Zero zeta functions

Proof of Theorem 14.2

By Theorem 11.14, the zeros of \({\mathsf D}_N\) have linear counting density, while the reduced zeros have counting order \(R^{1/N}\). Partial summation gives absolute convergence in the stated open half planes. The positive terms obtained after taking absolute values diverge on the boundary and to its left. Local uniform convergence follows by the Weierstrass test on compact subsets.

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Proof of Theorem 14.3

For \(N=2\), Corollary 11.10 and \({\mathsf D}_2(z)=\Delta_2(z^2)\) give

Thus the first formula holds with \(Q_2^{\mathrm{red}}=0\).

For \(N\geq3\), Corollary 11.13 writes all but finitely many reduced zeros as

where \(\lambda_m>0\), \(1+\varepsilon_m>0\), and

On every compact set of \(s\) values,

uniformly for large \(m\). The difference

therefore has a normally convergent sum on every compact subset of \({\mathbb C}\). Indeed, the model power has at most polynomial growth in \(m\) on a compact set, while \(|\varepsilon_m|\) decays exponentially. The finitely many exceptional terms \(|\lambda|^{-s}\) and the finitely many omitted model terms are entire functions of \(s\). Their addition to the normally convergent tail defines \(Q_N^{\mathrm{red}}\) and proves the first formula. Thus that formula is established for every \(N\geq2\).

In either case, the Hurwitz zeta function has a meromorphic continuation with one simple pole at its first argument equal to one and residue one. Composition with \(Ns\) gives the stated pole and residue.

For every \(N\geq2\), each nonzero zero \(\lambda\) of \(\Delta_N\) gives \(N\) zeros of \({\mathsf D}_N\), each of modulus \(|\lambda|^{1/N}\). Hence

Substitution gives the second formula with

which is entire.

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Proof of Corollary 14.4

For \(N=2\), the reduced zeros are

For \(N=4\), Example 11.12 gives the reduced zeros

Summing their negative powers proves the reduced formulas. The absolute formulas follow from

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Proof of Corollary 14.6

The tail consists of positive reduced zeros, so it has the same comparison with the Hurwitz zeta. Only finitely many terms depend on the branch choices.

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Proof of Theorem 14.8

For \(N\geq3\), Theorem 11.7 gives

for every sufficiently large simple zero. Let \(L\) be a compact subset of the \(s\) plane. Because the chosen arguments of the zeros lie in a fixed bounded interval, there are constants \(C_L\) and \(M_L\) such that

for every sufficiently large zero. Hence the absolute value of the corresponding summand is at most

The linear zero counting estimate makes this majorant summable. The Weierstrass test gives absolute local uniform convergence of the series of entire summands, so its sum is entire. The finite exceptional set causes no difficulty.

For \(N=2\), the derivative of the cosine has modulus one at every zero. The assertion then follows from the linear zero spacing.

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