On Montgomery’s pair correlation conjecture: A tale of three integrals

B.2.1 Existence of extremizers

The first step is to show that there exists f ∈ ℬ 2 ⁢ ( π ) that extremizes (B.2) (i.e. such that ∥ f ∥ L 2 ⁢ ( d ⁢ μ ) / ∥ f ∥ 2 = 𝐃 ). As we have argued in (B.2) and the remark thereafter, it is enough to find an extremizer in the class 𝒜 0 defined in Section 2.1 for

Let { g n } n ≥ 1 ⊂ 𝒜 0 be an extremizing sequence for (B.4), normalized so that ∥ g n ∥ 1 = 1 for all n. Hence,

as n → ∞ . Recall that supp ⁡ ( g n ^ ) ⊂ [ - 1 , 1 ] and that ∥ g n ^ ∥ ∞ = g n ^ ⁢ ( 0 ) = ∥ g n ∥ 1 = 1 . Therefore ∥ g n ^ ∥ 2 2 ≤ 2 ⁢ ∥ g n ^ ∥ ∞ 2 ≤ 2 , and we see that { g n } n ≥ 1 is a bounded sequence in ℬ 2 ⁢ ( 2 ⁢ π ) . By reflexivity, passing to a subsequence if necessary, we may assume that g n converges weakly to a certain g ♯ ∈ ℬ 2 ⁢ ( 2 ⁢ π ) . In particular,

and hence g ♯ ≠ 0 . Since ℬ 2 ⁢ ( 2 ⁢ π ) is a reproducing kernel Hilbert space, we also have the pointwise convergence

for all y ∈ ℝ . Hence g ♯ is even and non-negative on ℝ . Moreover, by Fatou’s lemma, it follows that

which implies that g ♯ ∈ 𝒜 0 . From (B.5) and (B.6), we see that this particular g ♯ is an extremizer for (B.4).

B.2.2 Solving the Euler–Lagrange equation

For a generic 0 ≠ h ∈ ℬ 2 ⁢ ( π ) let us write

For instance, for h ⁢ ( x ) = sin ⁡ π ⁢ x π ⁢ x , we have Φ ⁢ ( h ) = 2 3 . Let 0 ≠ f ∈ ℬ 2 ⁢ ( π ) be a maximizer for (B.7), normalized so that ∥ f ∥ 2 = 1 . That is,

In what follows let us write K 1 ⁢ ( x ) = ( sin ⁡ π ⁢ x π ⁢ x ) 2 , recalling our notation (2.23). For any function h ∈ ℬ 2 ⁢ ( π ) with ∥ h ∥ 2 = 1 and h ⊥ f , we have Φ ⁢ ( f + ε ⁢ h ) ≤ Φ ⁢ ( f ) for any ε ∈ ℝ , with equality if ε = 0 . Therefore

Similarly, for ε ∈ ℝ ,

We then conclude that

Since this holds for any function h with h ⊥ f in ℬ 2 ⁢ ( π ) , the function f ^ must verify the following Euler–Lagrange equation:

as functions in L 2 ⁢ [ - 1 2 , 1 2 ] , for some η ∈ ℂ . At this point observe that (B.8) yields

Hence η ∈ ℝ and we have seen that 1 > η ≥ 2 3 .

Since the left-hand side of (B.8) is continuous in [ - 1 2 , 1 2 ] , we may assume that f ^ is continuous in [ - 1 2 , 1 2 ] and hence

for all α ∈ [ - 1 2 , 1 2 ] . Since K 1 ^ is a Lipschitz function, the integral in (B.9) (as a function of α) is differentiable for all α ∈ ( - 1 2 , 1 2 ) . Recalling that ( K 1 ^ ) ′ ⁢ ( α ) = χ ( - 1 , 0 ) ⁢ ( α ) - χ ( 0 , 1 ) ⁢ ( α ) , and that supp ⁢ ( f ^ ) ⊂ [ - 1 2 , 1 2 ] , we have

The left-hand side of (B.10) is again differentiable in α, and an application of the fundamental theorem of calculus now yields

The general solution of this linear differential equation is

where A , B ∈ ℂ . Plugging this back into (B.10), we find the relation

Since η ≥ 2 3 > 2 π 2 , we have cos ⁡ ( 1 2 ⁢ η ) > 0 and therefore A = B ≠ 0 . Evaluating (B.9) at α = 0 , we arrive at the condition

that determines our η uniquely ( η = 0.67551 ⁢ … ). Finally, the normalization ∥ f ^ ∥ L 2 ⁢ [ - 1 2 , 1 2 ] = 1 together with (B.11) yields the value | A | = ( ( 2 ⁢ η + 1 ) / ( 8 ⁢ η + 2 ) ) 1 2 .

In sum, our extremal function is unique (up to multiplication by a complex number) and its Fourier transform, with the substitution θ = ( 2 ⁢ π 2 ⁢ η ) - 1 2 , is given by

which, by Fourier inversion, leads us to (B.3).