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

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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.

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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≤xa(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

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

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

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|ng(d) if and only if g(n) = ∑d|nf(d)μ(n/d) Thus, we get

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

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

and so

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

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

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

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:

(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

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

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

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

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≤x1 = [x] = x + O(1) Since F(x) = O(√x), there is some constant B > 0 such that

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

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

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

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