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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Every L1 approximate identity converges to the identity in Lp for 1p<

Statement

Assume the Axiom of Countable Choice.

Let (Kε)ε>0 be an L1 approximate identity on Rn. If 1p< and fLp(Rn), then

fKεfp0(ε0+).

Facts & Assumptions

Given: The Axiom of Countable Choice, an L1 approximate identity, an exponent 1p<, and fLp(Rn).

[L1]

Approximate identities are defined in An L1 approximate identity on Rn.

[L3]

Minkowski's integral inequality and Young's inequality are available (Minkowski's integral inequality, Young's convolution inequality).

Proof

technique · direct
1.1

Because Kε=1, one may write [L1, L3, given, algebra] fKεf=RnKε(y)(τyff)dy. Applying [L3] gives fKεfpKε(y)τyffpdy.

L1L3givenalgebra
2.1

Let η>0 be arbitrary. If fp=0, then f=0 in Lp and step [L1, L2, L3, step 1.1, choose, algebra] 1.1 gives fKεfp=0<η for every ε>0. Assume now fp>0, and put M:=supεKε1<. By [L2], choose δ>0 so that τyffp<η/(2M) whenever y<δ. By [L1], choose ε0>0 so that yδKε(y)dy<η4fp(0<ε<ε0). For such ε, split the integral from step 1.1 into y<δ and yδ. The near part is at most η/2. For the far part, [L3] gives τyffp2fp, so yδKε(y)τyffpdy2fpyδKε(y)dy<η/2. Hence fKεfp<η whenever 0<ε<ε0.

L1L2L3step 1.1choosealgebra
3.1

Because η>0 was arbitrary, fKεfp0 as [step 2.1] ε0+, proving the convergence in Lp.

step 2.1

Depends on

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