Joint Approximation by the Riemann and Hurwitz Zeta-Functions in Short Intervals

Let $K\in \mathcal{K}$ and $f\left(s\right)\in H_0\left(K\right)$. Then, for every $\epsilon >0$,  Moreover, the limit  exists and is positive for all but at most countably many $\epsilon >0$.

Suppose that the parameter α is transcendental, or rational $\ne 1,1/2$. Let $K\in \mathcal{K}$ and $f\left(s\right)\in H\left(K\right)$. Then, for every  $\epsilon >0$,  Moreover, the limit  exists and is positive for all but at most countably many $\epsilon >0$.

([15]). Suppose that the set $L({\alpha }_1,\dots ,{\alpha }_r)$ is linearly independent over $\mathbb{Q}$. For $j=1,\dots ,r$, let $K_j\in \mathcal{K}$ and $f_j\left(s\right)\in H\left(K_j\right)$. Then, for every $\epsilon >0$,

([23]). Suppose that the parameter α is transcendental. Let $K_1,K_2\in \mathcal{K}$ and $f_1\left(s\right)\in H_0\left(K_1\right)$, $f_2\left(s\right)\in H\left(K_2\right)$. Then, for every $\epsilon >0$,

([33]). Suppose that $T^{1/3}{(logT)}^{26/15}⩽H⩽T$. Let $K\in \mathcal{K}$, $f\left(s\right)\in H_0\left(K\right)$. Then, for every $\epsilon >0$,Moreover, “lim inf” can be replaced by “lim” for all but at most countably many $\epsilon >0$.

([35]). Suppose that $T^{27/82}⩽H⩽T^{1/2}$, and the parameter α is transcendental. Let $K\in \mathcal{K}$ and $f\left(s\right)\in H\left(K\right)$. Then, for every $\epsilon >0$,Moreover, the lower limit can be replaced by a limit for all but at most countably many $\epsilon >0$.

The lemma follows from Lemma 1 by taking the exponential pair $(11/30,16/30)$. □

The lemma is Theorem 2 from [24], where its proof is presented. □

Let We use the following integral representation [10]: which is a result of the classical Mellin formula that yields Let $K\subset D$ be a fixed compact set. Then, K is closed and bounded; hence, there exists a positive number $\delta <1/12$ such that $1/2+2\delta ⩽\sigma ⩽1-\delta$ for all $s=\sigma +it\in K$. We take $\theta =1/2+\delta$ and $\widehat{\theta }=1/2+\delta -\sigma$. Then, $\widehat{\theta }<0$ and $\widehat{\theta }⩾1/2+\delta -1+\delta =2\delta -1/2>-1/2-\delta =-\theta$. The integrand in (1) has only two simple poles in the strip $\widehat{\theta }<\mathrm{Re}z<\theta$, i.e., a pole at the point $z=0$ of the function $\mathrm{\Gamma }(s/\theta )$ and a pole at the point $z=1-s$ of the function $\zeta (s+z)$. Therefore, using the well-known estimate which is uniform in $\sigma \in ({\sigma }_1,{\sigma }_2)$ with every ${\sigma }_1<{\sigma }_2$, and replacing $\theta$ by $\widehat{\theta }$ in the line of integration in (1), via the residue theorem, we obtain, for $s\in K$, This gives, for $s\in K$, Hence, for $s\in K$, and, after change $t+u$ by u, we obtain In view of (2), we have, for $s\in K$, with $c_1>0$. Moreover, it is well known that, for $\sigma ⩾1/2$, This and (4) imply that Therefore, via (3), we find Using the Cauchy–Schwarz inequality gives For $\left|u\right|⩽{log}^{2}T$ and large T, Therefore, (6) and Lemma 2 show that, for $\left|u\right|⩽{log}^{2}T$, This and (4) give the estimate Similarly to (4), we obtain that, for $s\in K$, Thus, It is easily seen that This, (5), (7), and (8) lead to the following estimate: Taking $T\to \infty$, and then $n\to \infty$, gives the equality of the lemma. □

The lemma is Lemma 10 from [35], where its proof is given. □

Suppose that $T^{27/82}⩽H⩽T^{1/2}$, and $\alpha \ne 1/2$ or 1. Then

We use similar arguments as in the case $\tau \in [0,T]$. Let $F_{T,H,\alpha }({\underline{k}}_1,{\underline{k}}_2)$, ${\underline{k}}_1=(k_{1p}:k_{1p}\in \mathbb{Z},p\in \mathbb{P})$, ${\underline{k}}_2=(k_{2m}:k_{2m}\in \mathbb{Z},m\in {\mathbb{N}}_0)$, be the Fourier transform of $P_{T,H,\alpha }^{\mathsf{\Omega }}$, i. e., where the star ∗ shows that only a finite number of integers $k_{1p}$ and $k_{2m}$ are not zeroes. Thus, taking into account the definition of the measure $P_{T,H,\alpha }^{\mathsf{\Omega }}$, we have We have to show that where $\underline{0}=(0,0,\dots )$. Obviously, by (9), Therefore, only the case $({\underline{k}}_1,{\underline{k}}_2)\ne (\underline{0},\underline{0})$ remains for consideration. Since $\alpha$ is transcendental, the set $L\left(\alpha \right)=\{log(m+\alpha ):m\in {\mathbb{N}}_0\}$ is linearly independent over $\mathbb{Q}$. The set $\{logp:p\in \mathbb{P}\}$ is also linearly independent over $\mathbb{Q}$. The linear independence over $\mathbb{Q}$ for the set $L(\mathbb{P},\alpha )\stackrel{\mathrm{def}}{=}\{(logp:p\in \mathbb{P}),(log(m+\alpha ):m\in {\mathbb{N}}_0)\}$ is easily seen. Actually, if, for some non-zeroes $k_{1p_1},\dots ,k_{1p_r},k_{2m_1},\dots ,k_{2m_v}$, then From this, it follows that there exists a polynomial $p\left(s\right)$ with rational coefficients such that $p\left(\alpha \right)=0$, and this contradicts the transcendence of $\alpha$.

The linear independence of the set $L(\mathbb{P},\alpha )$ shows that, in the case $({\underline{k}}_1,{\underline{k}}_2)\ne (\underline{0},\underline{0})$, Hence, in view of (9), Thus, for $({\underline{k}}_1,{\underline{k}}_2)\ne (\underline{0},\underline{0})$, and with (11), we obtain (10). The lemma is proven to be true. □

The definition of $u_{n,\alpha }$ yields Therefore, by the definitions of $P_{T,H,\alpha }^{\mathsf{\Omega }}$ and $P_{T,H,n,\alpha }$, we have for all $A\in \mathcal{B}\left(H^2\left(D\right)\right)$. Hence, This, the continuity of $u_{n,\alpha }$, Lemma 8, and Theorem 5.1 of [38] prove that $P_{T,H,n,\alpha }$ converges weakly to $U_{n,\alpha }$. □