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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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The Fejer kernel is a positive approximate identity

Statement

For every N0, the Fejer kernel satisfies

FN(t)=1N+1j=0Nej(t)2.

Hence, for tZ,

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],

δ1δFN(t)dt1(N+1)sin2(πδ).

In particular,

δ1δFN(t)dt0(N).

Facts & Assumptions

Given: An integer N0 and a real δ(0,1/2].

[L1]

The Fejer kernel is FN(t)=1N+1m=0NDm(t), where Dm(t)=kmek(t) and 01Dm(t)dt=1 (Dirichlet and Fejer kernels).

Proof

technique · direct
1.1

Expanding the average in [L1] gives m=0NDm(t)=m=0Nkmek(t)=kN(N+1k)ek(t). On the other hand, j=0Nej(t)2=j=0N=0Nej(t)=kN(N+1k)ek(t). Therefore FN(t)=1N+1j=0Nej(t)2.

L1algebra
2.1

If tZ, 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 tZ, and at integers the same formula extends by continuity to FN(0)=N+1. Also [L1] gives 01FN(t)dt=1N+1m=0N01Dm(t)dt=1.

L1step 1.1algebra
3.1

For t[δ,1δ], one has sin(πt)sin(πδ). Using step 2.1 and sin((N+1)πt)1 therefore gives FN(t)1(N+1)sin2(πδ). Integrating over an interval of length at most 1 yields δ1δFN(t)dt1(N+1)sin2(πδ)0.

step 2.1algebra

Depends on

Used by

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