A conductor-uniform quadratic prime-harmonic bound

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Statement and role

The following claim from Section 4 of the independent record is checked here. For every nonprincipal real Dirichlet character presented modulo $q$,

$$\sup_{z\ge2}\left|\sum_{p\le z}\frac{\chi(p)}p\right| \ll\log(2q), \tag{1}$$

with an absolute constant. Differences of endpoints give the same bound on any prime interval. The estimates after fixed-character nonvanishing at $1$ are elementary; an elementary proof of that nonvanishing is included below. No prime number theorem, Siegel bound, or zero-free hypothesis is needed.

The existing quartic Euler majorant needs only upper bounds for $L$-functions, so (1) is additional input, not a missing step of that proof. Its two-sided character is needed by the random omission construction.

A weak polynomial lower bound at one

First suppose $\chi$ is primitive. We use $L(1,\chi)>0$. This follows, for example, from the fixed-character case of Ghidelli, Lemma 5.4, whose passage has been checked: partial summation gives convergence of $\sum_p\chi(p)/p$, and the Euler logarithm then has a finite real limit as $\sigma\downarrow1$. Constants in that fixed-character assertion need not be uniform and are not used in the estimates below.

Alternatively, Appendix E of third independent record supplies a valid elementary nonvanishing argument. For real nonprincipal $\chi$, the Dirichlet coefficients

$$\zeta(s)L(s,\chi)=\sum_{n\ge1}a(n)n^{-s},\qquad a(n)=\sum_{d\mid n}\chi(d)$$

are nonnegative, and $a(k^2)\ge1$. Their convergence abscissa is therefore in $[1/2,1]$. Periodic cancellation continues $L(s,\chi)$ holomorphically to $\Re s>0$, while

$$\zeta(s)=\frac{s}{s-1} -s\int_1^\infty\{t\}\,t^{-s-1}\,dt$$

continues zeta there with only its pole at one. If $L(1,\chi)=0$, the product would be holomorphic throughout that half-plane. This contradicts the elementary Landau lemma: a Dirichlet series with nonnegative coefficients is singular at its finite convergence abscissa. Indeed, a Taylor expansion about a real point just to its right, in the leftward direction, has nonnegative coefficient sums. If it extended past the abscissa, Tonelli would sum that Taylor series as the original Dirichlet series at a smaller real argument, forcing convergence there. Finally the positive Euler product for $s>1$ gives $L(1,\chi)>0$. Thus this argument removes the fixed-character prime-distribution input without changing any of the uniform bounds below.

Put $\zeta=e^{2\pi i/q}$ and $\tau=\sum_{a=1}^q\chi(a)\zeta^a$, so $|\tau|=\sqrt q$ and $\sum_a\chi(a)\zeta^{an}=\chi(n)\tau$ for every integer $n$. If $\chi(-1)=-1$, multiply the sine Fourier series

$$\sum_{n\ge1}\frac{\sin(2\pi na/q)}n =\pi\left(\frac12-\frac aq\right)\qquad(0<a<q)$$

by $\chi(a)$ and sum. The Gauss identity gives

$$|L(1,\chi)|=\pi q^{-3/2}\left|\sum_{a=1}^q a\chi(a)\right| \ge\pi q^{-3/2}. \tag{2}$$

The integer sum is nonzero by nonvanishing. The finite Fourier manipulations can equivalently be made with Abel factors and then passed to the limit.

If $\chi(-1)=1$, set

$$U=\prod_{(a,q)=1}(1-\zeta^a)^{\chi(a)}.$$

This is an algebraic unit. Indeed, $(1-\zeta^a)/(1-\zeta)$ and its reciprocal are algebraic integers by geometric sums, using an inverse of $a\bmod q$ for the reciprocal. The common factors cancel because $\sum_a\chi(a)=0$. Pairing $a$ and $-a$ shows $U>0$. The automorphism $\zeta\mapsto\zeta^b$ sends $U$ to $U^{\chi(b)}$, so $U+U^{-1}$ is a rational algebraic integer.

The cosine series $\log|1-e^{2\pi it}|=-\sum_{n\ge1}\cos(2\pi nt)/n$ and the Gauss identity yield

$$|\log U|=\sqrt q\,L(1,\chi).$$

Thus $U\ne1$, and $U+U^{-1}$ is an integer greater than two. It is at least three, whence

$$L(1,\chi)\ge q^{-1/2}\log\frac{3+\sqrt5}{2}. \tag{3}$$

Both parities give the absolute effective bound $L(1,\chi)\gg q^{-3/2}$.

Bounds just to the right of one

Periodicity and zero mean give $|\sum_{n\le x}\chi(n)|\le q$. Splitting at $q$ and applying partial summation to the tails, including the differentiated series, gives uniformly for $1\le\sigma\le2$

$$|L(\sigma,\chi)|\ll\log(2q),\qquad |L'(\sigma,\chi)|\ll\log^2(2q). \tag{4}$$

For $0\le\sigma-1\le c q^{-3/2}\log^{-2}(2q)$, (2)–(4) preserve half the lower bound at one if $c>0$ is sufficiently small. Outside that interval, the Euler factors give

$$L(\sigma,\chi)\ge\prod_p(1+p^{-\sigma})^{-1} =\frac{\zeta(2\sigma)}{\zeta(\sigma)}\gg\sigma-1.$$

Consequently, throughout $1<\sigma\le2$,

$$L(\sigma,\chi)\gg q^{-3/2}\log^{-2}(2q). \tag{5}$$

From the Euler logarithm to sharp cutoffs

For $z\ge3$, put $\sigma=1+1/\log z$. Uniformly for every real Dirichlet character,

$$\sum_{p\le z}\frac{\chi(p)}p=\log L(\sigma,\chi)+O(1). \tag{6}$$

Prime powers of exponent at least two have bounded total. Chebyshev's prime bounds give

$$\sum_{p\le z}\frac{1-p^{-1/\log z}}p \le\frac1{\log z}\sum_{p\le z}\frac{\log p}p\ll1, \qquad \sum_{p>z}p^{-1-1/\log z}\ll1.$$

For the last estimate, partial summation reduces to a constant multiple of $\int_1^\infty e^{-u}\,du/u$. This proves (6). Equations (4)–(6) prove (1) for primitive characters. Passing to a character presented modulo $q$ removes only Euler factors at primes dividing $q$; their contribution to a sharp prime sum is at most $\sum_{p\mid q}1/p\ll\log(2q)$. Thus (1) holds as stated.

The identical smoothing argument for the principal character yields $\sum_{p\le z}1/p=\log\log z+O(1)$, since $\zeta(1+1/\log z)=\log z+O(1)$. Applying (1) to $\chi_{-4}$ gives, uniformly for $3\le Y<Z$,

$$\sum_{\substack{Y<p\le Z\\p\equiv3\ (4)}}\frac1p =\frac12\log\frac{\log Z}{\log Y}+O(1). \tag{7}$$

Why the small multiplier matters

Exponentiating (1) with a fixed large coefficient can cost a polynomial in the conductor that is not summable against canonical coefficient weights. In the omission proof it is multiplied only by $O(\delta)$, where $\delta>0$ can be chosen arbitrarily small. The resulting loss $R^{C\delta}$ is summable. The complete quartic baseline still uses only the positive $L$-powers of the established Euler majorant.