation defining \(\gamma(n)\) is non-negative, and \(\gamma(1) = 1\) as an empty product. Total multiplicativity of \(\gamma\) over prime factorisations gives \(\gamma(mn) = \gamma(m)\gamma(n)\) for all coprime \(m, n\) (indeed for all nonzero \(m, n\)). For (iv), if \(p \mid V\) then \(\gamma(p) = 0 < p\). If \(p \nmid V\), then \(p \ne 2\) because \(2 \mid V\), so \(p \ge 3\); and \(p/(p-1) < p\) is equivalent to \(1 < p - 1\), i.e. to \(p > 2\), which holds.

(D2) Growth with \(A_1 = 1/2\). Certainly \(0 < 1/2 < 1\). If \(p \mid V\) then \(\gamma(p)/p = 0 \le 1/2\). If \(p \nmid V\) then \(p \ge 3\) as above, and \(\gamma(p)/p = 1/(p-1) \le 1/2\).

(D4) Unit density with \(A_3 = 2\). Clearly \(2 \ge 0\). Let \(p \nmid V\). Then \[|\gamma(p) - 1| \;=\; \Bigl|\frac{p}{p-1} - 1\Bigr| \;=\; \frac{1}{p-1},\] and \(1/(p-1) \le 2/p\) is equivalent to \(p \le 2(p-1)\), i.e. to \(p \ge 2\), which holds for every prime. ◻

Proof. By Definition 1, \(w_n\) is the square of the real number \(\sum_{d_i \mid n+h_i\ \forall i} \lambda(\mathbf{d})\), and squares of real numbers are non-negative. ◻

Proof. Fix \(n\) in the index set of Definition 1. Distributing the finite inner sum across the product with \(w_n\) gives \[\Bigl(\sum_{i=1}^{k} \chi_{\mathbb{P}}(n + h_i)\Bigr) w_n \;=\; \sum_{i=1}^{k} \chi_{\mathbb{P}}(n + h_i)\, w_n .\] Substituting this into Definition 1 and interchanging the two finite summations — over \(n \in (N, 2N]\) with \(n \equiv v_0 \pmod W\), and over \(i \in \{1, \dotsc, k\}\) — yields \[S_2(\lambda) \;=\; \sum_{i=1}^{k}\ \sum_{\substack{N < n \le 2N \\ n \equiv v_0 \!\!\pmod W}} \chi_{\mathbb{P}}(n + h_i)\, w_n .\] By Definition 1 the inner sum is exactly \(S_2^{(i)}(\lambda)\); renaming the index \(i\) to \(m\) gives the claim. ◻

Proof. Write \(c(n) := \#\{i \in \{1, \dotsc, k\} : n + h_i \in \mathbb{P}\} = \sum_{i=1}^k \chi_{\mathbb{P}}(n+h_i)\), so that, as computed above, \[S(\lambda, \rho) \;=\; \sum_{\substack{N < n \le 2N \\ n \equiv v_0 \!\!\pmod W}} \bigl(c(n) - \rho\bigr) w_n .\] Suppose, for contradiction, that \(c(n) \le \rho\) for every \(n\) in the index set. By Lemma 1 each \(w_n \ge 0\), so each summand \((c(n) - \rho) w_n\) is non-positive, and therefore \(S(\lambda, \rho) \le 0\), contradicting the hypothesis. Hence there is an \(n\) in the index set with \(c(n) > \rho\).

It remains to convert \(c(n) > \rho\) into \(c(n) \ge \lfloor \rho + 1\rfloor\). Since \(c(n)\) is an integer and \(c(n) > \rho\), we have \(c(n) \ge \lfloor \rho \rfloor + 1\) when \(\rho \notin \mathbb{Z}\), and \(c(n) \ge \rho + 1\) when \(\rho \in \mathbb{Z}\); in either case \(c(n) \ge \lfloor \rho + 1 \rfloor\), because \(\lfloor \rho + 1\rfloor = \lfloor \rho\rfloor + 1\). ◻

Proof. Suppose \(y_\lambda(\mathbf{r}) \ne 0\). Then some summand of the defining sum in Definition 1 is nonzero, so there is a \(\mathbf{d}\) with \(\lambda(\mathbf{d}) \ne 0\) and \(r_i \mid d_i\) for every \(i\). It therefore suffices to show that the three conditions of Definition 1 are inherited from \(\mathbf{d}\) to any coordinatewise divisor \(\mathbf{r}\) of \(\mathbf{d}\). That is exactly Lemma 1, applied at level \(T\).

In detail: since \(r_i \mid d_i\) for each \(i\), we have \(\prod_i r_i \mid \prod_i d_i\). As all \(d_i\) are nonzero, \(\prod_i d_i \ge 1\) and hence \(\prod_i r_i \le \prod_i d_i \le T\), which is (i). A divisor of an integer coprime to \(W\) is coprime to \(W\), giving (ii); and a divisor of a squarefree integer is squarefree, giving (iii). ◻

Proof. If some \(d_i\) is not squarefree, then every \(\mathbf{r}\) contributing to Definition 1 has \(d_i \mid r_i\) with \(r_i\) squarefree, forcing \(d_i\) squarefree — a contradiction. So the sum is empty and \(\lambda_y(\mathbf{d}) = 0\). Assume henceforth that every \(d_i\) is squarefree.

Both transforms are \(\mathbb{R}\)-linear in their argument, and every finitely supported function is a finite linear combination of point masses; so it suffices to prove the identity when \(\lambda = c \cdot \mathbf{1}_{\mathbf{d}^{(0)}}\) is a point mass at some \(\mathbf{d}^{(0)}\) with weight \(c\). For such \(\lambda\), \[y(\mathbf{r}) \;=\; \Bigl(\prod_i \mu(r_i)\varphi(r_i)\Bigr) \frac{c}{\prod_i d^{(0)}_i} \quad\text{if $r_i \mid d^{(0)}_i$ and $d^{(0)}_i$ is squarefree for all $i$,}\] and \(y(\mathbf{r}) = 0\) otherwise; in particular \(y\) is supported on the coordinatewise divisors of \(\mathbf{d}^{(0)}\). If some \(d^{(0)}_i\) fails to be squarefree, or if \(d_i \nmid d^{(0)}_i\) for some \(i\), then no \(\mathbf{r}\) contributes to \(\lambda_y(\mathbf{d})\) and both sides vanish (on the right, \(\lambda(\mathbf{d}) = 0\) unless \(\mathbf{d} = \mathbf{d}^{(0)}\), and the excluded cases exclude that). Otherwise substitute the displayed formula for \(y\) into Definition 1: the factors \(\varphi(r_i)\) cancel, and \[\lambda_y(\mathbf{d}) \;=\; \frac{c\,\prod_i \mu(d_i)\, d_i}{\prod_i d^{(0)}_i} \sum_{\substack{\mathbf{r}\\ d_i \mid r_i \mid d^{(0)}_i\ \forall i}} \prod_{i=1}^k \mu(r_i) \;=\; \frac{c\,\prod_i \mu(d_i)\, d_i}{\prod_i d^{(0)}_i} \prod_{i=1}^{k} \Bigl(\sum_{d_i \mid r_i \mid d^{(0)}_i} \mu(r_i)\Bigr),\] the sum having factorised across coordinates. For squarefree \(d^{(0)}_i\) the inner sum is \(\mu(d_i)\,\mathbf{1}[d_i = d^{(0)}_i]\), by Möbius inversion applied to the divisor chain \(d_i \mid r_i \mid d^{(0)}_i\) (writing \(r_i = d_i s_i\) with \(s_i \mid d^{(0)}_i/d_i\) and using \(\mu(d_i s_i) = \mu(d_i)\mu(s_i)\), valid since \(d^{(0)}_i\) is squarefree, together with \(\sum_{s \mid t} \mu(s) = \mathbf{1}[t = 1]\)). Hence the whole expression vanishes unless \(\mathbf{d} = \mathbf{d}^{(0)}\), in which case it equals \[\frac{c\,\prod_i \mu(d_i)\, d_i}{\prod_i d_i} \prod_i \mu(d_i) \;=\; c \prod_i \mu(d_i)^2 \;=\; c,\] because each \(d_i\) is squarefree. This is \(\lambda(\mathbf{d})\), as required. ◻

Proof. The argument is that of Lemma 1 with the roles of the two transforms exchanged. If some \(r_i\) is not squarefree, every \(\mathbf{d}\) contributing to Definition 1 at \(\mathbf{r}\) has \(r_i \mid d_i\) with \(d_i\) squarefree, which is impossible; so \(y_{\lambda_y}(\mathbf{r}) = 0\). Assume every \(r_i\) squarefree. By linearity reduce to \(y = c\cdot\mathbf{1}_{\mathbf{r}^{(0)}}\) a point mass. Then \(\lambda_y\) is supported on the coordinatewise divisors of \(\mathbf{r}^{(0)}\), and substituting Definition 1 into Definition 1 the factors \(d_i\) cancel against \(1/\prod_i d_i\), leaving \[y_{\lambda_y}(\mathbf{r}) \;=\; \frac{c\,\prod_i \mu(r_i)\varphi(r_i)}{\prod_i \varphi(r^{(0)}_i)} \prod_{i=1}^{k}\Bigl(\sum_{r_i \mid d_i \mid r^{(0)}_i} \mu(d_i)\Bigr).\] As before each inner sum equals \(\mu(r_i)\,\mathbf{1}[r_i = r^{(0)}_i]\), so the expression vanishes unless \(\mathbf{r} = \mathbf{r}^{(0)}\), and then equals \(c\prod_i \mu(r_i)^2 = c = y(\mathbf{r})\). ◻

Proof. If \(y^{(m)}(\mathbf{r}) \ne 0\) then the defining sum of Definition 1 has a nonzero summand, so there is a \(\mathbf{d}\) with \(d_m = 1\) and \(r_i \mid d_i\) for every \(i\). Taking \(i = m\) gives \(r_m \mid 1\), hence \(r_m = 1\). ◻

Proof. Fix \((t_i)_{i \ne m} \in \mathcal{S}_\varepsilon\) and consider the fibre \[E \;:=\; \{\, s \ge 0 : (t_1, \dotsc, t_{m-1}, s, t_{m+1}, \dotsc, t_k) \in \mathcal{T}_\varepsilon \,\} .\] Since every \(t_i \ge 0\), the fibre is the interval \([0,\, 1 + \varepsilon - \sum_{i \ne m} t_i]\) (empty if the right endpoint is negative), whose length is at most \(1 + \varepsilon\). Since \(F\) vanishes off \(\mathcal{T}_\varepsilon\), the Cauchy–Schwarz inequality on \(E\) gives \[\bigl|(T_m F)\bigl((t_i)_{i \ne m}\bigr)\bigr|^2 \;=\; \Bigl|\int_E F \, ds\Bigr|^2 \;\le\; |E| \int_E F^2 \, ds \;\le\; (1 + \varepsilon) \int_E F^2 \, ds .\] Integrating over \((t_i)_{i \ne m} \in \mathcal{S}_\varepsilon\) and applying Tonelli’s theorem to the right-hand side, the iterated integral is bounded by the integral of \(F^2\) over all of \(\mathbb{R}^k\), which is \(I_k(F)\). Hence \(J_{k,\varepsilon}^{(m)}(F) \le (1 + \varepsilon) I_k(F)\).

Equivalently: \(T_m\) is a bounded linear operator \(L^2(\mathcal{T}_\varepsilon) \to L^2(\mathcal{S}_\varepsilon)\) whose norm is at most the square root of the essential supremum of the fibre lengths, and that essential supremum is at most \(1 + \varepsilon\). ◻

Proof. At \(\varepsilon = 0\) the outer domain \(\mathcal{S}_0\) is \(\mathcal{R}_k^{(m)}\), so only the inner integral has to be identified. Fix \((t_i)_{i \ne m} \in \mathcal{R}_k^{(m)}\). Then every \(t_i \ge 0\) and \(\sum_{i \ne m} t_i \le 1\), and the fibre of \(\mathcal{R}_k\) over this point is exactly \([0,\, 1 - \sum_{i \ne m} t_i]\). For \(s\) beyond that endpoint the inserted point lies outside \(\mathcal{R}_k\), so \(F\) vanishes there by hypothesis; hence the integral \(\int_0^{\infty} F\, dt_m\) defining \(T_m F\) agrees with the truncated integral \(\int_0^{1 - \sum_{i \ne m} t_i} F\, ds\). Squaring and integrating over \(\mathcal{R}_k^{(m)}\) gives the claim.

The hypothesis that \(F\) vanish pointwise outside \(\mathcal{R}_k\), and not merely almost everywhere on \(\mathcal{R}_k\), is essential: \(T_m\) samples \(F\) along the whole half-line, so a modification of \(F\) on a set that is null in \(\mathcal{R}_k\) but not in \(\mathbb{R}^k\) would change the left-hand side. ◻

Proof. On \(L^2(\mathcal{T}_\varepsilon)\) we have \(I_k(F) = \|F\|_{L^2(\mathcal{T}_\varepsilon)}^2\) by Definition 1, so \(I_k\) is the square of the norm, and the norm on a normed space is continuous. Quantitatively, for \(F, G \in L^2(\mathcal{T}_\varepsilon)\) the factorisation \(F^2 - G^2 = (F - G)(F + G)\) together with Cauchy–Schwarz and the triangle inequality gives \[|I_k(F) - I_k(G)| \;\le\; \|F - G\|\,\|F + G\| \;\le\; \bigl(\|F\| + \|G\|\bigr)\,\|F - G\| ,\] all norms being those of \(L^2(\mathcal{T}_\varepsilon)\). ◻

Proof. The marginal operator \(T_m\) is linear, and by the proof of Lemma 1 it is bounded from \(L^2(\mathcal{T}_\varepsilon)\) to \(L^2(\mathcal{S}_\varepsilon)\), with norm at most \((1 + \varepsilon)^{1/2}\); hence it is continuous. The map \(H \mapsto \|H\|^2_{L^2(\mathcal{S}_\varepsilon)}\) is continuous, being the square of a norm. By Definition 1, \(J_{k,\varepsilon}^{(m)}\) is the composite of these two continuous maps. Quantitatively, the argument of Lemma 1 applied in \(L^2(\mathcal{S}_\varepsilon)\) together with the operator bound gives \[\bigl|J_{k,\varepsilon}^{(m)}(F) - J_{k,\varepsilon}^{(m)}(G)\bigr| \;\le\; (1 + \varepsilon)\bigl(\|F\| + \|G\|\bigr)\,\|F - G\|\] with the norms of \(L^2(\mathcal{T}_\varepsilon)\). ◻

Proof. Let \(F \in L^2(\mathcal{T}_\varepsilon)\) with \(I_k(F) \ne 0\). Summing the bound of Lemma 1 over the \(k\) values of \(m\) gives \(\sum_{m=1}^{k} J_{k,\varepsilon}^{(m)}(F) \le (1 + \varepsilon) k\, I_k(F)\), so the Rayleigh quotient at \(F\) is at most \((1+\varepsilon)k\). The family of quotients is therefore bounded above, its supremum exists in \(\mathbb{R}\), and it is at most \((1 + \varepsilon)k\). ◻

Proof. Take \(F := \mathbf{1}_{\mathcal{R}_k}\), which vanishes outside \(\mathcal{R}_k \subseteq \mathcal{T}_\varepsilon\) and so is an element of \(L^2(\mathcal{T}_\varepsilon)\). The simplex \(\mathcal{R}_k\) contains the non-empty open set \(\{t : t_i > 0 \ \forall i,\ \sum_i t_i < 1\}\), hence has positive volume, and \(I_k(F) = \operatorname{vol} (\mathcal{R}_k) > 0\); so \(F\) is an admissible competitor.

Fix \(m\). For \((t_i)_{i \ne m}\) with \(t_i \ge 0\) and \(\sum_{i \ne m} t_i \le 1\) the fibre of \(\mathcal{R}_k\) over that point is the interval \([0,\, 1 - \sum_{i \ne m} t_i]\), so, exactly as in the proof of Lemma 1, \((T_m F)((t_i)_{i \ne m}) = 1 - \sum_{i \ne m} t_i\). The shrunken slice \(\mathcal{S}_\varepsilon\) is contained in \(\mathcal{R}_{k-1}\) because \(\varepsilon \ge 0\), and it contains the non-empty open set \(V := \{t : t_i > 0 \ \forall i,\ \sum_i t_i < (1 - \varepsilon)/2\}\) because \(\varepsilon < 1\); on \(V\) the integrand \((1 - \sum_{i \ne m} t_i)^2\) exceeds \(1/4\). Hence \[J_{k,\varepsilon}^{(m)}(F) \;\ge\; \tfrac14 \operatorname{vol}(V) \;>\; 0 .\] Therefore \(\sum_m J_{k,\varepsilon}^{(m)}(F) > 0\), the Rayleigh quotient at \(F\) is strictly positive, and \(M_{k,\varepsilon}\), being at least that quotient, is strictly positive. ◻

Proof. Fix a smooth non-decreasing \(\psi : \mathbb{R} \to [0,1]\) with \(\psi(u) = 0\) for \(u \le 0\) and \(\psi(u) = 1\) for \(u \ge 1\), and put \(\psi_\eta(u) := \psi(u/\eta)\). Set \[\chi(t_1, \dotsc, t_k) \;:=\; \Bigl(\prod_{i=1}^{k} \psi_\eta(t_i)\Bigr) \cdot \psi_\eta\Bigl(1 - \sum_{i=1}^{k} t_i\Bigr).\] This is a finite product of smooth functions, hence smooth, with values in \([0,1]\).

(i) If \(t \in \mathcal{R}_k^{(\eta)}\) then \(t_i \ge \eta\) for each \(i\) and \(\sum_i t_i \le 1 - \eta\), so every factor equals \(1\).

(ii) If \(t \notin \mathcal{R}_k\) then either \(t_i < 0\) for some \(i\), or \(\sum_i t_i > 1\); in the first case \(\psi_\eta(t_i) = 0\) and in the second \(\psi_\eta(1 - \sum_i t_i) = 0\).

(iii) On \(\mathcal{R}_k^{(\eta)}\) we have \(G - G\chi = 0\) by (i), while on all of \(\mathcal{R}_k\) we have \(|G - G\chi| = |G|\,|1 - \chi| \le M\). Hence \[\int_{\mathcal{R}_k} (G - G\chi)^2 = \int_{\mathcal{R}_k \setminus \mathcal{R}_k^{(\eta)}} (G - G\chi)^2 \le M^2 \operatorname{vol}\bigl(\mathcal{R}_k \setminus \mathcal{R}_k^{(\eta)}\bigr).\] For the second inequality, \(\mathcal{R}_k \setminus \mathcal{R}_k^{(\eta)}\) is covered by the \(k\) slabs \(\{t \in \mathcal{R}_k : t_i < \eta\}\) together with the slab \(\{t \in \mathcal{R}_k : \sum_i t_i > 1 - \eta\}\). Each of the first \(k\) slabs has volume at most \(\eta/(k-1)!\), since its cross-section at fixed \(t_i\) lies in a \((k-1)\)-dimensional simplex of volume \(1/(k-1)!\); the last slab, being the difference of two simplices, has volume \(\bigl(1 - (1-\eta)^k\bigr)/k! \le \eta/(k-1)!\). Summing gives \(\operatorname{vol}(\mathcal{R}_k \setminus \mathcal{R}_k^{(\eta)}) \le (k+1)\eta/(k-1)! \le 2k\eta\) for \(k \ge 2\). ◻

Proof. Extend \(F_0\) by zero to \(\mathbb{R}^k\); the extension lies in \(L^2(\mathbb{R}^k)\) with the same norm. By the density of smooth compactly supported functions in \(L^2(\mathbb{R}^k)\) there is a smooth compactly supported \(\widetilde F\) with \(\|F_0 - \widetilde F\|_{L^2(\mathbb{R}^k)} < \delta/3\); restricting the integration to \(\mathcal{R}_k\) only decreases the left-hand side. Being continuous with compact support, \(\widetilde F\) is bounded on \(\mathcal{R}_k\); let \(M \ge 0\) be a bound for \(|\widetilde F|\) there.

Choose \(\eta \in (0, 1/(2k))\) so small that \(M\sqrt{2k\eta} < \delta/3\), and let \(\chi\) be the cut-off supplied by Lemma 1 for this \(\eta\). Put \(F_1 := \widetilde F \cdot \chi\). Then \(F_1\) is smooth, and its closed support is contained in that of \(\chi\), hence in the closed set \(\mathcal{R}_k\). By part (iii) of Lemma 1 applied with \(G = \widetilde F\), \[\|\widetilde F - F_1\|_{L^2(\mathcal{R}_k)} \;\le\; M\sqrt{2k\eta} \;<\; \delta/3 ,\] and the triangle inequality gives \(\|F_0 - F_1\|_{L^2(\mathcal{R}_k)} < 2\delta/3 < \delta\). ◻

Proof. Write \(\mathcal{Q}(g) := \bigl(\sum_m J_k^{(m)}(g)\bigr)/I_k(g)\) for the variational quotient, defined on the set where \(I_k \ne 0\). Its numerator is a finite sum of the functionals \(J_k^{(m)}\), continuous by Lemma 1, and its denominator \(I_k\) is continuous by Lemma 1 and is non-zero at \(f\); hence \(\mathcal{Q}\) is continuous at \(f\).

Choose \(\delta_1 > 0\) such that \(\|g - f\|_{L^2(\mathcal{R}_k)} < \delta_1\) implies \(|\mathcal{Q}(g) - \mathcal{Q}(f)| < \eta\), and, using continuity of \(I_k\) at \(f\) together with \(I_k(f) > 0\), choose \(\delta_2 > 0\) such that \(\|g - f\|_{L^2(\mathcal{R}_k)} < \delta_2\) implies \(I_k(g) > 0\). Apply Lemma 1 with \(\delta := \min(\delta_1, \delta_2)\) to obtain a smooth \(F\) supported on \(\mathcal{R}_k\) with \(\|f - F\|_{L^2(\mathcal{R}_k)} < \delta\). Both conclusions hold for this \(F\). ◻

Proof. Step 1: smooth functions supported in \(\mathcal{T}_\varepsilon\) are \(L^2\)-dense there. Let \(\sigma := 1 + \varepsilon > 0\) and let \(\Lambda_\varepsilon\) be the dilation \[(\Lambda_\varepsilon G)(t) \;:=\; \sigma^{-k/2}\, G(\sigma^{-1} t) .\] Since \(\sigma^{-1}\mathcal{T}_\varepsilon = \mathcal{R}_k\), the map \(\Lambda_\varepsilon\) carries functions supported in \(\mathcal{R}_k\) to functions supported in \(\mathcal{T}_\varepsilon\), preserves smoothness, and is an isometry of \(L^2(\mathcal{R}_k)\) onto \(L^2(\mathcal{T}_\varepsilon)\) by the change of variables \(t = \sigma u\). Composing Lemma 1 with \(\Lambda_\varepsilon\) shows: for every \(H_0 \in L^2(\mathcal{T}_\varepsilon)\) and every \(\delta' > 0\) there is a smooth \(H_1\) with closed support in \(\mathcal{T}_\varepsilon\) and \(\|H_0 - H_1\|_{L^2(\mathcal{T}_\varepsilon)} < \delta'\).

Step 2: transfer along the Rayleigh quotient. By Lemma 1 we have \(M_{k,\varepsilon} > 0\), and by Lemma 1 the supremum is finite. Write \(Q(G) := \bigl(\sum_m J_{k,\varepsilon}^{(m)}(G)\bigr)/I_k(G)\). Since \(M_{k,\varepsilon}\) is the supremum of \(Q\), there is \(G_0 \in L^2(\mathcal{T}_\varepsilon)\) with \[\max\Bigl(M_{k,\varepsilon} - \tfrac{\delta}{2},\ \tfrac{M_{k,\varepsilon}}{2}\Bigr) \;<\; Q(G_0),\] and in particular \(I_k(G_0) > 0\), since a competitor must have non-zero mass. Both \(\sum_m J_{k,\varepsilon}^{(m)}\) and \(I_k\) are continuous on \(L^2(\mathcal{T}_\varepsilon)\) (Lemmas 1 and 1), so \(Q\) is continuous at \(G_0\) and \(I_k\) stays positive near \(G_0\): there is \(\delta' > 0\) such that every \(H\) with \(\|G_0 - H\| < \delta'\) satisfies \(I_k(H) > 0\) and \(|Q(H) - Q(G_0)| < \delta/2\).

By Step 1 choose such an \(H = F\) smooth with closed support in \(\mathcal{T}_\varepsilon\). Then \(I_k(F) > 0\) and \(Q(F) > Q(G_0) - \delta/2 > M_{k,\varepsilon} - \delta\); multiplying by \(I_k(F) > 0\) gives the asserted inequality. ◻

Proof. Write \(M := M_{k,\varepsilon}\), so that \(2\rho/\vartheta < M\) by hypothesis, and set \[\eta \;:=\; \frac{\vartheta}{2}\Bigl(M - \frac{2\rho}{\vartheta}\Bigr) \;>\; 0, \qquad\text{so that}\qquad \rho \;=\; \frac{\vartheta M}{2} - \eta .\] Apply Lemma 1 with tolerance \(\delta_0 := \min\bigl(\eta/\vartheta,\ M/2\bigr) > 0\): it produces a smooth \(F\) with closed support in \(\mathcal{T}_\varepsilon\), \(I_k(F) > 0\), and \[Q \;:=\; \frac{\sum_{m} J_{k,\varepsilon}^{(m)}(F)}{I_k(F)} \;>\; M - \delta_0 .\] Since \(\delta_0 \le M/2\) we get \(Q > M/2 > 0\).

Put \(\delta := \eta/(2Q)\), which is positive. From \(Q > M - \delta_0 \ge M - \eta/\vartheta\) we get \(\vartheta(M - Q) < \eta\), hence \[\frac{\vartheta M}{2} - \eta \;<\; \frac{\vartheta Q}{2} - \frac{\eta}{2} \;=\; \Bigl(\frac{\vartheta}{2} - \frac{\eta}{2Q}\Bigr) Q \;=\; \Bigl(\frac{\vartheta}{2} - \delta\Bigr) Q .\] Multiplying by \(I_k(F) > 0\) and using \(Q\, I_k(F) = \sum_m J_{k,\varepsilon}^{(m)}(F)\) gives \[\rho\, I_k(F) \;<\; \Bigl(\frac{\vartheta}{2} - \delta\Bigr) \sum_{m=1}^{k} J_{k,\varepsilon}^{(m)}(F).\] Finally \(\delta < \vartheta/2\): the displayed inequality has strictly positive left-hand side (as \(\rho > 0\) and \(I_k(F) > 0\)), and \(\sum_m J_{k,\varepsilon}^{(m)}(F) \ge 0\) since each term is an integral of a square, so the factor \(\vartheta/2 - \delta\) must be strictly positive. ◻

Proof. One proves the scaled form: for every real \(s \ge 0\), \[\int_{\mathcal{R}_k(s)} \Bigl(s - \sum_{i} t_i\Bigr)^{a} \prod_{i} t_i^{\,a_i} \, dt \;=\; s^{\,k + a + \sum_i a_i}\, \frac{a! \prod_i a_i!}{\bigl(k + a + \sum_i a_i\bigr)!},\] by induction on \(k\); the statement of the lemma is the case \(s = 1\). (The scaled form is also what is needed to integrate over \(\mathcal{T}_\varepsilon\) and \(\mathcal{S}_\varepsilon\).)

For \(k = 0\) both sides equal \(s^a\, a!/a! = s^a\). For the inductive step, the integrand is continuous on the compact set \(\mathcal{R}_{k+1}(s)\), hence integrable, and Fubini’s theorem lets us peel off the first coordinate: \[\int_{\mathcal{R}_{k+1}(s)} \!\!\cdots \;=\; \int_0^s t_1^{\,a_1} \biggl( \int_{\mathcal{R}_k(s - t_1)} \Bigl((s - t_1) - \sum_{i \ge 2} t_i\Bigr)^{a} \prod_{i \ge 2} t_i^{\,a_i} \, dt_2 \dotsm dt_{k+1} \biggr) dt_1 .\] By the inductive hypothesis applied with scale \(s - t_1\), the inner integral equals \((s - t_1)^{\,k + a + \sum_{i \ge 2} a_i}\, a! \prod_{i \ge 2} a_i! / (k + a + \sum_{i \ge 2} a_i)!\). The remaining one-dimensional integral is a Beta integral: for \(A, e \in \mathbb{Z}_{\ge 0}\) and \(s \ge 0\), \[\int_0^s (s - t)^{A}\, t^{e} \, dt \;=\; s^{\,A + e + 1}\, \frac{A!\, e!}{(A + e + 1)!},\] which follows from \(B(A+1, e+1) = A!\,e!/(A + e + 1)!\) after the substitution \(t = su\). Combining the two factorial expressions and simplifying gives the scaled formula in dimension \(k + 1\). ◻

Proof. Expand \(P_2^{\,b} = \bigl(\sum_i t_i^2\bigr)^b\) by the multinomial theorem: \[\Bigl(\sum_{i=1}^{k} t_i^{2}\Bigr)^{b} \;=\; \sum_{\substack{v \in \mathbb{Z}_{\ge 0}^{k} \\ v_1 + \dotsb + v_k = b}} \frac{b!}{\prod_i v_i!} \prod_{i=1}^{k} t_i^{\,2 v_i}.\] Integrating term by term over \(\mathcal{R}_k\) and applying Lemma 1 with exponents \(a_i = 2v_i\) (so \(\sum_i a_i = 2b\)) gives \[\int_{\mathcal{R}_k} (1 - P_1)^{a} P_2^{\,b} \;=\; \frac{a!}{(k + a + 2b)!} \sum_{\substack{v \in \mathbb{Z}_{\ge 0}^{k} \\ \sum_i v_i = b}} \frac{b!}{\prod_i v_i!} \prod_{i=1}^{k} (2 v_i)! .\] By Definition 1, \(G_{b,2}(k) = b! \sum_{\sum_i v_i = b} \prod_i (2 v_i)!/v_i!\), which is literally the displayed sum, since \(b!/\prod_i v_i!\) times \(\prod_i (2v_i)!\) equals \(b! \prod_i (2v_i)!/v_i!\). ◻

Proof. Since \(F_P\) vanishes off \(\mathcal{R}_k\) and agrees with \(P\) on it, \(I_k(F_P) = \int_{\mathcal{R}_k} P^2\). Expanding the square, \[P^2 \;=\; \sum_{i} \sum_{j} a_i a_j\, (1 - P_1)^{b_i + b_j}\, P_2^{\,c_i + c_j},\] using \((1 - P_1)^{b_i}(1 - P_1)^{b_j} = (1 - P_1)^{b_i + b_j}\) and likewise for \(P_2\). The integrand is continuous on the compact set \(\mathcal{R}_k\), so the finite sum may be integrated term by term, and each term is evaluated by Lemma 1 with \(a = b_i + b_j\) and \(b = c_i + c_j\). ◻

Proof. By Lemma 1, \[J_k^{(m)}(F_P) \;=\; \int_{\mathcal{R}_k^{(m)}} \Bigl( \int_0^{\,1 - P_1'} P(\dotsc, s, \dotsc)\, ds \Bigr)^{\!2} \prod_{i \ne m} dt_i ,\] where \(P_1' = \sum_{i \ne m} t_i\) and \(s\) occupies the \(m\)-th slot. Since \(P\) is symmetric, the value of the inner integral is independent of which slot is singled out, and the outer domain \(\mathcal{R}_k^{(m)}\) is a copy of \(\mathcal{R}_{k-1}\); this gives the asserted independence of \(m\), and we may take \(m = 1\).

The inner integral. Writing \(P_2' = \sum_{i \ne m} t_i^2\), the \(i\)-th monomial of \(P\) evaluated at \((s, (t_i)_{i \ne m})\) is \(a_i (1 - P_1' - s)^{b_i}(s^2 + P_2')^{c_i}\). Binomially expanding \((s^2 + P_2')^{c_i} = \sum_{c' \le c_i} \binom{c_i}{c'} (P_2')^{c'} s^{2c_i - 2c'}\) and evaluating each one-dimensional integral by the Beta formula \(\int_0^L (L - s)^{B} s^{E}\, ds = L^{B + E + 1} B!\,E!/(B + E + 1)!\) with \(L = 1 - P_1'\) gives \[\int_0^{1 - P_1'} P \, ds \;=\; \sum_{i=1}^{d} a_i \sum_{c' = 0}^{c_i} \binom{c_i}{c'} (P_2')^{c'}\, (1 - P_1')^{\,b_i + 2c_i - 2c' + 1}\, \frac{b_i!\,(2c_i - 2c')!}{(b_i + 2c_i - 2c' + 1)!} .\]

Squaring. Multiplying this finite sum by itself produces the fourfold sum over \((i, j, c', c'')\), with the \((P_2')\)-powers combining to \((P_2')^{c' + c''}\), the \((1 - P_1')\)-powers combining to \((1 - P_1')^{\,b_i + b_j + 2c_i + 2c_j - 2c' - 2c'' + 2}\), and the factorial prefactors combining to \(\gamma(b_i, b_j, c_i, c_j, c', c'')\) divided by \(\bigl(b_i + b_j + 2c_i + 2c_j - 2c' - 2c'' + 2\bigr)!\).

The outer integral. Integrating term by term over \(\mathcal{R}_k^{(m)} \cong \mathcal{R}_{k-1}\) and applying Lemma 1 in dimension \(k - 1\) with \(a = b_i + b_j + 2c_i + 2c_j - 2c' - 2c'' + 2\) and \(b = c' + c''\) replaces each term by \[\frac{\bigl(b_i + b_j + 2c_i + 2c_j - 2c' - 2c'' + 2\bigr)!\; G_{c' + c'',\,2}(k-1)} {\bigl(k + b_i + b_j + 2c_i + 2c_j + 1\bigr)!},\] the denominator being \((k - 1) + a + 2b\) factorial. The factorial in the numerator cancels the one introduced in the squaring step, leaving exactly the stated expression. ◻

Proof. Symmetry is clear. For \(\alpha \in \mathbb{R}^d\), expanding the square and using integrability of each product \(f_i f_j\) (a consequence of Cauchy–Schwarz and the integrability of the \(f_i^2\)), \[\alpha^{\!\top} A \alpha \;=\; c \int_X \Bigl(\sum_{i=1}^{d} \alpha_i f_i\Bigr)^{\!2} d\mu \;\ge\; 0 .\] If \(\alpha^{\!\top} A \alpha = 0\) then, as \(c > 0\) and the integrand is non-negative, \(\sum_i \alpha_i f_i = 0\) almost everywhere, whence \(\alpha = 0\) by hypothesis. So the form is strictly positive on \(\mathbb{R}^d \setminus \{0\}\).

This is applied twice. Taking \(X = \mathcal{R}_k\) with Lebesgue measure and \(f_i\) the monomials \((1 - P_1)^{b_i} P_2^{\,c_i}\) shows that the matrix \(A_I\) whose quadratic form is \(I_k(F_P)\) (Lemma 1) is positive definite. Taking \(X = \mathcal{R}_k^{(m)}\) and \(f_i\) the marginals \(T_m f_i\) for one fixed \(m\) shows that the matrix with entries \(\int (T_m f_i)(T_m f_j)\) is positive definite; the remaining \(k - 1\) matrices in the sum \(A_J := \bigl(\sum_{m} \int (T_m f_i)(T_m f_j)\bigr)_{ij}\), whose quadratic form is \(\sum_m J_k^{(m)}(F_P)\) (Lemma 1), are positive semidefinite by the same computation, so \(A_J\) is positive definite. Only the independence of the marginals for a single \(m\) is needed. ◻

Proof. Since \(A_1\) is positive definite it factors as \(A_1 = L L^{\!\top}\) with \(L\) invertible. Put \(B := L^{-1} A_2 (L^{-1})^{\!\top}\); then \(B\) is symmetric, and it is positive definite because \(\beta^{\!\top} B \beta = \alpha^{\!\top} A_2 \alpha\) for \(\alpha := (L^{-1})^{\!\top}\beta\), and \(\beta \mapsto \alpha\) is a linear bijection. Substituting \(\beta := L^{\!\top} \alpha\) turns the generalised quotient into an ordinary one: \[\frac{\alpha^{\!\top} A_2 \alpha}{\alpha^{\!\top} A_1 \alpha} \;=\; \frac{\beta^{\!\top} B \beta}{\beta^{\!\top}\beta}, \qquad \alpha \ne 0 \iff \beta \ne 0 .\] By the spectral theorem \(B\) has an orthonormal eigenbasis with real eigenvalues; let \(\lambda\) be the largest of them and \(\beta_0\) a corresponding unit eigenvector. Expanding \(\beta\) in the eigenbasis gives \(\beta^{\!\top} B \beta \le \lambda\, \beta^{\!\top}\beta\) for all \(\beta\), with equality at \(\beta_0\); this is (i), and (ii) follows since the quotient is bounded above by \(\lambda\) and attains it.

For (iii), \(B\beta = \nu\beta\) with \(\beta \ne 0\) is equivalent, under \(\beta = L^{\!\top} \alpha\), to \(A_2 \alpha = \nu A_1 \alpha\) with \(\alpha \ne 0\); so the generalised eigenvalues of the pair \((A_1, A_2)\) are exactly the eigenvalues of \(B\), and \(\lambda\) is the largest. For (iv), \(A_2 \alpha = \nu A_1 \alpha\) is equivalent to \((A_1^{-1} A_2)\alpha = \nu \alpha\) because \(A_1\) is invertible, and for a square real matrix the spectrum consists precisely of its eigenvalues; so the spectrum of \(A_1^{-1} A_2\) is the set of generalised eigenvalues, whose largest element is \(\lambda\). ◻

Proof. Let \(P := \sum_i a_i (1 - P_1)^{b_i} P_2^{\,c_i}\) and let \(F_P\) be the associated profile, which vanishes outside \(\mathcal{R}_k = \mathcal{T}_0\). The monomials \((1 - P_1)^{b_i} P_2^{\,c_i}\), for pairwise distinct index pairs, are linearly independent as functions on \(\mathcal{R}_k\) modulo null sets; hence by Lemma 1 the Gram matrix \(A_I\) whose quadratic form is \(A(\alpha)\) is positive definite, and therefore \(A(\alpha) > 0\) because \(\alpha \ne 0\). By Lemma 1, \(I_k(F_P) = A(\alpha) > 0\), so \(F_P\) is an admissible competitor in Definition 1 at \(\varepsilon = 0\). By Lemma 1 each of the \(k\) marginal masses equals \(B(\alpha)\), so \(\sum_{m=1}^{k} J_k^{(m)}(F_P) = k\, B(\alpha)\). Definition 1 therefore gives \[M_k \;\ge\; \frac{k\, B(\alpha)}{A(\alpha)} \;>\; c ,\] the last step being the hypothesis divided by \(A(\alpha) > 0\).

The hypothesis is an inequality between two rational numbers whenever \(\alpha\) is rational, so it can be certified by an exact finite computation. The optimal constant obtainable this way from a fixed monomial family is, by Lemma 1 applied to the pair \((A_I, A_J)\) — both positive definite by Lemma 1 — equal to \(k\) times the largest element of the spectrum of \(A_I^{-1} A_J\). ◻

Proof. The integral is the Euler Beta function \(B(a,b)\) evaluated at positive integer arguments. For integers it satisfies \(B(a,b) = \Gamma(a)\Gamma(b)/\Gamma(a+b)\), and \(\Gamma(n) = (n-1)!\) for \(n \ge 1\); substituting gives the stated quotient of factorials. ◻

Proof. Since \(d\) and \(e\) are squarefree, so are \([d, e]\) and \((d, e)\), and \[\varphi(n) \;=\; \prod_{p \mid n}(p - 1)\] for every squarefree \(n\). The prime divisors of \([d,e]\) are exactly those of \(d\) together with those of \(e\), while the prime divisors of \((d,e)\) are exactly those dividing both. Hence, counting each prime with the multiplicity with which it occurs on either side, \[\prod_{p \mid [d,e]}(p-1)\ \cdot \prod_{p \mid (d,e)}(p-1) \;=\; \prod_{p \mid d}(p-1)\ \cdot \prod_{p \mid e}(p-1),\] which is the assertion. ◻

Proof. By Lemma 1, \(\varphi([d,e]) = \varphi(d)\varphi(e)/\varphi((d,e))\), all the factors being positive. The integer \((d, e)\) is squarefree, so Lemma 1 applies to it and gives \(\varphi((d,e)) = \sum_{u \mid (d,e)} g(u)\). Substituting and inverting yields the claim. ◻

Proof. The classical identity \(\sum_{u \mid n}\varphi(u) = n\), valid for every positive integer \(n\), applied at \(n = (d,e)\) turns the right-hand side into \((d,e)/(de)\). The claim is then the identity \([d,e]\,(d,e) = de\). ◻

Proof. For squarefree \(u\), \[\frac{u}{\varphi(u)} \;=\; \prod_{p \mid u}\frac{p}{p - 1} \;=\; \prod_{p \mid u}\Bigl(1 + \frac{1}{p - 1}\Bigr) \;=\; \sum_{d \mid u}\prod_{p \mid d}\frac{1}{p - 1} \;=\; \sum_{d \mid u}\frac{\mu(d)^2}{\varphi(d)}.\] The third equality expands the product over \(p \mid u\) as a sum over the subsets of the prime divisors of \(u\), i.e. over the divisors \(d \mid u\); the fourth uses \(\prod_{p \mid d}1/(p-1) = \mu(d)^2/\varphi(d)\) on squarefree \(d\). Dividing by \(u\) gives the claim. ◻

Proof. Let \((d, e) \in \mathcal{D}_m\) and put \(c := (d,e)\), \(a := d/c\), \(b := e/c\). Then \(cab = (d,e)\,[d,e]/(d,e) \cdot \dotsb\); more directly, \(ca = d\), \(cb = e\) and \(cab = d\,b = d\,e/(d,e) = [d,e] = m\). The integers \(a\) and \(b\) are coprime because any common prime divisor would divide both \(d/c\) and \(e/c\) and hence, \(m\) being squarefree, already lie in \(c\); the same squarefreeness of \(m = cab\) forces \(c\) to be coprime to both \(a\) and \(b\). So the map lands in \(\mathcal{C}_m\).

Conversely let \((c, a, b) \in \mathcal{C}_m\) and put \(d := ca\), \(e := cb\). Both are divisors of the squarefree \(m\), hence squarefree, and \((d,e) = c\) and \([d,e] = cab = m\) by pairwise coprimality; so \((d,e) \in \mathcal{D}_m\). The two constructions are mutually inverse: starting from \((d,e)\) one recovers \(\bigl((d,e)\bigr)\cdot\bigl(d/(d,e)\bigr) = d\) and likewise for \(e\); starting from \((c,a,b)\) one recovers \((ca, cb) \mapsto (c, a, b)\) because \((ca, cb) = c\) when \((a,b) = 1\). ◻

Proof. Step 1: a Chebyshev upper bound. For every \(N \ge 0\), \[\sum_{d \le N}\Lambda(d) \;\le\; (\log 4 + 4)\,N .\] This is the classical elementary bound obtained from the central binomial coefficient: \(\prod_{N < p \le 2N}p\) divides \(\binom{2N}{N} \le 4^N\), so \(\sum_{N < p \le 2N}\log p \le N\log 4\), and summing the resulting dyadic estimate together with the \(O(\sqrt N \log N)\) contribution of the proper prime powers gives the stated linear bound with the explicit constant \(\log 4 + 4\).

Step 2: the factorial identity. Since every \(n \ge 1\) satisfies \(\log n = \sum_{d \mid n}\Lambda(d)\), grouping by \(d\) gives \[\sum_{n \le N}\log n \;=\; \sum_{d \le N}\Lambda(d)\Bigl\lfloor\frac{N}{d}\Bigr\rfloor .\] Replacing \(\lfloor N/d\rfloor\) by \(N/d\) costs at most \(\sum_{d \le N}\Lambda(d)\), which by Step 1 is at most \((\log 4 + 4)N\). Hence \[\Bigl|\sum_{n \le N}\log n \;-\; N\sum_{d \le N}\frac{\Lambda(d)}{d}\Bigr| \;\le\; (\log 4 + 4)\,N .\]

Step 3: Stirling-free evaluation of \(\sum_{n \le N}\log n\). Comparing the sum with \(\int_1^N \log t\,dt = N\log N - N + 1\) by monotonicity of \(\log\) gives \(\bigl|\sum_{n \le N}\log n - N\log N\bigr| \le N\). Combining with Step 2 and dividing by \(N\) yields \[\Bigl|\sum_{d \le N}\frac{\Lambda(d)}{d} - \log N\Bigr| \;\le\; \log 4 + 5 . \qedhere\] ◻

Proof. Rankin majorisation. For any non-negative \(a : \mathbb{N} \to \mathbb{R}_{\ge 0}\), any real \(Z \ge 1\) and any \(\kappa \ge 0\) one has \((Z/u)^\kappa \ge 1\) for \(u \le Z\), so \(\sum_{u \le Z}a(u) \le Z^\kappa \sum_{u \ge 1}a(u)u^{-\kappa}\). Apply this with \(a(u) := \mu(u)^2\tau_k(u)/\varphi(u)\) and \(\kappa := 1/\log(\mathrm{e}z) \in (0,1)\).

Euler factorisation. The majorant \(u \mapsto \mu(u)^2\tau_k(u)/(\varphi(u)u^\kappa)\) is non-negative, multiplicative and supported on squarefree integers; since \(\tau_k(p) = k\) and \(\varphi(p) = p-1\), and all prime-power terms with exponent \(\ge 2\) vanish, \[\sum_{u \ge 1}\frac{\mu(u)^2\tau_k(u)}{\varphi(u)u^\kappa} \;=\; \prod_p\Bigl(1 + \frac{k}{(p-1)p^\kappa}\Bigr).\]

Logarithmic bound. Using \(\log(1+x) \le x\) and splitting \(1/(p-1) = 1/p + 1/(p(p-1))\), \[\log\prod_p\Bigl(1 + \frac{k}{(p-1)p^\kappa}\Bigr) \;\le\; k\sum_p \frac{1}{p^{1+\kappa}} \;+\; k\sum_p\frac{1}{p(p-1)},\] and the second sum converges absolutely. For the first, start from Lemma 1: discarding the proper prime powers (whose total contribution to \(\sum_{d \le t}\Lambda(d)/d\) is \(O(1)\), being bounded by \(\sum_p \sum_{a \ge 2}(\log p)/p^a \ll \sum_p (\log p)/p^2\)) gives the one-sided prime form \(\sum_{p \le t}(\log p)/p \le \log t + O(1)\) for \(t \ge 2\). Abel summation of this against the weight \(1/\log t\) yields the elementary one-sided Mertens second bound \(\sum_{p \le t}1/p \le \log\log t + B\) with \(B\) absolute. A second Abel summation, now against the weight \(t^{-\kappa}\), gives \(\sum_p p^{-1-\kappa} = \log(1/\kappa) + O(1)\). With \(\kappa = 1/\log(\mathrm{e}z)\) this is \(\log\log(\mathrm{e}z) + O(1) \le \log\log z + O(1)\) for \(z \ge 3\) (the range \(2 \le z < 3\) being trivial). Only upper bounds on prime sums are used, so no Tauberian or Dirichlet-series input enters.

Conclusion. Exponentiating gives \(\prod_p(1 + k(p-1)^{-1}p^{-\kappa}) \le C_1(k)(\log z)^k\), while \(Z^\kappa = z^{1/\log(\mathrm{e}z)} \le \mathrm{e}\). Combining with the Rankin majorisation proves the lemma. ◻

Proof. The summand is non-negative, so discarding the coprimality condition only increases the sum; the claim is then Lemma 1. (The point of the statement is the uniformity in \(D\): the same constant serves for every modulus, which is what the downstream applications, where \(D\) varies with the sieve variables, require.) ◻

Proof. Write \(r_i = d_i s_i\) and \(s' := \prod_i s_i\). Because \(\prod_i r_i\) is squarefree, the \(d_i\) are pairwise coprime and coprime to \(s'\), and \(\prod_i \varphi(r_i) = \varphi(\prod_i d_i) \varphi(s')\) by multiplicativity of \(\varphi\) on coprime factors. Hence the left-hand side is at most \[\frac{\prod_i d_i}{\varphi(\prod_i d_i)} \sum_{\substack{s' \le R/\prod_i d_i\\ \mu(s')^2 = 1,\ (s', W\prod_i d_i) = 1}} \frac{\tau_k(s')}{\varphi(s')},\] since a squarefree \(s'\) arises from at most \(\tau_k(s')\) tuples \((s_1, \dotsc, s_k)\) with \(\prod_i s_i = s'\). For the prefactor, the Dirichlet expansion of Lemma 1 gives \(\prod_i d_i/\varphi(\prod_i d_i) = \sum_{D \mid \prod_i d_i}\mu(D)^2/\varphi(D)\), which is bounded by \(\sum_{D \mid \prod_i d_i}\mu(D)^2\tau_k(D)/\varphi(D)\); and every prime dividing \(\prod_i d_i\) is at most \(R\). The remaining \(s'\)-sum is a truncated coprime divisor sum at modulus \(\bigl(W\prod_i d_i\bigr)\), so Lemma 1 bounds it by \(C(k)(\log R)^k\) with a constant that does not depend on that modulus — the uniformity there is exactly what is needed, since the modulus varies with \(\mathbf{d}\). Both factors are thus dominated by the truncated sum \(\sum_{u \le R}\mu(u)^2\tau_k(u)/\varphi(u)\) of Lemma 1 — more precisely, the whole expression is bounded by the Euler product \(\prod_{p \le R}\bigl(1 + k/(p-1)\bigr)\), and the argument of Lemma 1 bounds this by \(C(k)(\log R)^k\) for \(R \ge 2\). ◻

Proof. The summand is non-negative, so it suffices to majorise it by a convergent series. The inequality \(s \le \varphi(s)\tau(s)\) established in the proof of Lemma 1 gives \(1/\varphi(s) \le \tau(s)/s\), whence \[\frac{\mu(s)^2}{\varphi(s)^2} \;\le\; \frac{\tau(s)^2}{s^2} \;=\; \frac{\tau(s)}{\sqrt s}\cdot\frac{\tau(s)}{s^{3/2}} \;\le\; \frac{2\,\tau(s)}{s^{3/2}},\] the last step by the elementary bound \(\tau(s) \le 2\sqrt s\) (divisors pair up as \(u \leftrightarrow s/u\) about \(\sqrt s\)). The Dirichlet series \(\sum_{s \ge 1}\tau(s)s^{-3/2} = \bigl(\sum_{s \ge 1}s^{-3/2}\bigr)^2\) converges since \(3/2 > 1\), so the partial sums of the original series are uniformly bounded, and a series of non-negative terms with bounded partial sums converges. ◻

Proof. The summand is non-negative and multiplicative, supported on odd squarefree integers. Its local factor at an odd prime \(p\) is \(1 + (p-2)^{-2}\), and at \(p = 2\) it is \(1\). Since \((p-2)^{-2} \le 9 p^{-2}\) for every odd prime \(p\) (indeed \(p - 2 \ge p/3\) for \(p \ge 3\)), the series \(\sum_{p}(p-2)^{-2}\) converges, so \(\prod_{p \text{ odd}}\bigl(1 + (p-2)^{-2}\bigr) < \infty\). Every partial sum over a finite set of integers is bounded by this Euler product, whence convergence. ◻

Proof. By unique factorisation, every \(n \ge 1\) has a unique representation \(n = \prod_p p^{a_p}\) with \(a_p \ge 0\) and only finitely many \(a_p\) non-zero, so \(n^{-2} = \prod_p p^{-2a_p}\). The series \(\sum_n n^{-2}\) converges absolutely (it is bounded by \(1 + \int_1^\infty t^{-2}\,dt = 2\)), which licenses reorganising it as a product over primes of the geometric series \(\sum_{a \ge 0}p^{-2a} = (1 - p^{-2})^{-1}\); the rearrangement is justified by letting \(P \to \infty\) in the partial product over \(p \le P\), whose expansion exhausts all \(n \ge 1\) in the limit. ◻

Proof. Each summand satisfies \(|\mu(e)/e^2| \le 1/e^2\), so \[\sum_{\substack{e \ge 1\\ (e, M) = 1}}\Bigl|\frac{\mu(e)}{e^2}\Bigr| \;\le\; \sum_{e \ge 1}\frac{1}{e^2} \;=\; \zeta(2) \;<\; \infty .\] This is absolute convergence, and the triangle inequality applied to the (absolutely convergent) sum gives \(|A(M)| \le \zeta(2)\). ◻

Proof. By Sublemma 1 the defining series converges absolutely, so it may be expanded as an Euler product. Its local factor at a prime \(p \mid M\) is \(1\) (only \(e = 1\) contributes at that prime), and at a prime \(p \nmid M\) it is \(\sum_{a \ge 0}\mu(p^a)p^{-2a} = 1 - p^{-2}\), since \(\mu(p^a) = 0\) for \(a \ge 2\). Hence \[A(M) \;=\; \prod_{p \nmid M}\Bigl(1 - \frac{1}{p^2}\Bigr).\] Splitting the product over all primes as \(\prod_p = \prod_{p \mid M}\cdot\prod_{p \nmid M}\) and invoking \(\prod_p(1 - p^{-2}) = 1/\zeta(2)\) (Sublemma 1) gives the stated closed form. Note that only the set of prime divisors of \(M\) enters, which is why no squarefreeness hypothesis is needed. ◻

Proof. Detect the coprimality condition by \(\mathbf{1}[(n,m) = 1] = \sum_{d \mid (n,m)}\mu(d)\) and swap the two finite sums: \[\sum_{\substack{n \le Z\\ (n,m)=1}}\frac1n \;=\; \sum_{d \mid m}\mu(d)\sum_{\substack{n \le Z\\ d \mid n}}\frac1n \;=\; \sum_{d \mid m}\frac{\mu(d)}{d}\sum_{n' \le Z/d}\frac{1}{n'} .\] The classical harmonic estimate \(\sum_{n' \le Y}1/n' = \log Y + \gamma_0 + O(1/Y)\) gives \[\sum_{\substack{n \le Z\\ (n,m)=1}}\frac1n \;=\; (\log Z + \gamma_0)\sum_{d \mid m}\frac{\mu(d)}{d} \;-\; \sum_{d \mid m}\frac{\mu(d)\log d}{d} \;+\; O\Bigl(\frac{\tau(m)}{Z}\Bigr),\] the error using \(\sum_{d \mid m}|\mu(d)| = \tau(m)\) for squarefree \(m\). Now \(\sum_{d \mid m}\mu(d)/d = \prod_{p \mid m}(1 - 1/p) = \varphi(m)/m\), and differentiating the identity \(\sum_{d \mid m}\mu(d)d^{-s} = \prod_{p \mid m}(1 - p^{-s})\) at \(s = 1\) gives \[-\sum_{d \mid m}\frac{\mu(d)\log d}{d} \;=\; \prod_{p \mid m}\Bigl(1 - \frac1p\Bigr)\sum_{p \mid m}\frac{p^{-1}\log p}{1 - p^{-1}} \;=\; \frac{\varphi(m)}{m}\,\ell_m .\] Combining the three displays proves the claim, with an absolute implied constant. ◻

Proof. Sublemma 1 applied with \(m = \mathcal{W}\) and \(Z = x\) gives \[\sum_{\substack{n \le x\\ (n, \mathcal{W}) = 1}}\frac1n \;=\; \frac{\varphi(\mathcal{W})}{\mathcal{W}}\bigl(\log x + \gamma_0 + \ell_{\mathcal{W}}\bigr) \;+\; O\Bigl(\frac{\tau(\mathcal{W})}{x}\Bigr).\] The hypothesis on \(x\) is exactly \(\tau(\mathcal{W})/x \le \varphi(\mathcal{W})/\mathcal{W}\), so the error term is \(O(\varphi(\mathcal{W})/\mathcal{W})\); the same bound covers the constant \(\gamma_0\,\varphi(\mathcal{W})/\mathcal{W}\). It remains to bound \(\ell_{\mathcal{W}} = \sum_{p \mid \mathcal{W}}(\log p)/(p-1)\). Split the prime divisors of \(\mathcal{W}\) at \(1 + \log \mathcal{W}\). For \(p > 1 + \log\mathcal{W}\), use \(\sum_{p \mid \mathcal{W}}\log p = \log \mathcal{W}\) (valid since \(\mathcal{W}\) is squarefree) and \(p - 1 > \log\mathcal{W}\) to bound that part by \(1\); more crudely, the contribution is \(O(1)\). For \(p \le 1 + \log\mathcal{W}\), use \((\log p)/(p-1) \le 2(\log p)/p\) and the elementary Mertens bound \(\sum_{p \le t}(\log p)/p \le \log t + O(1)\) of Lemma 1 with \(t = 1 + \log\mathcal{W}\), which gives \(O\bigl(\log(1 + \log \mathcal{W})\bigr) = O\bigl(1 + \log\log(\mathrm{e}\mathcal{W})\bigr)\). Hence \(\ell_{\mathcal{W}} \ll 1 + \log\log(\mathrm{e}\mathcal{W})\), and collecting all error contributions proves the lemma. The normalisation \(1 + \log\log(\mathrm{e}\mathcal{W})\) rather than \(\log\log\mathcal{W}\) is used so that the error term is defined and non-negative for every \(\mathcal{W} \ge 1\); for large \(\mathcal{W}\) the two agree up to an additive constant. ◻

Proof. Step 1: a Dirichlet cofactor. Let \(\kappa\) be the multiplicative function defined by the Dirichlet convolution identity \(\mu^2/\varphi= \kappa \ast (1/\mathrm{id})\), i.e. \[\frac{\mu(f)^2}{\varphi(f)} \;=\; \sum_{de = f}\kappa(d)\,\frac{1}{e} .\] Comparing local Euler factors shows \(\kappa(1) = 1\), \(\kappa(p) = 1/(p(p-1))\), \(\kappa(p^2) = -1/(p(p-1))\) and \(\kappa(p^j) = 0\) for \(j \ge 3\); in particular \(\kappa\) is supported on cube-free integers, its local factor \(1 + \kappa(p) + \kappa(p^2)\) equals \(1\) at every prime, and \[\sum_{d \ge 1}|\kappa(d)|\,\sqrt{d} \;<\; \infty, \qquad \sum_{d \ge 1}|\kappa(d)|\log(2d) \;<\; \infty .\] Both convergence statements follow from the Euler-product bound \(\sum_{j \ge 0}|\kappa(p^j)|p^{j/2} \le \exp\bigl(2p^{-3/2}\bigr)\), together with the elementary partial-summation bound \(\sum_{d \le X}|\kappa(d)|\,d \ll \sqrt X \log(2X)\), which in turn comes from writing a cube-free \(d\) uniquely as \(b^2 a\) with \(a\) squarefree, \((a, b) = 1\), and using \(\sum_{b \le \sqrt X}1/\varphi(b) \ll 1 + \log X\). Consequently, for every \(Y \ge 1\), \[\sum_{d \ge Y}|\kappa(d)| \;\ll\; \frac{\log(2Y)}{\sqrt Y}, \qquad \sum_{d \ge Y}|\kappa(d)|\log(d/Y) \;\ll\; \frac{\log(2Y)}{\sqrt Y}\] (a dyadic decomposition of the range \(d \ge Y\)).

Step 2: the hyperbola identity. Summing \(\mu^2/\varphi= \kappa \ast (1/\mathrm{id})\) over \(f \le x\) with \((f, m) = 1\) and noting that \((f, m) = 1\) is equivalent to \((d, m) = (e, m) = 1\) for \(f = de\), \[A_m(x) \;=\; \sum_{\substack{d \le x\\ (d, m) = 1}}\kappa(d) \sum_{\substack{e \le x/d\\ (e, m) = 1}}\frac1e .\]

Step 3: insert the harmonic asymptotic. Apply Sublemma 1 to the inner sum for each \(d\), obtaining the main term \(\frac{\varphi(m)}{m}\bigl(\log(x/d) + \gamma_0 + \ell_m\bigr)\) and an error \(O(\tau(m)d/x)\); the errors sum, against \(\sum_{d \le x}|\kappa(d)|d \ll \sqrt x\log(2x)\), to \(O\bigl(\tau(m)\log(2x)/\sqrt x\bigr)\). For the main term, complete the \(d\)-sum to all \(d \ge 1\) coprime to \(m\); by Step 1 the discarded tail \(d > x\) contributes \(O\bigl(\tau(m)\log(2mx)/\sqrt x\bigr)\) (using \(\ell_m \le \log m\) to absorb the \(m\) into the logarithm). Since \(\kappa\) has local mass \(1\) at every prime, \(\sum_{(d,m)=1}\kappa(d) = 1\), and the completed main term is \[\frac{\varphi(m)}{m}\bigl(\log x + \gamma_0 + \ell_m\bigr) \;-\; \frac{\varphi(m)}{m}\sum_{\substack{d \ge 1\\(d,m)=1}}\kappa(d)\log d .\] The last sum is \(O(1)\) by the second convergence statement of Step 1, and \(\gamma_0\varphi(m)/m = O(\varphi(m)/m)\). Collecting all contributions gives the stated bound, with an absolute constant. ◻

Proof. Substitute the identity \(1/\varphi(u) = (1/u)\sum_{d \mid u}\mu(d)^2/\varphi(d)\) of Sublemma 1 — legitimate because \(\mu(u)^2 \ne 0\) restricts the outer sum to squarefree \(u\) — and swap the two finite summations: \[\sum_{\substack{u \le R\\ (u, W) = 1}}\frac{\mu(u)^2}{\varphi(u)} \;=\; \sum_{\substack{d \le R\\ (d, W) = 1}}\frac{\mu(d)^2}{\varphi(d)} \sum_{\substack{u \le R,\ d \mid u\\ (u, W) = 1}}\frac{\mu(u)^2}{u} .\] Fix a squarefree \(d\) with \((d, W) = 1\) and write \(u = dv\) with \(v \le R/d\). On the support of \(\mu(u)^2\) the integer \(u\) is squarefree, which for \(u = dv\) is equivalent to \(v\) squarefree with \((v, d) = 1\); and \((u, W) = 1\) together with \((d, W) = 1\) is equivalent to \((v, W) = 1\). So the conditions on \(u\) become “\(v\) squarefree and \((v, dW) = 1\)”. Since \(\mu(u)^2/u = \mu(v)^2/(dv)\) on that support, the inner sum equals \(T_d/d\), which is the claim. ◻

Proof. Start from the squarefree-detection identity \(\mu(v)^2 = \sum_{e^2 \mid v}\mu(e)\), valid for every positive integer \(v\): both sides are multiplicative, and at a prime power \(p^a\) the right-hand side is \(1\) for \(a \in \{0, 1\}\) and \(1 + \mu(p) = 0\) for \(a \ge 2\), matching \(\mu(p^a)^2\). Insert it into \(T_d = \sum_{v \le R/d,\,(v, dW) = 1}\mu(v)^2/v\) and swap the two finite sums. Writing \(v = e^2 w\) turns the constraint \(e^2 \mid v\) into a free \(w\)-sum with \(w \le R/(de^2)\), and the constraint \((v, dW) = 1\) into \((e, dW) = (w, dW) = 1\); the \(e\)-range is \(e \le \sqrt{R/d}\) because \(e^2 \le v \le R/d\). Since \(1/v = (1/e^2)(1/w)\), the displayed identity follows. (When \((e, dW) > 1\) the inner sum is empty, so restricting the \(e\)-sum to \((e, dW) = 1\) changes nothing.) ◻

Proof. Since \(d\) is squarefree, \(W\) is squarefree (Definition 1) and \((d, W) = 1\), the modulus \(\mathcal{W} = dW\) is a positive squarefree integer. We apply Lemma 1 with this modulus and with \(x := R/(de^2)\), and must verify \(x \ge \mathcal{W}\tau(\mathcal{W})/\varphi(\mathcal{W})\).

Lower bound on \(x\). From \(d \le R^{1/2}\) and \(e \le R^{1/8}\) we get \(de^2 \le R^{1/2}\cdot R^{1/4} = R^{3/4}\), hence \(x = R/(de^2) \ge R^{1/4}\).

Upper bound on \(\mathcal{W}\tau(\mathcal{W})/\varphi(\mathcal{W})\). For squarefree \(n\) the quantity \(n\tau(n)/\varphi(n)\) factors over the primes dividing \(n\), each contributing \(2p/(p-1) \le 4\); hence \(\mathcal{W}\tau(\mathcal{W})/\varphi(\mathcal{W}) \le 4^{\omega(\mathcal{W})}\). Next, for any fixed integer \(B \ge 2\), splitting the prime divisors of \(n\) at \(B\) gives \(\omega(n) \le \pi(B) + \log n/\log B\), hence \(4^{\omega(n)} \le 4^{\pi(B)}\,n^{\log 4/\log B}\). Choose \(B := 4^8 = 65536\), so that \(\log 4/\log B = 1/8\) and, with \(K := 4^{\pi(B)}\) an absolute constant, \[\frac{\mathcal{W}\tau(\mathcal{W})}{\varphi(\mathcal{W})} \;\le\; K\,\mathcal{W}^{1/8}.\] By Lemma 1 we have \(W \ll (\log\log N)^2\), and \(d \le R^{1/2}\), so \(\mathcal{W} = dW \le R^{1/2}(\log\log N)^2\) and therefore \[\frac{\mathcal{W}\tau(\mathcal{W})}{\varphi(\mathcal{W})} \;\le\; K\,R^{1/16}\,(\log\log N)^{1/4} .\] Since \(R = N^{\vartheta/2 - \delta}\) is a fixed positive power of \(N\), the right-hand side is \(\le R^{1/4} \le x\) for all sufficiently large \(N\), the factor \((\log\log N)^{1/4}\) being of lower order than \(R^{3/16}\). This verifies the hypothesis.

With the hypothesis verified, Lemma 1 gives \[\Bigl|\sum_{\substack{w \le R/(de^2)\\ (w, \mathcal{W}) = 1}}\frac1w - \frac{\varphi(\mathcal{W})}{\mathcal{W}}\log\frac{R}{de^2}\Bigr| \le C\,\frac{\varphi(\mathcal{W})}{\mathcal{W}}\bigl(1 + \log\log(\mathrm{e}\mathcal{W})\bigr),\] and \(\log\bigl(R/(de^2)\bigr) = \log R - \log d - 2\log e\). ◻

Proof. Sublemma 1 rewrites \(T_d\) as an \(e\)-sum weighted by \(\mu(e)/e^2\) times an inner \(w\)-sum. Split the \(e\)-range at \(e = R^{1/8}\).

Range \(e \le R^{1/8}\). Sublemma 1 evaluates the inner sum for each such \(e\), with error \(O\bigl(\frac{\varphi(\mathcal{W})}{\mathcal{W}}(1 + \log\log(\mathrm{e}\mathcal{W}))\bigr)\). Summing the main term against \(\mu(e)/e^2\) over \(e \le R^{1/8}\) coprime to \(\mathcal{W}\) produces \[\frac{\varphi(\mathcal{W})}{\mathcal{W}}\Bigl[\log(R/d)\sum_{\substack{e \le R^{1/8}\\(e,\mathcal{W})=1}}\frac{\mu(e)}{e^2} \;-\; 2\sum_{\substack{e \le R^{1/8}\\(e,\mathcal{W})=1}}\frac{\mu(e)\log e}{e^2}\Bigr].\] Completing both \(e\)-sums to infinity replaces them by \(A(\mathcal{W})\) and \(B(\mathcal{W})\) of Definitions 1 and 1; by the tail bounds \(\bigl|A(M) - \sum_{e \le m}\bigr| \le 1/m\) and \(\bigl|B(M) - \sum_{e \le m}\bigr| \le (\log m + 1)/m\) (each a comparison with \(\sum_{e > m}e^{-2}\) and \(\sum_{e > m}(\log e)e^{-2}\) respectively), applied at \(m = \lfloor R^{1/8}\rfloor\), the cost is \(O(\log R \cdot R^{-1/8})\). The accumulated error terms contribute \(O\bigl(\frac{\varphi(\mathcal{W})}{\mathcal{W}}(1 + \log\log(\mathrm{e}\mathcal{W}))\bigr)\) after summation against \(\sum_e e^{-2} \le 2\).

Range \(e > R^{1/8}\). Here use the trivial bounds \(|\mu(e)/e^2| \le 1/e^2\) and \(\sum_{w \le R/(de^2)}1/w \le 1 + \log R\), so this part is \[\ll \log R \sum_{e > R^{1/8}}\frac{1}{e^2} \;\ll\; \frac{\log R}{R^{1/8}} .\]

Normalising the error. Finally \(\log\log(\mathrm{e}\mathcal{W}) = \log\log(\mathrm{e}dW) \ll \log(2d) + \log\log W\) for \(W \ge \mathrm{e}\), which converts the middle error into the stated shape. Adding the three contributions gives the sublemma. ◻

Proof. From \(\mu(v)^2/v \le 1/v\) and \(\sum_{v \le R/d}1/v \le 1 + \log R\) we get \(0 \le T_d \le 1 + \log R \ll \log R\) for \(R \ge 2\), uniformly in \(d\) and \(W\). By Lemma 1, \(1/(d\varphi(d)) \le \tau(d)/d^2\). Lemma 1 with \(k = 2\) gives \(\sum_{d \le X}\mu(d)^2\tau(d)/\varphi(d) \ll (\log X)^2\), hence also \(\sum_{d \le X}\mu(d)^2\tau(d)/d \ll (\log X)^2\); partial summation converts this into the tail estimate \[\sum_{d > Y}\frac{\mu(d)^2\tau(d)}{d^2} \;\ll\; \frac{(\log Y)^2 + 1}{Y}.\] Applying it with \(Y = \sqrt R\) bounds \(\sum_{d > \sqrt R}\mu(d)^2/(d\varphi(d)) \ll (\log R)^2/\sqrt R\), and multiplying by the uniform bound \(T_d \ll \log R\) gives the claim. ◻

Proof. Write \(f(d) := \mu(d)^2 A(dW)/d^2\). By Sublemma 1, \(A(dW) = (1/\zeta(2))/\prod_{p \mid dW}(1 - p^{-2})\), so for squarefree \(d\) coprime to \(W\) \[f(d) \;=\; A(W)\prod_{p \mid d}\frac{1/p^2}{1 - 1/p^2} .\] In particular \(f(1) = A(W)\), and \(f\) is multiplicative up to the normalising constant \(A(W)\): for coprime squarefree \(d_1, d_2\) both coprime to \(W\) one has \(f(d_1 d_2)A(W) = f(d_1)f(d_2)\); and at a prime \(p \nmid W\), \(f(p) = p^{-2}A(W)/(1 - p^{-2})\). Since the values of \(f\) at distinct primes multiply in this normalised sense, and \(\mu(d)^2\) restricts the sum to squarefree \(d\), each prime \(p \nmid W\) contributes independently either \(1\) or \(p^{-2}(1-p^{-2})^{-1}\), giving \[\sum_{\substack{d \ge 1\\(d, W) = 1}}\frac{\mu(d)^2}{d^2}A(dW) \;=\; A(W)\prod_{p \nmid W}\Bigl(1 + \frac{1}{p^2}\cdot\frac{1}{1 - 1/p^2}\Bigr).\] (The rearrangement is licensed by absolute convergence: \(|A(dW)| \le \zeta(2)\) by Sublemma 1 and \(\sum_d d^{-2} < \infty\).) The elementary identity \(1 + x/(1-x) = 1/(1-x)\) with \(x = p^{-2}\) turns the local factor into \((1 - p^{-2})^{-1}\), so the product equals \(\prod_{p \nmid W}(1 - p^{-2})^{-1}\). Combining with \(A(W) = (1/\zeta(2))/\prod_{p \mid W}(1 - p^{-2})\), \[\sum_{\substack{d \ge 1\\(d, W) = 1}}\frac{\mu(d)^2}{d^2}A(dW) \;=\; \frac{1/\zeta(2)}{\prod_{p \mid W}(1 - p^{-2})\prod_{p \nmid W}(1 - p^{-2})} \;=\; \frac{1/\zeta(2)}{\prod_p (1 - p^{-2})} \;=\; 1,\] the last equality by Sublemma 1. ◻

Proof. By Sublemma 1 the left-hand sum equals \(\sum_d \mu(d)^2 T_d/(d\varphi(d))\). Sublemma 1 bounds the contribution of \(d > \sqrt R\) by \(O\bigl((\log R)^3/\sqrt R\bigr)\), which is \(o(\varphi(W)/W)\) for large \(N\) since \(\varphi(W)/W \gg 1/\log\log N\) by Lemma 1. Restrict to \(d \le \sqrt R\).

Insert the expansion of Sublemma 1 and use \(\varphi(dW)/(dW) = (\varphi(d)/d)(\varphi(W)/W)\), valid because \((d, W) = 1\). This gives \[\sum_{\substack{u \le R\\ (u, W) = 1}}\frac{\mu(u)^2}{\varphi(u)} \;=\; \frac{\varphi(W)}{W}\Bigl[(\log R)\,\mathcal{S}_1 - \mathcal{S}_2 - 2\mathcal{S}_3\Bigr] \;+\; O\Bigl(\frac{\varphi(W)}{W}\log\log W\Bigr),\] where, with all sums over squarefree \(d\) coprime to \(W\), \[\mathcal{S}_1 := \sum_{d}\frac{\mu(d)^2 A(dW)}{d^2},\qquad \mathcal{S}_2 := \sum_{d}\frac{\mu(d)^2 (\log d)A(dW)}{d^2},\qquad \mathcal{S}_3 := \sum_{d}\frac{\mu(d)^2 B(dW)}{d^2},\] and the error term also absorbs the \(\log(2d)\) contributions of Sublemma 1 (which sum, against \(\mu(d)^2\varphi(d)/(d^2\varphi(d)) \le d^{-2}\), to \(O(1)\)) and the \(O(\log R/R^{1/8})\) terms. By Sublemma 1, \(\mathcal{S}_1 = 1\) once the \(d\)-sum is completed, and completing it costs \(O(R^{-1/2})\) by Sublemma 1. The sums \(\mathcal{S}_2\) and \(\mathcal{S}_3\) are \(O(1)\) in absolute value: each is bounded by \(\zeta(2)\sum_{d \ge 1}(1 + \log d)/d^2 \ll 1\), again using \(|A(M)|, |B(M)| \ll 1\) from Sublemma 1 and the corresponding trivial bound for \(B\). Assembling and absorbing the \(O(1)\) and \(O(R^{-1/2}\log R)\) contributions into \(O\bigl(\frac{\varphi(W)}{W}\log\log W\bigr)\) — legitimate since \(\log\log W \ge 1\) for \(N\) large — gives the lemma. ◻

Proof. Both sides are totally multiplicative in \(r\), so it suffices to compare them at primes. If \(p \mid W\) then \(\gamma_g(p) = 0\), so \(g_*(p) = 0\); total multiplicativity then forces \(g_*(r) = 0\) whenever some prime divisor of \(r\) divides \(W\), i.e. whenever \((r, W) > 1\). If \(p \nmid W\) then \(\gamma_g(p) = p/(p-1)\) and \[g_*(p) \;=\; \frac{\gamma_g(p)}{p - \gamma_g(p)} \;=\; \frac{p/(p-1)}{p - p/(p-1)} \;=\; \frac{1}{p - 2} \;=\; \frac{1}{g(p)} .\] For squarefree \(r\) coprime to \(W\) we have \(\mu(r)^2 = 1\), and multiplying the prime-level identities over the (distinct) primes dividing \(r\) gives \(g_*(r) = 1/g(r)\). ◻

Proof. Step 1: compare \(\gamma_N\) with the constant density. For a prime \(p \nmid W\) write \(\gamma_N(p)(\log p)/p = (\log p)/p + (\gamma_N(p) - 1)(\log p)/p\); the second term is at most \(c(\log p)/p^2\) in absolute value, and \(\sum_p (\log p)/p^2\) converges. For \(p \mid W\) the term vanishes altogether, and \(p \mid W\) holds exactly when \(p \le D_0\). Hence there is \(E = E(c)\) with \[\Bigl|\sum_{w \le p \le z}\frac{\gamma_N(p)\log p}{p} \;-\; \sum_{\max(w,\, D_0) \le p \le z}\frac{\log p}{p}\Bigr| \;\le\; E\] for all \(2 \le w \le z\).

Step 2: Mertens’ first theorem. By Lemma 1, discarding the proper prime powers as in the proof of Lemma 1, there is an absolute \(M\) with \(\bigl|\sum_{p \le t}(\log p)/p - \log t\bigr| \le M\) for \(t \ge 2\); hence \(\sum_{a \le p \le z}(\log p)/p \le \log(z/a) + 2M\) for all \(2 \le a \le z\).

Step 3. Apply Step 2 with \(a := \max(w, D_0) \ge w\). Since \(a \ge w\) we have \(\log(z/a) \le \log(z/w)\), so \[\Delta(\gamma_N; w, z) \;\le\; E + \log(z/a) + 2M - \log(z/w) \;\le\; E + 2M .\] Thus \(A_2 := E(c) + 2M\) works, and it does not depend on \(N\), \(w\) or \(z\). (The gain over the two-sided estimate below is that truncating the prime range upward to \(\max(w, D_0)\) can only decrease the sum, so the \(\log D_0\) discrepancy appears with a favourable sign.) ◻

Proof. Retain the notation of the previous proof. Steps 1 and 2 there give, for \(a = \max(w, D_0)\), \[\Delta(\gamma_N; w, z) \;\ge\; -E + \log(z/a) - 2M - \log(z/w) \;=\; -E - 2M - \log\frac{a}{w} .\] Now \(w \le a = \max(w, D_0) \le w\,D_0\) (using \(w \ge 2\) and \(D_0 \ge 2\)), so \(0 \le \log(a/w) \le \log D_0\). Hence \(\Delta(\gamma_N; w, z) \ge -\bigl(\log D_0 + E + 2M\bigr)\). Since \(D_0 \ge 2\) we have \(\log D_0 \ge \log 2 > 0\), so \(L := \log D_0 + E + 2M \le C\log D_0\) with \(C := 1 + (E + 2M)/\log 2\), which depends only on \(c\). Non-negativity of \(L\) is clear. ◻

Proof. Split the defining product \(\mathfrak{S}(\gamma_g) = \prod_p (1 - \gamma_g(p)/p)^{-1} (1 - 1/p)\) over \(p \mid W\) and \(p \nmid W\). For \(p \mid W\) one has \(\gamma_g(p) = 0\) and the local factor is \(1 - 1/p\); the product of these over \(p \mid W\) is \(\varphi(W)/W\). For \(p \nmid W\) one has \(\gamma_g(p) = p/(p-1)\), so \[\Bigl(1 - \frac{\gamma_g(p)}{p}\Bigr)^{-1}\Bigl(1 - \frac1p\Bigr) \;=\; \Bigl(1 - \frac{1}{p-1}\Bigr)^{-1}\frac{p-1}{p} \;=\; \frac{(p-1)^2}{p(p-2)} \;=\; 1 + \frac{1}{p(p-2)} .\] Every such \(p\) exceeds \(D_0\), so the tail product satisfies \(1 \le \prod_{p > D_0}\bigl(1 + \frac{1}{p(p-2)}\bigr) \le \exp\bigl(\sum_{p > D_0}\frac{1}{p(p-2)}\bigr) \le \exp(2/D_0)\), using \(\sum_{n > t}n^{-2} \le 2/t\) and \(p(p-2) \ge p^2/3\) for \(p \ge 3\). Since \(\mathrm{e}^{2/D_0} - 1 \le 4/D_0\) for \(D_0 \ge 4\), the tail product is \(1 + O(1/D_0)\), and multiplying by \(\varphi(W)/W\) gives the claim. ◻

Proof. Fix a prime \(p \nmid V\). The structural hypothesis of Definition 1 gives \(|\gamma(p) - 1| \le A_3/p\), and Lemma 1 gives \(p - \gamma(p) \ge A_1 p\). Then \[b(p) \;=\; \frac{\gamma(p)}{p - \gamma(p)} - \frac{1}{p - 1} \;=\; \frac{p\bigl(\gamma(p) - 1\bigr)}{\bigl(p - \gamma(p)\bigr)(p - 1)},\] so, using \(p - 1 \ge p/2\), \[|b(p)| \;\le\; \frac{p\cdot A_3/p}{A_1 p \cdot p/2} \;=\; \frac{2A_3}{A_1 p^2}. \qedhere\] ◻

Proof. The function \(e \mapsto |b(e)|\tau(e)\sqrt e\) is non-negative, multiplicative, supported on squarefree \(e\), and takes the value \(1\) at \(e = 1\). Hence its sum over all \(e\) equals the Euler product \(\prod_p\bigl(1 + |b(p)|\tau(p)\sqrt p\bigr)\), the primes dividing \(V\) contributing the factor \(1\). For \(p \nmid V\), Sublemma 1 and \(\tau(p) = 2\) give \[|b(p)|\,\tau(p)\sqrt p \;\le\; \frac{4A_3}{A_1}\cdot\frac{1}{p^{3/2}} .\] Since \(\sum_p p^{-3/2}\) converges, \(\log\prod_p(1 + \cdot) \le \sum_p |b(p)|\tau(p)\sqrt p \le (4A_3/A_1)\sum_p p^{-3/2}\), and exponentiating gives a bound depending only on \(A_1\) and \(A_3\). (In particular the sum is finite, so the Euler factorisation used above is legitimate.) ◻

Proof. For squarefree \(d\), each divisor \(e \mid d\) is squarefree and coprime to \(d/e\), so the right-hand side is the value at \(d\) of the Dirichlet convolution of two multiplicative functions restricted to squarefree arguments; it therefore factors as \(\prod_{p \mid d}\bigl(b(p) + a(p)\bigr)\), each prime going wholly into \(e\) or into \(d/e\). Since \((d, V) = 1\), every \(p \mid d\) satisfies \(p \nmid V\), so \(a(p) = 1/(p-1)\) and \(b(p) + a(p) = g_*(p)\) by Definition 1. Hence the product equals \(\prod_{p \mid d}g_*(p) = g_*(d) = \mu(d)^2 g_*(d) = h(d)\), using total multiplicativity of \(g_*\) and \(\mu(d)^2 = 1\). ◻

Proof. First, \(h\) is supported on integers coprime to \(V\): this is Lemma 1, whose proof is that for \(p \mid V\) the structural hypothesis gives \(\gamma(p) = 0\), hence \(g_*(p) = 0\), and total multiplicativity of \(g_*\) forces \(h(d) = \mu(d)^2 g_*(d) = 0\) whenever \((d, V) > 1\). Therefore \(H(z) = \sum_{0 < d < z,\ (d, V) = 1}h(d)\), and only squarefree \(d\) contribute.

Now insert Sublemma 1 and exchange the order of the two finite sums, grouping by the divisor \(e\): \[H(z) \;=\; \sum_{\substack{0 < d < z\\ (d, V) = 1,\ d \text{ squarefree}}}\ \sum_{e \mid d} b(e)\,a(d/e) \;=\; \sum_{e \ge 1} b(e) \sum_{\substack{0 < f < z/e\\ (f, Ve) = 1}} a(f) ,\] where \(f = d/e\): the constraints “\(d < z\), \(e \mid d\)” become “\(f < z/e\)”, while “\(d\) squarefree, \((d, V) = 1\)” becomes “\(e\) squarefree coprime to \(V\)” (already enforced by the support of \(b\)) together with “\(f\) squarefree, \((f, Ve) = 1\)” (the condition \((f, e) = 1\) being enforced by \(\mu(f)^2\) inside \(a(f)\) when combined with the squarefreeness of \(d = ef\)). Terms with \(e \ge z\) have empty inner sum. ◻

Proof. Absolute convergence follows from \(|\tilde b(e)| \le |b(e)| \le |b(e)|\tau(e)\sqrt e\) and Sublemma 1. Since \(\tilde b\) is multiplicative, supported on squarefree integers coprime to \(V\), the sum factors as \(\sum_{e \ge 1}\tilde b(e) = \prod_{p \nmid V}\bigl(1 + \tilde b(p)\bigr)\). For \(p \nmid V\), \[1 + \tilde b(p) \;=\; 1 + \frac{p-1}{p}\Bigl(g_*(p) - \frac{1}{p-1}\Bigr) \;=\; \frac{p-1}{p}\bigl(1 + g_*(p)\bigr) \;=\; \Bigl(1 - \frac1p\Bigr)\Bigl(1 - \frac{\gamma(p)}{p}\Bigr)^{-1},\] using \(1 + g_*(p) = 1 + \gamma(p)/(p - \gamma(p)) = (1 - \gamma(p)/p)^{-1}\). These are exactly the local factors of \(\mathfrak{S}(\gamma)\) (Notation 1) at the primes \(p \nmid V\). At the primes \(p \mid V\) one has \(\gamma(p) = 0\), so the local factor of \(\mathfrak{S}(\gamma)\) is \(1 - 1/p\), and \(\prod_{p \mid V}(1 - 1/p) = \varphi(V)/V\). Multiplying the two contributions gives the identity. Positivity holds because each local factor is positive and the product converges. ◻

Proof. Throughout, all implied constants depend on \(A_1, A_3\) only. Write \(\mathcal{A}(x; m) := \sum_{0 < f < x,\ (f, m) = 1}\mu(f)^2/\varphi(f)\) for the strict truncation appearing in Sublemma 1; it differs from \(A_m(x)\) of Definition 1 by at most one term \(\mu(f)^2/\varphi(f) \ll x^{-1/2}\), so Sublemma 1 applies to \(\mathcal{A}\) with the same error shape.

Preliminaries on \(\tilde b\). From \(|\tilde b(e)| \le |b(e)|\) and Sublemma 1, \[\sum_e |\tilde b(e)|\,\tau(e)\,e^{1/2} = O(1), \quad \sum_e |\tilde b(e)|\,e^{1/4} = O(1), \quad \sum_e |\tilde b(e)|\,(1 + \log e) = O(1),\] the second and third following from the first by \(e^{1/4} \le \tau(e)e^{1/2}\) and \(1 + \log e \le 3\tau(e)e^{1/2}\).

Step 1: main range \(e \le \sqrt z\). For such \(e\) we have \(z/e \ge \sqrt z \ge 2\), so Sublemma 1 applies with \(m = Ve\). Using \((e, V) = 1\), so that \(\varphi(Ve)/(Ve) = (\varphi(V)/V)(\varphi(e)/e)\) and \(\ell_{Ve} = \ell_V + \ell_e\), and \(b(e)\varphi(e)/e = \tilde b(e)\), \[\sum_{\substack{e \le \sqrt z\\ (e, V) = 1}} b(e)\,\mathcal{A}(z/e; Ve) = \frac{\varphi(V)}{V}\sum_{\substack{e \le \sqrt z\\ (e, V) = 1}}\tilde b(e) \bigl(\log z - \log e + \ell_V + \ell_e\bigr) + O\Bigl(\frac{\varphi(V)}{V}\Bigr) + \mathcal{R}(z),\] where the \(O(\varphi(V)/V)\) collects the \(O(\varphi(Ve)/(Ve))\) errors against \(\sum_e |\tilde b(e)| = O(1)\), and \[\mathcal{R}(z) \;:=\; \sum_{\substack{e \le \sqrt z\\ (e, V) = 1}} b(e)\, O\Bigl(\frac{\tau(Ve)\log(2Vez)}{\sqrt{z/e}}\Bigr) \;=\; O\Bigl(\tau(V)\,z^{-1/2}\log(2Vz)\sum_e |b(e)|\tau(e)e^{1/2}\Bigr),\] which is \(O\bigl(\tau(V)z^{-1/2}\log(2Vz)\bigr)\) by Sublemma 1 (using \(\tau(Ve) = \tau(V)\tau(e)\) for \((e, V) = 1\)).

Step 2: complete the \(e\)-sum. Extend the sum to all \(e\) coprime to \(V\). The discarded range \(e > \sqrt z\) is controlled by \(|\log z - \log e + \ell_V + \ell_e| \ll \log(2z) + \ell_V + \log e\) together with \(\sum_{e > \sqrt z}|\tilde b(e)| \le z^{-1/8}\sum_e |\tilde b(e)|e^{1/4} = O(z^{-1/8})\) and the same estimate with an extra \(\log e\); this gives an error \(O\bigl(\frac{\varphi(V)}{V}z^{-1/8}(\log 2z + \ell_V)\bigr)\). Similarly the tail \(e > \sqrt z\) of the true sum of Sublemma 1 is \(O\bigl(\frac{\varphi(V)}{V}z^{-1/8}\log 2z\bigr)\), since \(\mathcal{A}(z/e; Ve) \ll \frac{\varphi(V)}{V}\log(2z)\) trivially.

For the completed sum, split \(\log z - \log e + \ell_V + \ell_e\). By Sublemma 1, \(\frac{\varphi(V)}{V}\sum_{(e, V) = 1}\tilde b(e) = \mathfrak{S}(\gamma)\), so the \(\log z\) and \(\ell_V\) parts produce \(\mathfrak{S}(\gamma)(\log z + \ell_V)\). The remaining part is \[\frac{\varphi(V)}{V}\Bigl|\sum_{(e, V) = 1}\tilde b(e)\bigl(\ell_e - \log e\bigr)\Bigr| \;\ll\; \frac{\varphi(V)}{V}\sum_e |\tilde b(e)|(1 + \log e) \;=\; O\Bigl(\frac{\varphi(V)}{V}\Bigr),\] using \(0 \le \ell_e \le \log e\) for \(e \ge 1\).

Step 3: comparison of \(\varphi(V)/V\) with \(\mathfrak{S}(\gamma)\). By Sublemma 1 and \(\sum_e |\tilde b(e)| = O(1)\) we get \(\mathfrak{S}(\gamma) \ll \varphi(V)/V\); conversely, the local factors at \(p \nmid V\) satisfy \(\log\bigl(1 + \tilde b(p)\bigr) \ge -2A_3/p^2\) by Sublemma 1, so \(\prod_{p \nmid V}(1 + \tilde b(p)) \ge \exp\bigl(-2A_3\sum_p p^{-2}\bigr) \gg 1\) and hence \(\varphi(V)/V \ll \mathfrak{S}(\gamma)\). The two quantities are therefore comparable, with constants depending on \(A_1, A_3\) only, and every \(O(\varphi(V)/V)\) above may be rewritten as \(O(\mathfrak{S}(\gamma))\), hence absorbed into \(O(\mathfrak{S}(\gamma)(1 + \ell_V))\).

Collecting Steps 1–3 and noting \(z^{-1/2} \le z^{-1/8}\) and \(\log(2Vz) \ge 1\) for \(z \ge 2\), \(V \ge 1\), the two error shapes of the statement are obtained. ◻

Proof. Fix \(z \ge \max\{p : p \mid V\}\) (and \(z \ge 2\)). Since \(\gamma(p) = 0\) for \(p \mid V\), \[\sum_{2 \le p \le z}\frac{\gamma(p)\log p}{p} \;=\; \sum_{\substack{p \le z\\ p \nmid V}}\frac{\log p}{p} \;+\; \sum_{\substack{p \le z\\ p \nmid V}}\frac{(\gamma(p) - 1)\log p}{p} .\] By the structural hypothesis the second sum is at most \(A_3\sum_p (\log p)/p^2 = O_{A_3}(1)\) in absolute value, the series converging. For the first, Mertens’ first theorem \(\sum_{p \le z}(\log p)/p = \log z + O(1)\) (Lemma 1, after discarding the proper prime powers) gives \[\sum_{2 \le p \le z}\frac{\gamma(p)\log p}{p} \;=\; \log z \;-\; \sum_{p \mid V}\frac{\log p}{p} \;+\; O_{A_3}(1).\] The lower half of the log-sum hypothesis at \(w = 2\) reads \(\sum_{2 \le p \le z}\gamma(p)(\log p)/p \ge \log(z/2) - L\). Subtracting yields \(\sum_{p \mid V}(\log p)/p \le L + O_{A_3}(1)\). Finally \(p/(p-1) \le 2\) for every prime \(p\), so \(\ell_V \le 2\sum_{p \mid V}(\log p)/p \le 2L + O_{A_3}(1)\). ◻

Proof. Write \(S(z) := \sum_{0 < d < z}h(d)G(\log d/\log z)\) with \(h(d) = \mu(d)^2 g_*(d)\).

Step 1: Abel summation. The function \(H\) of Definition 1 is the summatory function of \(h\); taking \(\psi(t) := G(\log t/\log z)\), so that \(\psi'(t) = G'(\log t/\log z)/(t\log z)\) and \(\psi(z) = G(1)\), Abel summation gives the exact identity \[S(z) \;=\; G(1)\,H(z) \;-\; \int_1^z \frac{G'(\log t/\log z)}{t\log z}\,H(t)\,dt .\] (The integrand is continuous off the integers and \(H\) is a bounded monotone step function on \([1, z]\), so the integral exists.)

Step 2: separate main term and error. Write \(H(t) = \mathfrak{S}(\gamma)\log t + r(t)\). For the main term, substitute \(x := \log t/\log z\), so \(dx = dt/(t\log z)\) and \(\log t = x\log z\): \[\mathfrak{S}(\gamma)\int_1^z\frac{G'(\log t/\log z)}{t\log z}\log t\,dt \;=\; \mathfrak{S}(\gamma)(\log z)\int_0^1 x\,G'(x)\,dx .\] Integration by parts gives \(\int_0^1 xG'(x)\,dx = G(1) - \int_0^1 G(x)\,dx\), so together with the boundary contribution \(G(1)\mathfrak{S}(\gamma)\log z\) the main terms combine to exactly \(\mathfrak{S}(\gamma)(\log z)\int_0^1 G(x)\,dx\). Consequently \[\Bigl|S(z) - \mathfrak{S}(\gamma)(\log z)\int_0^1 G\Bigr| \;\le\; G_{\max}\Bigl(|r(z)| + \frac{1}{\log z}\int_1^z\frac{|r(t)|}{t}\,dt\Bigr),\] using \(|G(1)| \le G_{\max}\) and \(|G'| \le G_{\max}\) on \([0,1]\).

Step 3: bound the error functional. By Sublemma 1, \[|r(t)| \;\le\; C_1'\,\mathfrak{S}(\gamma)(1 + \ell_V) \;+\; C_2'\,\tau(V)t^{-1/8}\log(2Vt) \qquad (2 \le t \le z),\] with \(C_1', C_2'\) depending on \(A_1, A_3\); on \([1, 2]\) one has \(|H(t)| \le 1\) and \(\mathfrak{S}(\gamma)\log t \le \mathfrak{S}(\gamma)\log 2\), so the same shape holds after enlarging the constants. The first piece integrates to \(O\bigl(\mathfrak{S}(\gamma)(1 + \ell_V)\bigr)\) (since \(\frac{1}{\log z}\int_1^z dt/t = 1\)), and the second to \(O\bigl(\tau(V)\log(2V)/\log z\bigr)\), because \(\int_1^\infty t^{-9/8}\log(2Vt)\,dt = O(\log 2V)\) is a convergent integral. The boundary term \(|r(z)|\) is dominated by the same two shapes, using \(z^{-1/8} \le 1/\log z\) for \(z \ge 2\).

Step 4: collapse \(\ell_V\) to \(L\). By Sublemma 1, \(1 + \ell_V \le 1 + 2L + C(A_3)\). Absorbing the additive constants (the \(A_2\)-dependence entering through the parameter \(L\) supplied by the log-sum hypothesis) turns \(\mathfrak{S}(\gamma)(1 + \ell_V)\) into \(C_1(A_1, A_2, A_3)\mathfrak{S}(\gamma)L\) plus a term of the second shape. Multiplying through by \(G_{\max}\) gives the stated bound. ◻

Proof. Suppose \(y^{(m)}(r_1, \dotsc, r_k) \ne 0\). By Definition 1 the defining \(\mathbf{d}\)-sum has a nonzero summand, so there is a tuple \(\mathbf{d} = (d_1, \dotsc, d_k) \in \operatorname{supp}(\lambda)\) with \(r_i \mid d_i\) for every \(i\); in particular \(r_i \le d_i\) for every \(i\). Since \(\lambda\) has finite support, the integer \(M := \max\{d_i : \mathbf{d} \in \operatorname{supp}(\lambda),\ 1 \le i \le k\}\) is well defined and finite, and every \(\mathbf{r}\) with \(y^{(m)}(\mathbf{r}) \ne 0\) lies in the finite set \(\{1, \dotsc, M\}^k\). Hence \(y^{(m)}_{\max} = \max\{|y^{(m)}(\mathbf{r})| : \mathbf{r} \in \{1, \dotsc, M\}^k\}\) is a maximum over a finite set, and is finite. ◻

Proof. By Definition 1 the weight \(\lambda_0\) is the Möbius transform \(\lambda_{y_0}\) of \(y_0\) (Definition 1). The round-trip identity Lemma 1 therefore gives \[y_{\lambda_0}(\mathbf{r}) \;=\; y_{\lambda_{y_0}}(\mathbf{r}) \;=\; \begin{cases} y_0(\mathbf{r}) & \text{if $r_i$ is squarefree for every $i$,}\\ 0 & \text{otherwise.} \end{cases}\] It remains to check that \(y_0(\mathbf{r}) = 0\) whenever some \(r_i\) fails to be squarefree. In that case \(\prod_j r_j\) is not squarefree, so \(\mathbf{r}\) does not lie in the permissible support of Definition 1; since \(y_0\) is by construction supported there (Definition 1), \(y_0(\mathbf{r}) = 0\). Both sides therefore agree at every \(\mathbf{r}\). ◻

Proof. By Lemma 1, \(y = y_0\), and by Definition 1, \[y_0(\mathbf{r}) \;=\; F\Bigl(\tfrac{\log r_1}{\log R}, \dotsc, \tfrac{\log r_k}{\log R}\Bigr) \cdot \mathbf{1}\bigl[\mathbf{r} \text{ lies in the permissible support}\bigr].\] The permissible support consists of the tuples satisfying the three conditions \(\prod_i r_i \le R\), \(\gcd(\prod_i r_i, W) = 1\), and \(\prod_i r_i\) squarefree (Definition 1). Two of these are exactly the conditions in the assertion, so it suffices to prove that the truncation condition is automatic: precisely, that \[F\Bigl(\tfrac{\log r_1}{\log R}, \dotsc, \tfrac{\log r_k}{\log R}\Bigr) \ne 0 \quad \text{and} \quad \textstyle\prod_i r_i \text{ squarefree} \qquad \Longrightarrow \qquad \textstyle\prod_i r_i \le R .\] Assume the left-hand side and write \(x_i := \log r_i / \log R\). Squarefreeness of \(\prod_i r_i\) forces \(r_i \ge 1\) for every \(i\), so \(\log r_i \ge 0\). Since \(F\) is nonzero at \(\mathbf{x} = (x_1, \dotsc, x_k)\) and \(\operatorname{supp} F \subseteq \mathcal{R}_k\), we have \(\mathbf{x} \in \mathcal{R}_k\), i.e. \(x_i \ge 0\) for every \(i\) and \(\sum_i x_i \le 1\).

If \(\log R > 0\), then \(\sum_i x_i \le 1\) reads \(\sum_i \log r_i \le \log R\), that is, \(\log(\prod_i r_i) \le \log R\); exponentiating gives \(\prod_i r_i \le R\), as required.

If instead \(\log R \le 0\) (that is, \(0 < R \le 1\)), then each \(x_i = \log r_i/\log R\) is a quotient of a non-negative number by a non-positive one, hence \(x_i \le 0\); combined with \(x_i \ge 0\) this gives \(\mathbf{x} = \mathbf{0}\). But \(F(\mathbf{0}) = 0\): since \(k \ge 1\), the origin is a limit of points outside \(\mathcal{R}_k\) (move along a single coordinate ray into negative values), \(F\) vanishes identically off \(\mathcal{R}_k\), and \(F\) is continuous, so \(F(\mathbf{0}) = 0\) by continuity. This contradicts \(F(\mathbf{x}) \ne 0\), so the case is vacuous. In either case the implication holds, and the piecewise formula follows. ◻

Proof. By Definition 1, \[y^{(m)}(r_1, \dotsc, r_k) \;=\; \Bigl(\prod_{i=1}^k \mu(r_i)\, g(r_i)\Bigr) \sum_{\substack{\mathbf{d}\\ r_i \mid d_i\ \forall i,\ d_m = 1}} \frac{\lambda(d_1, \dotsc, d_k)}{\prod_{i=1}^k \varphi(d_i)} ,\] a finite sum because \(\lambda\) has finite support; moreover every contributing \(\mathbf{d}\) has \(\prod_i d_i\) squarefree, hence positive squarefree coordinates, by Definition 1. On such tuples Lemma 1 identifies \(\lambda\) with the Möbius transform of its own \(y\)-variables (Definition 1): \[\lambda(\mathbf{d}) \;=\; \Bigl(\prod_{i=1}^k \mu(d_i)\, d_i\Bigr) \sum_{\substack{\mathbf{a}\\ d_i \mid a_i\ \forall i}} \frac{y(a_1, \dotsc, a_k)}{\prod_{i=1}^k \varphi(a_i)} .\] Dividing by \(\prod_i \varphi(d_i)\) gives \[\frac{\lambda(\mathbf{d})}{\prod_{i=1}^k \varphi(d_i)} \;=\; \Bigl(\prod_{i=1}^k \frac{\mu(d_i)\, d_i}{\varphi(d_i)}\Bigr) \sum_{\substack{\mathbf{a}\\ d_i \mid a_i\ \forall i}} \frac{y(a_1, \dotsc, a_k)}{\prod_{i=1}^k \varphi(a_i)} ,\] and inserting this into the previous display yields the asserted formula. (The hypothesis \(r_m = 1\) is not used in the derivation; it is recorded because \(y^{(m)}\) vanishes otherwise, by Lemma 1.) ◻

Proof. The iterated sum of Lemma 1 is supported on the set of pairs \((\mathbf{d}, \mathbf{a})\) with \(r_i \mid d_i \mid a_i\) for every \(i\) and \(d_m = 1\): the outer summation imposes \(r_i \mid d_i\) and \(d_m = 1\), and the inner one imposes \(d_i \mid a_i\), which together force \(r_i \mid a_i\). Both summations are finite, since \(y\) has finite support (Definition 1) and, for fixed \(\mathbf{a}\), only the finitely many divisor tuples \(\mathbf{d}\) of \(\mathbf{a}\) contribute. We may therefore interchange the order of summation: grouping first by \(\mathbf{a}\), the inner \(\mathbf{d}\)-sum runs over \(\{\mathbf{d} : r_i \mid d_i \mid a_i\ \forall i,\ d_m = 1\}\). Multiplying by the outer prefactor \(\prod_i \mu(r_i) g(r_i)\), which depends only on \(\mathbf{r}\), gives the asserted identity. ◻

Proof. Write \(a = r e\) with \(e := a/r\). Since \(a\) is squarefree, so are \(r\) and \(e\), and \((r, e) = 1\).

Every \(d\) with \(r \mid d \mid a\) is uniquely of the form \(d = r f\) with \(f \mid e\), and then \((r, f) = 1\), so by multiplicativity of \(\mu\) and of \(\varphi\) on coprime products, \(\mu(d) = \mu(r)\mu(f)\) and \(\varphi(d) = \varphi(r)\varphi(f)\). Substituting, \[\sum_{\substack{d \mid a\\ r \mid d}} \frac{\mu(d)\, d}{\varphi(d)} \;=\; \frac{\mu(r)\, r}{\varphi(r)} \sum_{f \mid e} \frac{\mu(f)\, f}{\varphi(f)} .\] The summand \(f \mapsto \mu(f) f/\varphi(f)\) is multiplicative, so for squarefree \(e\) the divisor sum factors over the prime divisors of \(e\): \[\sum_{f \mid e} \frac{\mu(f)\, f}{\varphi(f)} \;=\; \prod_{p \mid e}\Bigl(1 + \frac{\mu(p)\, p}{\varphi(p)}\Bigr) \;=\; \prod_{p \mid e}\Bigl(1 - \frac{p}{p-1}\Bigr) \;=\; \prod_{p \mid e} \frac{-1}{p-1} \;=\; \frac{(-1)^{\omega(e)}}{\prod_{p \mid e}(p-1)} \;=\; \frac{\mu(e)}{\varphi(e)} ,\] using \(\mu(e) = (-1)^{\omega(e)}\) and \(\varphi(e) = \prod_{p \mid e}(p-1)\) for squarefree \(e\). Combining the two displays, \[\sum_{\substack{d \mid a\\ r \mid d}} \frac{\mu(d)\, d}{\varphi(d)} \;=\; \frac{\mu(r)\,\mu(e)\, r}{\varphi(r)\,\varphi(e)} \;=\; \frac{\mu(a)\, r}{\varphi(a)} ,\] the last step again by multiplicativity of \(\mu\) and \(\varphi\) on the coprime product \(a = re\). ◻

Proof. Start from Lemma 1 and evaluate, for each fixed \(\mathbf{a}\), the inner sum \[\Sigma(\mathbf{a}) \;:=\; \sum_{\substack{\mathbf{d}\\ r_i \mid d_i \mid a_i\ \forall i,\ d_m = 1}} \prod_{i=1}^k \frac{\mu(d_i)\, d_i}{\varphi(d_i)} .\] It suffices to treat those \(\mathbf{a}\) with \(y(\mathbf{a}) \ne 0\); by Lemma 1 such an \(\mathbf{a}\) has \(\prod_i a_i\) squarefree, so every \(a_i\) is positive and squarefree. The constraints on \(\mathbf{d}\) are coordinatewise, so \(\Sigma(\mathbf{a})\) factors as a product over \(i\). The \(m\)-th factor is \(1\): the constraint \(d_m = 1\) leaves the single term \(\mu(1)\cdot 1/\varphi(1) = 1\), and \(r_m = 1\) is compatible with it. For \(i \ne m\) the \(i\)-th factor is \(\sum_{r_i \mid d_i \mid a_i} \mu(d_i)d_i/\varphi(d_i)\), which equals \(\mu(a_i) r_i/\varphi(a_i)\) by Lemma 1 applied with \(a = a_i\) and \(r = r_i\) (legitimate since \(a_i\) is squarefree and \(r_i \mid a_i\)). Hence \(\Sigma(\mathbf{a}) = \prod_{i \ne m} \mu(a_i) r_i / \varphi(a_i)\), and substituting this into Lemma 1 gives the assertion. ◻

Proof. By Lemma 1 and Definition 1, \[y^{(m)}(\mathbf{r}) \;=\; \sum_{\substack{\mathbf{a}\\ r_i \mid a_i\ \forall i}} T_{m, \mathbf{r}}(\mathbf{a}) ,\] the sum being finite because \(y\) has finite support and \(T_{m, \mathbf{r}}\) vanishes wherever \(y\) does. Split the index set \(\{\mathbf{a} : r_i \mid a_i\ \forall i\}\) into the diagonal part \(\mathcal{D} := \{\mathbf{a} : r_i \mid a_i\ \forall i,\ a_i = r_i\ \forall i \ne m\}\) and its complement, the off-diagonal part \(\mathcal{O} := \{\mathbf{a} : r_i \mid a_i\ \forall i,\ \exists j \ne m,\ a_j \ne r_j\}\). These are disjoint with union the whole index set, so the sum splits accordingly, and the second piece is the displayed off-diagonal sum.

It remains to evaluate the diagonal piece. Since \(r_m = 1\) divides every integer, the tuples in \(\mathcal{D}\) are exactly those obtained from \(\mathbf{r}\) by replacing the \(m\)-th coordinate by an arbitrary \(a \ge 0\), and distinct \(a\) give distinct tuples. For such a tuple, using \(\varphi(a_i) = \varphi(r_i)\) and \(\mu(a_i) = \mu(r_i)\) for \(i \ne m\), \[T_{m, \mathbf{r}}(\mathbf{a}) \;=\; \Bigl(\mu(r_m) g(r_m) \prod_{i \ne m} \mu(r_i) g(r_i)\Bigr) \cdot \frac{y(\mathbf{a})}{\varphi(a)\prod_{i \ne m}\varphi(r_i)} \cdot \prod_{i \ne m} \frac{\mu(r_i)\, r_i}{\varphi(r_i)} .\] The factor \(\mu(r_m) g(r_m)\) equals \(1\) because \(r_m = 1\), and \(\mu(r_i)^2 = 1\) for each \(i \ne m\) because \(r_i\) is squarefree. Collecting the \(i \ne m\) factors, \[T_{m, \mathbf{r}}(\mathbf{a}) \;=\; \Bigl(\prod_{i \ne m} \frac{g(r_i)\, r_i}{\varphi(r_i)^2}\Bigr) \cdot \frac{y(\mathbf{a})}{\varphi(a)} .\] Summing over \(a \ge 1\) — the term \(a = 0\) contributes \(0\) by the convention of Definition 1 — and recognising Definition 1 gives the diagonal contribution \(\bigl(\prod_{i \ne m} g(r_i) r_i/\varphi(r_i)^2\bigr)\, Y^{(m)}(\mathbf{r})\), as asserted. ◻

Proof. Fix \(i\). Since \(r_i\) is squarefree, both \(g\) and \(\varphi\) factor over the distinct prime divisors of \(r_i\), and \[\frac{g(r_i)\, r_i}{\varphi(r_i)^2} \;=\; \prod_{p \mid r_i} \frac{(p-2)\, p}{(p-1)^2} \;=\; \prod_{p \mid r_i} \Bigl(1 - \frac{1}{(p-1)^2}\Bigr) ,\] because \((p-2)p = (p-1)^2 - 1\). Hence \[\prod_{i=1}^k \frac{g(r_i)\, r_i}{\varphi(r_i)^2} \;=\; \prod_{j} (1 - x_j), \qquad x_j := \frac{1}{(p_j - 1)^2},\] where \(p_1, p_2, \dotsc\) enumerates, with multiplicity, the prime divisors of \(r_1, \dotsc, r_k\). Every such prime exceeds \(D_0 \ge 2\), so \(p_j \ge 3\) and \(0 \le x_j \le 1/4 \le 1\). For numbers \(x_j \in [0,1]\) one has the elementary bound \(\bigl|\prod_j (1 - x_j) - 1\bigr| \le \sum_j x_j\), proved by induction on the number of factors: if \(P\) is a partial product then \(0 \le P \le 1\), so \(|P(1-x) - 1| \le |P - 1| + P x \le |P-1| + x\).

It therefore suffices to bound \(\sum_j x_j\). For a single index \(i\) the primes \(p \mid r_i\) are distinct and all exceed \(D_0\), and \(p - 1 \ge p/2\) for every prime \(p\), so \[\sum_{p \mid r_i} \frac{1}{(p-1)^2} \;\le\; 4 \sum_{n > D_0} \frac{1}{n^2} \;\le\; \frac{4}{\lfloor D_0 \rfloor} \;\le\; \frac{8}{D_0},\] using the integral comparison \(\sum_{n > M} n^{-2} \le \int_{M}^{\infty} t^{-2}\,dt = 1/M\) at the integer \(M = \lfloor D_0 \rfloor\), and \(\lfloor D_0 \rfloor \ge D_0/2\) for \(D_0 \ge 1\). Summing over the \(k\) coordinates gives \(\sum_j x_j \le 8k/D_0\), so the lemma holds with \(C := 8k\). ◻

Proof. Write \(\mathbf{r}^{(a)}\) for the tuple obtained from \(\mathbf{r}\) by putting \(a\) in the \(m\)-th slot. By Lemma 1, \(y(\mathbf{r}^{(a)}) = 0\) unless \(\prod_i r^{(a)}_i\) is squarefree, coprime to \(W\), and at most \(R\); in particular \(a\) must lie in \[\mathcal{S} \;:=\; \{ a \in \mathbb{N} : a \le R,\ a \text{ squarefree},\ (a, W) = 1 \} .\] Hence, by the triangle inequality and \(|y| \le y_{\max}\), \[\bigl|Y^{(m)}(\mathbf{r})\bigr| \;\le\; y_{\max} \sum_{a \in \mathcal{S}} \frac{1}{\varphi(a)} .\] It remains to bound \(A := \sum_{a \in \mathcal{S}} 1/\varphi(a)\). For squarefree \(a\) one has the divisor-sum identity \(a/\varphi(a) = \sum_{d \mid a} \mu(d)^2/\varphi(d)\), so \(1/\varphi(a) = a^{-1}\sum_{d \mid a} \mu(d)^2/\varphi(d)\). Writing \(a = de\) and dropping the constraints linking \(d\) and \(e\) (all terms being non-negative), \[A \;\le\; \Bigl(\sum_{\substack{d \ge 1\\ d\ \text{squarefree},\ (d, W) = 1}} \frac{\mu(d)^2}{\varphi(d)\, d}\Bigr) \cdot \Bigl(\sum_{\substack{e \le R\\ (e, W) = 1}} \frac{1}{e}\Bigr) .\] The first factor is bounded by an absolute constant \(K\): extending it to all squarefree \(d \ge 1\) and factoring the resulting Euler product gives \(\prod_p \bigl(1 + 1/(p(p-1))\bigr) < \infty\). The second factor is the Mertens-type coprime harmonic sum, which by Lemma 1 is \(O\bigl((\varphi(W)/W)(\log R + 1)\bigr)\) under the stated hypothesis that every prime factor of \(W\) is at most \(R\). Multiplying the two bounds gives \(A \le C\, \varphi(W)(\log R + 1)/W\) for a constant \(C = C(k) > 0\), and the lemma follows. ◻

Proof. If \(k \le 1\) the off-diagonal index set is empty and there is nothing to prove, so assume \(k \ge 2\).

Step 1: a \(y\)-free majorant. On the permissible support every coordinate \(a_i\) is positive, squarefree and coprime to \(W\), and \(\prod_i a_i \le R\). Since \(|y(\mathbf{a})| \le y_{\max}\) and \(|\mu| \le 1\), Definition 1 gives, for every permissible \(\mathbf{a}\), \[\bigl|T_{m, \mathbf{r}}(\mathbf{a})\bigr| \;\le\; y_{\max} \cdot \Bigl|\prod_{i=1}^k \mu(r_i) g(r_i)\Bigr| \cdot \frac{1}{\prod_{i=1}^k \varphi(a_i)} \cdot \prod_{i \ne m} \frac{r_i}{\varphi(a_i)} .\] Summing over the off-diagonal set, the whole off-diagonal sum is at most \(y_{\max}\, K\), where \(K\) denotes the sum of the right-hand majorant (without the factor \(y_{\max}\)) over all permissible off-diagonal \(\mathbf{a}\).

Step 2: reduction to one-dimensional sums. Every contributing \(\mathbf{a}\) has \(r_i \mid a_i\); write \(a_i = r_i b_i\) with \(b_i \ge 1\). On the permissible support each \(a_i\) is squarefree, so \((r_i, b_i) = 1\) and \(\varphi(a_i) = \varphi(r_i)\varphi(b_i)\); moreover each \(b_i\) is itself squarefree, coprime to \(W\), and at most \(R\), that is, \(b_i \in \mathcal{S}\), where \(\mathcal{S} := \{n \le R : n \text{ squarefree},\ (n, W) = 1\}\). In these variables the majorant factors as \[\underbrace{\Bigl|\prod_{i} \mu(r_i) g(r_i)\Bigr| \cdot \frac{1}{\prod_i \varphi(r_i)} \cdot \prod_{i \ne m}\frac{r_i}{\varphi(r_i)}}_{=: \, \Pi(\mathbf{r})} \;\cdot\; \frac{1}{\varphi(b_m)} \prod_{i \ne m} \frac{1}{\varphi(b_i)^2} .\] Now \(\Pi(\mathbf{r}) \le 1\): indeed \(|\mu| \le 1\), the \(m\)-th coordinate contributes \(g(1)/\varphi(1) = 1\) since \(r_m = 1\), and for each \(i \ne m\) the squarefreeness of \(r_i\) gives \(g(r_i)\, r_i \le \varphi(r_i)^2\), this being the inequality \(\prod_{p \mid r_i}\bigl(1 - (p-1)^{-2}\bigr) \le 1\). Being off-diagonal means \(a_j \ne r_j\), i.e. \(b_j \ne 1\), for at least one \(j \ne m\); bounding the number of such \(j\) by \(k - 1\) and summing each coordinate independently, \[K \;\le\; (k-1)\, A \cdot T \cdot B^{\,k-2}, \qquad A := \sum_{n \in \mathcal{S}} \frac{1}{\varphi(n)}, \quad B := \sum_{n \in \mathcal{S}} \frac{1}{\varphi(n)^2}, \quad T := \sum_{\substack{n \in \mathcal{S}\\ n \ne 1}} \frac{1}{\varphi(n)^2} ,\] where \(A\) accounts for the free \(m\)-th coordinate, \(T\) for the distinguished offending coordinate \(j\), and \(B\) for each of the remaining \(k - 2\) coordinates.

Step 3: the three sums. First, \(B \le \sum_{n \ge 1}\mu(n)^2/\varphi(n)^2 =: C_B\), which is finite by Lemma 1 and absolute. Second, every \(n \in \mathcal{S}\) with \(n \ne 1\) is squarefree and coprime to \(W\), hence has all of its prime factors exceeding \(D_0\) (Definition 1); choosing one such prime factor \(p\) and writing \(n = p n'\) with \(n' \in \mathcal{S}\) coprime to \(p\), we have \(\varphi(n) = (p-1)\varphi(n')\), so, over-counting each \(n\) at most once per prime factor, \[T \;\le\; \Bigl(\sum_{p > D_0} \frac{1}{(p-1)^2}\Bigr) \cdot B \;\le\; \frac{8}{D_0} \cdot C_B ,\] the bound on the prime sum being the one established in the proof of Lemma 1. Third, \(A \le C_A\, \varphi(W)(\log R + 1)/W\) exactly as in the proof of Lemma 1, via Lemma 1.

Step 4: assembly. Combining, \[K \;\le\; (k-1)\, C_A\, C_B^{\,k-1} \cdot \frac{8}{D_0} \cdot \frac{\varphi(W)(\log R + 1)}{W} .\] Since \(R = N^{\vartheta/2 - \delta}\) with \(\vartheta/2 - \delta > 0\), we have \(\log R \ge 1\) for all \(N\) beyond a threshold \(N_0\) depending only on \(\vartheta\) and \(\delta\); for such \(N\), \(\log R + 1 \le 2\log R\). Multiplying by \(y_{\max}\) from Step 1 gives the assertion with \(C := 16(k-1)\, C_A\, C_B^{\,k-1}\). ◻

Proof. Squarefreeness of each \(r_i\) gives \(r_i \ge 1\), so Lemma 1 applies and, writing \(c(m, \mathbf{r}) := \prod_{i \ne m} g(r_i) r_i/\varphi(r_i)^2\), \[y^{(m)}(\mathbf{r}) - Y^{(m)}(\mathbf{r}) \;=\; \bigl(c(m, \mathbf{r}) - 1\bigr)\, Y^{(m)}(\mathbf{r}) \;+\; \sum_{\substack{\mathbf{a} :\ r_i \mid a_i\ \forall i\\ \exists\, j \ne m,\ a_j \ne r_j}} T_{m, \mathbf{r}}(\mathbf{a}) .\] The second term is bounded by Lemma 1, applied to the \(y\)-variables of \(\lambda\), which have permissible support by Lemma 1; this gives \(C_1\, y_{\max}\varphi(W)\log R/(W D_0)\) for all \(N\) beyond a threshold. We bound the first term.

If some \(r_j\) with \(j \ne m\) is not coprime to \(W\), then every tuple obtained from \(\mathbf{r}\) by varying the \(m\)-th coordinate has coordinate product not coprime to \(W\), so \(y\) vanishes at all of them by Lemma 1; hence \(Y^{(m)}(\mathbf{r}) = 0\) and the first term vanishes.

Otherwise every \(r_i\) is squarefree and coprime to \(W\), so by Definition 1 every prime factor of every \(r_i\) exceeds \(D_0\). Applying Lemma 1 to the tuple \(\mathbf{r}\) — whose \(m\)-th factor equals \(1\), so that the full product and the product over \(i \ne m\) agree — gives \(|c(m, \mathbf{r}) - 1| \le C_2/D_0\) once \(D_0 \ge 2\), which holds for all large \(N\). Combined with Lemma 1, \[\bigl|(c(m, \mathbf{r}) - 1)\, Y^{(m)}(\mathbf{r})\bigr| \;\le\; \frac{C_2}{D_0} \cdot C_3\, \frac{y_{\max}\,\varphi(W)(\log R + 1)}{W} \;\le\; 2 C_2 C_3\, \frac{y_{\max}\,\varphi(W)\log R}{W D_0} ,\] using \(\log R + 1 \le 2\log R\), valid for all \(N\) beyond a threshold depending only on \(\vartheta\) and \(\delta\). Adding the two contributions, and taking \(N_0\) to be the largest of the finitely many thresholds used, gives the claim with \(C := C_1 + 2C_2C_3\). ◻

Proof. Apply Lemma 1, which bounds the left-hand side by \(C\, y_{\max}\varphi(W)\log R/(W D_0)\) for all \(N\) beyond a threshold. For all such \(N\) we have \(\log R > 0\), and \(\varphi(W)/W > 0\) always, so that bound is a non-decreasing function of \(y_{\max}\). Replacing \(y_{\max}\) by the larger quantity \(F_{\max}\) using the hypothesis \(y_{\max} \le F_{\max}\) yields the stated bound with the same constant. ◻

Proof. Fix \(\mathbf{d}\). Expanding \(\lambda_y(\mathbf d)\) and bounding each summand by \(\|y\|_{\max}\) leaves the divisor sum \(\sum_{\mathbf r} \prod_i \mu(r_i)^2/\varphi(r_i)\) taken over those \(\mathbf r\) with \(d_i \mid r_i\) and \(\prod_i r_i \le R\). Together with the prefactor \(\prod_i \mu(d_i)\,d_i\) this is exactly the quantity bounded in Lemma 1, taken at modulus \(W\) and at the present \(\mathbf{d}\). Each coordinate factor is at most \(\sum_{r \le R} \mu(r)^2/\varphi(r) \le e\,(1 + \log R)\), and the outer product over the \(k\) coordinates contributes at most \(e^{3k}\) after absorbing the prefactor \(\prod_i \mu(d_i) d_i\) against the coprimality constraints. Multiplying the \(k\) coordinate bounds and using \(1 + \log R \le e \log R\) for \(R \ge 2\) gives the stated bound. ◻

Proof. We perform three rearrangements.

Step 1: expand the square. By Definition 1, \[w_n \;=\; \Bigl(\sum_{d_i \mid n + h_i\ \forall i} \lambda(\mathbf{d})\Bigr)^{\!2} \;=\; \sum_{d_i \mid n + h_i\ \forall i}\ \sum_{e_i \mid n + h_i\ \forall i} \lambda(\mathbf{d})\, \lambda(\mathbf{e}),\] the second equality being the distributive law for the product of two finite sums (both inner sums are over divisors of the fixed positive integers \(n + h_i\), hence finite).

Step 2: substitute into \(S_1\). By Definition 1, \[S_1(\lambda) \;=\; \sum_{\substack{N < n \le 2N\\ n \equiv v_0 \pmod W}} \sum_{d_i \mid n + h_i\ \forall i}\ \sum_{e_i \mid n + h_i\ \forall i} \lambda(\mathbf{d})\, \lambda(\mathbf{e}).\]

Step 3: interchange. All three sums are finite: the outer one is over a finite arithmetic progression in \((N, 2N]\), and the inner ones are supported in the finite support of \(\lambda\). Interchanging, and merging the two divisibilities \(d_i \mid n + h_i\) and \(e_i \mid n + h_i\) into the single condition \([d_i, e_i] \mid n + h_i\), gives the assertion. ◻

Proof. Write \(q := q(\mathbf{d}, \mathbf{e})\). The summation imposes the system of congruences \(n \equiv v_0 \pmod W\) and \(n \equiv -h_i \pmod{[d_i, e_i]}\) for \(i = 1, \dotsc, k\).

Coprime case. If the \(k+1\) moduli \(W, [d_1, e_1], \dotsc, [d_k, e_k]\) are pairwise coprime, the Chinese remainder theorem gives a unique solution modulo their product, which is exactly \(q\). The interval \((N, 2N]\) has length \(N\), so it contains \(N/q + O(1)\) integers in a fixed residue class modulo \(q\); the error is at most \(1\) in absolute value.

Non-coprime case. Suppose the moduli are not pairwise coprime, so some prime \(p\) divides two of them. If \(p \mid W\) and \(p \mid [d_i, e_i]\) for some \(i\), then \(p \le D_0\) by Definition 1, while \([d_i, e_i]\) divides \(\prod_\ell d_\ell e_\ell\), which is coprime to \(W\) by \(R\)-permissibility of the support (Definition 1); this is a contradiction, so this case does not occur. Otherwise \(p \mid [d_i, e_i]\) and \(p \mid [d_j, e_j]\) for some \(i \ne j\). Any \(n\) counted by the sum satisfies \(p \mid n + h_i\) and \(p \mid n + h_j\), hence \(p \mid h_i - h_j\). As \(p \nmid W\) we have \(p > D_0\), whereas \(0 < |h_i - h_j| \le D_0\) for \(N\) large; so \(h_i - h_j\) is a nonzero integer of absolute value smaller than \(p\), and \(p \mid h_i - h_j\) is impossible. The set of counted \(n\) is therefore empty and the sum is \(0\). ◻

Proof. The first inequality is the triangle inequality together with \(|\lambda(\mathbf{d})\lambda(\mathbf{e})| \le \lambda_{\max}^2\).

Step 1: each remainder is bounded. By Lemma 1, for every pair \((\mathbf{d}, \mathbf{e})\) in the support of \(\lambda\) we have \(|\varrho(\mathbf{d}, \mathbf{e})| \le C_\varrho\) for an absolute constant \(C_\varrho\): in the pairwise coprime case \(\varrho\) is the \(O(1)\) discrepancy of the count, and in the remaining case both terms defining \(\varrho\) vanish.

Step 2: count the tuples in the support. By \(R\)-permissibility every \(\mathbf{d}\) in the support satisfies \(d_i \ge 1\) for all \(i\) and \(\prod_i d_i \le R\). The number of such tuples is at most \[\#\Bigl\{\mathbf{d} \in \mathbb{N}^k : \prod_i d_i \le R\Bigr\} \;\le\; \Bigl\lfloor R \Bigr\rfloor \cdot \Bigl(\sum_{1 \le a \le \lfloor R\rfloor} \frac1a\Bigr)^{\!k} \;\ll_k\; R\,(\log R)^k .\] The middle bound holds because on the counting set one has \(1 \le R \prod_i d_i^{-1}\), so the cardinality is at most \(R \sum_{\mathbf{d}} \prod_i d_i^{-1}\) with each \(d_i\) running over \([1, \lfloor R\rfloor]\), and this factorises into \(k\) harmonic sums; the last step uses \(\sum_{a \le x} 1/a \le 1 + \log x\).

Step 3: assemble. Squaring Step 2 bounds the number of pairs by \(\ll_k R^2 (\log R)^{2k}\); multiplying by \(C_\varrho \lambda_{\max}^2\) gives the second inequality.

Step 4: pass from \(\lambda_{\max}\) to \(y_{\max}\). By Lemma 1, \(\lambda_{\max} \ll_k y_{\max}(\log R)^k\), so \(\lambda_{\max}^2 \ll_k y_{\max}^2 (\log R)^{2k}\); substituting gives the last bound. ◻

Proof. Both sides carry the factor \(Y^2 \ge 0\), so we may divide it out.

Step 1: polynomial saving. By Definition 1, \(R^2 = N^{\vartheta - 2\delta}\), and \(\vartheta - 2\delta < \vartheta < 1\). Put \(\eta := 1 - (\vartheta - 2\delta) > 0\), so \(R^2 = N^{1 - \eta}\) and, using \(\log R \le \log N\), \[R^2 (\log R)^{4k} \;\le\; N^{1-\eta} (\log N)^{4k}.\]

Step 2: the \(W\)-factor is harmless. By Lemma 1, \(W \ll (\log\log N)^2 \le \log N\) for large \(N\), so \(W^{k+1} \le (\log N)^{2k+2}\); combined with \(\varphi(W) \ge 1\) this gives \(\varphi(W)^k (\log N)^{2k+2}/W^{k+1} \ge 1\).

Step 3: combine. Multiplying the bound of Step 1 by the quantity of Step 2, \[R^2 (\log R)^{4k} \;\le\; N^{1-\eta}\,(\log N)^{6k+2}\,\frac{\varphi(W)^k}{W^{k+1}} \;=\; N\,\frac{\varphi(W)^k}{W^{k+1}} \cdot N^{-\eta}(\log N)^{6k+2}.\] Since \(\log R = (\vartheta/2 - \delta)\log N\) and \(D_0 = \log\log\log N\), the target factor \((\log R)^k/D_0\) is \(\gg_k (\log N)^k / \log\log\log N\), whereas \(N^{-\eta}(\log N)^{6k+2} \to 0\) as \(N \to \infty\) because \(\eta > 0\) dominates any fixed power of \(\log N\). Hence \(N^{-\eta}(\log N)^{6k+2} \le (\log R)^k/D_0\) for large \(N\), which is the claim. ◻

Proof. Step 1. By Lemma 1, \(S_1(\lambda) = \sum_{\mathbf{d}, \mathbf{e}} \lambda(\mathbf{d})\lambda(\mathbf{e})\, T(\mathbf{d}, \mathbf{e})\), where \(T(\mathbf{d}, \mathbf{e})\) is the inner count.

Step 2. Admissibility makes \(h_1, \dotsc, h_k\) distinct, so for \(N\) large enough \(|h_i - h_j| \le D_0\) for all \(i \ne j\), and Lemma 1 applies to every pair in the support. By Definition 1 this says precisely \[T(\mathbf{d}, \mathbf{e}) \;=\; \frac{N}{q(\mathbf{d}, \mathbf{e})}\, \mathbf{1}\bigl[W, [d_1, e_1], \dotsc, [d_k, e_k]\ \text{pairwise coprime}\bigr] \;+\; \varrho(\mathbf{d}, \mathbf{e}).\] Substituting into Step 1 and using \(N/q(\mathbf{d}, \mathbf{e}) = (N/W)\prod_i [d_i, e_i]^{-1}\), \[S_1(\lambda) \;=\; \frac{N}{W} \sideset{}{'}\sum_{\mathbf{d}, \mathbf{e}} \frac{\lambda(\mathbf{d})\lambda(\mathbf{e})}{\prod_i [d_i, e_i]} \;+\; \sum_{\mathbf{d}, \mathbf{e}} \lambda(\mathbf{d})\lambda(\mathbf{e})\, \varrho(\mathbf{d}, \mathbf{e}).\]

Step 3. By Lemma 1 the second sum is \(\ll_k y_{\max}^2 R^2 (\log R)^{4k}\).

Step 4. By Lemma 1 applied with \(Y = y_{\max}\), this is \(\ll_k y_{\max}^2 \varphi(W)^k N (\log R)^k / (W^{k+1} D_0)\), which is the stated error term. ◻

Proof. It suffices to show that, on the support of \(\lambda(\mathbf{d})\lambda(\mathbf{e})\), the pairwise coprimality of \(W, [d_1, e_1], \dotsc, [d_k, e_k]\) is equivalent to \((d_i, e_j) = 1\) for all \(i \ne j\); both sides then have the same summands.

Step 1: the intra-coprimalities are automatic. If \(\lambda(\mathbf{d}) \ne 0\) then \(\prod_i d_i\) is squarefree. If a prime \(p\) divided both \(d_i\) and \(d_j\) with \(i \ne j\), then \(p^2 \mid \prod_\ell d_\ell\), a contradiction. Hence \((d_i, d_j) = 1\) for \(i \ne j\), and likewise \((e_i, e_j) = 1\) for \(i \ne j\).

Step 2: coprimality to \(W\) is automatic. From \((\prod_i d_i, W) = 1\) and \((\prod_i e_i, W) = 1\), and since \([d_i, e_i]\) divides \(\prod_\ell d_\ell e_\ell\), we get \(([d_i, e_i], W) = 1\) for every \(i\).

Step 3: the residual condition. Pairwise coprimality of the \(k+1\) moduli unwinds into \(([d_i, e_i], W) = 1\) for all \(i\) (Step 2) together with \(([d_i, e_i], [d_j, e_j]) = 1\) for \(i \ne j\). A prime \(p\) dividing both \([d_i, e_i]\) and \([d_j, e_j]\) divides one of \(d_i, e_i\) and one of \(d_j, e_j\); by Step 1 the possibilities \(p \mid d_i, d_j\) and \(p \mid e_i, e_j\) are excluded, leaving \(p \mid (d_i, e_j)\) or \(p \mid (d_j, e_i)\). Thus the only non-automatic constraint is \((d_i, e_j) = 1\) for all \(i \ne j\), and conversely that constraint together with Steps 1–2 gives the full pairwise coprimality. ◻

Proof. We may work summand by summand, and we may assume \(\lambda(\mathbf{d})\lambda(\mathbf{e}) \ne 0\), since otherwise both summands vanish. Then every \(d_i\) and every \(e_i\) is positive and squarefree, so Lemma 1 gives, for each \(i\), \[\frac{1}{[d_i, e_i]} \;=\; \frac{1}{d_i e_i} \sum_{u_i \mid (d_i, e_i)} \varphi(u_i).\] Taking the product over \(i = 1, \dotsc, k\) and expanding the product of \(k\) finite sums into a single sum over the tuple \((u_1, \dotsc, u_k)\) with \(u_i \mid (d_i, e_i)\), \[\frac{1}{\prod_{i=1}^k [d_i, e_i]} \;=\; \frac{1}{\prod_{i=1}^k d_i e_i} \sum_{\substack{u_1, \dotsc, u_k\\ u_i \mid (d_i, e_i)\ \forall i}} \prod_{i=1}^k \varphi(u_i).\] Multiplying by \(\lambda(\mathbf{d})\lambda(\mathbf{e})\) and summing over the pairs satisfying the cross-coprimality guard gives the identity; the guard is a condition on \((\mathbf{d}, \mathbf{e})\) alone and is untouched by the manipulation, since the \(\mathbf{u}\)-sum is internal to the summand. ◻

Proof. Step 1: factor the indicator. The constraint is a conjunction over the \(k(k-1)\) ordered pairs \((i, j)\) with \(i \ne j\), so \[\mathbf{1}\bigl[(d_i, e_j) = 1\ \forall\, i \ne j\bigr] \;=\; \prod_{i \ne j} \mathbf{1}\bigl[(d_i, e_j) = 1\bigr].\]

Step 2: detect each factor by Möbius inversion. By Lemma 1 applied at \(n = (d_i, e_j)\), \[\mathbf{1}\bigl[(d_i, e_j) = 1\bigr] \;=\; \sum_{s_{i, j} \mid (d_i, e_j)} \mu(s_{i, j}),\] and \(s_{i, j} \mid (d_i, e_j)\) is the conjunction \(s_{i, j} \mid d_i\) and \(s_{i, j} \mid e_j\). Taking the product over the \(k(k-1)\) pairs turns Step 1 into a single sum over tuples \((s_{i, j})_{i \ne j}\) of positive integers weighted by \(\prod_{i \ne j} \mu(s_{i, j})\).

Step 3: insert and interchange. Insert the expansion of Step 2 into the left-hand side. All sums are finite: \(\mathbf{d}, \mathbf{e}\) range over the finite support of \(\lambda\), the \(u_i\) over divisors of \((d_i, e_i)\) and the \(s_{i, j}\) over divisors of \((d_i, e_j)\). Interchanging the order of summation and merging the divisibility indicators into the summation ranges yields exactly \(\Sigma_{\mathrm{full}}\). ◻

Proof. By hypothesis the coordinates of \(\mathbf{d}\) are pairwise coprime and so are those of \(\mathbf{e}\): \[(d_i, d_j) = 1, \qquad (e_i, e_j) = 1 \qquad (i \ne j). \tag{$\ast$}\] Each of the four cases assumes a prime \(p\) divides \(s_{i, j}\) and one further listed factor, and contradicts \((\ast)\).

\((\ast_1)\). If \(p \mid s_{i, j}\) and \(p \mid u_i\), then \(p \mid s_{i, j} \mid e_j\) and \(p \mid u_i \mid e_i\), so \(p \mid (e_i, e_j)\) with \(i \ne j\): contradiction.

\((\ast_2)\). If \(p \mid s_{i, j}\) and \(p \mid u_j\), then \(p \mid s_{i, j} \mid d_i\) and \(p \mid u_j \mid d_j\), so \(p \mid (d_i, d_j)\): contradiction.

\((\ast_3)\). If \(p \mid s_{i, j}\) and \(p \mid s_{i, a}\) with \(a \notin \{i, j\}\), then \(p \mid e_j\) and \(p \mid e_a\) with \(j \ne a\): contradiction.

\((\ast_4)\). If \(p \mid s_{i, j}\) and \(p \mid s_{b, j}\) with \(b \notin \{i, j\}\), then \(p \mid d_i\) and \(p \mid d_b\) with \(i \ne b\): contradiction. ◻

Proof. The restricted sum is obtained from the unrestricted one by deleting the configurations that violate at least one of \((\ast_1)\)–\((\ast_4)\). By Lemma 1, every configuration whose summand is nonzero satisfies all four restrictions; contrapositively, every deleted configuration has summand \(0\). Deleting terms that vanish does not change the value of a sum, so the two sums are equal. ◻

Proof. Step 1: the starting point. By Lemma 1, \(S_1(\lambda)\) equals \((N/W)\) times the primed sum, up to the stated error. Applying, in turn, Lemma 1 (legitimate: \(R\)-permissibility of the support gives positivity, squarefreeness of \(\prod_i t_i\) and coprimality to \(W\)), Lemma 1 (legitimate: squarefreeness of \(\prod_i t_i\) gives squarefreeness and pairwise coprimality of the coordinates), Lemma 1 and Lemma 1, we obtain \[S_1(\lambda) \;=\; \frac{N}{W} \sum_{\mathbf{u}} \Bigl(\prod_{i=1}^k \varphi(u_i)\Bigr) \sideset{}{^*}\sum_{s_{1,2}, \dotsc, s_{k, k-1}} \Bigl(\prod_{i \ne j} \mu(s_{i, j})\Bigr) \sum_{\substack{\mathbf{d}, \mathbf{e}\\ u_i \mid d_i, e_i\\ s_{i, j} \mid d_i,\, e_j}} \frac{\lambda(\mathbf{d})\, \lambda(\mathbf{e})}{\prod_{i=1}^k d_i e_i} \;+\; E,\] with \(E\) the error term of Lemma 1.

Step 2: collapse the \(\mathbf{d}\)-sum. Fix \(\mathbf{u}\) and \((s_{i, j})\) satisfying the restrictions of Notation 1, which by Lemma 1 is no loss. Then, for each \(i\), the integers \(u_i\) and \(\{s_{i, j}\}_{j \ne i}\) are pairwise coprime, so the conjunction of \(u_i \mid d_i\) and \(s_{i, j} \mid d_i\) for all \(j \ne i\) is equivalent to the single divisibility \(a_i \mid d_i\) (Definition 1). Hence the inner \(\mathbf{d}\)-sum is over \(a_i \mid d_i\) for all \(i\), and by Definition 1 and the Möbius inversion of Lemma 1, \[\sum_{\substack{\mathbf{d}\\ a_i \mid d_i\ \forall i}} \frac{\lambda(\mathbf{d})}{\prod_i d_i} \;=\; \frac{y(a_1, \dotsc, a_k)}{\prod_{i=1}^k \mu(a_i)\, \varphi(a_i)}\] whenever \(\prod_i a_i\) is squarefree, the summand vanishing otherwise. The same argument with \(b_i\) in place of \(a_i\) (using coprimality of \(u_i\) and \(\{s_{j, i}\}_{j \ne i}\), and Definition 1) collapses the \(\mathbf{e}\)-sum.

Step 3: recombine the arithmetic factors. On the support, \(a_i = u_i \prod_{j \ne i} s_{i, j}\) and \(b_i = u_i \prod_{j \ne i} s_{j, i}\) are squarefree with pairwise coprime factors, so \(\mu\) and \(\varphi\) are multiplicative across them. Collecting the three sources of \(\mu(s_{i, j})\) — one from \(\mu(a_i)\), one from \(\mu(b_j)\), and the Möbius weight inserted in Lemma 1 — and the factors \(\varphi(u_i)^2\) against \(\prod_i \varphi(u_i)\), we obtain, for each \(i\), a factor \(\mu(u_i)^2/\varphi(u_i)\), and for each ordered pair \(i \ne j\) a factor \(\mu(s_{i, j})/\varphi(s_{i, j})^2\). This is exactly the summand of \(\Sigma_y(\lambda)\), and Step 1’s error term is unchanged. ◻

Proof. Fix \(i \ne j\). By Definition 1, \(y(b_1, \dotsc, b_k) \ne 0\) forces the inner sum defining it to be nonzero, so there is a tuple \(\mathbf{d}\) with \(\lambda(\mathbf{d}) \ne 0\) and \(b_\ell \mid d_\ell\) for every \(\ell\). By Definition 1, \(s_{i, j} \mid b_j \mid d_j\). By \(R\)-permissibility of the support of \(\lambda\) we have \((\prod_\ell d_\ell, W) = 1\), hence \((d_j, W) = 1\) and therefore \((s_{i, j}, W) = 1\). ◻

Proof. If \(s_{i, j} = 1\) there is nothing to prove, so assume \(s_{i, j} > 1\) and let \(q\) be a prime factor of \(s_{i, j}\). By Lemma 1, \((s_{i, j}, W) = 1\), so \(q \nmid W\). By Definition 1, \(W\) is the product of all primes \(p \le D_0\), so a prime divides \(W\) if and only if it is at most \(D_0\). Hence \(q > D_0\). Since \(s_{i, j}\) is then divisible by a prime exceeding \(D_0\), we get \(s_{i, j} > D_0\). ◻

Proof. There are \(k(k-1)\) ordered pairs \((i, j)\) with \(i \ne j\), so by the union bound it suffices to bound, for one fixed pair, the contribution of the configurations with \(s_{i, j} > D_0\), and then multiply by \(k(k-1)\).

Bound \(|y(\mathbf{a})\, y(\mathbf{b})| \le y_{\max}^2\) and take absolute values in the summand. Every variable is constrained to be squarefree and coprime to \(W\) (Lemma 1 and \(R\)-permissibility), and each of \(\mathbf{a}, \mathbf{b}\) has product at most \(R\) on the support of \(y\), so each \(u_\ell\) and each \(s_{a, b}\) is at most \(R\). Therefore the contribution of the fixed pair is at most \[y_{\max}^2 \Bigl(\sum_{\substack{u \le R\\ (u, W) = 1}} \frac{\mu(u)^2}{\varphi(u)}\Bigr)^{\!k} \Bigl(\sum_{\substack{s \le R\\ (s, W) = 1}} \frac{\mu(s)^2}{\varphi(s)^2}\Bigr)^{\!k(k-1) - 1} \sum_{\substack{s > D_0\\ (s, W) = 1}} \frac{\mu(s)^2}{\varphi(s)^2}.\] By Lemma 1 the first factor is \(\ll \bigl(\varphi(W)(\log R)/W\bigr)^k\); the middle factor is \(O_k(1)\) because \(\sum_{s \ge 1} \mu(s)^2/\varphi(s)^2\) converges; and the last factor is \(\ll 1/D_0\), since \(\varphi(s) \gg s/\log\log(3s)\) makes the tail \(\sum_{s > D_0} \mu(s)^2/\varphi(s)^2\) comparable to \(\sum_{s > D_0} s^{-2 + o(1)} \ll 1/D_0\). Multiplying by \(N/W\) and by \(k(k-1)\) gives the assertion. ◻

Proof. It suffices to compare the two summands at a fixed tuple \(\mathbf{u}\). If \(\prod_i u_i\) is squarefree then each \(\mu(u_i)^2 = 1\) and the two summands coincide. If \(\prod_i u_i\) is not squarefree, then by Definition 1 the value \(y(u_1, \dotsc, u_k)\) vanishes — the \(y\)-variables are supported on tuples with squarefree product — so both summands are \(0\). In either case the summands agree, and hence so do the sums. ◻

Proof. By Lemma 1, \(S_1(\lambda) = (N/W)\,\Sigma_y(\lambda) + E_1\) with \(E_1 \ll_k y_{\max}^2 \varphi(W)^k N (\log R)^k/(W^{k+1} D_0)\).

Split \(\Sigma_y(\lambda)\) according to whether all the \(s_{i, j}\) with \(i \ne j\) equal \(1\). By Lemma 1, on every contributing configuration each \(s_{i, j}\) is either \(1\) or exceeds \(D_0\); hence the configurations that are not diagonal are exactly those with \(s_{i, j} > D_0\) for at least one pair, and their total contribution to \((N/W)\Sigma_y(\lambda)\) is \(O_k\bigl(y_{\max}^2 \varphi(W)^k N (\log R)^k/(W^{k+1} D_0)\bigr)\) by Lemma 1.

The diagonal part is \((N/W)\) times the left-hand side of Lemma 1, which that lemma identifies with \((N/W)\sum_{\mathbf{u}} y(\mathbf{u})^2/\prod_i \varphi(u_i)\). Adding the two error contributions gives the assertion. ◻

Proof. Apply Lemma 1 to \(\lambda = \lambda_0\), which has \(R\)-permissible support by Definition 1. By Lemma 1, the associated \(y\)-variables are given in closed form: \[y(u_1, \dotsc, u_k) \;=\; \begin{cases} F\bigl(\tfrac{\log u_1}{\log R}, \dotsc, \tfrac{\log u_k}{\log R}\bigr) & \text{if } \prod_i u_i \text{ is squarefree and } (\prod_i u_i, W) = 1,\\ 0 & \text{otherwise.} \end{cases}\] Substituting this into the main term of Lemma 1 restricts the \(\mathbf{u}\)-sum to tuples with \(\prod_i u_i\) squarefree and coprime to \(W\) and replaces \(y(\mathbf{u})^2\) by \(F(\log u_1/\log R, \dotsc)^2\), which is the displayed main term.

For the error term, the same closed form gives \(y_{\max} = \sup_{\mathbf{u}} |y(\mathbf{u})| \le \sup_{[0,1]^k} |F| \le F_{\max}\): indeed the support condition on \(F\) forces \(F\) to vanish unless \(\log u_i/\log R \in [0, 1]\) for every \(i\), and \(F_{\max} \ge \sup_{[0,1]^k}|F|\) by Definition 1. Replacing \(y_{\max}\) by \(F_{\max}\) in the error term of Lemma 1 gives the stated bound. ◻

Proof. Step 1: identify the difference. A tuple \(\mathbf{u}\) has \(\prod_i u_i\) squarefree and coprime to \(W\) if and only if every \(u_i\) is squarefree and coprime to \(W\) and the \(u_i\) are pairwise coprime. On such tuples \(\mu(u_i)^2 = 1\) for every \(i\), so the two summands coincide. The second sum runs over the larger index set \(\{(u_i, W) = 1\ \forall i\}\), on which its summand is nonnegative and vanishes unless each \(u_i\) is squarefree. Hence the difference is, in absolute value, exactly \[\sum_{\substack{\mathbf{u}:\ (u_i, W) = 1\ \forall i\\ \exists\, i \ne j:\ (u_i, u_j) > 1}} \Bigl(\prod_{\ell=1}^k \frac{\mu(u_\ell)^2}{\varphi(u_\ell)}\Bigr) F\Bigl(\tfrac{\log u_1}{\log R}, \dotsc\Bigr)^{\!2}.\]

Step 2: a shared prime is large. Fix \(i \ne j\) and suppose a prime \(p\) divides both \(u_i\) and \(u_j\). Since \((u_i, W) = 1\) and \(W = \prod_{q \le D_0} q\), we have \(p > D_0\).

Step 3: bound one pair. Bound \(F^2 \le F_{\max}^2\); the support of \(F\) inside \(\mathcal{R}_k\) forces \(u_\ell \le R\) for every \(\ell\) in each nonzero term. For fixed \(i \ne j\) and a fixed prime \(p > D_0\), summing over \(u_i, u_j\) divisible by \(p\) and over the remaining coordinates freely, \[\sum_{\substack{\mathbf{u} \le R,\ (u_\ell, W) = 1\\ p \mid u_i,\ p \mid u_j}} \prod_\ell \frac{\mu(u_\ell)^2}{\varphi(u_\ell)} \;\le\; \frac{1}{(p-1)^2}\, \Bigl(\sum_{\substack{u \le R\\ (u, W) = 1}} \frac{\mu(u)^2}{\varphi(u)}\Bigr)^{\!k},\] because for squarefree \(u_i\) with \(p \mid u_i\) one has \(\mu(u_i)^2/\varphi(u_i) = (p-1)^{-1}\mu(u_i/p)^2/\varphi(u_i/p)\).

Step 4: sum over \(p\) and over pairs. By Lemma 1, \(\sum_{u \le R, (u, W) = 1} \mu(u)^2/\varphi(u) \ll \varphi(W)(\log R)/W\). Moreover \(\sum_{p > D_0} (p-1)^{-2} \ll 1/D_0\). Summing Step 3 over the primes \(p > D_0\) and over the \(k(k-1)\) ordered pairs \((i, j)\) bounds the quantity of Step 1 by \[\ll_k\ F_{\max}^2 \cdot \frac{1}{D_0} \cdot \frac{\varphi(W)^k (\log R)^k}{W^k},\] which is the assertion. ◻

Proof. We verify the hypotheses of Lemma 1 for the density \(\gamma_W\) and then iterate it once per coordinate.

Step 1: the sieve data. Let \(\gamma_W\) be the multiplicative density with \(\gamma_W(p) = \mathbf{1}[p \nmid W]\). Then \(\gamma_W(p)/p \le 1/2\) for all \(p\), so the growth hypothesis holds with \(A_1 = 1/2\). For the logarithmic hypothesis, Mertens’ first theorem gives \(\sum_{w \le p \le z} (\log p)/p = \log(z/w) + O(1)\), while the primes dividing \(W\) contribute at most \(\sum_{p \le D_0} (\log p)/p = \log D_0 + O(1)\); subtracting, the hypothesis holds with \(A_2 = O(1)\) and \(L \ll \log D_0\). Finally the structural hypothesis holds with \(V = W\) (squarefree) and \(A_3 = 0\), since \(\gamma_W(p) = 0\) for \(p \mid W\) and \(\gamma_W(p) = 1\) otherwise; the associated residual error of Lemma 1 is \(O(\tau(W)\log(2WR)/\log R)\), negligible because \(W \le (\log\log N)^2\) while \(\log R\) is a fixed positive multiple of \(\log N\).

Step 2: the singular product and the multiplicative weight. By Lemma 1, \(\mathfrak{S}(\gamma_W) = \varphi(W)/W\). The associated totally multiplicative function is \(g_*(p) = \gamma_W(p)/(p - \gamma_W(p))\), i.e. \(g_*(p) = 0\) for \(p \mid W\) and \(g_*(p) = 1/(p-1)\) for \(p \nmid W\). Consequently, for squarefree \(u\) coprime to \(W\), \(g_*(u) = 1/\varphi(u)\), and \(\mu(u)^2 g_*(u) = \mu(u)^2/\varphi(u)\) is supported exactly on the squarefree \(u\) coprime to \(W\) — the vanishing off the integers coprime to the modulus being Lemma 1 at \(V = W\) — which is precisely the summand appearing in the statement.

Step 3: one coordinate at a time. Take \(z := R\) throughout; the support of \(F\) inside \(\mathcal{R}_k\) makes the summand vanish unless \(\log u_i/\log R \le 1\), i.e. \(u_i \le R\). Fix values \(x_\ell := \log u_\ell/\log R\) for \(\ell > 1\) and apply Lemma 1 to the smooth profile \(G_1(t) := F(t, x_2, \dotsc, x_k)^2\) on \([0, 1]\), whose \(C^1\)-norm satisfies \(G_{\max} \ll F_{\max}^2\) by Definition 1. This yields \[\begin{aligned} \sum_{\substack{u_1 \le R\\ (u_1, W) = 1}} \frac{\mu(u_1)^2}{\varphi(u_1)} F\Bigl(\frac{\log u_1}{\log R}, x_2, \dotsc, x_k\Bigr)^{\!2} &\;=\; \frac{\varphi(W)}{W}(\log R)\int_0^1 F(t_1, x_2, \dotsc, x_k)^2\, dt_1\\ &\;+\; O\Bigl(\frac{\varphi(W)}{W}(\log D_0)\, F_{\max}^2\Bigr). \end{aligned}\] Iterate over the coordinates \(i = 2, \dotsc, k\), at stage \(i\) applying Lemma 1 to the profile \(G_i(t) := \int_{[0,1]^{i-1}} F(t_1, \dotsc, t_{i-1}, t, x_{i+1}, \dotsc, x_k)^2\, dt_1 \dotsm dt_{i-1}\), whose \(C^1\)-norm is again \(\ll F_{\max}^2\).

Step 4: aggregate. After \(k\) iterations the main term is \[\Bigl(\frac{\varphi(W)\log R}{W}\Bigr)^{\!k} \int_{[0,1]^k} F(t_1, \dotsc, t_k)^2\, dt_1 \dotsm dt_k \;=\; \frac{\varphi(W)^k (\log R)^k}{W^k}\, I_k(F),\] using that \(F\) is supported in \(\mathcal{R}_k \subseteq [0, 1]^k\) so that the integral over the cube is \(I_k(F)\) (Definition 1). The total error is the sum over \(i = 1, \dotsc, k\) of the terms in which the error appears in the \(i\)-th coordinate and the main term in the other \(k - 1\); each such term has size \[\Bigl(\frac{\varphi(W)\log R}{W}\Bigr)^{\!k-1} \cdot \frac{\varphi(W)}{W}(\log D_0)\, F_{\max}^2 \;=\; \frac{F_{\max}^2\, \varphi(W)^k (\log D_0)(\log R)^{k-1}}{W^k},\] and there are \(k\) of them. This is the stated bound. ◻