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A mean value theorem and logarithmic phases at large height

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The zero-free region of A zero-free strip at large height needs cancellation in \(\sum n^{iv}\) over progressions of length about \(\exp(L/(\log L)^2)\) at heights \(|v|\) up to \(x^3\). Then \(\log|v|/\log(\text{length})\) is of order \((\log L)^2\), and repeated differencing (Lesson 3) gives no useful saving. Vinogradov's method expands the phase into a polynomial of degree \(k\approx4\log|v|/\log(\text{length})\) and controls the resulting polynomial sums on average by a mean value theorem. What matters here is that every constant depends on \(k\) only through a factor \(\exp(Ck^{C})\). This lesson proves such a mean value theorem by Linnik's \(p\)-adic iteration and deduces the required cancellation.

Prerequisites. Primes, divisors, mean squares and rough numbers, Lemma 1.3; Logarithmic phases at polynomial scales, Lemma 1.1; Measure and Hilbert space tools for Haar integration, Theorem 3.1 (Hölder's inequality).

1. A mean value theorem with uniform constants

For integers \(k\ge2\), \(s\ge1\) and real \(M\ge1\), let \(J_{s,k}(M)\) be the number of pairs of \(s\)-tuples \(\mathbf u,\mathbf v\in\{1,\dots,\lfloor M\rfloor\}^s\) with

\[ \begin{gathered} \sum_{i=1}^su_i^j=\sum_{i=1}^sv_i^j\qquad(1\le j\le k), \end{gathered} \]

and put \(\kappa=k(k+1)/2\). For \(\boldsymbol\alpha\in\mathbb R^k\) let

\[ \begin{gathered} f(\boldsymbol\alpha)=\sum_{n\le M}e\Bigl(\sum_{j=1}^k\alpha_jn^j\Bigr). \end{gathered} \]

Lemma 1.1 (basic properties).

  1. \(J_{s,k}(M)=\int_{[0,1]^k}|f(\boldsymbol\alpha)|^{2s}\,d\boldsymbol\alpha\).
  2. The system \(\sum_iu_i^j=\sum_iv_i^j\) \((j\le k)\) is invariant under the translation of all variables by one integer, so the number of its solutions with all variables in any set of \(\lfloor M\rfloor\) consecutive integers is \(J_{s,k}(M)\).
  3. For every integer vector \(\mathbf c\) and every set \(I\) of \(\lfloor M\rfloor\) consecutive integers, the number of \(\mathbf u,\mathbf v\in I^s\) with \(\sum_i(u_i^j-v_i^j)=c_j\) for all \(j\le k\) is at most \(J_{s,k}(M)\).

Proof. 1. Expand \(|f|^{2s}=f^s\overline f^{\,s}\) and use \(\int_0^1e(m\alpha)\,d\alpha=\mathbf 1_{m=0}\) for integers \(m\). 2. By the binomial theorem, \(\sum_i(u_i+b)^j-\sum_i(v_i+b)^j\) is a combination of the differences \(\sum_iu_i^l-\sum_iv_i^l\), \(l\le j\), because both sides have \(s\) terms. 3. With \(g\) the exponential sum over \(I\), the count is \(\int|g|^{2s}e(-\boldsymbol\alpha\cdot\mathbf c)\,d\boldsymbol\alpha\le\int|g|^{2s}\), which is \(J_{s,k}(M)\) by parts 1 and 2. \(\square\)

Lemma 1.2 (power sums modulo prime powers). Let \(r>k\) be prime and \(h\ge1\). The \(k\)-tuples \(\mathbf z\in(\mathbb Z/r^h\mathbb Z)^k\) whose entries are pairwise distinct modulo \(r\) and whose power sums \(\sum_iz_i^j\) \((1\le j\le k)\) take prescribed values modulo \(r^h\) are at most \(k!\) in number; any two of them are permutations of each other.

Proof. Let \(e_j\) be the elementary symmetric functions of \(\mathbf z\) and \(p_j=\sum_iz_i^j\). Newton's identities \(je_j=\sum_{l=1}^j(-1)^{l-1}e_{j-l}p_l\), with \(e_0=1\), determine \(e_1,\dots,e_k\) modulo \(r^h\) from the \(p_j\), since \(j\le k<r\) is invertible. So all such tuples share the polynomial \(F(X)=\prod_i(X-z_i)\) modulo \(r^h\). Let \(\mathbf z,\mathbf z'\) be two of them. Modulo \(r\), \(F\) splits over the field \(\mathbb F_r\) with the roots \(z_1,\dots,z_k\), so each \(z_i'\) is congruent modulo \(r\) to exactly one \(z_{\sigma(i)}\), and \(\sigma\) is a permutation because the \(z_i'\) are distinct modulo \(r\). In \(0\equiv F(z_i')=\prod_l(z_i'-z_l)\pmod{r^h}\), every factor with \(l\ne\sigma(i)\) is a unit modulo \(r\), so \(z_i'\equiv z_{\sigma(i)}\pmod{r^h}\). \(\square\)

Lemma 1.3 (the congruence step). Let \(k\ge2\), \(s\ge1\), let \(r>k\) be a prime with \(r^k\ge M\) and \(r\le M\), and let \(N_r\) be the number of solutions counted by \(J_{s+k,k}(M)\) in which the first \(k\) coordinates of \(\mathbf u\) are pairwise distinct modulo \(r\), and likewise for \(\mathbf v\). Then

\[ \begin{gathered} N_r\le k!\,r^{2s+\kappa-k}M^k\,J_{s,k}(\lceil M/r\rceil).\quad(1.1) \end{gathered} \]

Proof. Let \(F_r(\boldsymbol\alpha)\) be the sum of \(e(\sum_j\alpha_j\sum_{i\le k}z_i^j)\) over \(\mathbf z\in\{1,\dots,\lfloor M\rfloor\}^k\) with entries distinct modulo \(r\), and \(f_a\) the part of \(f\) with \(n\equiv a\ (r)\). Then \(N_r=\int|F_r|^2|f|^{2s}\). Since \(f=\sum_{a\bmod r}f_a\), Hölder's inequality for the counting measure on the \(r\) residues (Theorem 3.1 of Measure and Hilbert space tools for Haar integration) gives \(|f|^{2s}\le r^{2s-1}\sum_a|f_a|^{2s}\), so \(N_r\le r^{2s}\max_a\int|F_r|^2|f_a|^{2s}\).

Fix \(a\). The integral counts solutions in which the last \(s\) coordinates on each side are \(\equiv a\ (r)\). Translate every variable by \(-a\) (Lemma 1.1(2)). The last coordinates become \(r\tilde u_i\), \(r\tilde v_i\), so the first parts \(\mathbf z=(u_i-a)_{i\le k}\) and \(\mathbf z'=(v_i-a)_{i\le k}\) satisfy

\[ \begin{gathered} \sum_{i\le k}z_i^j\equiv\sum_{i\le k}{z'_i}^{\,j}\pmod{r^j}\\ (1\le j\le k). \end{gathered} \]

There are at most \(M^k\) choices of \(\mathbf z\). Given \(\mathbf z\), the power sums of \(\mathbf z'\) are prescribed modulo \(r^j\); lifting them to residues modulo \(r^k\) can be done in \(\prod_{j\le k}r^{k-j}=r^{\kappa-k}\) ways, and for each lift Lemma 1.2 leaves at most \(k!\) residue vectors \(\mathbf z'\bmod r^k\). Since \(M\le r^k\), each residue class modulo \(r^k\) has at most one representative among the \(\lfloor M\rfloor\) admissible values. Given both first parts, the variables \(\tilde u_i,\tilde v_i\) run over a set of at most \(\lceil M/r\rceil\) consecutive integers and satisfy \(\sum_i(\tilde u_i^j-\tilde v_i^j)=c_j\) for integers \(c_j\) determined by \(\mathbf z,\mathbf z'\); Lemma 1.1(3) bounds their number by \(J_{s,k}(\lceil M/r\rceil)\). Multiplying the three counts proves (1.1). \(\square\)

Theorem 1.4 (mean value with uniform constants). There are absolute constants \(C_1,C_2\) such that for every \(k\ge2\) some integer \(s\) with \(k\le s\le C_1k^3\) satisfies

\[ \begin{gathered} J_{s,k}(M)\le\exp(C_1k^{C_2})\\ \times M^{2s-\kappa+1/100}\quad(M\ge1).\quad(1.2) \end{gathered} \]

Proof. We prove by induction along \(s=k,2k,3k,\dots\) bounds

\[ \begin{gathered} J_{s,k}(M)\le C_sM^{E_s}\quad(M\ge1),\\ E_s=2s-\kappa+\epsilon_s,\\ \epsilon_{s+k}=\Bigl(1-\frac1k\Bigr)\epsilon_s,\quad(1.3) \end{gathered} \]

starting from \(C_k=1\), \(\epsilon_k=\kappa\), which is the trivial bound \(J_{k,k}(M)\le M^{2k}\). Note that \(E_k=2k\) and \(E_{s+k}-E_s=2k-\epsilon_s/k\)\(\ge2k-(k+1)/2>0\), so \(E_s\ge2k\) throughout.

By Lemma 1.3 of Lesson 2 there is an absolute \(A_1\) such that for every \(k\ge2\) and every \(y\ge A_1k^6\) the interval \([y,2y]\) contains at least \(k^3\) primes, all larger than \(k\). Fix \(s\) and let \(M\ge1\).

Small \(M\). If \(M^{1/k}<A_1k^6\), the trivial bound \(J_{s+k,k}(M)\le M^{2(s+k)}=M^{E_{s+k}}M^{\kappa-\epsilon_{s+k}}\) gives (1.3) with a constant at most \(M^\kappa\le(A_1k^6)^{k\kappa}=\exp(O(k^3\log2k))\).

Large \(M\). Let \(M^{1/k}\ge A_1k^6\) and choose primes \(r_1,\dots,r_{k^3}\in[M^{1/k},2M^{1/k}]\), all larger than \(k\); then \(r^k\ge M\) and \(r\le2M^{1/k}\le M\). Call an \((s+k)\)-tuple degenerate if it has fewer than \(k\) distinct coordinates. Such a tuple is described by a list of \(k-1\) values and an assignment of each coordinate to one of them, so there are at most \(M^{k-1}k^{s+k}\) of them. The solutions with a degenerate \(\mathbf u\) number \(\int G\overline f^{\,s+k}\), where \(G\) is the exponential sum over degenerate tuples, and

\[ \begin{gathered} \Bigl|\int G\overline f^{\,s+k}\Bigr|\le M^{k-1}k^{s+k}\int|f|^{s+k}\\ \le M^{k-1}k^{s+k}J_{s+k,k}(M)^{1/2}. \end{gathered} \]

The same holds for degenerate \(\mathbf v\). In a solution with both sides nondegenerate, choose \(k\) positions with distinct entries on each side and permute them to the front; this shows that such solutions number at most \((s+k)^{2k}\) times those whose first \(k\) entries are distinct on both sides. The product of the two Vandermonde determinants of these entries is a nonzero integer of modulus less than \(M^{k(k-1)}\), so at most \(k^2(k-1)<k^3\) of the primes \(r_l\ge M^{1/k}\) divide it, and some \(r_l\) separates both sets of entries. By Lemma 1.3,

\[ \begin{gathered} J_{s+k,k}(M)\\ \le2M^{k-1}k^{s+k}J_{s+k,k}(M)^{1/2}\\ +k^3(s+k)^{2k}k!\max_lr_l^{2s+\kappa-k}\\ \times M^kJ_{s,k}(\lceil M/r_l\rceil). \end{gathered} \]

Insert (1.3) for \(s\), with \(\lceil M/r\rceil\le2M/r\). Since \(2s+\kappa-k-E_s=k^2-\epsilon_s\ge0\) and \(r\le2M^{1/k}\),

\[ \begin{aligned} &r^{2s+\kappa-k}M^k\Bigl(\frac{2M}r\Bigr)^{E_s}\\ &\quad\le2^{E_s+k^2}M^{k+E_s+k-\epsilon_s/k}\\ &\quad=2^{E_s+k^2}M^{E_{s+k}}. \end{aligned} \]

An inequality \(J\le a\sqrt J+b\) implies \(J\le a^2+2b\). Here \(a^2=4k^{2(s+k)}M^{2k-2}\le4k^{2(s+k)}M^{E_{s+k}}\), because \(E_{s+k}\ge2k\). Thus (1.3) holds for \(s+k\) with

\[ \begin{gathered} \log C_{s+k}\le\log(1+C_s)\\ +O\bigl((s+k^3)\log2k\bigr), \end{gathered} \]

which also covers the small-\(M\) constant; here \(E_s\le2s+\kappa\) was used for the factor \(2^{E_s}\).

After \(n\) steps, \(\epsilon=\kappa(1-1/k)^n\le\kappa e^{-n/k}\), which is at most \(1/100\) once \(n=\lceil k\log(100\kappa)\rceil\). Then \(s=k+nk=O(k^2\log2k)\le C_1k^3\), and the constants satisfy \(\log C_s\)\(=O(n(k^2\log2k+k^3)\log2k)\)\(=O(k^5)\). Since \(M\ge1\) and \(E_s\le2s-\kappa+1/100\), this proves (1.2). \(\square\)

The exponent \(2s-\kappa\) cannot be lowered: the vector of power sums takes at most \(\prod_j(sM^j+1)\) values, and Cauchy's inequality gives \(J_{s,k}(M)\ge\lfloor M\rfloor^{2s}/\prod_j(sM^j+1)\) (Exercise 4.2). Theorem 1.4 loses only \(1/100\) in the exponent, at a cost \(\exp(C_1k^{C_2})\) that is polynomial in \(k\) after taking logarithms. Much sharper results are known, but this weak form with explicit uniformity is all that is needed below.

2. Logarithmic phases on short progressions

Put \(T=\log L=\log\log x\).

Theorem 2.1 (logarithmic phases at large height). Fix \(C_0>0\). There is an absolute constant \(C_3\) such that, for \(x\) large in terms of \(C_0\), uniformly for

\[ \begin{gathered} \exp(L/T^2)\le N\le2x^5,\qquad q\le L^{C_0},\\ \exp\bigl(L/(2T^2)\bigr)\le|v|\le4x^3, \end{gathered} \]

every residue \(a\) modulo \(q\) and every interval \(I\subseteq[N,2N]\),

\[ \begin{gathered} \Bigl|\sum_{\substack{n\in I\\ n\equiv a\ (q)}}n^{iv}\Bigr|\le\frac Nq\exp\Bigl(-\frac L{T^{C_3}}\Bigr).\quad(2.1) \end{gathered} \]

One can take \(C_3=10\).

Proof. Let \(a_0\in[1,q]\) with \(a_0\equiv a\ (q)\), and write \(n=qb+a_0\), \(u=a_0/q\in(0,1]\) and \(H=N/q\). Then \(n^{iv}=q^{iv}(b+u)^{iv}\), and \(b\) runs over the integers of an interval \(\mathcal I\) on which \(b+u\in[H,2H]\). Let \(S=\sum_{b\in\mathcal I}e(F(b))\) with \(F(y)=(v/2\pi)\log(y+u)\), and

\[ \begin{gathered} y_*=\frac{\log|v|}{\log H}. \end{gathered} \]

For large \(x\), \(\log H\ge L/T^2-C_0T\ge L/(2T^2)\) and \(\log|v|\le3L+2\), so \(y_*\le7T^2\).

The case \(y_*<4/5\). Then \(|v|<H^{4/5}<\pi H\), and the proof of Lemma 1.1 of Logarithmic phases at polynomial scales, which uses only \(b+u\in[H,2H]\), gives \(|S|\ll H/|v|\le H\exp(-L/(2T^2))\).

The case \(y_*\ge4/5\). Put

\[ \begin{gathered} M=\lfloor H^{3/4}\rfloor,\qquad k=\lceil4y_*+8\rceil, \end{gathered} \]

so that \(12\le k\le28T^2+9\). Shifting the summation by \(h\in[1,M]\) changes \(S\) by at most \(2h\) terms, so

\[ \begin{gathered} \Bigl|S-\frac1M\sum_{h=1}^M\sum_{b\in\mathcal I}e\bigl(F(b+h)\bigr)\Bigr|\le2M. \end{gathered} \]

Taylor's formula with \(\alpha_j(b)=F^{(j)}(b)/j!\)\(=(-1)^{j-1}v/(2\pi j(b+u)^j)\) gives \(F(b+h)=F(b)+\sum_{j\le k}\alpha_j(b)h^j+R\), where

\[ \begin{gathered} |R|\le\frac{|v|}{2\pi}\Bigl(\frac MH\Bigr)^{k+1}\\ \le H^{y_*-(k+1)/4}\le H^{-9/4}, \end{gathered} \]

because \(M\le H^{3/4}\) and \((k+1)/4\ge y_*+9/4\). With \(S_k(\boldsymbol\alpha)=\sum_{h\le M}e(\sum_{j\le k}\alpha_jh^j)\) this yields

\[ \begin{gathered} |S|\le\frac1M\sum_{b\in\mathcal I}\bigl|S_k(\boldsymbol\alpha(b))\bigr|\\ +2M+1.\quad(2.2) \end{gathered} \]

Occupancy. Divide \([0,1)^k\) into the \(M^\kappa\) boxes that are products of intervals of length \(M^{-j}\) in the \(j\)th coordinate. We claim that each box contains \(\boldsymbol\alpha(b)\bmod1\) for at most \(e^{O(k)}H^{4/5}\) values of \(b\). Let \(j_0=\lceil y_*+3/20\rceil\), so \(1\le j_0\le y_*+23/20\le k\). The coordinate \(\alpha_{j_0}(b)\) has modulus at most \(H^{y_*-j_0}\le H^{-3/20}<\frac12\) and is monotone in \(b\), with

\[ \begin{gathered} \Bigl|\frac{d\alpha_{j_0}}{db}\Bigr|=\frac{|v|}{2\pi(b+u)^{j_0+1}}\ge\frac{H^{y_*-j_0-1}}{2\pi\,2^{j_0+1}} . \end{gathered} \]

An interval of length \(M^{-j_0}\le2^{j_0}H^{-3j_0/4}\) on the circle meets \((-\frac12,\frac12)\) in at most two intervals of total length \(M^{-j_0}\), whose preimage under \(\alpha_{j_0}\) has total length at most \(e^{O(k)}H^{1+j_0/4-y_*}\). Since \(y_*\ge4/5\),

\[ \begin{gathered} 1+\frac{j_0}4-y_*\le1-\frac{3y_*}4+\frac{23}{80}\\ \le1-\frac5{16}, \end{gathered} \]

and the preimage contains at most \(e^{O(k)}H^{11/16}+2\) integers, which proves the claim.

Suprema over boxes. For every \(s\ge1\),

\[ \begin{gathered} \sum_Q\sup_{\boldsymbol\alpha\in Q}|S_k(\boldsymbol\alpha)|^{2s}\\ \le(4\pi)^{2sk}M^\kappa J_{s,k}(M).\quad(2.3) \end{gathered} \]

To see this, rescale a box \(Q\) to the unit cube by \(\alpha_j=\beta_j+\theta_jM^{-j}\) and let \(\Phi(\boldsymbol\theta)\) be \(S_k\) in these coordinates. In one variable, \(\phi(\theta)=\phi(\theta')+\int_{\theta'}^\theta\phi'\) and averaging over \(\theta'\in[0,1]\) give \(|\phi(\theta)|\le\int_0^1|\phi|+\int_0^1|\phi'|\). Applying this in each coordinate, \(\sup|\Phi|\le\sum_{\boldsymbol\epsilon\in\{0,1\}^k}\int_{[0,1]^k}|\partial^{\boldsymbol\epsilon}\Phi|\), and by Hölder's inequality

\[ \begin{gathered} \sup|\Phi|^{2s}\le2^{k(2s-1)}\sum_{\boldsymbol\epsilon}\int_{[0,1]^k}|\partial^{\boldsymbol\epsilon}\Phi|^{2s}. \end{gathered} \]

Each \(\partial^{\boldsymbol\epsilon}\Phi\) is an exponential sum of the same shape as \(S_k\) whose coefficients \(\prod_{j:\epsilon_j=1}2\pi i(h/M)^j\) have modulus at most \((2\pi)^k\). Summing over the boxes undoes the rescaling at the cost of the factor \(M^\kappa\), and the full-torus moment of such a sum is at most \((2\pi)^{2sk}J_{s,k}(M)\) by Lemma 1.1(1). This proves (2.3).

Conclusion. Take \(s\) from Theorem 1.4. By Hölder's inequality, the occupancy bound, (2.3) and (1.2),

\[ \begin{aligned} &\frac1M\sum_{b\in\mathcal I}|S_k(\boldsymbol\alpha(b))|\\ &\quad\le\frac{(2H)^{1-1/(2s)}}M\\ &\qquad\times\Bigl(\sum_b|S_k(\boldsymbol\alpha(b))|^{2s}\Bigr)^{1/(2s)}\\ &\quad\le2H\bigl(e^{O(sk+k^{C_2})}H^{-1/5}M^{1/100}\bigr)^{1/(2s)} . \end{aligned} \]

Here \(H^{-1/5}M^{1/100}\le H^{-77/400}\), \(sk+k^{C_2}\ll T^{2\max(4,C_2)}\), \(2s\ll T^6\), and \(\log H\ge L/(2T^2)\). Since \(L/T^2\) exceeds every fixed power of \(T\) for large \(x\), the right side is at most \(H\exp(-L/T^9)\). The other terms in (2.2) are \(O(H^{3/4})\), which is smaller. Hence \(|S|\le H\exp(-L/T^{10})\) for large \(x\), and \(|\sum_{n\in I,\,n\equiv a}n^{iv}|=|S|\). \(\square\)

3. Exercises

4.1. Show that \(J_{k,k}(M)\le k!\lfloor M\rfloor^k\). (Two \(k\)-tuples of integers with the same first \(k\) power sums are permutations of each other.)

4.2. Show that \(J_{s,k}(M)\ge\lfloor M\rfloor^{2s}/\prod_{j\le k}(s\lfloor M\rfloor^j+1)\), and deduce that the exponent \(2s-\kappa\) in (1.2) cannot be lowered.

4.3. Show that Lemma 1.2 fails for \(r=k=2\) and \(h=2\): find two pairs with distinct residues modulo \(2\), equal power sums modulo \(4\), that are not permutations of each other modulo \(4\).

4.4. In the proof of Theorem 2.1, take \(y_*=1\). Compute \(k\) and \(j_0\), and check the occupancy exponent \(1+j_0/4-y_*\).

4.5. Deduce from Theorem 2.1 and partial summation that, under its hypotheses, \(\sum_{n\in I,\,n\equiv a}n^{-\sigma+iv}\)\(\ll N^{1-\sigma}q^{-1}\exp(-L/T^{C_3})\) uniformly for \(|\sigma-1|\le1/L\).

4. Solutions

4.1. If \(\mathbf u\) and \(\mathbf v\) have the same power sums \(p_1,\dots,p_k\), Newton's identities over \(\mathbb Q\) give the same elementary symmetric functions, hence the same polynomial \(\prod(X-u_i)=\prod(X-v_i)\), so \(\mathbf v\) is a permutation of \(\mathbf u\). For each of the \(\lfloor M\rfloor^k\) choices of \(\mathbf u\) there are at most \(k!\) choices of \(\mathbf v\).

4.2. Let \(R(\mathbf c)\) be the number of \(\mathbf u\in\{1,\dots,\lfloor M\rfloor\}^s\) with power-sum vector \(\mathbf c\). Then \(\sum R=\lfloor M\rfloor^s\) and \(J_{s,k}(M)=\sum R^2\). The vector \(\mathbf c\) takes at most \(\prod_j(s\lfloor M\rfloor^j+1)\) values, so Cauchy's inequality gives the bound. The denominator is \(\asymp_{s,k}M^\kappa\), so \(J_{s,k}(M)\gg_{s,k}M^{2s-\kappa}\), and a bound \(M^{2s-\kappa-\eta}\) with \(\eta>0\) is impossible for large \(M\).

4.3. Take \(\mathbf z=(0,1)\) and \(\mathbf z'=(2,3)\). Both have entries \(0,1\) modulo \(2\). Their sums are \(1\) and \(5\equiv1\) modulo \(4\), and their sums of squares are \(1\) and \(13\equiv1\) modulo \(4\). But \(\{2,3\}\ne\{0,1\}\) modulo \(4\). The proof breaks at Newton's identity \(2e_2=e_1p_1-p_2\), which cannot be divided by \(2\).

4.4. For \(y_*=1\), \(k=\lceil12\rceil=12\) and \(j_0=\lceil1.15\rceil=2\). Then \(1+j_0/4-y_*=1/2\le11/16\), so each box contains \(e^{O(k)}H^{1/2}\) points.

4.5. By Lemma 4.2 of Lesson 3 with \(g(n)=n^{-\sigma}\), whose supremum plus variation on \([N,2N]\) is at most \(2N^{-\sigma}\), the sum is at most \(2N^{-\sigma}\) times the maximum of (2.1) over initial segments of \(I\).

References