Abstract
As demonstrated in Hayman’s proof of the classical Borel lemma, every continuous increasing function \(T(r)\geq 1\) satisfies \(T\big{(}r+\frac{1}{T(r)}\big{)}<2T(r)\) outside a possible exceptional set of linear measure \(2\) . In this work, we prove that \(T(r)\) satisfies \(T\big{(}r+\frac{1}{T(r)}\big{)}<\big{(}\sqrt{T(r)}+1\big{)}^{2}\) outside a possible exceptional set of linear measure \(\zeta(2)=\frac{\pi^{2}}{6}<2\) for the Riemann zeta function \(\zeta(s)\) , which is sharper if \(T(r)\geq\big{(}\sqrt{2}+1\big{)}^{2}\) . Observe \(T\big{(}r+\frac{1}{T(r)}\big{)}<\big{(}\sqrt{T(r)}+1\big{)}^{2}\leq 2T(r)\) outside a possible exceptional set of linear measure \(\zeta\big{(}2,\sqrt{2}+1\big{)}\leq 0.52<2\) for the Hurwitz zeta function \(\zeta(s,a)\) . This is noteworthy, provided the set of \(r\) with \(1\leq T(r)<\big{(}\sqrt{2}+1\big{)}^{2}\) has linear measure less than \(1.48\) . Focusing solely on meromorphic functions of infinite order, we apply Hinkkanen’s refined version of the second main theorem, draw comparisons with the classical results of Borel, Nevanlinna, and Hayman, and finally extend an earlier work of Fernández Árias.
