## A zero-density theorem for the Riemann zeta-function.(English)Zbl 0548.10024

Let $$N(\sigma$$,T) denote as usual the number of zeros of $$\zeta$$ (s) in the box $$\sigma \leq Re s\leq 1$$, $$| Im s|\leq T$$. The paper considers estimates $$N(\sigma,T)<<T^{A(\sigma)(1-\sigma)+\epsilon},$$ and shows that one may take $$A(\sigma)\leq 3/(7\sigma -4)$$ for 3/4$$\leq\sigma \leq 10/13$$ and $$A(\sigma)\leq 9/(8\sigma -2)$$ for 10/13$$\leq\sigma \leq 1$$. These improve on the bound $$A(\sigma)\leq 3/(3\sigma -1)$$ due to M. N. Huxley [Invent. Math. 15, 164-170 (1972; Zbl 0241.10026)], throughout the range 3/4$$\leq\sigma \leq 1$$. It is claimed that these are the first bounds to do so. However M. Jutila [Acta Arith. 32, 55-62 (1977; Zbl 0307.10045)] showed that one may take $$A(\sigma)\leq 3k/((3k-2)\sigma +2-k)$$ if $(*)\quad (5k-4)/(6k- 4)\leq\sigma \leq (3k^ 2-2k+1)/(4k^ 2-3k+1)$ and if $$k\geq 2$$ is an integer. The author comments that for any fixed k Huxley’s bound is better, if $$\sigma$$ is sufficiently close to 3/4. None the less by choosing $$k=[(11\sigma -8)/(4\sigma -3)]$$ (in which case $$\sigma$$ lies in the range (*)) one obtains $A(\sigma)\leq 3(11\sigma -8)/(25\sigma^ 2- 21\sigma +2)\quad for\quad 3/4\leq\sigma \leq 10/13;$ this is superior to the bound 3/(7$$\sigma$$ -4) given in the present paper. The case $$k=5$$ of Jutila’s result yields $$A(\sigma)\leq 15/(13\sigma -3) (<9/(8\sigma -2))$$ for 10/13$$\leq\sigma \leq 21/26$$; and the reviewer’s estimate $$A(\sigma)\leq 9/(7\sigma -1)$$ for 11/14$$\leq\sigma \leq 1$$ [J. Lond. Math. Soc., II. Ser. 19, 221-232 (1979; Zbl 0393.10043)] is superior to $$9/(8\sigma -2)$$ for the remaining range 21/26$$\leq\sigma \leq 1$$. Thus the author’s results are inferior to those already known, for the whole range 3/4$$\leq\sigma \leq 1$$.
Reviewer: D.R.Heath-Brown

### MSC:

 11M06 $$\zeta (s)$$ and $$L(s, \chi)$$