Advanced Number Theory Note #6: Reformulation of the prime number theorem using the Möbius function 7 August 2012 at 00:43

Transcription

1 Advanced Number Theory Note #6: Reformulation of the prime number theorem using the Möbius function 7 August 2012 at 00:43 Public One of the more intricate and tricky proofs concerning reformulations of the prime number theorem is the one showing that the prime number theorem is equivalent to a certain asymptotic formula involving the Möbius function. In this note, which is intended mainly as a technical memo for myself, I provide a detailed 'blow-by-blow' account of this proof. Specifically, for x 1, define the partial sum function of the Möbius function as Then the task is to prove that the prime number theorem is equivalent to The process will be in three stages. First, it is necessary to relate M(x) to another weighted average of the Möbius function. Next, this result will be used to prove that the prime number theorem implies (1). Finally, the proof will be completed by proving that (1) implies the prime number theorem

2 Relating M(x) to another weighted average of μ(n) For x 1, define Then we can show that if one of M(x)/x or H(x)/xlogx tends to a limit, so does the other, and the two limits must be equal. That is, we have To see this, we use Abel's identity in Advanced Number Theory Note #4, equation (6), which says where A(x) = n x a(n) and a(n) is any arithmetical function. In the present context, we set a(n) = μ(n) A(x) = M(x) f(t) = logt and observe that the term in Abel's identity corresponding to A(y)f(y) is zero with y = 1, since log1 = 0. Then we have

3 Dividing through by xlogx (valid if x > 1) and rearranging we get Therefore to prove (2) we need to show that But we have and thus

4 Therefore we have from which we get (3) and hence (2). QED The prime number theorem implies (1) To prove that the prime number theorem implies (1), we use the prime number theorem in the form and show that this implies The result will then follow from (2). In order to achieve this, we first need to establish the identity

5 where is Chebyshev's ψ-function. To prove (4), we use the result obtained in Advanced Number Theory Note #1 (section on the Mangoldt function). We apply Möbius inversion to this, discussed in Advanced Number Theory Note #1 (section on the Möbius function). Möbius inversion says f = g u if and only if g = f μ or, writing these in full, f(n) = d n g(d) if and only if g(n) = d n f(d)μ(n/d) Thus, we get

6 Now we need to sum this over all n x using the formula for the partial sums of a Dirichlet convolution in Advanced Number Theory Note #3 (equation 2). This says that for h = f g, Therefore taking f = μ, g = Λ, h = μ Λ, and G(x/n) = ψ(x/n), we get (4). // Given ψ(x) ~ x, and also some ϵ > 0, there is a constant A > 0 such that

7 This is equivalent to saying that we have Choose x > A and split the sum on the right of (4) into two parts, where y = [x/a]. In the first sum we have n y n x/a x/n A Therefore we can use (5) to write Then we have

8 and so

9 where the term circled in red follows from Asymptotic formula 1 in Advanced Number Theory Note #2. In the second sum we have y < n x n y + 1 (since y = [x/a] must be an integer). Therefore

10 because The inequality (x/n) < A implies that ψ(x/n) ψ(a). Therefore the second sum is dominated by xψ(a) Therefore if ϵ < 1, the full sum on the right hand side of (4) is dominated by In other words, given any ϵ such that 0 < ϵ < 1, we have or Note that the first term on the right of this inequality is decreasing in x, so we can choose B > A such that (2 + ψ(a))/logx < ϵ when x > B. Then for this x > B we have

11 which shows that H(x)/xlogx 0 as x. Therefore it follows from (2) that the prime number theorem (in the form ψ(x) ~ x) implies (1). QED The asymptotic relation (1) implies the prime number theorem We prove this by showing that implies (Here, the 'little oh' notation has its usual meaning, so (7) means the same as (1)). To prove this, we first show that the Chebyshev ψ-function obeys the formula

12 and then use (7) to show that the sum in (8) is o(x) as x. The function f in (8) is given by where C is Euler's constant and σ(n) is the number of divisors of n. To obtain (8) we begin with the identities We now express each of the summands in these identities as a Dirichlet convolution involving the Möbius function, as follows:

13 (This follows from the Möbius inversion formula. We have σ = u u, so u = μ σ). (This was obtained in Advanced Number Theory Note #1, the section on the Mangoldt function). (This was obtained in Advanced Number Theory Note #1, the section on the Möbius function). With these, we can now write

14 where the third equality follows by noting that d n n = qd. This, upon rearranging, gives (8). Therefore the proof will be complete if we can show that

15 For this purpose we can use the general identity for the partial sums of a Dirichlet convolution obtained in Advanced Number Theory Note #3, equation (12). This result involved the functions

16

17 We can use this result in the present context to write where the first equality follows from the commutativity of Dirichlet convolution. We next show that F(x) = O( x) by using equation (9) in Advanced Number Theory Note #2, which says We use this together with the relation

18 where the right hand side is from equation (9) in Advanced Number Theory Note #3. Using these two results we get where the last two terms on the third line follow from the fact that n x 1 = [x] = x + O(1) Since F(x) = O( x), there is some constant B > 0 such that

19 Using this in the second sum on the right hand side of (10) we get for some constant A > B > 0. Now let ϵ > 0 be arbitrary and choose a > 1 to be such that Then (11) becomes

20 for all x 1. (Note that a depends on ϵ and not on x). Since M(x) = o(x) as x, for the same ϵ there exists some c > 0 (again depending only on ϵ) such that where K is any positive number. (K will be specified shortly). The first sum on the right of (10) satisfies provided that x/n > c for all n a. Therefore (13) holds if x > ac (because n a, so x/n > c is guaranteed if x/a > c). Now take

21 Then (13) implies The last term on the right of (10) is dominated by since ab = x. Combining this with (12) and (14), we conclude that (10) implies provided that x > ac, where a and c depend only on ϵ. This proves (9), which completes the overall proof. QED