proof of Ingham Inequality



If k(x) is a integrable function in (-,), let


It is easy to prove the equalities.

-k(x)|f(x)|2𝑑x = j,h=-nncjch¯K(tj-th).
-k(x)f(x)e-ixth𝑑x = j=-nncjK(tj-th).

For the rest of the proof we make the following choices:


Then, after computation, we have


Let T=π and γ=1+δ (δ=ε/π>0). Since |tj+1-tj|γ, we have that |tj-th||j-h|γ. If we suppose that h is fixed, we have

jh|K(tj-th)| = jh|4cosπ(tj-th)1-4(tj-th)2|jh44(j-h)2γ2-1
= 82γ2r=1(12r-1-12r+1)=4γ2=K(0)γ2.

But, since 2|cjch¯||cj|2+|ch|2 and K(tj-th)=K(th-tj), we find

-k(x)|f(x)|2𝑑x = j,h=-nncjch¯K(tj-th)=
= j|cj|2K(0)+j,h;jh𝒪(j,h)|cj|2+|ch|22|K(tj-th)|
= j|cj|2(K(0)+hj𝒪(j,h)|K(tj-th)|),

where |𝒪(j,h)|1. Using the definition of k and the previous inequality, we have


and we have obtained the conclusion.

Title proof of Ingham Inequality
Canonical name ProofOfInghamInequality
Date of creation 2013-03-22 15:55:18
Last modified on 2013-03-22 15:55:18
Owner ncrom (8997)
Last modified by ncrom (8997)
Numerical id 8
Author ncrom (8997)
Entry type Proof
Classification msc 42B05