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单词 ProofOfInghamInequality
释义

proof of Ingham Inequality


Let

f(x)=12πj=-nncjeitjx.

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

K(u)=-k(x)eixu𝑑x.

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:

k(x)={cos(x/2),|x|π0,|x|>π

Then, after computation, we have

K(u)=4cos(πu)1-4u2.

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
4γ2jh14(j-h)2-1<8γ2r=114r2-1=
=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

-ππ|f(x)|2𝑑xj|cj|2K(0)(1-1γ2),

and we have obtained the conclusion.

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更新时间:2025/5/4 3:43:11