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The Fejer kernel is a positive approximate identity

Statement

For every N≥0, the Fejer kernel satisfies

FN(t)=1N+1∣∑j=0Nej(t)∣2.

Hence, for t∉Z,

FN(t)=1N+1(sin⁡((N+1)πt)sin⁡(πt))2,

so FN(t)≥0 for all t, ∫01FN(t) dt=1, and for every δ∈(0,1/2],

∫δ1−δFN(t) dt≤1(N+1)sin⁡2(πδ).

In particular,

∫δ1−δFN(t) dt⟶0(N→∞).

Facts & Assumptions

Given: An integer N≥0 and a real δ∈(0,1/2].

[L1]

The Fejer kernel is FN(t)=1N+1∑m=0NDm(t), where Dm(t)=∑∣k∣≤mek(t) and ∫01Dm(t) dt=1 (Dirichlet and Fejer kernels).

Proof

technique · direct
1.1L1algebra

Expanding the average in [L1] gives ∑m=0NDm(t)=∑m=0N∑∣k∣≤mek(t)=∑∣k∣≤N(N+1−∣k∣)ek(t). On the other hand, ∣∑j=0Nej(t)∣2=∑j=0N∑ℓ=0Nej−ℓ(t)=∑∣k∣≤N(N+1−∣k∣)ek(t). Therefore FN(t)=1N+1∣∑j=0Nej(t)∣2.

2.1L1step 1.1algebra

If t∉Z, the finite geometric-series formula gives ∑j=0Nej(t)=∑j=0Ne2πijt=eπiNtsin⁡((N+1)πt)sin⁡(πt), so step 1.1 yields the displayed square formula. This proves FN(t)≥0 for every t∉Z, and at integers the same formula extends by continuity to FN(0)=N+1. Also [L1] gives ∫01FN(t) dt=1N+1∑m=0N∫01Dm(t) dt=1.

3.1step 2.1algebra∎

For t∈[δ,1−δ], one has ∣sin⁡(πt)∣≥sin⁡(πδ). Using step 2.1 and ∣sin⁡((N+1)πt)∣≤1 therefore gives FN(t)≤1(N+1)sin⁡2(πδ). Integrating over an interval of length at most 1 yields ∫δ1−δFN(t) dt≤1(N+1)sin⁡2(πδ)→0.

Depends on

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Sources