(Almost) Real Proof of the Prime Number Theorem

Transcription

1 (Almost Real Proof of the Prime Number Theorem Michael Müger Institute for Mathematics, Astrophysics and Particle Physics Radboud University Nijmegen, The Netherlands April 2, 27 Abstract We explain a fairly simple proof of the Prime Number Theorem that uses only basic real analysis and the elementary arithmetic of complex numbers. This includes the ζ-function (as realdifferentiable function and the Fourier transform on R, but neither Fourier inversion nor anything from complex analysis. Introduction The first proof of the Prime Number Theorem (PNT, found in 896 by Hadamard and de la Vallée-Poussin (independently, made massive use of complex analysis, well beyond relying on the Riemann function ζ(z for complex z. Its crucial ingredient is the non-vanishing of ζ on the line Re(z =, but it requires additional estimates on ζ to ensure the existence of certain integrals. Landau gave a more conceptual proof of the PNT in terms of a Tauberian theorem, cf. [, 24], which however still needed a growth condition. In the 93s, Ikehara [5] used Wiener s general Tauberian theory [7, 8] to einate the growth condition from Landau s Tauberian theorem, thereby deducing the PNT from ζ( + it alone. Bochner [] finally gave a simple proof of the Landau-Ikehara theorem without appealing to Wiener s Tauberian theory. The point of this note is just to make plain that this approach to the PNT can be formulated in a way that, while keeping the complex numbers for notational convenience and using ζ, avoids all notions and results from complex analysis, like holomorphicity, analytic continuation, Cauchy s theorem, etc. The resulting proof probably is the simplest imaginable. (There is, of course, nothing new about this. Cf. e.g. [9]. 2 The ζ-function: Re z > For z C with Re(z > we define ζ(z = n z. (2. In view of n z = n Re(z and the convergence of /nα for α >, it is clear that the above sum converges absolutely for Re(z > and uniformly for Re(z α >. It therefore defines a nice function on the open half plane Re(z >. For our purposes it will be sufficient to read nice as C. 2. Remark As explained in the Introduction, the proof given below makes no use of complex analysis but does involve complex numbers. Writing z = s + it, we have ζ(z = e z log n = it log n e n s = cos(t log n n s i sin(t log n n s.

2 Since the dependence of ζ(s + it on the real variables s and t is quite different, and since we do not need the holomorphicity of ζ, we could interpret ζ as a function from an open half-plane in R 2 to R 2. But since a complex number just is a pair of real numbers, this seems pointless, in particular since multiplying complex numbers, which is inconvenient in terms of pairs of reals, is essential for the proof, cf. e.g. Lemma 2.2. Furthermore, the Fourier transform, which we use in Section 4, is much more natural for complex-valued than for real-valued functions since R C, t e it is a homomorphism, which is not true for sin t, cos t. It really does not seem that there is a reasonable way of writing proofs of the PNT involving the ζ-function in purely real terms. (As opposed to the elementary proofs, cf. e.g. [3]. 2.2 Lemma Denoting by P the set of primes, for Re(z > we have ζ(z = p P. (2.2 p z Proof. By the definition of infinite products, the r.h.s. of the above equation is the it x of p z = ( p z = p P p P k= k p x p x = k l (p k = pk l l = z n z, P + (n x where {p,, p l } = P [, x] and P + (n is the largest prime factor of n. The last identity is due to the fundamental theorem of arithmetic, i.e. the existence and uniqueness of the decomposition of n into primes. Now, as x the summation in the last term runs over all of N, and the expression tends to (the absolutely convergent expression n n z = ζ(z. We now want a function H such that ζ(z = e H(z for Re(z >. Here some care is required since the exponential function is not injective ( the logarithm is multi-valued. 2.3 Lemma (i The series F (z = ( z k k= k converges whenever z <. (ii If z (, 2 then F (z = log z, the real logarithm of z. (iii For every z C such that z <, we have e F (z = z. Proof. (i is obvious, and (ii is well-known. Now (ii implies e F (z = z for z (, 2. The exponential function and F being given by power series, the power series identity e F (z = z (which essentially is an algebraic fact continues to hold for all complex z satisfying z <. 2.4 Proposition Write H(z = p P k=. (2.3 kpkz (i The sum in (2.3 converges absolutely if Re(z > and defines a nice function on this open half-plane. (ii For real z >, H(z is the (real logarithm of ζ(z. (iii For all z with Re(z > we have ζ(z = exp(h(z and thus ζ(z. Proof. (i It is obvious from (2. that ζ(s > for real s >. Since also ( p s >, equation (2.2, together with continuity of the logarithm (for positive argument gives log ζ(s = p P ( log p s = p P log( p s = p P k=, (2.4 kpks where the last identity follows the fact that p s < and parts (i and (ii of Lemma

3 For s = Re(z > we have kp kz p P k= = p P k= = log ζ(s <, kpks using (2.4. Thus the double sum in (2.3 converges absolutely and defines a nice function on {z Re(z > }. (ii This is just (2.4. (iii With p z = p s <, part (iii of Lemma 2.3 gives exp[ k (kpkz ] = ( p z whenever Re(z >. Now e H(z = x exp( p x k= = kp kz x p x ( p z = ζ(z by continuity of exp. This clearly implies ζ(z for Re(z >. 2.5 Lemma Defining the von Mangoldt function by { log p if n = p Λ(n = k, p P, k N else and ψ(x = n x Λ(n, we have for Re(z > : ζ (z ζ(z = Λ(n n z Proof. Using Proposition 2.4, we have ζ (z ζ(z = H (z = p P = z kp kz k, e zx ψ(e x dx (2.5 = p P k log p p kz = Λ(n n z, where the differentiation under the sum is justified by the uniform convergence of the sum of the derivatives. This is the first half of (2.5, proving also the absolute convergence of the middle sum. Convergence of the integral follows from ψ(x = O(x log x = O(x +ε. Now z e zx ψ(e x dx = z proves the second half of (2.5. = z e zx Λ(n dx = z n e x ] Λ(n = n z [ Λ(n z e zx log n Λ(n e zx dx log n 3 The ζ-function: Re z 3. Lemma There is a C -function g : {z Re(z > } C such that ζ(z = g(z + z s = Re(z >. Thus ζ has a C -extension, which we also denote ζ, to {z Re(z > }\{}. Proof. For Re(z > we have t z dt = z. Thus Now, g(z = ζ(z z = n+ n ( n z t z dt = z ( n z n+ n n+ t n n du u t z dt z+ dt = sup n+ n u [n,n+] so that (3. converges, and thus defines g(z as a C -function on Re(z >. when ( n z t z dt. (3. z z = u z+ n s+, 3

4 3.2 Remark The above argument of course also shows that ζ(z extends to a meromorphic function on Re(z > with a simple pole of residue one at z =. This extension clearly is unique. We will have no use for this, since we will only need that ζ(z /(z has a C -extension. Extensions of differentiable functions are, of course, not unique since there exist compactly supported smooth functions. This is not a problem since we only need a continuous extension of ζ(z /(z to the closed half plane Re(z, telling us that s ζ(s + it s+it exists for all t. This extension (whose existence follows from the lemma still is unique for simple reasons of point set topology (the Hausdorff property of R Proposition We have ζ(z whenever Re(z and z. Proof. We already know that ζ(z when Re(z >. It remains to prove ζ( + it for t. With Proposition 2.4 and z = s + it we obtain log ζ(z = log e H(z = Re(H(z = p P k= ( Re kp kz = p P k= Re ( ikt log e p kp ks. (3.2 Using Re(3 + 4e iθ + e 2iθ = 3 + 2e iθ + 2e iθ + e2iθ + e 2iθ 2 with θ = kt log p and (3.2 we have = (2 + eiθ + e iθ 2 2 log ζ 3 (sζ 4 (s + itζ(s + 2it = p k= Re(3 + 4e ikt log p + e 2ikt log p kp ks, thus ζ 3 (sζ 4 (s + itζ(s + 2it. This can be restated as (s ζ(s 3 ζ(s + it s 4 ζ(s + 2it s. (3.3 If t, the function ζ(z = ζ(s + it as extended in Lemma 3. clearly has the partial derivative s ζ(s + it ζ at z = + it. Assuming that ζ( + it =, it follows that the it ζ(s + it ζ(s + it ζ( + it = = ζ ( + it s s s s exists. Together with s (s ζ(s =, this implies that the l.h.s. of (3.3 tends to the finite it ζ ( + it 4 ζ( + 2it as s, whereas the r.h.s. of (3.3 diverges. This contradiction shows that ζ( + it for all t. From now on, ζ (z, where z = s + it, will always denote the partial derivative s ζ(s + it, as in the above proof. (Since all functions considered are holomorphic, ζ (z of course coincides with the complex derivative, but this will not be needed. 3.4 Corollary The function z ζ (z ζ(z z, well-defined on Re(z > by Proposition 2.4(iii, has a continuous extension to Re(z. Proof. By Lemma 3., there is a C -function g on Re(z > such that ζ(z = g(z + z on Re(z >. Since ζ, ζ are nice functions and ζ(z on {Re(z }\{} by Proposition 3.3, the claim is clear away from z =. To see that this also holds near z =, we compute ζ (z ζ(z = g (z (z 2 g(z + z = (z g (z (z g(z + + z = (z g (z + z (z g(z + ( (z g(z (z g(z +. 4

5 Thus ζ (z ζ(z = z + k(z, where k = (z g (z g(z. (z g(z + The claim follows, since k is continuous on Re(z. 4 Ikehara s Tauberian theorem The aim of this section is to prove the following theorem (following [3] quite closely: 4. Theorem Let f : R + R + be non-decreasing. If there are a, c such that the integral F (z = f(xe zx dx, converges for Re(z > a [thus automatically defines a continuous function there] and F (z has a continuous extension to the half-plane Re(z a then f(x ce ax as x +. The proof will be given using the Fourier transform on R, concerning which we only need the definition and an easy special case of the Riemann-Lebesgue Lemma: 4.2 Lemma Let f : R C be continuous with compact support. Then the Fourier transform f(ξ := f(xe iξx dx (4. c z a (integrals without bounds always extend over R satisfies f(ξ as ξ. Proof. For ξ, we have f(ξ = f(xe iξ(x π ξ dx = f(x + π ξ e iξx dx, which leads to f(ξ = 2 (f(x f(x + πξ e iξx dx. (4.2 Since f is continuous with compact support, it is uniformly continuous, thus sup x f(x f(x+ε as ε. Inserting this in (4.2 and keeping in mind that supp(f is bounded, the claim follows. Proof of Theorem 4.. As a preparatory step, pick any continuous, compactly supported even function g : R R such that g( = and ĝ(ξ ξ, where ĝ is as in (4.. (A possible choice is the tent function g(x = max( x,. By an easy computation, ĝ(ξ = 2( cos ξ/ξ 2, so that all assumptions are satisfied. Then the Fourier inversion theorem applies, thus g(x = (2π ĝ(ξe iξx dξ, and in particular we have ĝ(ξdξ = 2πg( = 2π. (This is our only use of Fourier inversion, and it can be avoided at the expense of showing ĝ(ξdξ = 2π by hand, which can be done for the above choice of g, if somewhat arduously. Defining g n (x = g(x/n, it is immediate that ĝ n (ξ = nĝ(nξ and ĝ n = 2π for all n. For every δ > we have ξ δ ĝn(ξ = u nδ ĝ(udu, which tends to zero as n since ĝ n is integrable and non-negative. Thus the normalized functions h n = ĝ n /2π constitute an approximation of unity. (I.e. n fh n f( for sufficiently nice f. Let a, b >. Twofold partial integration gives e ax a cos bx dx =. Applying c db gives a 2 +b 2 ax sin cx e dx = arctan c. Applying f dc gives ax cos fx e dx = f arctan f a log ( + ( f 2. (The exchanges of integration order are justified by absolute convergence and Fubini. Taking f = and the it a we x a x 2 a 2 a obtain cos x dx = π. x 2 2 5

6 Let G be the continuous extension (clearly unique of F to Re(z a. Defining φ(x = e ax f(x, we have for all t R, ε > : G(a + ε + it = F (a + ε + it c ε + it = (φ(x ce (ε+itx dx = ψ ε (xe itx dx, (4.3 where we abbreviate ψ ε (x = e εx (φ(x c to make the following computation readable. Multiplying (4.3 by e iyt g n (t and integrating over t, we have ( e iyt g n (tg(a + ε + itdt = e iyt g n (t ψ ε (xe itx dx dt ( = ψ ε (x g n (te it(y x dt dx = = ψ ε (xĝ n (y xdx e εx φ(xĝ n (y xdx c e εx ĝ n (y xdx. (4.4 (The second equality holds by Fubini s theorem, since ψ ε and g n are integrable. The third identity uses that g n is even. We now consider the it ε of (4.4. Since G by assumption is continuous on Re(z a, we have G(a + ε + it G(a + it, uniformly for t in the compact support of g n. Thus ε e iyt g n (tg(a + ε + itdt = e iyt g n (tg(a + itdt. Since φ and ĝ n are non-negative, the monotone convergence theorem gives ε ε e εx φ(xĝ n (y xdx = e εx ĝ n (y xdx = φ(xĝ n (y xdx, ĝ n (y xdx = y ĝ n (xdx. Thus the ε it of (4.4 is e iyt g n (tg(a + itdt = y φ(xĝ n (y xdx c ĝ n (xdx. (4.5 Since t g n (tg(a + it is continuous and compactly supported, Lemma 4.2 gives e iyt g n (tg(a + itdt =, y + with which the complex numbers leave the stage. Obviously, y ĝ n (xdx = ĝ n (xdx = 2π. y + Thus for y +, (4.5 becomes (in terms of h n = ĝ n /2π y + φ(xh n (y xdx = c. (4.6 It remains to show that this implies x + φ(x = c, using that {h n } is an approximate unit 6

7 and that f(x = e ax φ(x is non-decreasing. For δ > we have φ(x + δ yh n (ydy = f(x + δ ye a(x+δ y h n (ydy δ δ f(x + δ ye a(x+δ y h n (ydy f(xe a(x+2δ h n (ydy δ = φ(xe 2aδ h n (ydy = φ(xe 2aδ I n (δ, where we write I n (δ = δ h n(ydy. In view of (4.6, we have sup φ(x c e2aδ x + I n (δ. Recalling that I n (δ n for each δ >, taking n gives sup x + φ(x ce 2aδ, and for δ we obtain sup x + φ(x c. This also implies that φ is bounded above: φ(x M for some M >. On the other hand, δ φ(x δ yh n (ydy φ(x δ yh n (ydy + M h n (ydy = δ y δ f(x δ ye a(x y h n (ydy + M φ(xe 2aδ I n (δ + M( I n (δ. ( δ h n (ydy This gives With (4.6 we obtain φ(x δ yhn (ydy M( I n (δ φ(x e 2aδ. I n (δ inf x φ(x c M( I n(δ e 2aδ. I n (δ Taking n we find inf x φ(x c, and δ gives inf e 2aδ x φ(x c. We have thus proven φ(x c as x +, which is equivalent to f(x ce ax. 5 Proof of the PNT 5. Theorem (Prime Number Theorem (i ψ(x x as x. (ii With π(x = #(P [, x] we have π(x (iii If p n is the n-th prime, we have p n n log n. x log x. Proof. (i By Lemma 2.5, the integral e zx ψ(e x dx converges for Re(z > and equals ζ (z zζ(z. By Corollary 3.4, the function z ζ (z ζ(z z has a continuous extension to Re(z, thus also ζ (z zζ(z z = z ( ζ (z ζ(z z. In view of Λ, the function ψ(x = n x Λ(n is non-decreasing. Now Theorem 4. gives ψ(e x e x as x, and therefore ψ(x x. (ii We compute ψ(x = Λ(n = log p = log x log p π(x log x. log p n x p x p k x 7

8 If < y < x then π(x π(y = y<p x y<p x Thus π(x y + ψ(x/ log y. Taking y = x/ log 2 x this gives ψ(x x π(x log x ψ(x x x log p log y ψ(x log y. log x log(x/ log 2 x + log x, thus ψ(x π(x log x. Together with (i, this gives π(x x/ log x. (iii Taking logarithms of π(x x/ log x, we have log π(x log x log log x log x and thus π(x log π(x x. Taking x = p n and using π(p n = n gives n log n p n. 6 Comments Should one prefer complex over harmonic analysis (why??, a weak version of Ikehara s Tauberian theorem, sufficient for obtaining the PNT, can be proven using complex analysis, as shown by Newman [4] (and presented efficiently in [9]. The proof is not simpler than the one given here and gives rather less insight. Without too much more work, cf. [], the proof of the PNT given above can be strengthened to the more quantitative statement ( x ψ(x = x + O log n n N, x which implies ( x π(x = Li(x + O log n x n, where Li(x = x 2 dt log t. In order to do so, one needs some estimates on ζ(+it and its derivatives, but no information on ζ(z for Re(z <. In order to prove stronger results on the error ψ(x x (or π(x Li(x one needs information on the non-vanishing of ζ(z for Re(z <. It is known that where α (,, is equivalent to ζ(s + it when s > ψ(x x = O(xe d logβ x, where β = α + <. c log α ( t +, (6. With β = this would become ψ(x x = O(xe d log x = x d. It is known that ζ(s + it when s > γ ψ(x x = O(x γ+ε ε >. Note that for any β, γ (, and n N we have x γ xe and xe d log β x d logβ x x log n x. Since there are (infinitely many t R such that ζ( 2 + it =, the best possible result is ψ(x x = O(x 2 +ε (which then actually implies ψ(x x = O( x log 2 x. However, the best current proven result is (6. with α = 2/3, implying an error estimate of the form O(xe d log3/5 x. 8

9 Most proofs of the PNT do not work directly with π(x, but rather with ψ(x or with the (only superficially simpler function θ(x = p x log p. See however [8, 2] for a deviation from this rule, following a Fourier analytic approach similar to the present one (but more sophisticated due to the use of distributional language and less Tauberian, thus perhaps less conceptual. The approach to the PNT discussed above combines Tauberian theorems that have no numbertheoretic content themselves with the number-theoretic information contained in the nonvanishing of ζ( + it. A beautiful alternative approach due to Landau [, 6] and Ingham [6] uses information about ζ to prove strong Tauberian theorems which, while not looking particularly number theoretical, actually contain the PNT in the sense of implying it with very little additional work, not appealing to the zeta-function again. Similarly to the history of the Landau-Ikehara theorem, Landau s proof used information on ζ stronger than ζ(+it, which Ingham einated using Wiener s theory. However, a simpler early result of Wiener [6], cf. also [2], is sufficient. The latter has much in common with the proof of Ikehara s theorem given here. In 948, 5 years after the first proofs of the PNT, Selberg and Erdös found a proof of the PNT that avoids not only complex analysis, but also harmonic analysis and every use of C. Not much later, Karamata showed that this approach leads to an elementary proof of a version of Ingham s (first Tauberian theorem, sufficient for the PNT. [Despite the fact that different elementary proofs of the PNT have appeared since, this author thinks that the one of Karamata still is the simplest and most conceptual to date. Cf. [3] for an exposition.] For some insights on the relationship between the traditional ζ-based and elementary proofs of the PNT cf. [7], [4, Chap. 3, 4] and [5]. But there still is much room for a better understanding. References [] S. Bochner: Ein Satz von Laudau und Ikehara. Math. Zeit. 37, -9 (933. [2] J.-B. Bost: Le théorème des nombres premiers et la transformation de Fourier. In: N. Berline, C. Sabbah (eds.: La fonction zeta. Ecole Polytechnique, 23. [3] D. Choimet, H. Queffelec: Twelve landmarks of twentieth-century analysis. Cambridge University Press, 25. [4] W. & F. Ellison: Prime numbers. Wiley, Hermann, 985. (Original French version: W. Ellison, M. Mendès France: Les nombres premiers. Hermann, 975. [5] S. Ikehara: An extension of Landau s Theorem in the analytic theory of numbers. Journ. Math. Phys. M.I.T., -2 (93. [6] A. E. Ingham: Some Tauberian theorems connected with the prime number theorem. J. Lond. Math. Soc. 2, 7-8 (945. [7] A. E. Ingham: Review of the papers of Selberg and Erdös. Mathem. Reviews, MR294 (,595c [8] J.-P. Kahane: Une formule de Fourier sur les nombres premiers. Gaz. Math. 67, 3-9 (996. [9] A. Kienast: Über die Unabhängigkeit des Beweises des Primzahlsatzes vom Begriff der analytischen Funktion einer komplexen Variable. Comm. Math. Helv. 8, 3-4 (935 and 9, 4-4 (936 (erratum. [] A. Kienast: Beweis des Satzes x log q x(ψ(x x ohne Überschreitung der Geraden σ =. Math. Zeit. 43, 3-9 (938. [] E. Landau: Handbuch der Lehre von der Verteilung der Primzahlen. Teubner, 99. [2] N. Levinson: On the elementary character of Wiener s general tauberian theorem. J. Math. Anal. Appl. 42, (973. 9

10 [3] M. Müger: On Karamata s proof of the Landau-Ingham Tauberian theorem. L Ens. Math. 6, (25. [4] D. J. Newman: Simple analytic proof of the prime number theorem. Amer. Math. Monthly 87, (98. [5] T. Tao: Fourier-analytic proofs of the prime number theorem. tao/preprints/expository/prime.dvi [6] N. Wiener: A new method in Tauberian theorems. Journ. Math. Phys. M.I.T. 7, 6-84 (928. [7] N. Wiener: Tauberian theorems. Ann. Math. 33, - (932. [8] N. Wiener: The Fourier integral and certain of its applications. Cambridge University Press, 933. [9] D. Zagier: Newman s short proof of the prime number theorem. Amer. Math. Monthly 4, (997.