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Under the Riemann hypothesis, every $\beta =1/2$, so for $\sigma >1/2$ the absolutely convergent formula givesThe zero set is nonempty: otherwise Hadamard factorization would make $\xi$ an exponential of a linear polynomial, and its functional symmetry would force it to be constant, contrary to gamma growth on the positive real axis. Thus the inequality is strict in the open right half-plane. Continuity at $\sigma =1/2$, including at boundary zeros, proves the claimed increasing modulus on the closed half-line.

Conversely, suppose the modulus is nondecreasing for every fixed $t$. If a zero $\beta +i\gamma$ had $\beta >1/2$, then nonnegativity and monotonicity would force $|\xi (\sigma +i\gamma )|=0$ throughout $1/2\le \sigma \le \beta$. The identity theorem would make $\xi$ identically zero, a contradiction. A zero left of the line reflects to one right of the line by the functional equation and complex conjugation symmetry. Hence every zero lies on the critical line. This proves the xi modulus criterion for the Riemann hypothesis. The two following roman headers refer to supplied asymptotic assumptions, not further questions, and require no Solution sections.

Removing the completion's poles produces an entire function with $\xi (0)=\xi (1)=1/2$ and $\xi (s)=\xi (1-s)$. Its zeros are the nontrivial zeros of the Riemann zeta function. Its finite order allows Hadamard factorization, while the xi modulus criterion for the Riemann hypothesis characterizes their horizontal location.