Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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τhffp0 in Lp(Rn) as h0, for 1p<

Statement

Assume the Axiom of Countable Choice.

Let 1p< and fLp(Rn). Then

τhffp0(h0).

Facts & Assumptions

Given: The Axiom of Countable Choice, 1p<, fLp(Rn), and ε>0.

[L1]

Cc(Rn) is dense in Lp(Rn) (Cc(Rn) is dense in Lp(Rn) for 1p<).

[L2]

Compactly supported continuous functions are translation-continuous in Lp (Continuous compactly supported functions are translation-continuous in Lp).

[L3]

Lebesgue measure is translation invariant, so τhup=up (Translation of a function on Rn, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[L4]

Minkowski's inequality is available (Minkowski's inequality for integrals, including p=).

Proof

technique · direct
1.1

By [L1], choose gCc(Rn) with [L1, L2, given, choose] fgp<ε/3. By [L2], choose δ>0 such that h<δ implies τhggp<ε/3.

L1L2givenchoose
2.1

For h<δ, [L3] and [L4] give [L3, L4, step 1.1, algebra] τhffpτh(fg)p+τhggp+gfp=2fgp+τhggp<ε.

L3L4step 1.1algebra
3.1

Since ε>0 was arbitrary, τhffp0 as [step 2.1] h0.

step 2.1

Depends on

Used by

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