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Assume Riemann hypothesis. First fix $0<\delta <1/4$ and work on ${\sigma }_0=1/2+\delta$. We prove the subpower zeta bound to the right of the critical line, then move back to the line by the functional equation and Phragmén–Lindelöf principle.

Put $x=\log t$ for large positive $t$ and integrate the supplied smoothed logarithmic derivative identity horizontally from ${\sigma }_0$ to two. There are no zeros on this path under Riemann hypothesis, so the Euler-product logarithm at $2+it$ continues along it. Each prime-power term contributes at most $\Lambda (n)n^{-{\sigma }_0}/\log n\le n^{-{\sigma }_0}$, and the smoothing weights are at most one. Hence the integrated prime terms are bounded byAll zeros have $\rho =1/2+i\gamma$. The local zero-count estimate from Question 3, with its reflected version for negative ordinates, gives uniformly for ${\sigma }_0\le \sigma \le 2$To see the uniformity, sum the $O(\log (2+|\gamma |))$ zeros in successive unit ordinate intervals against $({\delta }^2+|t-\gamma {|}^2{)}^{-1}$; the distant dyadic tails are summable. The zero-term numerator has modulus at most $x^{-2\delta }+x^{-\delta }$. Its integrated contribution is therefore at most $C_{\delta }{x}^{-\delta }\log t/\log x=o(\log t)$. The integrated supplied remainder is $O(x^{-1-\delta }\log t/\log x)$, also $o(\log t)$. Since $\log \zeta (2+it)$ is bounded, we obtainThus for every fixed $\delta >0$ and $\eta >0$, $|\zeta (1/2+\delta +it)|{\ll }_{\delta ,\eta }{t}^{\eta }$. Negative $t$ follow by complex conjugation.

The zeta functional equation and the gamma ratio give $|\zeta (1/2-\delta +it)|{\ll }_{\delta ,\eta }{t}^{\delta +\eta }$. Zeta is holomorphic throughout this strip, since its pole at one is outside it, and Euler summation supplies polynomial vertical growth. The strip convexity conclusion of Phragmén–Lindelöf principle therefore gives at the midpointFor a prescribed $\epsilon >0$, choose $\delta$ and $\eta$ with $\delta /2+\eta <\epsilon$; the bounded $t$ range is harmless. This provesThe explicit-formula estimate is deliberately first made a fixed distance to the right of the critical line. No divergent zero bound at $\sigma =1/2$ is used.

A normalization is important here. I use the covolume-one theta function of a fractional ideal, for which the requested limit is $1$, and also give the formula for the unnormalized Gaussian theta sum.

Let $n=[K:\mathbb{Q}]$, let $d_v=[K_v:\mathbb{R}]\in \{1,2\}$, and choose one field embedding ${\sigma }_v$ for each Archimedean place. The Minkowski embedding of a number field identifieswith a real inner product space havingIts Euclidean Lebesgue measure is $d\mu ={\prod }_{v\text{ real}}d{x}_v{\prod }_{v\text{ complex}}2d\mathrm{Re}{x}_{v}d\mathrm{Im}{x}_v$. With this convention, the covolume of a fractional ideal lattice $j(\mathfrak{b})$ iswhere $D_K$ is the field discriminant and $N(\mathfrak{b})$ is the positive absolute norm of a fractional ideal. Thus the Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C_{\mathfrak{b}}^{-1/n}j(\mathfrak{b})$ has covolume $1$. DefineThis Gaussian theta sum converges absolutely whenever every $y_v>0$.

The trace dual of a fractional ideal isHere ${\mathfrak{D}}_{K/\mathbb{Q}}^{-1}$ is the inverse different. Since $N({\mathfrak{D}}_{K/\mathbb{Q}})=|D_K|$, we have $C_{{\mathfrak{b}}^{\vee }}=C_{\mathfrak{b}}^{-1}$. There is a subtle distinction between the trace pairing and the positive inner product: the dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,Consequently ${\Lambda }_{\mathfrak{b}}^{\ast }=\overline{{\Lambda }_{{\mathfrak{b}}^{\vee }}}$. Coordinatewise complex conjugation preserves the weighted squared lengths in the Gaussian theta sum.

Here are the precise analytic formulas used in the proof. For a Schwartz function on $K_{\infty }$, take the Fourier transform to beFor a full Euclidean lattice $\Lambda$ of covolume $C$, the Poisson summation formula for a Euclidean lattice isFor $f_y(x)=\exp (-\pi {\sum }_v{d}_v{y}_v|x_v{|}^2)$, the Gaussian Fourier transform, applied in orthonormal real coordinates, givesMore generally, $\widehat{\exp (-\pi x^{T}Ax)}(z)=(\det A{)}^{-1/2}\exp (-\pi z^T{A}^{-1}z)$ for a real positive-definite matrix $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.

Apply the Poisson summation formula for a Euclidean lattice to ${\Lambda }_{\mathfrak{b}}$, whose covolume is $1$, and use the preceding description of its dual lattice. The anisotropic theta functional equation isFor comparison, the unnormalized Gaussian theta sumhas the functional equationand its corresponding limit is $C_{\mathfrak{b}}^{-1}$ rather than $1$. Explicitly, our normalization is $\Theta (y,\mathfrak{b})=\theta (C_{\mathfrak{b}}^{-2/n}y,\mathfrak{b})$.

For the small-parameter asymptotic of a lattice theta sum, $y\to 0$ means that every coordinate tends to zero. In $\Theta (y^{-1},{\mathfrak{b}}^{\vee })$, the zero lattice vector contributes $1$. Every nonzero vector contributes a term tending to zero. Once every $y_v\le 1$, all these terms are bounded by the summable Gaussian theta sum ${\sum }_{\lambda \in {\Lambda }_{{\mathfrak{b}}^{\vee }}}{e}^{-\pi |\lambda {|}^2}$. The dominated convergence theorem therefore gives $\Theta (y^{-1},{\mathfrak{b}}^{\vee })\to 1$. Using the anisotropic theta functional equation,The condition that every coordinate tends to zero matters; $\Vert y\Vert \to 0$ alone would not justify this argument.

The subpower zeta bound to the right of the critical line gives an arbitrarily small exponent at $1/2+\delta$. The functional equation gives exponent $\delta +\epsilon$ at $1/2-\delta$. The Phragmén–Lindelöf principle gives exponent $\delta /2+\epsilon$ at the midpoint. Letting these fixed positive parameters be arbitrarily small proves the Lindelöf hypothesis.

For two level-one weight-$k$ cusp forms, $F(s)=\sum a_n{\overline{b}}_n{n}^{-s}$. Unfolding against the nonholomorphic lattice sum normalized by ${\pi }^{-w}\Gamma (w)$ gives the displayed completion as the integral of $f\overline{g}{y}^k$ against that Eisenstein series. Its functional equation is $\mathcal{R}(s)=\mathcal{R}(2k-1-s)$, and its only possible simple poles are at $k-1,k$. The raw series is obtained by dividing out the gamma and zeta factors, so it does not inherit the same unqualified pole statement; zeros of the denominator zeta factor can yield additional possible poles. The raw series is regular at $k-1$ and can have a simple pole at $k$.