TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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$∥ τ_{h} f - f ∥_{p} \to 0$ in $L^{p} (R^{n})$ as $h \to 0$ , for $1 \leq p < \infty$

Statement

Assume the Axiom of Countable Choice.

Let $1 \leq p < \infty$ and $f \in L^{p} (R^{n})$ . Then

$∥ τ_{h} f - f ∥_{p} ⟶ 0 (h \to 0) .$

Facts & Assumptions

Given: The Axiom of Countable Choice, $1 \leq p < \infty$ , $f \in L^{p} (R^{n})$ , and $ε > 0$ .

[L1]

$C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ ( $C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ ).

[L2]

Compactly supported continuous functions are translation-continuous in $L^{p}$ (Continuous compactly supported functions are translation-continuous in $L^{p}$ ).

[L3]

Lebesgue measure is translation invariant, so $∥ τ_{h} u ∥_{p} = ∥ u ∥_{p}$ (Translation of a function on $R^{n}$ , Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[L4]

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

Proof

technique · direct

1.1L1L2givenchoose

By [L1], choose $g \in C_{c} (R^{n})$ with [L1, L2, given, choose] $∥ f - g ∥_{p} < ε / 3$ . By [L2], choose $δ > 0$ such that $∣ h ∣ < δ$ implies $∥ τ_{h} g - g ∥_{p} < ε / 3$ .

2.1L3L4step 1.1algebra

For $∣ h ∣ < δ$ , [L3] and [L4] give [L3, L4, step 1.1, algebra] $∥ τ_{h} f - f ∥_{p} \leq ∥ τ_{h} (f - g) ∥_{p} + ∥ τ_{h} g - g ∥_{p} + ∥ g - f ∥_{p} = 2 ∥ f - g ∥_{p} + ∥ τ_{h} g - g ∥_{p} < ε .$

3.1step 2.1∎

Since $ε > 0$ was arbitrary, $∥ τ_{h} f - f ∥_{p} \to 0$ as [step 2.1] $h \to 0$ .

$∥ τ_{h} f - f ∥_{p} \to 0$ in $L^{p} (R^{n})$ as $h \to 0$ , for $1 \leq p < \infty$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

∥τhf−f∥p→0∥τh​f−f∥p​→0 in Lp(Rn)Lp(Rn) as h→0h→0, for 1≤p<∞1≤p<∞

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

$∥ τ_{h} f - f ∥_{p} \to 0$ in $L^{p} (R^{n})$ as $h \to 0$ , for $1 \leq p < \infty$