Zeros of the Riemann Zeta-Function on the Critical Line

Transcription

1 Zeros of the Riemann Zeta-Function on the Critical Line D.R. Heath-Brown Magdalen College, Oxford It was shown by Selberg [3] that the Riemann Zeta-function has at least c log zeros on the critical line up to height, for some positive absolute constant c. Indeed Selberg s method counts only zeros of odd order, and counts each such zero once only, regardless of its multiplicity. With this in mind we shall write ˆγ i for the distinct ordinates of zeros of ζ(s) on the critical line of odd multiplicity. We shall number the points ˆγ i so that 0 < ˆγ < ˆγ <.... he purpose of the present note is to extract a little more from Selberg s argument, by obtaining further information on the distribution of the ˆγ i. his is given in the following result. heorem For any constant µ (0, ) we have (ˆγ i+ ˆγ i ) µ µ (log ) µ. ˆγ i In particular, if f( ) is any function which tends to infinity with, then almost all intervals [, + f( )(log ) ] contain a point ˆγ i. Clearly this result includes Selberg s. Moreover it is apparent that the second statement of the theorem follows from the first. We also remark that, if one merely sums over ordinates γ i of the zeros in the usual sense, not restricting to those zeros which are on the critical line, then one has (γ i+ γ i ) µ µ (log ) µ γ i for any µ > 0, as was shown by Fujii []. In giving the proof of our result we shall refer to the version of Selberg s argument presented by itchmarsh [4: ]. he proof uses a mollifier φ(s) = ν X β ν ν s,

2 in which the numbers β ν are defined in terms of the coefficients α ν in the expansion ζ(s) / = α ν ν s σ >. itchmarsh takes ν= β ν = α ν ( log ν log X ), but for our purpose the choice α ν, ν X /, β ν = log X/ν α ν log X, X/ ν X, is required. One then defines F (t) = Ξ(t) t + φ( + it) e ( 4 π δ)t, 4 where δ is small and positive. In fact we shall take δ =, where [, ] is the interval in which we are looking for zeros. With the above definition of F (t) it follows (itchmarsh [4: Lemma 0.7]) that t h F (u)du dt δ / log X. () his is subject to the conditions X = δ c and h = (a log X), where a, c are positive and satisfy (a + )c 4. itchmarsh takes a to be constant, but this is unnecessary. We shall set c = 6 so that any value a (0, ) is permissable. We see from () that t h F (u)du dt / log, () on changing h into h and substituting t h for t. he bound () is subject to the conditions X = /6 and h = (a log ), where a (0, 4 ). he proof of (), and hence of (), depends on the definition of β ν, so we must check that our modification does not materially alter the estimates. It is only Lemma 0. of itchmarsh [4] which needs any change. It is shown that κ X/d α κ κ θ log X dκ (X d )θ (log X d )/ p ρ ( + p ) / (3)

3 uniformly for 0 < θ, where κ is restricted to integers coprime to ρ. We shall require a corresponding estimate in which the function { log X/dκ, dκ X, f(x, d, κ) = 0, dκ X, on the left is replaced by log X, dκ X /, g(x, d, κ) = log X/dκ, X / dκ X, 0, dκ X. However, since g(x, d, κ) = f(x, d, κ) f(x /, d, κ), one sees that (3) remains true with g in place of f. We shall also require the estimate F (t) dt log /δ δ / log X / (4) given by Lemma 0.8 of itchmarsh [4]. he proof of this requires no modification. We now establish a lower bound for t h F (u) du = J(t), say, on the interval t. itchmarsh does this only on average, while we shall, in effect, obtain a lower bound for almost all t. We begin by choosing a large constant integer K, and writing so that We now consider the integral /+i / i w(z) = ( sin z z )K, e iλt w(t)dt = 0 for λ K. ζ(s + it)φ(s + it) w( s )ds = I, (5) i say, where log 3/4. (6) 3

4 he integral will converge if K is chosen large enough. We now move the line of integration to σ =, producing a residue φ() w( it i ) (log ) ( e/ t / )K (log ). On the line σ = we may integrate termwise. We have +i i n s w( s )ds = i n / it w(t)dt i on moving the line of integration back to σ = /, so that terms for which log n K make no contribution. Since ζ(s)φ(s) = a n n s n= with a = and a n = 0 for n X /, we now see that where providing that I = i C K + O( log ), C K = w(t)dt > 0, At this point we observe that if h 3/4 then 64K log. (7) /4 J(t) ζ( t h + iu)φ( + iu) du ζ( t h + iu)φ( + iu) w( u )du I + O{ ζ( + i(u + t))φ( + i(u + du t)) ( u / ) u h K }, whence /4 J(t) + u h ζ( + i(u + t))φ( + i(u + du t)). ( u / ) K Since ζ( + i(u + t))φ( + i(u + t)) ( + u ) /4 X 4

5 for t, it follows that the range u / will contribute only O( / ), say. Here we use the facts that 3/4, by (6), and that K is sufficiently large. Moreover ζ( + i(u + t))φ( + i(u + t)) ( u / ) K du h u / say. It follows that We now observe that K(t)dt = /4 = /4 K(t), h u / F (t + u) ( u / ) K du J(t) + K(t) /4. (8) h u / { } +u ( u / ) K F (v) dv du, +u and Cauchy s inequality, in conjunction with (4) yields +u +u We therefore see that since F (v) dv / { We shall write (8) as h u / 5/ / K(t)dt ( u / ) K du J(t) + K(t) C /4, F (v) dv} / 3/4. 3/4, (9) (h/ ) K h (h/ ) K. and define R h = {t [, ] : J(t) C /4 }. hen K(t) /4 on R h, whence /4 mes(r h ) 3/4 (h/ ) K, by (9). It follows that mes(r h ). (0) (h/ ) K 5

6 We can now complete the proof of the theorem. We consider the set S h of t [, ] for which the interval t h u t + h contains no sign change of the function F (u), or equivalently no zero of odd order of ζ( + iu). hus J(t) = t h F (u)du for t S h. If t S h \ R h then J(t) C /4, whence () yields hus On choosing ( C /4 ) mes(s h \ R h ) mes(s h \ R h ) we therefore deduce from (0) that log. = h(h log ) /(K+) / log. mes(s h ) (h log ) +/(K+). Our choice of will satisfy the conditions (6) and (7) if h = (a log ) with h 3/4 and 0 < a a (K), say. We are now ready to estimate ˆγ i (ˆγ i+ ˆγ i ) µ for ˆγ i. We shall choose K to be a fixed integer such that K + < µ. According to a result of Hardy and Littlewood [] we have ˆγ i+ ˆγ i θ for any θ > 4, so that it suffices to prove that (ˆγ i+ ˆγ i ) µ (log ) µ () ˆγ i for ˆγ i < ˆγ i+. Moreover, summands for which ˆγ i+ ˆγ i 8 a (K) log clearly make a satisfactory contribution. We shall classify the remaining terms according to the value of h = H, H =,,..., for which 4h < ˆγ i+ ˆγ i 8h. 6

7 hus we may assume that a (K) log h 3/4. Since t S h for ˆγ i + h t ˆγ i + h, we see that the number of points ˆγ i corresponding to any such h is at most h mes(s h ) h +/(K+) (log ) +/(K+). he corresponding contribution to () is therefore h µ +/(K+) (log ) +/(K+), and summing over h = H (log ) yields the required result. References [] Fujii, A.: On the distribution of the zeros of the Riemann Zeta function in short intervals. Bull. Amer. Math. Soc. 8, 39-4 (975) [] Hardy, G.H., Littlewood, J.E.: Contributions to the theory of the Rieman zeta-function and the theory of the distribution of primes. Acta Math. 4, 9-96 (98) [3] Selberg, A.: On the zeros of Riemann s zeta-function. Skr. Norske Vid. Akad. Oslo no. 0, (94) [4] itchmarsh, E.C.: he theory of the Riemann Zeta-function, nd Edition, revised by D.R. Heath-Brown. Oxford: Clarendon Press 986 7