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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Cc∞(Rn) is dense in Lp(Rn) for 1≤p<∞

Statement

Assume the Axiom of Countable Choice.

Let 1≤p<∞. Then Cc∞(Rn) is dense in Lp(Rn).

Facts & Assumptions

Given: The Axiom of Countable Choice, 1≤p<∞, ε>0, and f∈Lp(Rn).

[L1]

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

[L2]

Mollifier families are L1 approximate identities, and convolution with a mollifier is smooth (A unit-mass smooth bump generates an L1 approximate identity, Convolution with a mollifier is smooth, and derivatives pass under the integral sign).

[L3]

Approximate identities converge in Lp, and for bounded continuous functions the convergence is uniform on compacta (Every L1 approximate identity converges to the identity in Lp for 1≤p<∞, L1 approximate identities converge uniformly on compacta for bounded continuous functions).

[L4]

The support of a convolution lies in the closure of the support sumset (The support of a convolution lies in the closure of the support sumset).

Proof

technique · direct
1.1L1L2givenchoose

By [L1], choose g∈Cc(Rn) with ∥f−g∥p<ε/2. [L1, L2, given, choose] Let (φδ) be a mollifier family as in [L2].

2.1L2L3L4step 1.1choose

By [L3], choose δ>0 so small that [L2, L3, L4, step 1.1, choose] ∥g∗φδ−g∥p<ε/2. The function g∗φδ is smooth by [L2]. Because g has compact support and φδ has compact support, [L4] gives compact support for g∗φδ, so g∗φδ∈Cc∞(Rn).

3.1step 1.1step 2.1algebra∎

Therefore [step 1.1, step 2.1, algebra] ∥f−g∗φδ∥p≤∥f−g∥p+∥g−g∗φδ∥p<ε. Hence Cc∞(Rn) is dense in Lp(Rn).

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Sources