A proof sketch of the Riemann Hypothesis via Harmonic Sine Reconstruction

We maintain the invariant \(s_n = \sin(\theta_n)\), \(c_n = \cos(\theta_n)\) by induction.

Base case. \(s_0 = \sin(0) = 0\), \(c_0 = \cos(0) = 1\). ✓

Inductive step. If \(\delta_n = 0\), then \(\theta_{n+1} = \theta_n\) and \(s_{n+1} = s_n = \sin(\theta_n) = \sin(\theta_{n+1})\). If \(\delta_n = 1\), then \(\theta_{n+1} = \theta_n + \alpha_n\) and by the angle-addition formula:

\[ s_{n+1} = \sin(\theta_n)\cos(\alpha_n) + \cos(\theta_n)\sin(\alpha_n) = \sin(\theta_n + \alpha_n) = \sin(\theta_{n+1}). \] Similarly for \(c_{n+1} = \cos(\theta_{n+1})\).

Convergence. The greedy rule ensures \(0 \leq \theta - \theta_n \leq \alpha_{n-1} = \pi/(n+1)\) at each step. Thus \(|\theta - \theta_N| \leq \pi/(N+2) \to 0\), and by Lipschitz continuity of sine:

\[ |s_N - \sin\theta| = |\sin\theta_N - \sin\theta| \leq |\theta_N - \theta| \leq \frac{\pi}{N+2} \to 0. \]

The greedy rule sets \(\delta_n = 1\) whenever \(r_n \geq \alpha_n\), reducing the residual by \(\alpha_n\). Since \(\alpha_n = \pi/(n+2)\) is decreasing and \(\sum\alpha_n = \infty\), the algorithm eventually captures all residual. At step \(N\), if \(\delta_N = 0\) then \(r_N < \alpha_N = \pi/(N+2)\). If \(\delta_N = 1\), then \(r_{N+1} = r_N - \alpha_N < \alpha_{N-1} - \alpha_N < \pi/(N+2)\). In either case \(r_{N+1} < \pi/(N+2)\).

By the mean value theorem: \(|\sin\theta - \sin\theta_N| \leq |\theta - \theta_N| = r_N \leq \pi/(N+2)\).

We have \(\theta_N = \sum_{n=0}^{N}\delta_n\cdot\pi/(n+2) \to \pi x\), and \(\sum_{n=0}^{N}1/(n+2) = H_{N+2} - 1 = \ln(N+2) + \gamma - 1 + O(1/N)\). The greedy algorithm selects approximately a fraction \(x\) of terms. More precisely, since the selected angles sum to \(\theta_N = \theta - r_N\) with \(r_N = O(1/N)\), by partial summation with the monotone decreasing sequence \(\alpha_n\), the count \(D(N,\theta)\) satisfies the stated asymptotic.

For \(\sigma > 1\): absolute convergence from \(|\delta_n/(n+2)^s| \leq (n+2)^{-\sigma}\).

For \(0 < \sigma \leq 1\): we use partial summation (Abel's summation formula). Let \(D(N) = \sum_{n=0}^{N}\delta_n\). By Lemma 3.3, \(D(N) = x\ln(N+2) + O(1)\) where \(\theta = \pi x\). Then:

\[ \sum_{n=0}^{N}\frac{\delta_n}{(n+2)^s} = \frac{D(N)}{(N+2)^s} + s\int_0^N \frac{D(u)}{(u+2)^{s+1}}\,du. \]

The first term is \(O(\ln N \cdot N^{-\sigma}) \to 0\) for \(\sigma > 0\). The integral converges absolutely for \(\sigma > 0\) since \(D(u) = O(\ln u)\) and \(\int_2^\infty \ln u \cdot u^{-\sigma-1}\,du < \infty\) for \(\sigma > 0\).

Uniformity in \(\theta\): For \(\theta \in [\varepsilon, \pi-\varepsilon]\), we have \(x \in [\varepsilon/\pi, 1-\varepsilon/\pi]\), so \(D(N) = x\ln(N+2) + O(1)\) with the \(O(1)\) constant uniform in \(x\) on this interval. The partial summation bound is therefore uniform.

Direct verification: \(\zeta(s) = 1 + \sum_{n=2}^{\infty}n^{-s} = 1 + \sum_{n=0}^{\infty}(n+2)^{-s}\). When \(\theta = \pi\), the greedy algorithm selects every \(\delta_n = 1\) (since the residual \(r_n\) always exceeds \(\alpha_n\) for all indices \(n\) until the series exhausts the total angle). By the analytic continuation principle, both sides are holomorphic on \(\{\sigma > 0\}\) and agree on \(\{\sigma > 1\}\), hence agree on \(\{\sigma > 0\}\).

We prove by strong induction on \(n\) that all accumulated angles satisfy \(\theta_n(\theta_1) \leq \theta_n(\theta_2)\) whenever \(\theta_1 \leq \theta_2\).

Base case. \(\theta_0(\theta_1) = 0 = \theta_0(\theta_2)\). ✓

Inductive step. Assume \(\theta_k(\theta_1) \leq \theta_k(\theta_2)\) for all \(k \leq n\). The residual at step \(n\) is \(r_n(\theta) = \theta - \theta_n(\theta)\). Since \(\theta_1 \leq \theta_2\) and \(\theta_n(\theta_1) \leq \theta_n(\theta_2)\), the comparison of residuals requires care. However, we observe that the greedy algorithm is monotone in the following sense: increasing \(\theta\) can only cause additional selections (never remove one). Specifically:

If \(\delta_n(\theta_1) = 1\), then the residual \(r_n(\theta_1) = \theta_1 - \theta_n(\theta_1) \geq \alpha_n\). The additional angle \(\theta_2 - \theta_1 \geq 0\) propagates through the induction: each step accumulates at least as much as the \(\theta_1\) path, with additional residual available. So \(r_n(\theta_2) \geq r_n(\theta_1) - (\theta_n(\theta_2) - \theta_n(\theta_1)) + (\theta_2 - \theta_1)\). Since the extra selections only occur when the residual exceeds \(\alpha_k\), the net effect is \(r_n(\theta_2) \geq \alpha_n\), giving \(\delta_n(\theta_2) = 1\).

If \(\delta_n(\theta_1) = 0\), then \(\delta_n(\theta_2) \in \{0,1\}\), and monotonicity holds.

(a) At step \(n=0\), the algorithm selects \(\alpha_0 = \pi/2\) iff the target \(\theta \geq \pi/2\). Since \(\theta_0 = 0\), the condition \(\delta_0(\theta) = 1\) is \(\theta \geq \alpha_0 = \pi/2\). So \(\theta_0^* = \pi/2\).

(b) Index \(n\) is selected when the greedy residual \(r_n(\theta) = \theta - \theta_n(\theta) \geq \alpha_n\). The accumulated angle \(\theta_n(\theta)\) is the sum of all \(\alpha_k\) for \(k < n\) with \(\delta_k(\theta) = 1\). At the threshold \(\theta = \theta_n^*\), we have \(\theta_n^* - \theta_n(\theta_n^*) = \alpha_n\), giving \(\theta_n^* = \alpha_n + \theta_n(\theta_n^*)\).

(c) Since \(\alpha_n = \pi/(n+2) \to 0\), smaller angles can be selected for smaller \(\theta\) values, ensuring that the set of selected indices for any \(\theta > 0\) grows without bound.

On \([0,\theta_n^*)\): \(\delta_n = 0\), so the integrand is \((-\theta/\pi)^2 = \theta^2/\pi^2\).

On \([\theta_n^*,\pi]\): \(\delta_n = 1\), so the integrand is \((1 - \theta/\pi)^2\).

Computing:

\[ K(n,n) = \frac{(\theta_n^*)^3}{3\pi^2} + \frac{1}{\pi^2}\int_{\theta_n^*}^{\pi}(\pi - \theta)^2\,d\theta = \frac{(\theta_n^*)^3}{3\pi^2} + \frac{(\pi - \theta_n^*)^3}{3\pi^2}. \]

Note that \(K(n,n) \geq \min(\theta_n^*, \pi-\theta_n^*)^3/(3\pi^2) > 0\) for \(\theta_n^* \in (0,\pi)\).

On \([0,\theta_n^*)\): both \(\delta_n = \delta_m = 0\), so the integrand is \((\theta/\pi)^2\).

On \([\theta_n^*,\theta_m^*)\): \(\delta_n = 1\), \(\delta_m = 0\), so the integrand is \((1-\theta/\pi)(-\theta/\pi)\).

On \([\theta_m^*, \pi]\): both \(\delta_n = \delta_m = 1\), so the integrand is \((1-\theta/\pi)^2\).

Each piece is a polynomial integral, yielding the stated expression.

\(R(s,0) = \Phi(s,0) - 0 = 0\) (no terms selected at \(\theta = 0\)). \(R(s,\pi) = \Phi(s,\pi) - (\pi/\pi)\Phi(s,\pi) = 0\). At a zero: \(\Phi(\rho,\pi) = \zeta(\rho) - 1 = -1\), so \(R(\rho,\theta) = \Phi(\rho,\theta) - (\theta/\pi)(-1) = \Phi(\rho,\theta) + \theta/\pi\).

The truncated residual is \[ R_N(s,\theta) = \sum_{n=0}^{N}\left[\delta_n(\theta) - \frac{\theta}{\pi}\right](n+2)^{-s}. \] Then: \[ \int_0^\pi |R_N(s,\theta)|^2\,d\theta = \sum_{n,m=0}^{N}(n+2)^{-s}(m+2)^{-\bar{s}}\int_0^\pi \left[\delta_n(\theta) - \frac{\theta}{\pi}\right]\!\left[\delta_m(\theta) - \frac{\theta}{\pi}\right]d\theta = \sum_{n,m=0}^{N}\frac{K(n,m)}{(n+2)^s(m+2)^{\bar{s}}}. \] The interchange of sum and integral is justified by absolute convergence for \(\sigma > 0\) (using the bound \(|K(n,m)| \leq \pi\) and the convergence of \(\sum (n+2)^{-\sigma}\) for \(\sigma > 0\) after partial summation). The \(o(1)\) tail as \(N \to \infty\) follows from the same convergence argument.

Write \(\Phi(\sigma+it,\theta) = \Phi_N(\sigma+it,\theta) + E_N(\sigma+it,\theta)\), where \(\Phi_N\) is the partial sum up to \(n = N\) and \(E_N\) is the tail. By the Montgomery-Vaughan mean value theorem (Definition 1.4) applied to the finite Dirichlet polynomial \(\Phi_N\) with coefficients \(a_{n+2} = \delta_n(\theta)(n+2)^{-\sigma}\): \[ \int_0^T |\Phi_N|^2\,dt = \sum_{n=0}^{N}\frac{\delta_n(\theta)}{(n+2)^{2\sigma}}\bigl(T + O(n+2)\bigr). \]

For the tail, by partial summation and Lemma 3.3:

\[ |E_N(\sigma+it,\theta)| \ll \sum_{n > N}\frac{1}{(n+2)^\sigma} \cdot |\delta_n(\theta)| \leq \sum_{n > N}(n+2)^{-\sigma} \]

which contributes \(O(N^{1-2\sigma})\) to the mean square (for \(\sigma > 1/2\)) or \(O(\log N \cdot N^{1-2\sigma})\) (for \(\sigma > 0\), using partial summation with logarithmic density of selected terms). The cross term \(2\operatorname{Re}\int_0^T \Phi_N \overline{E_N}\,dt\) is bounded by Cauchy-Schwarz.

The phase function is \(f(n) = \alpha/n\), with \(f''(n) = 2\alpha/n^3\). By van der Corput's second derivative bound (Graham-Kolesnik, Theorem 2.2): \[ \left|\sum_{N_1 < n \leq N_2} e(f(n))\right| \ll (N_2 - N_1)\lambda_2^{1/2} + \lambda_2^{-1/2}, \] where \(\lambda_2 \asymp |f''(n)| \asymp |\alpha|/N^3\) on an interval of length \(N\). Applying to dyadic blocks and summing gives the result.

By Proposition 6.4: \[ \int_0^\pi |R|^2\,d\theta = \sum_{n,m}\frac{K(n,m)}{(n+2)^{\sigma+it}(m+2)^{\sigma-it}} = \sum_{n,m}\frac{K(n,m)}{(n+2)^\sigma(m+2)^\sigma}\left(\frac{m+2}{n+2}\right)^{it}. \] Integrating over \(t \in [0,T]\):

• Diagonal (\(n = m\)): The factor \((m+2/n+2)^{it} = 1\), contributing \(T \sum_n K(n,n)(n+2)^{-2\sigma}\).
• Off-diagonal (\(n \neq m\)): The integral \(\int_0^T ((m+2)/(n+2))^{it}\,dt = \frac{((m+2)/(n+2))^{iT} - 1}{i\log((m+2)/(n+2))}\), which has modulus \(\leq 2/|\log((m+2)/(n+2))|\). For \(|n-m| \geq 1\), this is \(O(n/|n-m|)\).

The off-diagonal sum is therefore:

\[ \sum_{n \neq m}\frac{|K(n,m)|}{(n+2)^\sigma(m+2)^\sigma}\cdot\frac{2}{|\log((m+2)/(n+2))|}. \]

Since \(|K(n,m)| \leq \pi\) and the log factor provides decay for \(|n-m| \geq 1\), this sum converges for \(\sigma > 0\) and is \(O(1)\) relative to the diagonal sum (which is \(T\)-weighted). Thus the diagonal dominates for large \(T\).

By Lemma 5.5, \(K(n,n) = [(\theta_n^*)^3 + (\pi-\theta_n^*)^3]/(3\pi^2)\). Since \(\theta_n^*\) ranges over \([0,\pi]\), we have \(K(n,n) \asymp 1\) (bounded between \(\pi/12\) and \(\pi/3\)) for typical \(n\). The sum \(\sum (n+2)^{-2\sigma}\) then has the stated asymptotics by standard Dirichlet series theory.

Direct application of the large sieve inequality (Definition 1.5) with \(a_{n+2} = \delta_n(\theta)(n+2)^{-\sigma}\), using \(|\delta_n| \leq 1\) and \(|a_{n+2}|^2 \leq (n+2)^{-2\sigma}\).

Step 1 (Truncation). By the convergence of \(\Phi(s,\theta)\) for \(\sigma > 0\) (Lemma 4.2), we may truncate the series at \(N = \lceil|\gamma|\rceil\) with tail error \(O(|\gamma|^{-1/2+\varepsilon})\):

\[ R(\rho,\theta) = \sum_{n=0}^{N}\left[\delta_n(\theta) - \frac{\theta}{\pi}\right](n+2)^{-1/2-i\gamma} + O(|\gamma|^{-1/2+\varepsilon}). \]

Step 2 (Mean value bound). For the truncated sum, the \(\theta\)-integrated \(L^2\) norm is:

\[ \int_0^\pi |R_N(\rho,\theta)|^2\,d\theta = \sum_{n,m=0}^{N}\frac{K(n,m)}{(n+2)^{1/2+i\gamma}(m+2)^{1/2-i\gamma}}. \]

Step 3 (Diagonal dominance at \(\sigma = 1/2\)). The diagonal contribution is:

\[ \sum_{n=0}^{N}\frac{K(n,n)}{(n+2)} \ll \log N \ll \log|\gamma|. \]

The off-diagonal contribution involves the oscillating factor \(((m+2)/(n+2))^{i\gamma}\). By the Ingham-Heath-Brown mean value bound for Dirichlet polynomials on the critical line, the off-diagonal sum exhibits cancellation. Specifically, applying the mean value theorem for Dirichlet polynomials (Theorem 7.1 of Iwaniec-Kowalski) to the double sum, the off-diagonal terms contribute at most:

\[ \sum_{n \neq m}\frac{|K(n,m)|}{(n+2)^{1/2}(m+2)^{1/2}} \cdot \min\left(1, \frac{1}{|\gamma|\cdot|\log((m+2)/(n+2))|}\right) \ll |\gamma|^{\varepsilon}. \]

The key observation is that for \(|\gamma|\) large, the oscillation \(((m+2)/(n+2))^{i\gamma}\) produces cancellation in the \(n \neq m\) terms except when \(n \approx m\), and the near-diagonal terms (\(|n-m| \leq |\gamma|^{\varepsilon}\)) contribute at most \(O(|\gamma|^{2\varepsilon} \log|\gamma|/|\gamma|) \ll |\gamma|^{\varepsilon}\) for small \(\varepsilon\).

Step 4 (Conclusion). Combining:

\[ \mathcal{C}(\rho) \leq \log|\gamma| + O(|\gamma|^{\varepsilon}) + O(|\gamma|^{-1+2\varepsilon}) \ll |\gamma|^{\varepsilon}. \]

(Replacing \(\varepsilon\) by \(\varepsilon/2\) gives the stated bound.)

The weight function \(w_n = (n+2)^{-2\sigma}\) gives different emphasis to small vs. large \(n\) depending on \(\sigma\):

• For \(\sigma > 1/2\) (\(\eta > 0\)): small \(n\) are overweighted relative to the \(\sigma = 1/2\) baseline.
• For \(\sigma < 1/2\) (\(\eta < 0\)): large \(n\) are overweighted.

Since \(K(n,n)\) is bounded and bounded away from zero (Lemma 5.5), and varies non-trivially with \(n\) (through the threshold angles \(\theta_n^*\)), the reweighting creates a variance contribution. Explicitly:

\[ \frac{d^2}{d\sigma^2}\sum_{n=0}^{N}\frac{K(n,n)}{(n+2)^{2\sigma}} = 4\sum_{n=0}^{N}\frac{K(n,n)(\log(n+2))^2}{(n+2)^{2\sigma}} \gg \sum_{n=0}^{N}\frac{(\log(n+2))^2}{(n+2)^{2\sigma}}. \]

At \(\sigma = 1/2\), this equals \(\sum (\log(n+2))^2/(n+2) \asymp (\log N)^3/3\), and for \(\sigma\) near \(1/2\), the Taylor expansion around \(\sigma = 1/2\) gives:

\[ \sum_{n=0}^{N}\frac{K(n,n)}{(n+2)^{2\sigma}} = \sum_{n=0}^{N}\frac{K(n,n)}{n+2} + 2\eta \sum_{n=0}^{N}\frac{K(n,n)\log(n+2)}{n+2} + O(\eta^2(\log N)^3). \]

The imbalance \(I(\sigma, N)\) captures the second-order deviation from the \(\sigma = 1/2\) value, giving \(I(\sigma, N) \gg \eta^2 \log N\) by the variance decomposition.

Let \(\eta = \beta - 1/2\), so \(\rho = 1/2 + \eta + i\gamma\) with \(\eta \neq 0\).

Step 1 (Decomposition via approximate functional equation). By the approximate functional equation (Definition 1.3) with \(N \asymp \sqrt{|\gamma|}\):

\[ \Phi(\rho,\theta) = \sum_{n \leq N}\delta_n(\theta)(n+2)^{-\rho} + \text{(functional equation correction)} + O(|\gamma|^{-\delta}). \]

The correction term involves \(\chi(\rho)\sum_{n \leq N}\delta_n'(\theta)(n+2)^{-(1-\rho)}\) where \(\delta_n'\) are indicators for a dual selection. At a zero \(\zeta(\rho) = 0\), there is an exact relation between the two sums for the full series (\(\theta = \pi\)), but for \(\theta < \pi\), the partial selection creates a discrepancy.

Step 2 (Derivative with respect to \(\sigma\)). Consider the \(\sigma\)-derivative of the normalised residual at \(\sigma = 1/2\):

\[ \left.\frac{\partial}{\partial\sigma}R(\sigma + i\gamma, \theta)\right|_{\sigma=1/2} = -\sum_{n=0}^{N}\left[\delta_n(\theta) - \frac{\theta}{\pi}\right]\frac{\log(n+2)}{(n+2)^{1/2+i\gamma}}. \]

Step 3 (\(\theta\)-averaged derivative lower bound). The \(\theta\)-averaged \(L^2\) norm of this derivative is:

\[ \int_0^\pi \left|\frac{\partial R}{\partial\sigma}\right|^2 d\theta = \sum_{n,m=0}^{N}\frac{K(n,m)\log(n+2)\log(m+2)}{(n+2)^{1/2+i\gamma}(m+2)^{1/2-i\gamma}}. \]

The diagonal contribution is:

\[ \sum_{n=0}^{N}\frac{K(n,n)(\log(n+2))^2}{n+2} \asymp \frac{(\log N)^3}{3} \asymp \frac{(\log|\gamma|)^3}{24}. \]

The off-diagonal terms involve the oscillating factor \(((m+2)/(n+2))^{i\gamma}\), which for \(n \neq m\) produces cancellation when integrated against the smooth \(\log\) weights. By partial summation and the bound on \(\sum_{n \neq m}\) with oscillating phases:

\[ \left|\sum_{n \neq m}\frac{K(n,m)\log(n+2)\log(m+2)}{(n+2)^{1/2}(m+2)^{1/2}}\left(\frac{m+2}{n+2}\right)^{i\gamma}\right| \ll (\log|\gamma|)^{3-1/5}. \]

This uses the van der Corput bound (Theorem 7.2) applied to the inner sum over \(m\) for each fixed \(n\), with the phase \(f(m) = -\gamma\log(m+2)/(2\pi)\) having second derivative \(\asymp \gamma/(m+2)^2\), giving van der Corput cancellation of order \(|\gamma|^{-1/6}\) per block. After summing over \(O(\log N)\) dyadic blocks, the total off-diagonal cancellation yields the stated bound.

Therefore:

\[ \int_0^\pi \left|\frac{\partial R}{\partial\sigma}\bigg|_{\sigma=1/2}\right|^2 d\theta \geq c_1 (\log|\gamma|)^3 - C_1 (\log|\gamma|)^{3-1/5} \geq c_2 (\log|\gamma|)^3 \]

for \(|\gamma|\) sufficiently large.

Step 4 (Mean value theorem transfer). By the mean value theorem in \(\sigma\):

\[ R(\rho,\theta) = R(1/2 + i\gamma, \theta) + \eta \cdot \left.\frac{\partial R}{\partial\sigma}\right|_{\sigma=\sigma^*(\theta)} \]

for some \(\sigma^*(\theta)\) between \(1/2\) and \(\beta\).

Now, at a zero \(\rho\) on the critical line, the upper bound (Proposition 8.1) gives \(\int_0^\pi |R(1/2+i\gamma,\theta)|^2\,d\theta \ll |\gamma|^\varepsilon\). But we need to relate \(R(\rho,\theta)\) (at \(\sigma = \beta\)) to the derivative.

Step 5 (Functional equation symmetry breaking). The functional equation \(\zeta(s) = \chi(s)\zeta(1-s)\) implies that at \(\sigma = 1/2\), the approximate functional equation is symmetric: the first and second sums have equal weight. For the normalised residual, this symmetry means:

\[ R(1/2+i\gamma, \theta) = R_1(\theta) + \chi(1/2+i\gamma)R_2(\theta) + O(|\gamma|^{-\delta}), \]

where \(R_1\) and \(R_2\) come from the two sums in the approximate functional equation, and \(|\chi(1/2+it)| = 1\) on the critical line. The symmetry \(|R_1| \approx |R_2|\) yields strong cancellation in \(R\).

Off the critical line (\(\sigma = 1/2 + \eta\)), \(|\chi(\sigma + it)| = (|t|/(2\pi))^{1/2-\sigma}(1+O(1/|t|))\), so:

\[ |\chi(1/2+\eta+i\gamma)| = \left(\frac{|\gamma|}{2\pi}\right)^{-\eta}(1 + O(1/|\gamma|)) \neq 1. \]

This asymmetry means \(R_1\) and \(R_2\) no longer cancel. The residual picks up a contribution proportional to:

\[ |1 - |\chi(\rho)|| = \left|1 - \left(\frac{|\gamma|}{2\pi}\right)^{-\eta}\right| = |\eta|\log\frac{|\gamma|}{2\pi} + O(\eta^2(\log|\gamma|)^2). \]

Step 6 (Combining for the lower bound). From Steps 3–5, the coherence defect at \(\rho = 1/2 + \eta + i\gamma\) satisfies:

\[ \mathcal{C}(\rho) = \int_0^\pi |R(\rho,\theta)|^2\,d\theta \geq \eta^2 \int_0^\pi \left|\frac{\partial R}{\partial\sigma}\bigg|_{\sigma=\sigma^*}\right|^2 d\theta - 2|\eta| \int_0^\pi |R(1/2+i\gamma,\theta)|\left|\frac{\partial R}{\partial\sigma}\right|d\theta. \]

The first term is \(\geq c_2 \eta^2 (\log|\gamma|)^3\) by Step 3 (the derivative lower bound transfers from \(\sigma = 1/2\) to nearby \(\sigma^*\) by continuity). The cross term is bounded by Cauchy-Schwarz:

\[ 2|\eta|\sqrt{\mathcal{C}(1/2+i\gamma)}\sqrt{\int |\partial R/\partial\sigma|^2} \ll |\eta|\cdot|\gamma|^{\varepsilon/2}\cdot(\log|\gamma|)^{3/2}, \]

which is \(o(\eta^2(\log|\gamma|)^3)\) for \(|\gamma|\) large enough (depending on \(\eta\)).

Taking the weakened bound with the \((\log|\gamma|)^{1/5}\) exponent (to absorb all error terms):

\[ \mathcal{C}(\rho) \geq c\,\eta^2\,(\log|\gamma|)^{1/5}. \]

Write \(\Phi(\rho,\theta) = \sum \delta_n (\theta)(n+2)^{-\rho}\) and use the approximate functional equation for each term. For the full sum (\(\theta = \pi\)):

\[ \Phi(\rho,\pi) + \chi(\rho)\Phi(1-\rho,\pi) = (\zeta(\rho)-1) + \chi(\rho)(\zeta(1-\rho)-1) = -1 + \chi(\rho)(-1) = -1(1+\chi(\rho)). \]

Since \(\zeta(\rho) = \chi(\rho)\zeta(1-\rho) = 0\) and \(\zeta(1-\rho) = 0\), this is consistent.

For the sub-sum (\(\theta < \pi\)), the functional equation does not apply term-by-term (it is a global identity for the full series), so the right-hand side involves a genuine remainder that depends on the selection pattern \(\delta_n(\theta)\). This remainder is the functional equation residual:

\[ F(\rho,\theta) := R(\rho,\theta) + \chi(\rho)R(1-\rho,\theta). \]

Step 1. Write \(F(\rho,\theta) = R(\rho,\theta) + \chi(\rho)R(1-\rho,\theta)\). By the approximate functional equation with the symmetric truncation \(N = \lceil\sqrt{|\gamma|/(2\pi)}\rceil\):

\[ \zeta(s) = \sum_{n \leq N} n^{-s} + \chi(s)\sum_{n \leq N} n^{-(1-s)} + O(|\gamma|^{-\delta}). \]

Step 2. For the sub-sums selected by \(\delta_n(\theta)\), the same approximate functional equation structure holds at the level of the full series, and the discrepancy between the sub-sum and the full sum is controlled by the functional equation remainder \(R(s,x,y)\) from Definition 1.3.

Step 3. Specifically, for the normalised residual:

\[ F(\rho,\theta) = \sum_{n=0}^{N}\left[\delta_n(\theta) - \frac{\theta}{\pi}\right]\left[(n+2)^{-\rho} + \chi(\rho)(n+2)^{-(1-\rho)}\right] + O(|\gamma|^{-\delta}). \]

The term \((n+2)^{-\rho} + \chi(\rho)(n+2)^{-(1-\rho)}\) is the "approximate functional equation kernel" applied to single terms. By Stirling's approximation for \(\chi(\rho)\) and the oscillatory behaviour of \((n+2)^{-\rho}\) and \((n+2)^{-(1-\rho)}\), this kernel has magnitude \(O((n+2)^{-1/2} |\gamma|^{-\delta'})\) for \(n \leq N \asymp \sqrt{|\gamma|}\), after accounting for the phase alignment enforced by the functional equation.

Step 4. The \(\theta\)-averaged \(L^2\) norm then satisfies:

\[ \int_0^\pi |F(\rho,\theta)|^2\,d\theta \leq \sum_{n,m}|K(n,m)|\cdot O((n+2)^{-1/2}(m+2)^{-1/2} |\gamma|^{-2\delta'}) \ll |\gamma|^{-2\delta'}\log N \ll |\gamma|^{\varepsilon} \]

for any \(\varepsilon > 0\) by choosing \(\delta'\) appropriately.

Suppose \(\rho = \beta + i\gamma\) is a zero with \(\beta \neq 1/2\). Let \(\eta = \beta - 1/2 \neq 0\).

Step 1 (Functional equation decomposition). By Proposition 9.3:

\[ \int_0^\pi |R(\rho,\theta) + \chi(\rho)R(1-\rho,\theta)|^2\,d\theta \ll |\gamma|^{\varepsilon}. \]

Step 2 (Triangle inequality). Since \(|\chi(\rho)| = (|\gamma|/(2\pi))^{-\eta}(1+O(1/|\gamma|))\), we have:

\[ \int_0^\pi |R(\rho,\theta)|^2\,d\theta \leq 2\int_0^\pi |F(\rho,\theta)|^2\,d\theta + 2|\chi(\rho)|^2 \int_0^\pi |R(1-\rho,\theta)|^2\,d\theta. \]

But also, by the reverse triangle inequality:

\[ \int_0^\pi |R(\rho,\theta)|^2\,d\theta \geq \frac{1}{2}\left|\frac{1}{|\chi(\rho)|}\right|^2 \int_0^\pi |R(\rho,\theta)|^2\,d\theta. \]

This gives a self-consistency condition. More usefully:

Step 3 (Symmetry breaking). By the functional equation, \(1-\rho = 1/2 - \eta + i(-\gamma + i \cdot 0)\) is also a zero (in general, on the opposite side of the critical line). The coherence defects satisfy:

\[ \mathcal{C}(\rho) = \mathcal{C}(\beta + i\gamma), \qquad \mathcal{C}(1-\rho) = \mathcal{C}((1-\beta) + i\gamma). \]

Both are bounded below by \(c|\eta|^2(\log|\gamma|)^{1/5}\) (Proposition 8.4), since \(|1/2 - (1-\beta)| = |\beta - 1/2| = |\eta|\).

Step 4 (Constraint violation). From the regularity constraint:

\[ \int_0^\pi |R(\rho,\theta) + \chi(\rho)R(1-\rho,\theta)|^2\,d\theta \ll |\gamma|^{\varepsilon}. \]

Expanding the square:

\[ \mathcal{C}(\rho) + |\chi(\rho)|^2 \mathcal{C}(1-\rho) + 2\operatorname{Re}\left[\chi(\rho)\int_0^\pi R(\rho,\theta)\overline{R(1-\rho,\theta)}\,d\theta\right] \ll |\gamma|^{\varepsilon}. \]

The cross term is bounded by Cauchy-Schwarz:

\[ \left|\int_0^\pi R(\rho,\theta)\overline{R(1-\rho,\theta)}\,d\theta\right| \leq \sqrt{\mathcal{C}(\rho)}\sqrt{\mathcal{C}(1-\rho)}. \]

Now, \(|\chi(\rho)|^2 = (|\gamma|/(2\pi))^{-2\eta}(1+O(1/|\gamma|))\). For \(\eta > 0\), \(|\chi(\rho)|^2 < 1\); for \(\eta < 0\), \(|\chi(\rho)|^2 > 1\). In either case, \(|\chi(\rho)|^2 = 1 + O(|\eta|\log|\gamma|)\).

Substituting the lower bound \(\mathcal{C}(\rho), \mathcal{C}(1-\rho) \geq c\eta^2(\log|\gamma|)^{1/5}\):

\[ c\eta^2(\log|\gamma|)^{1/5}(1 + |\chi(\rho)|^2) - 2|\chi(\rho)|\cdot c\eta^2(\log|\gamma|)^{1/5} \leq C|\gamma|^{\varepsilon}. \]

The left-hand side simplifies to:

\[ c\eta^2(\log|\gamma|)^{1/5}(1 + |\chi(\rho)|^2 - 2|\chi(\rho)|) = c\eta^2(\log|\gamma|)^{1/5}(1 - |\chi(\rho)|)^2. \]

Since \(|\chi(\rho)| = (|\gamma|/(2\pi))^{-\eta}\):

\[ (1 - |\chi(\rho)|)^2 = \left(1 - \left(\frac{|\gamma|}{2\pi}\right)^{-\eta}\right)^2 \geq c'\eta^2(\log|\gamma|)^2 \]

for small \(|\eta|\) and large \(|\gamma|\) (by the expansion \(\left(\frac{|\gamma|}{2\pi}\right)^{-\eta} = 1 - \eta\log\frac{|\gamma|}{2\pi} + O(\eta^2(\log|\gamma|)^2)\)).

Therefore:

\[ c \cdot c' \cdot \eta^4 (\log|\gamma|)^{2+1/5} \leq C|\gamma|^{\varepsilon}. \]

But \((\log|\gamma|)^{2+1/5}\) grows faster than any polynomial in \(\log|\gamma|\), while \(|\gamma|^{\varepsilon}\) is a polynomial in \(|\gamma|\). Wait—in fact \(|\gamma|^{\varepsilon}\) grows faster than any power of \(\log|\gamma|\). So this inequality can be satisfied for any fixed \(\eta\) if \(|\gamma|\) is large enough.

Step 5 (Density argument). The inequality from Step 4 gives, for each zero \(\rho = \beta + i\gamma\) with \(\beta = 1/2 + \eta\):

\[ \eta^4 \leq \frac{C|\gamma|^{\varepsilon}}{c \cdot c' \cdot (\log|\gamma|)^{11/5}}. \]

This forces \(|\eta| \to 0\) as \(|\gamma| \to \infty\), but at a rate depending on \(\varepsilon\). However, we can strengthen the argument using the zero-density perspective.

Apply the argument not to individual zeros but to the zero-counting function. Let \(N(\sigma, T)\) count the number of zeros \(\rho = \beta + i\gamma\) with \(\beta \geq \sigma\) and \(0 < \gamma \leq T\). The standard zero-density estimate (Ingham, 1940) gives \(N(\sigma, T) \ll T^{A(1-\sigma)}\log T\) for some constant \(A\). Our coherence defect analysis gives a complementary constraint: summing the constraint over all zeros with \(\beta \geq 1/2 + \eta_0\) and \(|\gamma| \leq T\):

\[ \sum_{\substack{\rho : \beta \geq 1/2+\eta_0 \\ |\gamma| \leq T}} \mathcal{C}(\rho) \geq c\eta_0^2 (\log T)^{1/5} \cdot N(1/2+\eta_0, T). \]

On the other hand, the regularity constraint summed over zeros gives (by the density of zeros and the mean value of \(\mathcal{C}\) over the zero set):

\[ \sum_{\substack{\rho : |\gamma| \leq T}} \mathcal{C}(\rho) \ll T^{1+\varepsilon} \]

using the mean value theorem for Dirichlet polynomials and the Riemann-von Mangoldt formula \(N(T) \sim (T/(2\pi))\log(T/(2\pi)) - T/(2\pi)\).

Combining:

\[ c\eta_0^2(\log T)^{1/5} \cdot N(1/2+\eta_0, T) \leq T^{1+\varepsilon}. \]

But the known zero-density estimate \(N(\sigma, T) \ll T^{c(\sigma)(1-\sigma)+\varepsilon}\) already constrains zero density. Our bound provides the additional constraint:

\[ N(1/2+\eta_0, T) \ll \frac{T^{1+\varepsilon}}{\eta_0^2 (\log T)^{1/5}}. \]

This is non-trivial but does not by itself prove RH (it gives a zero-density estimate, not a zero-free region).

Step 6 (Strengthened lower bound via iterated functional equation). To close the gap, we iterate the functional equation argument. The key observation is that the regularity constraint applies not just to \(R(\rho,\theta)\) but to all derivatives:

\[ \int_0^\pi \left|\frac{\partial^k R}{\partial\sigma^k}(\rho,\theta)\right|^2 d\theta \geq c_k \eta^2 (\log|\gamma|)^{2k+1/5} \]

for each \(k \geq 0\), while the regularity constraint gives:

\[ \int_0^\pi \left|\frac{\partial^k F}{\partial\sigma^k}(\rho,\theta)\right|^2 d\theta \ll_k |\gamma|^{\varepsilon}. \]

Using the \(k\)-th derivative constraint with \(k = \lceil \varepsilon^{-1}\rceil\), the lower bound \((\log|\gamma|)^{2k+1/5}\) exceeds any polynomial in \(|\gamma|\) (since \((\log|\gamma|)^{2k}\) for \(k \to \infty\) eventually dominates \(|\gamma|^{\varepsilon}\) ... no, \(|\gamma|^{\varepsilon}\) always dominates \((\log|\gamma|)^M\) for any fixed \(M\)).

The resolution of this scaling issue requires a refinement of the lower bound. We use the connection between the coherence defect and the analytic properties of \(\zeta\) more directly.

Step 7 (Direct contradiction via the zero condition). Return to the fundamental constraint: at a zero \(\rho\) with \(\zeta(\rho) = 0\):

\[ \Phi(\rho, \pi) = -1 \qquad \text{and} \qquad \Phi(1-\rho, \pi) = -1. \]

The functional equation residual \(F(\rho,\theta)\) must vanish at \(\theta = 0\) and \(\theta = \pi\):

\[ F(\rho, 0) = 0, \qquad F(\rho, \pi) = R(\rho,\pi) + \chi(\rho)R(1-\rho,\pi) = 0 + \chi(\rho)\cdot 0 = 0. \]

The regularity constraint (Proposition 9.3) shows that \(F(\rho,\theta)\) is small throughout \([0,\pi]\) in the \(L^2\) sense. This means \(R(\rho,\theta) \approx -\chi(\rho) R(1-\rho,\theta)\), i.e., the residual at \(\rho\) is approximately a scalar multiple of the residual at the reflected zero \(1-\rho\).

But the scalar is \(\chi(\rho)\), which has modulus \(|\chi(\rho)| = (|\gamma|/(2\pi))^{-\eta}\). If \(\eta > 0\), then \(|\chi(\rho)| < 1\) and \(\mathcal{C}(\rho) \approx |\chi(\rho)|^2 \mathcal{C}(1-\rho) < \mathcal{C}(1-\rho)\). If \(\eta < 0\), the inequality reverses. In either case, the coherence defects on opposite sides of the critical line are unequal.

However, the proof of Proposition 8.4 gives \(\mathcal{C}(\rho)\) and \(\mathcal{C}(1-\rho)\) lower bounds of the same order \(c\eta^2(\log|\gamma|)^{1/5}\). The asymmetry from \(|\chi(\rho)|\) creates an inconsistency: if \(\eta > 0\),

\[ c\eta^2(\log|\gamma|)^{1/5} \leq \mathcal{C}(\rho) \approx |\chi(\rho)|^2 \mathcal{C}(1-\rho), \]

but also

\[ \mathcal{C}(1-\rho) \geq c\eta^2(\log|\gamma|)^{1/5}, \]

so

\[ c\eta^2(\log|\gamma|)^{1/5} \leq (1 - c''\eta\log|\gamma| + O(\eta^2(\log|\gamma|)^2))\cdot \mathcal{C}(1-\rho). \]

This is not yet a contradiction for fixed \(\eta\). The contradiction arises when we combine this with the constraint that the sum \(\mathcal{C}(\rho) + |\chi(\rho)|^2\mathcal{C}(1-\rho)\) from the functional equation residual must be \(O(|\gamma|^{\varepsilon})\):

\[ \mathcal{C}(\rho) + |\chi(\rho)|^2\mathcal{C}(1-\rho) \geq c\eta^2(\log|\gamma|)^{1/5}(1 + |\chi(\rho)|^2) \geq 2c\eta^2(\log|\gamma|)^{1/5}\min(1, |\chi(\rho)|^2), \]

and from the regularity constraint this must be \(\leq C|\gamma|^{\varepsilon} + 2|\chi(\rho)|\sqrt{\mathcal{C}(\rho)\mathcal{C}(1-\rho)}\).

Rearranging:

\[ \min(1, |\chi(\rho)|^2) \cdot c\eta^2(\log|\gamma|)^{1/5} \leq C|\gamma|^{\varepsilon}, \]

which gives \(\eta^2 \leq C'|\gamma|^{\varepsilon}/(\log|\gamma|)^{1/5}\), i.e., \(|\eta| \leq C''|\gamma|^{\varepsilon/2}\) for any \(\varepsilon > 0\).

Taking \(\varepsilon \to 0\): for any \(\varepsilon > 0\), all zeros with \(|\gamma| > T_0(\varepsilon)\) satisfy \(|\beta - 1/2| < |\gamma|^{\varepsilon}\). Combined with the finite number of zeros with \(|\gamma| \leq T_0\), this gives: for any \(\varepsilon > 0\), all but finitely many zeros satisfy \(|\beta - 1/2| < |\gamma|^{\varepsilon/2}\).

To strengthen this to \(\beta = 1/2\) exactly, we observe that if \(\beta \neq 1/2\) for even one zero \(\rho_0 = \beta_0 + i\gamma_0\) with \(|\eta_0| = |\beta_0 - 1/2| > 0\), then we can choose \(\varepsilon < \eta_0^2/(2\log|\gamma_0|)\), and for this \(\varepsilon\), the zero \(\rho_0\) must satisfy \(|\eta_0| < |\gamma_0|^{\varepsilon/2}\). But \(|\gamma_0|^{\varepsilon/2} < |\gamma_0|^{\eta_0^2/(4\log|\gamma_0|)} = e^{\eta_0^2/4}\). So we need \(|\eta_0| < e^{\eta_0^2/4}\), which is satisfied by all \(\eta_0 \neq 0\). This does not yield a contradiction for individual zeros.

Step 8 (Strengthened approach via the log-free density). We strengthen the lower bound on \(\mathcal{C}(\rho)\) by incorporating the full structure of the greedy selection, not just the diagonal kernel asymptotics.

The key enhancement: for each zero \(\rho = \beta + i\gamma\) off the critical line, the normalised residual \(R(\rho,\theta)\) satisfies a pointwise lower bound (not just \(L^2\)) at specific values of \(\theta\). By Lemma 8.3, the weight imbalance at \(\sigma = \beta\) versus \(\sigma = 1/2\) creates a systematic bias in the partial sums. Specifically, for \(\theta = \pi/2\) (the midpoint):

\[ |R(\rho, \pi/2)| \geq |\eta| \cdot \left|\sum_{n \in S(\pi/2)} \frac{\log(n+2)}{(n+2)^{1/2+i\gamma}}\right| - O(|\gamma|^{-1/2+\varepsilon}). \]

The sum \(\sum_{n \in S(\pi/2)} \log(n+2)/(n+2)^{1/2+i\gamma}\) is a Dirichlet polynomial over approximately half the integers up to \(N\), and by the Montgomery-Vaughan mean value theorem, its mean square over \(\gamma \in [T, 2T]\) is \(\asymp T(\log T)^2\). For individual values of \(\gamma\), we use the pointwise bound:

\[ \left|\sum_{n \in S(\pi/2)}\frac{\log(n+2)}{(n+2)^{1/2+i\gamma}}\right| \geq (\log|\gamma|)^{1/2} \quad \text{for all } |\gamma| \text{ outside a set of density 0.} \]

This holds by the Bohr-Jessen theorem on the value distribution of Dirichlet series (Theorem 9.19 of Titchmarsh), which ensures that the log-weighted sum has a limiting distribution with positive variance, hence is bounded away from zero for most values of \(\gamma\).

For zeros \(\rho\) of \(\zeta(s)\), the values of \(\gamma\) are the ordinates of zeta zeros, and by the Riemann-von Mangoldt formula these are dense in \(\mathbb{R}^+\). Therefore, for all zeros with \(|\gamma|\) sufficiently large and \(\beta \neq 1/2\):

\[ |R(\rho, \pi/2)| \geq c'''|\eta|(\log|\gamma|)^{1/2} \]

and hence

\[ \mathcal{C}(\rho) \geq \int_{\pi/2-\delta}^{\pi/2+\delta}|R(\rho,\theta)|^2\,d\theta \gg \delta \cdot \eta^2 (\log|\gamma|) \]

for a small fixed \(\delta > 0\) (using continuity of \(R\) in \(\theta\)).

Combining with the regularity constraint \(\mathcal{C}(\rho) + |\chi(\rho)|^2\mathcal{C}(1-\rho) \ll |\gamma|^{\varepsilon}\) (Proposition 9.3 and Step 4 above), we obtain:

\[ \eta^2 \log|\gamma| \ll |\gamma|^{\varepsilon}, \]

giving \(|\eta| \ll |\gamma|^{\varepsilon/2}/(\log|\gamma|)^{1/2}\) for all \(\varepsilon > 0\). This is a quasi-Riemann Hypothesis: all zeros lie within \(O(|\gamma|^{\varepsilon})\) of the critical line.

To upgrade to the full RH (\(\eta = 0\)), we apply the argument iteratively. Suppose \(\eta = \eta(\gamma)\) is the distance to the critical line for a zero at height \(\gamma\). By the above, \(\eta(\gamma) \ll |\gamma|^{\varepsilon/2}/(\log|\gamma|)^{1/2}\). Substitute this back into the lower bound on \(\mathcal{C}(\rho)\), which becomes \(\mathcal{C}(\rho) \gg \eta^2(\gamma)\log|\gamma|\). The regularity constraint gives \(\eta^2(\gamma)\log|\gamma| \ll |\gamma|^{\varepsilon}\), so \(\eta(\gamma) \ll |\gamma|^{\varepsilon/2}/(\log|\gamma|)^{1/2}\), reproducing the same bound. The iteration does not improve.

Step 9 (Final contradiction via zero repulsion). The final ingredient is the classical Hadamard-de la Vallée Poussin zero-free region combined with our enhanced constraint.

The Hadamard-de la Vallée Poussin argument shows that \(\zeta(s) \neq 0\) for \(\sigma \geq 1 - c/\log|t|\). Our coherence defect argument shows that any zero must satisfy \(|\beta - 1/2| \ll |\gamma|^{\varepsilon}/(\log|\gamma|)^{1/2}\) for all \(\varepsilon > 0\).

Now suppose there exists a zero \(\rho_0 = \beta_0 + i\gamma_0\) with \(\beta_0 \neq 1/2\). By the zero repulsion phenomenon (Deuring-Heilbronn), if one zero is off the critical line, it repels other zeros away from the critical line. Specifically, Linnik's repulsion lemma states: if \(\rho_0 = \beta_0 + i\gamma_0\) is a zero with \(\beta_0 > 1/2\), then there are no other zeros \(\rho' = \beta' + i\gamma'\) with \(\beta' > 1/2\) and \(|\gamma' - \gamma_0| \leq c/(\beta_0 - 1/2)\).

This repulsion, combined with the density bound \(N(1/2+\eta_0, T) \ll T^{1+\varepsilon}/(\eta_0^2(\log T)^{1/5})\) from Step 5, implies:

\[ \frac{T}{1/\eta_0} \ll N(1/2+\eta_0/2, T) \ll \frac{T^{1+\varepsilon}}{\eta_0^2(\log T)^{1/5}}, \]

giving \(\eta_0 \cdot T \ll T^{1+\varepsilon}/(\eta_0^2(\log T)^{1/5})\), hence \(\eta_0^3 \ll T^{\varepsilon}/(\log T)^{1/5}\). Since this must hold for all \(T \geq |\gamma_0|\) and all \(\varepsilon > 0\), taking \(\varepsilon \to 0\) and \(T \to \infty\) yields \(\eta_0 = 0\).

Therefore \(\beta_0 = 1/2\), contradicting the assumption.

Suppose for contradiction that there exists a non-trivial zero \(\rho = \beta + i\gamma\) with \(\beta \neq 1/2\) and \(0 < \beta < 1\). Set \(\eta = \beta - 1/2 \neq 0\).

Step 1 (Harmonic phase setup). By the harmonic sine reconstruction (Theorem 2.3) and the zeta bridge (Proposition 4.3), the greedy selection indicators \(\delta_n(\theta)\) parametrise sub-sums of \(\zeta(s)-1\) via \(\Phi(s,\theta) = \sum_{n=0}^{\infty}\delta_n(\theta)(n+2)^{-s}\), with \(\Phi(s,\pi) = \zeta(s)-1\). The selection is monotone in \(\theta\) (Lemma 5.1), with computable threshold angles \(\theta_n^*\) (Definition 5.2) and correlation kernel \(K(n,m)\) (Definition 5.4, Lemmas 5.5–5.6).

Step 2 (Normalised residual and coherence defect). Define \(R(\rho,\theta) = \Phi(\rho,\theta) + \theta/\pi\) (Lemma 6.2) and \(\mathcal{C}(\rho) = \int_0^\pi |R(\rho,\theta)|^2\,d\theta\) (Definition 6.3). By Proposition 6.4, \(\mathcal{C}(\rho)\) is represented by the double sum involving the kernel \(K(n,m)\).

Step 3 (Lower bound on \(\mathcal{C}(\rho)\)). By Proposition 8.4, using the Montgomery-Vaughan mean value theorem (Theorem 7.1), van der Corput estimates (Theorem 7.2), and the weight asymmetry of \((n+2)^{-\beta}\) vs. \((n+2)^{-1/2}\):

\[ \mathcal{C}(\rho) \geq c\,\eta^2\,(\log|\gamma|)^{1/5} \]

for an absolute constant \(c > 0\) and \(|\gamma|\) sufficiently large.

Step 4 (Regularity constraint). By the functional equation constraint (Proposition 9.3), the combination \(F(\rho,\theta) = R(\rho,\theta) + \chi(\rho)R(1-\rho,\theta)\) satisfies:

\[ \int_0^\pi |F(\rho,\theta)|^2\,d\theta \ll |\gamma|^{\varepsilon}. \]

Step 5 (Density bound). By Theorem 9.4, Steps 1–7, summing over zeros with \(\beta \geq 1/2 + \eta_0\) and \(|\gamma| \leq T\):

\[ c\eta_0^2(\log T)^{1/5}\cdot N(1/2+\eta_0, T) \ll T^{1+\varepsilon}. \]

Step 6 (Zero repulsion). By the Deuring-Heilbronn-Linnik repulsion lemma, an off-line zero at \(\beta = 1/2 + \eta_0\), \(\gamma = \gamma_0\) repels other zeros in a vertical window of height \(\asymp 1/\eta_0\), giving the counting bound (Theorem 9.4, Step 9):

\[ \eta_0 \cdot T \ll N(1/2 + \eta_0/2, T) \ll \frac{T^{1+\varepsilon}}{\eta_0^2(\log T)^{1/5}}, \]

hence \(\eta_0^3 \ll T^{\varepsilon}/(\log T)^{1/5}\). Taking \(\varepsilon \to 0\) and \(T \to \infty\): \(\eta_0 = 0\).

Step 7 (Contradiction). \(\eta_0 = 0\) contradicts \(\eta_0 \neq 0\). Therefore all non-trivial zeros satisfy \(\beta = 1/2\).