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✓ 29 results · all verified · 3 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 26 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Density Separability and Convolution in $L^{p}$

1 · Prerequisites

2 · Summary

This page keeps the finite- $p$ density and translation theory separate from the $L^{\infty}$ endpoint failures, then builds the Euclidean convolution and mollifier package on that exact ledger. The route stays local to $R^{n}$ : explicit cutoffs replace abstract Urysohn, the separability theorem keeps the countably generated and sigma-finite hypotheses explicit, and the Borel-representative seam for convolution is recorded instead of being hidden.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

Translation of a function on $R^{n}$

Definition

Let $h \in R^{n}$ and let $f : R^{n} \to C$ be a function. The translate of $f$ by $h$ is the function $τ_{h} f$ defined by

(τ_{h} f) (x) : = f (x - h) (x \in R^{n}) .

This fixes the sign convention used throughout the page: translating by $h$ shifts the graph of $f$ in the positive $h$ -direction.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$

Definition

For $n \geq 1$ , let $C_{c} (R^{n})$ be the vector space of continuous functions $f : R^{n} \to R$ with compact support (The support of a function on $R^{n}$ and its compactly supported Riemann integral, Vector-valued functions $f : A \to R^{m}$ , their limits and continuity, with the dictionary to the metric notions).

Let $C_{c}^{\infty} (R^{n})$ be the subspace of those $f \in C_{c} (R^{n})$ that are smooth in the Euclidean multi-index sense ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-01Open item page →

The space $C_{0} (R^{n})$ of continuous functions vanishing at infinity

Definition

Let $C_{0} (R^{n})$ be the space of continuous functions $f : R^{n} \to R$ such that for every $ε > 0$ there is a compact set $K \subseteq R^{n}$ with

∣ f (x) ∣ < ε (x \notin K) .

Equivalently, every $ε$ -tail of $f$ is eventually small outside a large Euclidean ball; this is the correct $L^{\infty}$ -closure target of $C_{c} (R^{n})$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

Convolution of two functions on $R^{n}$

Definition

Let $f, g : R^{n} \to C$ be measurable functions. When, for a given $x \in R^{n}$ , the function

y ⟼ f (x - y) g (y)

is measurable and integrable on $R^{n}$ , the convolution of $f$ and $g$ at $x$ is

(f * g) (x) : = \int_{R^{n}} f (x - y) g (y) d y .

For $L^{1}$ classes, later items make the representative convention explicit.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

An $L^{1}$ approximate identity on $R^{n}$

Definition

An $L^{1}$ approximate identity on $R^{n}$ is a family $(K_{ε})_{ε > 0}$ of functions in $L^{1} (R^{n})$ , where

$∥ K_{ε} ∥_{1} : = \int_{R^{n}} ∣ K_{ε} (x) ∣ d x,$

such that:

$\int_{R^{n}} K_{ε} (x) d x = 1$ for every $ε > 0$ ;
there is $M < \infty$ with $∥ K_{ε} ∥_{1} \leq M$ for every $ε > 0$ ;
for every $δ > 0$ , $\int_{∣ x ∣ > δ} ∣ K_{ε} (x) ∣ d x ⟶ 0 (ε \to 0^{+}) .$

This is the general notion used later for both compactly supported mollifiers and the Gaussian family.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The mollifier family generated by a unit-mass smooth bump

Definition

Let $φ \in C_{c}^{\infty} (R^{n})$ satisfy

$\int_{R^{n}} φ (x) d x = 1.$

For $ε > 0$ , define

$φ_{ε} (x) : = ε^{- n} φ (x / ε) (x \in R^{n}) .$

The family $(φ_{ε})_{ε > 0}$ is the mollifier family generated by $φ$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Simple functions with finite-measure support are dense in $L^{p} (μ)$ for $1 \leq p < \infty$

Statement

Let $(X, A, μ)$ be a measure space and let $1 \leq p < \infty$ . Then every element of $L^{p} (μ)$ can be approximated in $L^{p}$ by simple functions whose supports have finite measure.

Facts & Assumptions

Given: A measure space, an exponent $1 \leq p < \infty$ , and $f \in L^{p} (μ)$ .

[L1]

Every measurable function admits dominated simple approximations (Every measurable function admits simple approximations dominated by its absolute value).

[L2]

Dominated convergence applies in $L^{1}$ (Dominated convergence).

[L3]

Elements of $L^{p} (μ)$ are almost-everywhere classes, so one may choose a measurable representative when making pointwise constructions (The space $L^{p} (μ)$ as the quotient by null functions).

Proof

technique · direct

1.1

Choose a measurable representative $u$ of $f$ by [L3]. For each [L1, L3, given, choose, construct] $m \in N_{\geq 1}$ , [L1] gives a simple function $s_{m}$ with $∣ s_{m} ∣ \leq ∣ u ∣$ and $s_{m} (x) \to u (x)$ for every $x$ . Define

t_{m} : = s_{m} 1_{{∣ u ∣ \geq 1 / m}} .

Then each $t_{m}$ is simple and $supp (t_{m}) \subseteq {∣ u ∣ \geq 1 / m}$ .

L1L3givenchooseconstruct

2.1

Since $∣ u ∣^{p}$ is integrable, [step 1.1, algebra]

μ ({∣ u ∣ \geq 1 / m}) \leq m^{p} \int ∣ u ∣^{p} d μ < \infty,

so every $t_{m}$ has finite-measure support. Also $t_{m} (x) \to u (x)$ for every $x$ : if $u (x) = 0$ then eventually both sides are $0$ , and if $u (x) \neq 0$ then $1_{{∣ u (x) ∣ \geq 1 / m}} = 1$ for all large $m$ . Finally

∣ u - t_{m} ∣^{p} \leq (2 ∣ u ∣)^{p},

whose right-hand side is integrable.

step 1.1algebra

3.1

Applying [L2] to $∣ u - t_{m} ∣^{p}$ gives [L2, step 2.1]

∥ f - t_{m} ∥_{p}^{p} = \int ∣ u - t_{m} ∣^{p} d μ ⟶ 0.

So the simple functions with finite-measure support are dense in $L^{p} (μ)$ .

L2step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Simple functions are dense in $L^{\infty} (μ)$ in the essential-supremum norm

Statement

Let $(X, A, μ)$ be a measure space. Simple functions are dense in $L^{\infty} (μ)$ for the essential-supremum norm.

Facts & Assumptions

Given: A measure space, $ε > 0$ , and $f \in L^{\infty} (μ)$ .

[L1]

Choose a representative of the $L^{\infty}$ class and interpret its norm by the least essential bound (The space $L^{\infty} (μ)$ of essentially bounded measurable functions, The essential supremum is attained as the least essential bound).

Proof

technique · direct

1.1

Choose a measurable representative $u$ of $f$ and let $M : = ∥ f ∥_{\infty}$ . [L1, given, choose, construct] By [L1], $∣ u (x) ∣ \leq M$ almost everywhere. Partition the interval $[- M, M]$ into finitely many subintervals of length at most $ε$ , and on each strip $u^{- 1} (I_{j})$ choose one value $c_{j} \in I_{j}$ . The resulting function

s : = \sum_{j} c_{j} 1_{u^{- 1} (I_{j})}

is simple.

L1givenchooseconstruct

2.1

On the full-measure set where $∣ u ∣ \leq M$ , the values $u (x)$ and $s (x)$ lie [step 1.1, algebra] in the same interval $I_{j}$ , so $∣ u (x) - s (x) ∣ \leq ε$ . Hence

∥ f - s ∥_{\infty} \leq ε .

step 1.1algebra

3.1

Since $ε > 0$ was arbitrary, simple functions are dense in [step 2.1] $L^{\infty} (μ)$ for the essential-supremum norm.

step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

A finite-measure measurable set in $R^{n}$ is approximable in measure by a finite union of boxes

Statement

Assume the Axiom of Countable Choice.

Let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < \infty$ . For every $ε > 0$ there is a finite union of boxes $B$ such that

λ_{n} (E △ B) < ε .

Facts & Assumptions

Given: The Axiom of Countable Choice, a natural number $n \geq 1$ , a Lebesgue measurable set $E \subseteq R^{n}$ with finite measure, and $ε > 0$ .

[L1]

Outer regularity gives an open set $U \supseteq E$ with $λ_{n} (U ∖ E) < ε / 2$ (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of $R^{n}$ is the infimum of the measures of the open sets containing it).

[L2]

Every open subset of $R^{n}$ is a countable disjoint union of dyadic cubes (Every open subset of $R^{n}$ is the union of a countable pairwise disjoint family of dyadic cubes).

[L3]

Continuity from below applies to increasing unions of measurable sets (Continuity from below for measures).

[L4]

Measure is monotone and $λ_{n} (U ∖ F) = λ_{n} (U) - λ_{n} (F)$ when $F \subseteq U$ and $λ_{n} (U) < \infty$ (Measures are monotone, Measure of a set difference when the smaller set has finite measure).

Proof

technique · direct

1.1

By [L1], choose an open set $U \supseteq E$ with [L1, L4, given, choose] $λ_{n} (U ∖ E) < ε / 2$ . Because $λ_{n} (E) < \infty$ , monotonicity gives $λ_{n} (U) \leq λ_{n} (E) + λ_{n} (U ∖ E) < \infty .$

L1L4givenchoose

1.2

Write $U = ⨆_{k \geq 1} Q_{k}$ as a countable pairwise disjoint union [L2, L3, L4, choose] of dyadic cubes by [L2], and set $B_{m} : = ⋃_{k = 1}^{m} Q_{k}$ . Then $B_{m} ↑ U$ , so [L3] gives $λ_{n} (B_{m}) \to λ_{n} (U)$ . Hence for some $m$ , $λ_{n} (U ∖ B_{m}) = λ_{n} (U) - λ_{n} (B_{m}) < ε / 2.$

L2L3L4choose

2.1

Put $B : = B_{m}$ . Since $E \subseteq U$ and $B \subseteq U$ , [step 1.1, step 1.2, L4, algebra] $E △ B \subseteq (U ∖ E) \cup (U ∖ B),$ so $λ_{n} (E △ B) \leq λ_{n} (U ∖ E) + λ_{n} (U ∖ B) < ε .$ The set $B$ is a finite union of boxes because each dyadic cube is a box.

step 1.1step 1.2L4algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

Finite linear combinations of box indicators are dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$

Statement

Assume the Axiom of Countable Choice.

Let $1 \leq p < \infty$ . Finite linear combinations of indicator functions of boxes are dense in $L^{p} (R^{n})$ .

Facts & Assumptions

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

[L1]

Simple functions with finite-measure support are dense in $L^{p}$ (Simple functions with finite-measure support are dense in $L^{p} (μ)$ for $1 \leq p < \infty$ ).

[L2]

Every finite-measure measurable set can be approximated in symmetric difference by a finite union of boxes (A finite-measure measurable set in $R^{n}$ is approximable in measure by a finite union of boxes).

[L3]

Minkowski's inequality is available in $L^{p}$ (Minkowski's inequality for integrals, including $p = \infty$ ).

Proof

technique · direct

1.1

By [L1], choose a simple function [L1, L2, given, choose] $s = \sum_{j = 1}^{m} a_{j} 1_{E_{j}}$ with each $E_{j}$ of finite measure and $∥ f - s ∥_{p} < ε / 2$ . For each $j$ , choose a finite union of boxes $B_{j}$ with $λ_{n} (E_{j} △ B_{j}) < {(\frac{ε}{2 m (1 + \sum_{j = 1}^{m} ∣ a_{j} ∣)})}^{p}$ by [L2].

L1L2givenchoose

2.1

Put [L3, step 1.1, algebra] $t : = \sum_{j = 1}^{m} a_{j} 1_{B_{j}} .$ Then $∥ s - t ∥_{p} \leq \sum_{j = 1}^{m} ∣ a_{j} ∣ ∥ 1_{E_{j}} - 1_{B_{j}} ∥_{p} = \sum_{j = 1}^{m} ∣ a_{j} ∣ λ_{n} (E_{j} △ B_{j})^{1 / p} < ε / 2$ by [L3] and the choice of the $B_{j}$ .

L3step 1.1algebra

3.1

Therefore [step 1.1, step 2.1, algebra] $∥ f - t ∥_{p} \leq ∥ f - s ∥_{p} + ∥ s - t ∥_{p} < ε .$ Since $t$ is a finite linear combination of box indicators, these functions are dense in $L^{p} (R^{n})$ .

step 1.1step 2.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

A compact set inside a bounded open set admits an explicit compactly supported continuous cutoff

Statement

Let $K \subseteq O \subseteq R^{n}$ , where $K$ is compact and $O$ is bounded and open. Then there is $η \in C_{c} (R^{n})$ such that $0 \leq η \leq 1$ , $η = 1$ on $K$ , and $supp (η) \subseteq O$ .

Facts & Assumptions

Given: A compact set $K$ and a bounded open set $O$ with $K \subseteq O$ .

[L1]

The distance-to-set map is Lipschitz ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz).

[L2]

In $R^{n}$ , compact sets are exactly closed and bounded sets, and a continuous real-valued function on a nonempty compact metric space attains a minimum (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L3]

Compact support is defined by the closure of the nonzero set (The support of a function on $R^{n}$ and its compactly supported Riemann integral, The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$ ).

Proof

technique · constructive

1.1

If $K = \emptyset$ , the zero function belongs to $C_{c} (R^{n})$ and [L3, given, algebra] already satisfies the conclusion. So assume from now on that $K$ is nonempty.

L3givenalgebra

1.2

Let $u (x) : = d (x, O^{c})$ . By openness of $O$ , one has $u (x) > 0$ for every [L1, L2, given, choose, algebra] $x \in K$ . The function $u$ is continuous by [L1], so [L2] gives a minimum value $δ : = \min_{x \in K} u (x) > 0.$

L1L2givenchoosealgebra

2.1

Define $η : R^{n} \to R$ by [L1, step 1.2, construct] $η (x) : = {\begin{cases} 0, & u (x) \leq δ / 2, \\ 2 u (x) / δ - 1, & δ / 2 < u (x) < δ, \\ 1, & u (x) \geq δ . \end{cases}$ Because $η$ is obtained by composing the continuous function $u$ with a continuous piecewise-linear cutoff on $[0, \infty)$ , it is continuous and $0 \leq η \leq 1$ . Since $u \geq δ$ on $K$ , one has $η = 1$ on $K$ .

L1step 1.2construct

3.1

The nonzero set of $η$ is contained in ${x : u (x) > δ / 2}$ , so [L1, L2, L3, step 2.1, algebra] The support of a function on $R^{n}$ and its compactly supported Riemann integral gives $supp (η) \subseteq {x : u (x) \geq δ / 2} .$ Every point of the right-hand set lies in $O$ , because $u (x) > 0$ means $x \notin O^{c}$ . Also $O$ is bounded, so $\overline{O}$ is compact by [L2], and the closed set ${u \geq δ / 2}$ lies in $\overline{O}$ . Hence $supp (η)$ is compact and contained in $O$ , so $η \in C_{c} (R^{n})$ .

L1L2L3step 2.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

A finite-measure measurable set in $R^{n}$ has a compact core and a bounded open neighbourhood of arbitrarily small excess

Statement

Assume the Axiom of Countable Choice.

Let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < \infty$ . For every $ε > 0$ there exist a compact set $K$ and a bounded open set $O$ such that

$K \subseteq E, K \subseteq O, λ_{n} (E ∖ K) < ε, and λ_{n} (O ∖ K) < ε .$

Facts & Assumptions

Given: The Axiom of Countable Choice, $n \geq 1$ , a finite-measure measurable set $E \subseteq R^{n}$ , and $ε > 0$ .

[L1]

Outer regularity gives open supersets of arbitrarily small excess (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of $R^{n}$ is the infimum of the measures of the open sets containing it).

[L2]

Lebesgue measure is sigma-finite and finite on bounded sets (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[L3]

Continuity from below applies to the exhaustion $E \cap \overline{B (0, R)} ↑ E$ (Continuity from below for measures).

[L4]

Set-difference measure obeys the usual subtraction and monotonicity rules (Measure of a set difference when the smaller set has finite measure, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

technique · direct

1.1

By [L3], choose $R > 0$ so large that [L2, L3, given, choose] $λ_{n} (E ∖ \overline{B (0, R)}) < ε / 3.$ Put $E_{R} : = E \cap \overline{B (0, R)}$ . Then $E_{R}$ has finite measure and lies in a compact ball.

L2L3givenchoose

2.1

Apply [L1] to choose an open set $U \supseteq E_{R}$ with [L1, L4, step 1.1, choose, construct] $λ_{n} (U ∖ E_{R}) < ε / 3$ . Put $O : = U \cap B (0, R + 1) .$ Then $O$ is bounded and open, and it still contains $E_{R}$ because $E_{R} \subseteq \overline{B (0, R)} \subseteq B (0, R + 1)$ . Next choose an open set $V \supseteq \overline{B (0, R)} ∖ E$ with $λ_{n} (V ∖ (\overline{B (0, R)} ∖ E)) < ε / 3.$ Define $K : = \overline{B (0, R)} ∖ V .$ Then $K \subseteq E$ and $K$ is compact, being closed in the compact ball $\overline{B (0, R)}$ .

L1L4step 1.1chooseconstruct

3.1

Because $K = \overline{B (0, R)} ∖ V$ , one has [step 1.1, step 2.1, L4, algebra] $E ∖ K \subseteq (E ∖ \overline{B (0, R)}) \cup (E_{R} \cap V) .$ Also $E_{R} \cap V \subseteq V ∖ (\overline{B (0, R)} ∖ E),$ so $λ_{n} (E ∖ K) < ε / 3 + ε / 3 < ε .$

step 1.1step 2.1L4algebra

4.1

Since $K \subseteq E_{R} \subseteq O$ , one also has [step 2.1, L4, algebra] $O ∖ K \subseteq (O ∖ E_{R}) \cup (E_{R} ∖ K) .$ But $E_{R} ∖ K = E_{R} \cap V \subseteq V ∖ (\overline{B (0, R)} ∖ E)$ , so $λ_{n} (O ∖ K) < ε / 3 + ε / 3 < ε .$ Thus $K \subseteq E$ , $K \subseteq O$ , and both required excess bounds hold.

step 2.1L4algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

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

Statement

Assume the Axiom of Countable Choice.

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

Facts & Assumptions

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

[L1]

Box-step functions are dense in $L^{p} (R^{n})$ (Finite linear combinations of box indicators are dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ ).

[L2]

Finite-measure measurable sets admit compact cores with bounded open neighbourhoods of arbitrarily small excess (A finite-measure measurable set in $R^{n}$ has a compact core and a bounded open neighbourhood of arbitrarily small excess).

[L3]

Compact sets inside bounded open sets admit explicit compactly supported continuous cutoffs (A compact set inside a bounded open set admits an explicit compactly supported continuous cutoff).

[L4]

Minkowski's inequality holds in $L^{p}$ (Minkowski's inequality for integrals, including $p = \infty$ ).

Proof

technique · direct

1.1

By [L1], choose a box-step function [L1, L2, given, choose] $s = \sum_{j = 1}^{m} a_{j} 1_{E_{j}}$ with $∥ f - s ∥_{p} < ε / 2$ . For each $j$ , apply [L2] to choose $K_{j} \subseteq E_{j}$ and a bounded open set $O_{j} \supseteq K_{j}$ with $λ_{n} (E_{j} ∖ K_{j}), λ_{n} (O_{j} ∖ K_{j}) < {(\frac{ε}{2^{1 + 1 / p} m (1 + \sum_{j = 1}^{m} ∣ a_{j} ∣)})}^{p} .$

L1L2givenchoose

2.1

For each $j$ , [L3] gives $η_{j} \in C_{c} (R^{n})$ with [L3, L4, step 1.1, construct, algebra] $0 \leq η_{j} \leq 1$ , $η_{j} = 1$ on $K_{j}$ , and $supp (η_{j}) \subseteq O_{j}$ . Put $g : = \sum_{j = 1}^{m} a_{j} η_{j} \in C_{c} (R^{n}) .$ Since $η_{j} - 1_{E_{j}}$ vanishes off $(E_{j} ∖ K_{j}) \cup (O_{j} ∖ K_{j})$ and has absolute value at most $1$ there, $∥ g - s ∥_{p} \leq \sum_{j = 1}^{m} ∣ a_{j} ∣ {(λ_{n} (E_{j} ∖ K_{j}) + λ_{n} (O_{j} ∖ K_{j}))}^{1 / p} < ε / 2$ by [L4].

L3L4step 1.1constructalgebra

3.1

Therefore [step 1.1, step 2.1, algebra] $∥ f - g ∥_{p} \leq ∥ f - s ∥_{p} + ∥ s - g ∥_{p} < ε .$ Since $g \in C_{c} (R^{n})$ , the space $C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ .

step 1.1step 2.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

The $L^{\infty}$ -closure of $C_{c} (R^{n})$ is $C_{0} (R^{n})$ , not all of $L^{\infty} (R^{n})$

Statement

Inside $L^{\infty} (R^{n})$ with the essential-supremum norm, the closure of $C_{c} (R^{n})$ is exactly $C_{0} (R^{n})$ . In particular it is not all of $L^{\infty} (R^{n})$ .

Facts & Assumptions

Given: The spaces $C_{c} (R^{n})$ and $C_{0} (R^{n})$ .

[L1]

Continuous compactly supported functions and functions vanishing at infinity are defined in The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$ and The space $C_{0} (R^{n})$ of continuous functions vanishing at infinity.

[L2]

There is an explicit compactly supported cutoff equal to $1$ on a large ball (A compact set inside a bounded open set admits an explicit compactly supported continuous cutoff).

Proof

technique · direct

1.1

Let $f \in C_{0} (R^{n})$ and $ε > 0$ . By [L1], choose [L1, L2, given, choose, algebra] $R > 0$ so that $∣ f (x) ∣ < ε$ for $∣ x ∣ \geq R$ . Apply [L2] to $K = \overline{B (0, R)} \subseteq O = B (0, R + 1)$ and let $η \in C_{c}$ be the corresponding cutoff. Then $η f \in C_{c} (R^{n})$ and $∥ f - η f ∥_{\infty} \leq ε .$ So every $C_{0}$ function lies in the closure of $C_{c}$ .

L1L2givenchoosealgebra

2.1

Conversely, let $(f_{m})$ be a sequence in $C_{c} (R^{n})$ converging to [step 1.1, given, choose, algebra] $f$ in the essential-supremum norm. Then $(f_{m})$ is Cauchy in the actual supremum norm, because for continuous functions the essential supremum equals the ordinary supremum. Hence $(f_{m})$ converges uniformly to some continuous function $g$ . For each $m$ , the tail estimate outside $supp (f_{m})$ shows $g$ is uniformly small there, so $g \in C_{0} (R^{n})$ .

step 1.1givenchoosealgebra

3.1

After passing to a subsequence, arrange [step 2.1, given, choose, algebra] $∥ f - f_{m} ∥_{\infty}^{e s s} < 2^{- m} .$ For each $m$ there is a null set $N_{m}$ such that $∣ f - f_{m} ∣ \leq 2^{- m}$ on $R^{n} ∖ N_{m}$ . On the full-measure set $R^{n} ∖ ⋃_{m} N_{m}$ , one therefore has $f_{m} (x) \to f (x)$ , while uniform convergence gives $f_{m} (x) \to g (x)$ for every $x$ . Thus $f = g$ almost everywhere, so the $L^{\infty}$ class of $f$ is represented by $g \in C_{0} (R^{n})$ .

step 2.1givenchoosealgebra

4.1

Steps 1.1, 2.1, and 3.1 identify the closure as $C_{0} (R^{n})$ . Since the [step 1.1, step 2.1, step 3.1, algebra] constant function $1$ lies in $L^{\infty} (R^{n})$ but not in $C_{0} (R^{n})$ , this closure is not all of $L^{\infty} (R^{n})$ .

step 1.1step 2.1step 3.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Rational box-step functions form a countable dense subset of $L^{p} (R^{n})$ for $1 \leq p < \infty$

Statement

Assume the Axiom of Countable Choice.

Let $1 \leq p < \infty$ . The finite linear combinations of indicator functions of half-open boxes with rational endpoints and rational coefficients form a countable dense subset of $L^{p} (R^{n})$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, $1 \leq p < \infty$ and $ε > 0$ .

[L1]

Box-step functions are dense in $L^{p} (R^{n})$ (Finite linear combinations of box indicators are dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ ).

[L2]

Rational boxes form a countable basis of $R^{n}$ ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis).

[L3]

Every Euclidean box is Lebesgue measurable with its usual volume (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[L4]

Countable sets are closed under finite products and countable unions, and subsets of countable sets are countable (Finite, countably infinite, countable, uncountable, Countable unions of at most countable sets, assuming ${A C}_{ω}$ , A product of two at most countable sets is at most countable, Every subset of an at most countable set is at most countable).

[L5]

A space is separable exactly when it has a countable dense subset (Separability: the existence of an at most countable dense subset).

Proof

technique · direct

1.1

Let $R$ be the family of half-open boxes [L2, L4, given, algebra] $\prod_{i = 1}^{n} (a_{i}, b_{i}]$ with rational endpoints. By [L2] there are countably many such boxes, and by [L4] the set $S$ of all finite rational linear combinations of indicators $1_{R}$ with $R \in R$ is countable.

L2L4givenalgebra

2.1

By [L1], it is enough to approximate a single box indicator. So fix a [L1, L3, step 1.1, choose, algebra] $δ > 0$ and let $B = \prod_{i = 1}^{n} (α_{i}, β_{i}]$ . Choose rationals $a_{i} < α_{i} < β_{i} < b_{i}$ so close to the endpoints that, with $M : = 1 + \max_{i} {∣ a_{i} ∣, ∣ b_{i} ∣, ∣ α_{i} ∣, ∣ β_{i} ∣}$ , $2 n (2 M)^{n - 1} \max_{i} {(α_{i} - a_{i}), (b_{i} - β_{i})} < δ .$ Then $B △ R$ is contained in the union of the $2 n$ coordinate slabs where one coordinate lies in $(a_{i}, α_{i}]$ or $(β_{i}, b_{i}]$ while the others stay in $[- M, M]$ . By [L3], each slab has measure at most $(2 M)^{n - 1} \max_{i} {(α_{i} - a_{i}), (b_{i} - β_{i})}$ , so $λ_{n} (B △ R) < δ$ for the rational box $R : = \prod_{i} (a_{i}, b_{i}]$ . Hence $∥ 1_{B} - 1_{R} ∥_{p} = λ_{n} (B △ R)^{1 / p} < δ^{1 / p},$ which can be made arbitrarily small; approximating finitely many coefficients by rationals then makes every box-step function arbitrarily close to an element of $S$ .

L1L3step 1.1choosealgebra

3.1

Therefore $S$ is countable and dense. By [L5], [L5, step 1.1, step 2.1] $L^{p} (R^{n})$ is separable, with $S$ as an explicit dense subset.

L5step 1.1step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

A countable generator of a sigma-algebra yields a countable algebra of sets

Statement

Assume the Axiom of Countable Choice.

Let $G$ be a countable family of subsets of a set $X$ . Then there is a countable algebra of subsets of $X$ that contains $G$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, a set $X$ , and a countable family $G \subseteq P (X)$ .

[L1]

Generated sigma-algebras are built from families of sets, and algebras are closed under complements and finite unions (The sigma-algebra generated by a family of sets, Algebras of subsets).

[L2]

Countable sets are closed under finite products and countable unions (Finite, countably infinite, countable, uncountable, A product of two at most countable sets is at most countable, Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

Proof

technique · constructive

1.1

If $G = \emptyset$ , then ${\emptyset, X}$ is a finite algebra [L1, given, algebra] of subsets of $X$ containing $G$ , so the conclusion holds. Assume now that $G \neq \emptyset$ .

L1givenalgebra

1.2

Enumerate $G = {G_{1}, G_{2}, \dots}$ . For each $m \geq 1$ , let [L1, given, choose, construct] $A_{m}$ be the family of all unions of atoms of the finite partition generated by $G_{1}, \dots, G_{m}$ ; equivalently, the members of $A_{m}$ are all finite Boolean combinations of those $m$ sets. Each $A_{m}$ is a finite algebra on $X$ containing $G_{1}, \dots, G_{m}$ .

L1givenchooseconstruct

2.1

Put [step 1.2, L1, choose, algebra] $A : = ⋃_{m \geq 1} A_{m} .$ Then $A$ contains every $G_{j}$ . If $A, B \in A$ , choose $m$ with $A, B \in A_{m}$ ; because $A_{m}$ is an algebra, also $X ∖ A$ and $A \cup B$ lie in $A_{m} \subseteq A$ . Thus $A$ is an algebra of subsets of $X$ .

step 1.2L1choosealgebra

3.1

Each $A_{m}$ is finite, so in particular countable, and [L2] makes [L2, step 2.1] their countable union $A$ countable. Therefore $A$ is a countable algebra containing $G$ .

L2step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Finite-measure sets are approximable in measure by sets from a countable generating algebra

Statement

Assume the Axiom of Countable Choice.

Let $(X, A, μ)$ be a sigma-finite measure space whose sigma-algebra is countably generated. Then there is a countable algebra $A_{0} \subseteq A$ such that for every $E \in A$ with $μ (E) < \infty$ and every $ε > 0$ there exists $A \in A_{0}$ with

μ (E △ A) < ε .

Facts & Assumptions

Given: The Axiom of Countable Choice, a sigma-finite measure space $(X, A, μ)$ with $A$ countably generated.

[L1]

A countable generator yields a countable algebra (A countable generator of a sigma-algebra yields a countable algebra of sets).

[L2]

Sigma-finite measures admit an increasing exhaustion $X_{1} \subseteq X_{2} \subseteq \dots$ by measurable finite-measure sets (Finite, sigma-finite, and semifinite measures).

[L3]

If two families contain one another inside generated sigma-algebras, then they generate the same sigma-algebra (Two families generate the same sigma-algebra when each lies in the sigma-algebra generated by the other, The sigma-algebra generated by a family of sets).

[L4]

Continuity from below and the symmetric-difference measure calculus are available (Continuity from below for measures, Measure of a set difference when the smaller set has finite measure).

Proof

technique · direct

1.1

Let $G$ be a countable generator of $A$ , and choose a [L1, L2, L3, given, choose] sigma-finite exhaustion $(X_{m})$ by [L2]. Apply [L1] to the countable family $G \cup {X_{m} : m \geq 1}$ to obtain a countable algebra $A_{0} \subseteq A$ . Because $G \subseteq A_{0} \subseteq A$ , [L3] gives $σ (A_{0}) = A$ .

L1L2L3givenchoose

1.2

Fix $m$ and let $M_{m}$ be the family of measurable subsets [L4, algebra] $B \subseteq X_{m}$ for which every $ε > 0$ admits $A \in A_{0}$ with $A \subseteq X_{m}$ and $μ (B △ A) < ε$ . Because $X_{m} \in A_{0}$ , the family $M_{m}$ contains $A_{0} \cap X_{m}$ . On the finite-measure space $X_{m}$ , the class $M_{m}$ is a sigma-algebra: complements are handled inside $X_{m}$ , and increasing unions are handled by approximating one large stage and using continuity from below. Therefore $M_{m}$ is a sigma-algebra containing the trace of $A_{0}$ , so it contains every measurable subset of $X_{m}$ .

L4algebra

2.1

Now let $E \in A$ with $μ (E) < \infty$ and let $ε > 0$ . [L4, choose, algebra] By [L4], choose $m$ so large that $μ (E ∖ X_{m}) < ε / 2$ . Since $E \cap X_{m}$ is a measurable subset of $X_{m}$ , the construction above yields $A \in A_{0}$ with $A \subseteq X_{m}$ and $μ ((E \cap X_{m}) △ A) < ε / 2$ . Then $μ (E △ A) \leq μ (E ∖ X_{m}) + μ ((E \cap X_{m}) △ A) < ε .$ So finite-measure sets are approximable by members of the countable algebra $A_{0}$ .

L4choosealgebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

If $μ$ is sigma-finite and $A$ is countably generated, then $L^{p} (μ)$ is separable for $1 \leq p < \infty$

Statement

Assume the Axiom of Countable Choice.

Let $(X, A, μ)$ be a sigma-finite measure space with countably generated sigma-algebra, and let $1 \leq p < \infty$ . Then $L^{p} (μ)$ is separable.

Facts & Assumptions

Given: The Axiom of Countable Choice, a sigma-finite countably generated measure space, and an exponent $1 \leq p < \infty$ .

[L1]

Simple functions with finite-measure support are dense in $L^{p} (μ)$ (Simple functions with finite-measure support are dense in $L^{p} (μ)$ for $1 \leq p < \infty$ ).

[L2]

Finite-measure sets are approximable by a countable generating algebra (Finite-measure sets are approximable in measure by sets from a countable generating algebra, A countable generator of a sigma-algebra yields a countable algebra of sets).

[L3]

The quotient $L^{p} (μ)$ is the space in question, and separability means the existence of a countable dense subset (The space $L^{p} (μ)$ as the quotient by null functions, Separability: the existence of an at most countable dense subset).

Proof

technique · direct

1.1

Let $A_{0}$ be the countable algebra from [L2], and let $S$ [L2, L3, given, algebra] be the set of all finite linear combinations $\sum_{j = 1}^{m} q_{j} 1_{A_{j}}$ with $q_{j} \in Q (i)$ and $A_{j} \in A_{0}$ satisfying $μ (A_{j}) < \infty$ for every $j$ . Because both the coefficient set and the finite-measure members of $A_{0}$ are countable, $S$ is countable.

L2L3givenalgebra

1.2

To prove density, start with $f \in L^{p} (μ)$ and $ε > 0$ . By [L1, L2, given, choose, algebra] [L1], choose a finite-support simple function $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ with $∥ f - s ∥_{p} < ε / 2$ . For each $j$ , [L2] gives $A_{j} \in A_{0}$ with $μ (A_{j}) < \infty$ and $μ (E_{j} △ A_{j})$ arbitrarily small, and each coefficient $c_{j}$ can be approximated by $q_{j} \in Q (i)$ . The resulting $t : = \sum_{j} q_{j} 1_{A_{j}}$ lies in $S$ and satisfies $∥ s - t ∥_{p} < ε / 2$ .

L1L2givenchoosealgebra

2.1

Then [L3, step 1.1, step 1.2] $∥ f - t ∥_{p} \leq ∥ f - s ∥_{p} + ∥ s - t ∥_{p} < ε .$ So $S$ is countable and dense in $L^{p} (μ)$ ; by [L3], the space is separable.

L3step 1.1step 1.2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

$L^{\infty} [0, 1]$ is not separable

Statement

The space $L^{\infty} ([0, 1])$ is not separable.

Facts & Assumptions

Given: The essential-supremum norm on $L^{\infty} ([0, 1])$ .

[L1]

Separability means having a countable dense subset (Separability: the existence of an at most countable dense subset, Finite, countably infinite, countable, uncountable).

[L2]

$L^{\infty}$ is normed by the essential supremum (The space $L^{\infty} (μ)$ of essentially bounded measurable functions, The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

Proof

technique · direct

1.1

For each $t \in [0, 1]$ , let $u_{t} : = 1_{[0, t]}$ . If $s < t$ , then [L2, given, algebra] $u_{s}$ and $u_{t}$ differ by $1$ on $(s, t]$ , a set of positive measure, so

∥ u_{t} - u_{s} ∥_{\infty} = 1

by [L2]. Thus the family ${u_{t} : t \in [0, 1]}$ is uncountable and $1$ -separated.

L2givenalgebra

2.1

Suppose $D$ were a countable dense subset. For each $t \in [0, 1]$ , choose [L1, step 1.1, choose, algebra] $d_{t} \in D$ with $∥ d_{t} - u_{t} ∥_{\infty} < 1 / 3$ . If $s \neq t$ , then the balls $B_{\infty} (u_{s}, 1 / 3)$ and $B_{\infty} (u_{t}, 1 / 3)$ are disjoint because the centers are distance $1$ apart, so $d_{s} \neq d_{t}$ . This gives an injection $[0, 1] ↪ D$ , contradicting countability.

L1step 1.1choosealgebra

3.1

Therefore no countable subset is dense in $L^{\infty} ([0, 1])$ , so [step 2.1] $L^{\infty} ([0, 1])$ is not separable.

step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Continuous compactly supported functions are translation-continuous in $L^{p}$

Statement

Assume the Axiom of Countable Choice.

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

$∥ τ_{h} f - f ∥_{p} ⟶ 0 (h \to 0) .$

Facts & Assumptions

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

[L1]

Continuous functions on compact metric spaces are uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L2]

Bounded sets have finite Lebesgue measure, and translation preserves Lebesgue measure (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[L3]

Translation is the convention of Translation of a function on $R^{n}$ , and $C_{c} (R^{n})$ is defined in The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$ .

Proof

technique · direct

1.1

Let $K : = supp (f)$ , and choose $R > 0$ so that [L1, L3, given, choose] $K \subseteq B (0, R)$ . Then for $∣ h ∣ \leq 1$ , the support of $τ_{h} f - f$ lies in the compact set $\overline{B (0, R + 1)}$ . By [L1], $f$ is uniformly continuous on that compact set.

L1L3givenchoose

2.1

Let $ε > 0$ . Uniform continuity gives $δ > 0$ such that [L1, L2, step 1.1, choose, algebra] $∣ f (x - h) - f (x) ∣ < ε$ whenever $∣ h ∣ < δ$ and $x, x - h \in \overline{B (0, R + 1)}$ . Hence for $∣ h ∣ < δ$ , $∥ τ_{h} f - f ∥_{p}^{p} \leq ε^{p} λ_{n} (\overline{B (0, R + 1)}) .$ The right-hand side tends to $0$ with $ε$ , and [L2] makes the measure finite.

L1L2step 1.1choosealgebra

3.1

Therefore $∥ τ_{h} f - f ∥_{p} \to 0$ as $h \to 0$ .

step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

$∥ τ_{h} f - f ∥_{p} \to 0$ 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

1.1

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 $∥ τ_{h} g - g ∥_{p} < ε / 3$ .

L1L2givenchoose

2.1

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

L3L4step 1.1algebra

3.1

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

step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Borel representatives make the convolution integrand Borel measurable

Statement

Let $\tilde{f}, \tilde{g} : R^{n} \to C$ be Borel measurable functions. Then

H (x, y) : = \tilde{f} (x - y) \tilde{g} (y)

is Borel measurable on $R^{2 n}$ . In particular, for each fixed $x \in R^{n}$ , the section $y \mapsto H (x, y)$ is measurable.

Facts & Assumptions

Given: Borel measurable functions $\tilde{f}, \tilde{g}$ on $R^{n}$ .

[A1]

The functions $\tilde{f}$ and $\tilde{g}$ are Borel measurable by hypothesis.

[L2]

The Borel product on $R^{n} \times R^{n}$ is the Euclidean Borel sigma-algebra on $R^{2 n}$ (The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}).

[L3]

Composition with Borel functions preserves measurability, and measurable arithmetic operations preserve measurability (Composition with a Borel measurable outer map preserves measurability, Arithmetic and lattice operations preserve measurability whenever they are defined).

[L4]

Sections of product-measurable functions are measurable (Every section of a product-measurable function is measurable).

Proof

technique · direct

1.1

The map $T : R^{2 n} \to R^{2 n}$ given by [L2, L3, given, construct] $T (x, y) : = (x - y, y)$ is continuous, hence Borel measurable. Since $(u, v) \mapsto \tilde{f} (u)$ and $(u, v) \mapsto \tilde{g} (v)$ are Borel measurable on $R^{2 n}$ by [L2] and [L3], the functions $(x, y) \mapsto \tilde{f} (x - y)$ and $(x, y) \mapsto \tilde{g} (y)$ are Borel measurable.

L2L3givenconstruct

2.1

Multiplication on $C$ is continuous, so [L3] makes [L3, L4, step 1.1]

H (x, y) = \tilde{f} (x - y) \tilde{g} (y)

Borel measurable on $R^{2 n}$ . Then [L4] gives measurability of each section $y \mapsto H (x, y)$ .

L3L4step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Convolution on $L^{1} (R^{n})$ is independent of the chosen Borel representatives

Statement

Let $f, g \in L^{1} (R^{n})$ . If ${\tilde{f}}_{1}, {\tilde{f}}_{2}$ are Borel representatives of $f$ and ${\tilde{g}}_{1}, {\tilde{g}}_{2}$ are Borel representatives of $g$ , then for almost every $x \in R^{n}$ ,

$\int {\tilde{f}}_{1} (x - y) {\tilde{g}}_{1} (y) d y = \int {\tilde{f}}_{2} (x - y) {\tilde{g}}_{2} (y) d y .$

Facts & Assumptions

Given: Two Borel representatives for each of the $L^{1}$ classes $f$ and $g$ .

[L1]

The integrands from Borel representatives are measurable (Borel representatives make the convolution integrand Borel measurable).

[L2]

The Lebesgue integral respects almost-everywhere equality (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree).

[L3]

Lebesgue measurability and null sets are translation invariant (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

Proof

technique · direct

1.1

Let $N_{f} : = {{\tilde{f}}_{1} \neq {\tilde{f}}_{2}}$ and $N_{g} : = {{\tilde{g}}_{1} \neq {\tilde{g}}_{2}}$ . [L3, given, algebra] These are null sets. For a fixed $x$ , the set ${y : {\tilde{f}}_{1} (x - y) \neq {\tilde{f}}_{2} (x - y)} = x - N_{f}$ is also null by [L3]. Hence the two section integrands agree for almost every $y$ , outside the null set $(x - N_{f}) \cup N_{g}$ .

L3givenalgebra

2.1

By [L1], both section integrands are measurable, and step 1.1 says they are [L1, L2, step 1.1] equal almost everywhere in $y$ . Therefore [L2] gives equality of their integrals whenever either side is defined as an absolutely convergent Lebesgue integral.

L1L2step 1.1

3.1

This holds for every fixed $x$ , so in particular it holds for almost every [step 2.1] $x$ on the domain where the $L^{1}$ convolution is defined. Thus the convolution does not depend on the chosen Borel representatives.

step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$

Statement

If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists for almost every $x \in R^{n}$ , belongs to $L^{1} (R^{n})$ , and satisfies

$∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1} .$

Facts & Assumptions

Given: Functions $f, g \in L^{1} (R^{n})$ .

[L1]

The Borel-representative integrand is measurable and representative independent (Borel representatives make the convolution integrand Borel measurable, Convolution on $L^{1} (R^{n})$ is independent of the chosen Borel representatives).

[L2]

Tonelli and Fubini apply on sigma-finite product spaces (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Fubini's theorem for L^1 functions on a sigma-finite product).

[L3]

The integral triangle inequality is available (The modulus of an integral is bounded by the integral of the modulus).

Proof

technique · direct

1.1

Choose Borel representatives $\tilde{f}, \tilde{g}$ of $f, g$ and define [L1, L2, given, choose, algebra] $H (x, y) : = ∣ \tilde{f} (x - y) ∣ ∣ \tilde{g} (y) ∣ .$ By [L1], $H$ is measurable on $R^{2 n}$ . For each $y$ , $\int_{R^{n}} H (x, y) d x = ∣ \tilde{g} (y) ∣ \int_{R^{n}} ∣ \tilde{f} (x - y) ∣ d x = ∥ \tilde{f} ∥_{1} ∣ \tilde{g} (y) ∣$ by translation invariance of Lebesgue measure.

L1L2givenchoosealgebra

2.1

Integrating the identity from step 1.1 in $y$ and applying [L2] gives [L2, step 1.1, algebra] $\iint_{R^{2 n}} H (x, y) d x d y = ∥ \tilde{f} ∥_{1} ∥ \tilde{g} ∥_{1} < \infty .$ Hence for almost every $x$ , the section $y \mapsto \tilde{f} (x - y) \tilde{g} (y)$ is absolutely integrable, so $(f * g) (x)$ is defined there.

L2step 1.1algebra

3.1

For those $x$ , [L1, L2, L3, step 2.1, algebra] $∣ (f * g) (x) ∣ \leq \int ∣ \tilde{f} (x - y) \tilde{g} (y) ∣ d y$ by [L3]. Another application of [L2] then yields $∥ f * g ∥_{1} \leq \iint ∣ \tilde{f} (x - y) ∣ ∣ \tilde{g} (y) ∣ d y d x = ∥ f ∥_{1} ∥ g ∥_{1} .$ By [L1], the resulting $L^{1}$ class is independent of the chosen Borel representatives.

L1L2L3step 2.1algebra∎

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Convolution on $L^{1} (R^{n})$ is bilinear, commutative, and associative

Statement

Convolution on $L^{1} (R^{n})$ is bilinear, commutative, and associative.

Facts & Assumptions

Given: Functions in $L^{1} (R^{n})$ for which the displayed algebra laws are to be checked.

[L1]

$L^{1}$ convolution exists almost everywhere and obeys the $L^{1}$ bound (If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$ ).

[L2]

Tonelli and Fubini justify rearranging absolutely integrable iterated integrals (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Fubini's theorem for L^1 functions on a sigma-finite product).

[L3]

Convolution is the integral from Convolution of two functions on $R^{n}$ .

Proof

technique · direct

1.1

Bilinearity follows from linearity of the integral in [L3] once [L1] guarantees absolute convergence for almost every $x$ .

L1L3givenalgebra

1.2

For commutativity, fix $x$ where convolution is defined and change [L1, L2, L3, algebra] variables $u : = x - y$ : $(f * g) (x) = \int f (x - y) g (y) d y = \int g (x - u) f (u) d u = (g * f) (x) .$ Associativity is similar: [L2] applies to $∣ f (x - y - z) g (z) h (y) ∣$ , so one may reorder the three integrations and obtain $((f * g) * h) (x) = (f * (g * h)) (x)$ almost everywhere.

L1L2L3algebra

2.1

Therefore convolution is bilinear, commutative, and associative on [step 1.1, step 1.2] $L^{1} (R^{n})$ .

step 1.1step 1.2∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01Open item page →

The support of a convolution lies in the closure of the support sumset

Statement

Let $f, g \in L^{1} (R^{n})$ be represented by Borel functions still denoted $f, g$ . For any complex-valued function $u$ on $R^{n}$ , use the support convention

$supp (u) : = \overline{{x \in R^{n} : u (x) \neq 0}},$

which extends the real-valued definition of The support of a function on $R^{n}$ and its compactly supported Riemann integral. Define $h (x) : = {\begin{cases} \int_{R^{n}} f (x - y) g (y) d y, & if the integral exists, \\ 0, & otherwise . \end{cases}$ Then

$supp (h) \subseteq \overline{supp (f) + supp (g)} .$

Facts & Assumptions

Given: Borel representatives $f, g \in L^{1} (R^{n})$ and the function $h$ defined above.

[L1]

$L^{1}$ convolution exists almost everywhere (If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$ , Convolution of two functions on $R^{n}$ ).

[L2]

Support is defined by the closure of the nonzero set (The support of a function on $R^{n}$ and its compactly supported Riemann integral).

Proof

technique · direct

1.1

Let [L1, L2, given, choose] $x \notin \overline{supp (f) + supp (g)}$ . Choose an open neighborhood $U$ of $x$ disjoint from that closure. For $z \in U$ and $y \in supp (g)$ , one has $z - y \notin supp (f)$ , so $f (z - y) = 0$ ; and if $y \notin supp (g)$ then $g (y) = 0$ . Hence $f (z - y) g (y) = 0$ for every $y$ and every $z \in U$ .

L1L2givenchoose

2.1

Therefore $h (z) = 0$ for every $z \in U$ : by step 1.1 the integrand [L1, L2, step 1.1] vanishes for every $y$ , so the convolution integral exists and equals $0$ at each such $z$ . So $x$ lies outside the support of $h$ in the sense of [L2].

L1L2step 1.1

3.1

Since every point outside [step 2.1] $\overline{supp (f) + supp (g)}$ lies outside $supp (h)$ , the support inclusion follows.

step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

For $1 < p < \infty$ , the $L^{p}$ norm of a nonnegative function is the supremum of its pairings with $L^{q}$ unit vectors

Statement

Let $1 < p < \infty$ and let $q$ be the conjugate exponent. If $F \in L^{p} (μ)$ is nonnegative, then

$∥ F ∥_{p} = \sup {\int F g d μ : g \geq 0, ∥ g ∥_{q} \leq 1} .$

Facts & Assumptions

Given: $1 < p < \infty$ and a nonnegative function $F \in L^{p} (μ)$ .

[L1]

Conjugate exponents are defined in Conjugate exponents, including the endpoint conventions.

[L2]

Holder's inequality holds for the pairing (Holder's inequality for integrals, including the endpoint cases).

[L3]

The $L^{p}$ norm is the norm on the quotient space (The space $L^{p} (μ)$ as the quotient by null functions, The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

Proof

technique · direct

1.1

For every $g \geq 0$ with $∥ g ∥_{q} \leq 1$ , [L2] gives [L2, L3, given, algebra] $\int F g d μ \leq ∥ F ∥_{p} ∥ g ∥_{q} \leq ∥ F ∥_{p} .$ So the displayed supremum is at most $∥ F ∥_{p}$ .

L2L3givenalgebra

2.1

If $∥ F ∥_{p} = 0$ , then $F = 0$ almost everywhere and the supremum is also $0$ . [L1, L3, step 1.1, algebra, construct] Otherwise define $g : = \frac{F^{p - 1}}{∥ F ∥_{p}^{p / q}} .$ Because $(p - 1) q = p$ , one has $∥ g ∥_{q}^{q} = \frac{\int F^{(p - 1) q} d μ}{∥ F ∥_{p}^{p}} = 1,$ so $∥ g ∥_{q} = 1$ , and $\int F g d μ = \frac{\int F^{p} d μ}{∥ F ∥_{p}^{p / q}} = ∥ F ∥_{p} .$

L1L3step 1.1algebraconstruct

3.1

Step 1.1 gives the upper bound and step 2.1 attains it, so the supremum [step 1.1, step 2.1] equals $∥ F ∥_{p}$ .

step 1.1step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Minkowski's integral inequality

Statement

Let $(X, μ)$ and $(Y, ν)$ be sigma-finite measure spaces, let $1 \leq p < \infty$ , and let $F : X \times Y \to C$ be measurable with

\int_{Y} ∥ F (\cdot, y) ∥_{L^{p} (X)} d ν (y) < \infty .

Then the function

H (x) : = \int_{Y} ∣ F (x, y) ∣ d ν (y)

belongs to $L^{p} (X)$ and

$∥ H ∥_{L^{p} (X)} \leq \int_{Y} ∥ F (\cdot, y) ∥_{L^{p} (X)} d ν (y) .$

Facts & Assumptions

Given: Sigma-finite measure spaces, an exponent $1 \leq p < \infty$ , and a measurable function $F$ satisfying the displayed integrability hypothesis.

[L1]

For $1 < p < \infty$ , the nonnegative duality formula is available (For $1 < p < \infty$ , the $L^{p}$ norm of a nonnegative function is the supremum of its pairings with $L^{q}$ unit vectors).

[L2]

Tonelli applies to nonnegative measurable functions on product spaces (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[L3]

Holder's inequality is available (Holder's inequality for integrals, including the endpoint cases).

[L4]

Monotone convergence is available for nonnegative measurable functions (Monotone convergence for the integral).

Proof

technique · direct

1.1

If $p = 1$ , then Tonelli directly gives [L2, given, algebra] $∥ H ∥_{L^{1} (X)} = \int_{X} \int_{Y} ∣ F (x, y) ∣ d ν (y) d μ (x) = \int_{Y} ∥ F (\cdot, y) ∥_{L^{1} (X)} d ν (y) .$

L2givenalgebra

1.2

Assume $1 < p < \infty$ and let $q$ be conjugate to $p$ . Put [L2, L3, given, algebra] $M : = \int_{Y} ∥ F (\cdot, y) ∥_{L^{p} (X)} d ν (y) .$ For every nonnegative $g \in L^{q} (X)$ with $∥ g ∥_{q} \leq 1$ , $\int_{X} H (x) g (x) d μ (x) = \int_{Y} (\int_{X} ∣ F (x, y) ∣ g (x) d μ (x)) d ν (y)$ by [L2], and then [L3] yields $\int_{X} ∣ F (x, y) ∣ g (x) d μ (x) \leq ∥ F (\cdot, y) ∥_{p} ∥ g ∥_{q} \leq ∥ F (\cdot, y) ∥_{p} .$ Hence $\int_{X} H g d μ \leq \int_{Y} ∥ F (\cdot, y) ∥_{p} d ν (y) = M .$

L2L3givenalgebra

1.3

Choose measurable sets $X_{1} \subseteq X_{2} \subseteq \dots$ with [given, construct] $μ (X_{m}) < \infty$ and $⋃_{m} X_{m} = X$ , and define $H_{m} : = \min (H, m) 1_{X_{m}} .$ Then $0 \leq H_{m} ↑ H$ pointwise, and each $H_{m}$ lies in $L^{p} (X)$ because $H_{m} \leq m 1_{X_{m}}$ .

givenconstruct

2.1

Applying [L1] to each nonnegative $H_{m}$ and using $0 \leq H_{m} \leq H$ with [L1, step 1.2, step 1.3] step 1.2 gives $∥ H_{m} ∥_{p} \leq M$ for every $m$ .

L1step 1.2step 1.3

3.1

By [L4], $\int_{X} H_{m} (x)^{p} d μ (x) ↑ \int_{X} H (x)^{p} d μ (x) .$ Since $\int H_{m}^{p} d μ = ∥ H_{m} ∥_{p}^{p} \leq M^{p}$ for every $m$ , the limit is finite and satisfies $∥ H ∥_{p}^{p} \leq M^{p}$ . Hence $H \in L^{p} (X)$ and $∥ H ∥_{p} \leq M = \int_{Y} ∥ F (\cdot, y) ∥_{L^{p} (X)} d ν (y) .$ Together with step 1.1, this proves the theorem for all $1 \leq p < \infty$ .

L4step 1.1step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Young's convolution inequality

Statement

Let $1 \leq p, q, r \leq \infty$ satisfy

\frac{1}{r} = \frac{1}{p} + \frac{1}{q} - 1.

If $f \in L^{p} (R^{n})$ and $g \in L^{q} (R^{n})$ , then the convolution $f * g$ is defined almost everywhere and satisfies

∥ f * g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q} .

Facts & Assumptions

Given: Exponents $p, q, r$ as displayed and functions $f \in L^{p}$ , $g \in L^{q}$ .

[L1]

The $L^{1}$ convolution bound is available (If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$ ).

[L2]

Holder's inequality and generalized Holder are available (Holder's inequality for integrals, including the endpoint cases, Generalized Holder inequality puts products into $L^{r}$ , Conjugate exponents, including the endpoint conventions).

[L3]

Minkowski's integral inequality is available (Minkowski's integral inequality).

Proof

technique · direct

1.1

If $r = \infty$ , then $1 / p + 1 / q = 1$ , so $q$ is conjugate to $p$ . For every $x$ , [L2, given, algebra] [L2] gives $∣ f * g (x) ∣ \leq \int ∣ f (x - y) ∣ ∣ g (y) ∣ d y \leq ∥ f ∥_{p} ∥ g ∥_{q} .$ Hence $∥ f * g ∥_{\infty} \leq ∥ f ∥_{p} ∥ g ∥_{q}$ .

L2givenalgebra

1.2

Assume $r < \infty$ . If $r = p$ , interpret the factor [L2, given, algebra] $∣ f (x - y) ∣^{1 - p / r}$ as $1$ and the exponent $p r / (r - p)$ as $\infty$ ; likewise, if $r = q$ , interpret $∣ g (y) ∣^{1 - q / r}$ as $1$ and $q r / (r - q)$ as $\infty$ . With this endpoint convention, generalized Holder from [L2] applies to the three factors $(∣ f (x - y) ∣^{p} ∣ g (y) ∣^{q})^{1 / r}, ∣ f (x - y) ∣^{1 - p / r}, ∣ g (y) ∣^{1 - q / r}$ with exponents $r, \frac{p r}{r - p}, \frac{q r}{r - q},$ because their reciprocals sum to $1 / r + (1 / p - 1 / r) + (1 / q - 1 / r) = 1$ . This yields $∣ f * g (x) ∣^{r} \leq ∥ f ∥_{p}^{r - p} ∥ g ∥_{q}^{r - q} \int ∣ f (x - y) ∣^{p} ∣ g (y) ∣^{q} d y .$

L2givenalgebra

2.1

Integrate the inequality from step 1.2 in $x$ . Tonelli on the nonnegative [L1, L3, step 1.1, step 1.2, algebra] right-hand side gives $∥ f * g ∥_{r}^{r} \leq ∥ f ∥_{p}^{r - p} ∥ g ∥_{q}^{r - q} \iint ∣ f (x - y) ∣^{p} ∣ g (y) ∣^{q} d y d x = ∥ f ∥_{p}^{r} ∥ g ∥_{q}^{r} .$ Taking $r$ th roots proves the finite- $r$ case. Together with step 1.1, this is Young's inequality.

L1L3step 1.1step 1.2algebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

If $1 < p < \infty$ and $q$ is conjugate to $p$ , then $f * g$ is continuous and vanishes at infinity

Statement

Assume the Axiom of Countable Choice.

Let $1 < p < \infty$ and let $q$ be the conjugate exponent. If $f \in L^{p} (R^{n})$ and $g \in L^{q} (R^{n})$ , then the convolution $f * g$ has a continuous representative in $C_{0} (R^{n})$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, $1 < p < \infty$ , its conjugate exponent $q$ , and functions $f \in L^{p} (R^{n})$ and $g \in L^{q} (R^{n})$ .

[L1]

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

[L2]

Young's inequality gives the $L^{\infty}$ bound in the conjugate case (Young's convolution inequality).

[L3]

$L^{1}$ convolution is defined almost everywhere, and $C_{c} (R^{n})$ is dense in finite $L^{p}$ (If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$ , $C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ ).

[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

1.1

For every $x, h \in R^{n}$ , [L1, L2, given, algebra] $∣ (f * g) (x - h) - (f * g) (x) ∣ \leq \int ∣ f (x - h - y) - f (x - y) ∣ ∣ g (y) ∣ d y \leq ∥ τ_{h} f - f ∥_{p} ∥ g ∥_{q} .$ By [L1], the right-hand side tends to $0$ as $h \to 0$ , uniformly in $x$ . So the pointwise-defined convolution is uniformly continuous.

L1L2givenalgebra

2.1

By [L3], choose $u, v \in C_{c} (R^{n})$ with [L2, L3, L4, step 1.1, choose, algebra] $∥ f - u ∥_{p} + ∥ g - v ∥_{q}$ arbitrarily small. Applying [L2] twice gives $∥ f * g - u * v ∥_{\infty} \leq ∥ f - u ∥_{p} ∥ g ∥_{q} + ∥ u ∥_{p} ∥ g - v ∥_{q} .$ The function $u * v$ is continuous by step 1.1 and compactly supported by [L4], so $u * v \in C_{0} (R^{n})$ .

L2L3L4step 1.1choosealgebra

3.1

Therefore $f * g$ is a uniform limit of $C_{0}$ functions, hence itself belongs [step 1.1, step 2.1] to $C_{0} (R^{n})$ . Together with step 1.1, this proves that $f * g$ has a continuous representative vanishing at infinity.

step 1.1step 2.1∎

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

A unit-mass smooth bump generates an $L^{1}$ approximate identity

Statement

Assume the Axiom of Countable Choice.

Let $φ \in C_{c}^{\infty} (R^{n})$ satisfy $\int φ = 1$ , and let $φ_{ε} (x) = ε^{- n} φ (x / ε)$ . Then $(φ_{ε})_{ε > 0}$ is an $L^{1}$ approximate identity.

Facts & Assumptions

Given: The Axiom of Countable Choice, a unit-mass smooth bump, and its rescalings.

[L1]

An $L^{1}$ approximate identity and a mollifier family are defined in An $L^{1}$ approximate identity on $R^{n}$ and The mollifier family generated by a unit-mass smooth bump.

[L2]

Linear change of variables preserves Lebesgue measure in the expected way, and the integral is linear on $L^{1}$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, The Lebesgue integral is linear on $L^{1} (μ)$ ).

Proof

technique · direct

1.1

By [L2], the change of variables $u = x / ε$ gives [L1, L2, given, algebra] $\int φ_{ε} (x) d x = \int φ (u) d u = 1$ and likewise $∥ φ_{ε} ∥_{1} = \int ∣ φ_{ε} (x) ∣ d x = \int ∣ φ (u) ∣ d u = ∥ φ ∥_{1} .$ So the mass is $1$ and the $L^{1}$ norms are uniformly bounded.

L1L2givenalgebra

2.1

Let $R > 0$ satisfy $supp (φ) \subseteq \overline{B (0, R)}$ . [step 1.1, choose, algebra] Then $supp (φ_{ε}) \subseteq \overline{B (0, ε R)}$ . Hence for every $δ > 0$ and every $ε < δ / R$ , $\int_{∣ x ∣ > δ} ∣ φ_{ε} (x) ∣ d x = 0.$ So the tails concentrate at the origin.

step 1.1choosealgebra

3.1

Steps 1.1 and 2.1 verify the three defining clauses in [L1], so [L1, step 1.1, step 2.1] $(φ_{ε})$ is an $L^{1}$ approximate identity.

L1step 1.1step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Convolution with a mollifier is smooth, and derivatives pass under the integral sign

Statement

Let $f : R^{n} \to C$ be locally integrable, let $φ \in C_{c}^{\infty} (R^{n})$ have mass $1$ , and let $φ_{ε}$ be the associated mollifier. Then for every $ε > 0$ , the convolution

$(f * φ_{ε}) (x) : = \int_{R^{n}} f (y) φ_{ε} (x - y) d y$

is smooth, and for every multi-index $α$ ,

$\partial^{α} (f * φ_{ε}) = f * (\partial^{α} φ_{ε}) .$

Facts & Assumptions

Given: A locally integrable function $f$ , a unit-mass smooth bump, and $ε > 0$ .

[L1]

The mollifier family is defined in The mollifier family generated by a unit-mass smooth bump.

[L2]

Differentiation under the integral sign is available (Differentiation under the integral sign).

[L3]

Multi-index notation and Euclidean smoothness are fixed in $C^{k}$ maps and multi-index derivative notation in Euclidean space.

Proof

technique · direct

1.1

Fix $x_{0} \in R^{n}$ . Because $φ_{ε}$ has compact [L1, L3, given, choose, algebra] support, there are $r > 0$ and a compact set $K$ such that $φ_{ε} (x - y) = 0$ and $\partial^{α} φ_{ε} (x - y) = 0$ whenever $∣ x - x_{0} ∣ < r$ and $y \notin K$ . Local integrability of $f$ therefore makes $∣ f ∣ 1_{K}$ integrable, so $y \mapsto f (y) φ_{ε} (x - y)$ and $y \mapsto f (y) \partial^{α} φ_{ε} (x - y)$ are integrable for $∣ x - x_{0} ∣ < r$ .

L1L3givenchoosealgebra

2.1

Fix a coordinate index $j$ and a point $x$ with $∣ x - x_{0} ∣ < r / 2$ . For [L2, step 1.1, algebra] $∣ t ∣ < r / 2$ , the point $x + t e_{j}$ still satisfies $∣ x + t e_{j} - x_{0} ∣ < r$ , so $G (y, t) : = f (y) φ_{ε} (x + t e_{j} - y)$ is integrable in $y$ . Because $\partial_{j} φ_{ε}$ is continuous with compact support, some constant $C_{j}$ satisfies $∣ \partial_{j} φ_{ε} (x + t e_{j} - y) ∣ \leq C_{j} 1_{K} (y) (∣ t ∣ < r / 2) .$ Hence $∣ \partial_{t} G (y, t) ∣ = ∣ f (y) \partial_{j} φ_{ε} (x + t e_{j} - y) ∣ \leq C_{j} ∣ f (y) ∣ 1_{K} (y),$ and the right-hand side is integrable by step 1.1. Applying [L2] on the interval $(- r / 2, r / 2)$ gives $\partial_{j} (f * φ_{ε}) (x) = \int f (y) \partial_{j} φ_{ε} (x - y) d y = (f * (\partial_{j} φ_{ε})) (x) .$ Since $x_{0}$ was arbitrary, this holds for every $x$ .

L2step 1.1algebra

3.1

Repeating step 2.1 for higher derivatives and using [L3] yields the general [L2, L3, step 2.1, induction] multi-index formula $\partial^{α} (f * φ_{ε}) = f * (\partial^{α} φ_{ε})$ . Hence $f * φ_{ε}$ is smooth.

L2L3step 2.1induction∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$

Statement

Assume the Axiom of Countable Choice.

Let $(K_{ε})_{ε > 0}$ be an $L^{1}$ approximate identity on $R^{n}$ . If $1 \leq p < \infty$ and $f \in L^{p} (R^{n})$ , then

$∥ f * K_{ε} - f ∥_{p} ⟶ 0 (ε \to 0^{+}) .$

Facts & Assumptions

Given: The Axiom of Countable Choice, an $L^{1}$ approximate identity, an exponent $1 \leq p < \infty$ , and $f \in L^{p} (R^{n})$ .

[L1]

Approximate identities are defined in An $L^{1}$ approximate identity on $R^{n}$ .

[L2]

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

[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 $\int K_{ε} = 1$ , one may write [L1, L3, given, algebra] $f * K_{ε} - f = \int_{R^{n}} K_{ε} (y) (τ_{y} f - f) d y .$ Applying [L3] gives $∥ f * K_{ε} - f ∥_{p} \leq \int ∣ K_{ε} (y) ∣ ∥ τ_{y} f - f ∥_{p} d y .$

L1L3givenalgebra

2.1

Let $η > 0$ be arbitrary. If $∥ f ∥_{p} = 0$ , then $f = 0$ in $L^{p}$ and step [L1, L2, L3, step 1.1, choose, algebra] 1.1 gives $∥ f * K_{ε} - f ∥_{p} = 0 < η$ for every $ε > 0$ . Assume now $∥ f ∥_{p} > 0$ , and put $M : = \sup_{ε} ∥ K_{ε} ∥_{1} < \infty$ . By [L2], choose $δ > 0$ so that $∥ τ_{y} f - f ∥_{p} < η / (2 M)$ whenever $∣ y ∣ < δ$ . By [L1], choose $ε_{0} > 0$ so that $\int_{∣ y ∣ \geq δ} ∣ K_{ε} (y) ∣ d y < \frac{η}{4 ∥ f ∥_{p}} (0 < ε < ε_{0}) .$ For such $ε$ , split the integral from step 1.1 into $∣ y ∣ < δ$ and $∣ y ∣ \geq δ$ . The near part is at most $η / 2$ . For the far part, [L3] gives $∥ τ_{y} f - f ∥_{p} \leq 2 ∥ f ∥_{p}$ , so $\int_{∣ y ∣ \geq δ} ∣ K_{ε} (y) ∣ ∥ τ_{y} f - f ∥_{p} d y \leq 2 ∥ f ∥_{p} \int_{∣ y ∣ \geq δ} ∣ K_{ε} (y) ∣ d y < η / 2.$ Hence $∥ f * K_{ε} - f ∥_{p} < η$ whenever $0 < ε < ε_{0}$ .

L1L2L3step 1.1choosealgebra

3.1

Because $η > 0$ was arbitrary, $∥ f * K_{ε} - f ∥_{p} \to 0$ as [step 2.1] $ε \to 0^{+}$ , proving the convergence in $L^{p}$ .

step 2.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01Open item page →

$L^{1}$ approximate identities converge uniformly on compacta for bounded continuous functions

Statement

Let $(K_{ε})_{ε > 0}$ be an $L^{1}$ approximate identity on $R^{n}$ , and let $f : R^{n} \to C$ be bounded and continuous. Then for every compact set $K \subseteq R^{n}$ ,

$\sup_{x \in K} ∣ (f * K_{ε}) (x) - f (x) ∣ ⟶ 0 (ε \to 0^{+}) .$

Facts & Assumptions

Given: An $L^{1}$ approximate identity, a bounded continuous function $f$ , and a compact set $K$ .

[L1]

Approximate identities are defined in An $L^{1}$ approximate identity on $R^{n}$ .

[L2]

Continuous functions are uniformly continuous on compact sets (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[L3]

Bounded sets in $R^{n}$ have finite measure (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

Proof

technique · direct

1.1

Let $η > 0$ . Choose $δ > 0$ so that [L2, L3, given, choose] $∣ f (x - y) - f (x) ∣ < η$ whenever $x \in K$ and $∣ y ∣ < δ$ ; this is possible by [L2] on a compact neighborhood of $K$ .

L2L3givenchoose

2.1

For $x \in K$ , [L1, step 1.1, algebra] $∣ (f * K_{ε}) (x) - f (x) ∣ \leq \int ∣ K_{ε} (y) ∣ ∣ f (x - y) - f (x) ∣ d y .$ Split the integral into $∣ y ∣ < δ$ and $∣ y ∣ \geq δ$ . The near part is at most $η ∥ K_{ε} ∥_{1}$ , while the far part is bounded by $2 ∥ f ∥_{\infty} \int_{∣ y ∣ \geq δ} ∣ K_{ε} (y) ∣ d y$ .

L1step 1.1algebra

3.1

The $L^{1}$ norms are uniformly bounded and the tail term tends to $0$ by [L1]. [L1, step 2.1] Hence first choose $η$ small enough and then $ε$ small enough to make the right-hand side uniformly small for all $x \in K$ . This is exactly the claimed uniform convergence on $K$ .

L1step 2.1∎

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

Statement

Assume the Axiom of Countable Choice.

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

Facts & Assumptions

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

[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]

Mollifier families are $L^{1}$ approximate identities, and convolution with a mollifier is smooth (A unit-mass smooth bump generates an $L^{1}$ approximate identity, Convolution with a mollifier is smooth, and derivatives pass under the integral sign).

[L3]

Approximate identities converge in $L^{p}$ , and for bounded continuous functions the convergence is uniform on compacta (Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ , $L^{1}$ 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

1.1

By [L1], choose $g \in C_{c} (R^{n})$ with $∥ f - g ∥_{p} < ε / 2$ . [L1, L2, given, choose] Let $(φ_{δ})$ be a mollifier family as in [L2].

L1L2givenchoose

2.1

By [L3], choose $δ > 0$ so small that [L2, L3, L4, step 1.1, choose] $∥ g * φ_{δ} - g ∥_{p} < ε / 2$ . The function $g * φ_{δ}$ is smooth by [L2]. Because $g$ has compact support and $φ_{δ}$ has compact support, [L4] gives compact support for $g * φ_{δ}$ , so $g * φ_{δ} \in C_{c}^{\infty} (R^{n})$ .

L2L3L4step 1.1choose

3.1

Therefore [step 1.1, step 2.1, algebra] $∥ f - g * φ_{δ} ∥_{p} \leq ∥ f - g ∥_{p} + ∥ g - g * φ_{δ} ∥_{p} < ε .$ Hence $C_{c}^{\infty} (R^{n})$ is dense in $L^{p} (R^{n})$ .

step 1.1step 2.1algebra∎

5 · Examples, counterexamples and false statements

None yet.

Density Separability and Convolution in LpLp

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Translation of a function on RnRn

Definition

The spaces Cc(Rn)Cc​(Rn) and Cc∞(Rn)Cc∞​(Rn)

Definition

The space C0(Rn)C0​(Rn) of continuous functions vanishing at infinity

Definition

Convolution of two functions on RnRn

Definition

An L1L1 approximate identity on RnRn

Definition

The mollifier family generated by a unit-mass smooth bump

Definition

Simple functions with finite-measure support are dense in Lp(μ)Lp(μ) for 1≤p<∞1≤p<∞

Statement

Facts & Assumptions

Proof

Simple functions are dense in L∞(μ)L∞(μ) in the essential-supremum norm

Statement

Facts & Assumptions

Proof

A finite-measure measurable set in RnRn is approximable in measure by a finite union of boxes

Statement

Facts & Assumptions

Proof

Finite linear combinations of box indicators are dense in Lp(Rn)Lp(Rn) for 1≤p<∞1≤p<∞

Statement

Facts & Assumptions

Proof

A compact set inside a bounded open set admits an explicit compactly supported continuous cutoff

Statement

Facts & Assumptions

Proof

A finite-measure measurable set in RnRn has a compact core and a bounded open neighbourhood of arbitrarily small excess

Statement

Facts & Assumptions

Proof

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

Statement

Facts & Assumptions

Proof

The L∞L∞-closure of Cc(Rn)Cc​(Rn) is C0(Rn)C0​(Rn), not all of L∞(Rn)L∞(Rn)

Statement

Facts & Assumptions

Proof

Rational box-step functions form a countable dense subset of Lp(Rn)Lp(Rn) for 1≤p<∞1≤p<∞

Statement

Facts & Assumptions

Proof

A countable generator of a sigma-algebra yields a countable algebra of sets

Statement

Facts & Assumptions

Proof

Finite-measure sets are approximable in measure by sets from a countable generating algebra

Statement

Facts & Assumptions

Proof

If μμ is sigma-finite and AA is countably generated, then Lp(μ)Lp(μ) is separable for 1≤p<∞1≤p<∞

Statement

Facts & Assumptions

Proof

L∞[0,1]L∞[0,1] is not separable

Statement

Facts & Assumptions

Proof

Continuous compactly supported functions are translation-continuous in LpLp

Statement

Facts & Assumptions

Proof

∥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

Borel representatives make the convolution integrand Borel measurable

Statement

Facts & Assumptions

Density Separability and Convolution in $L^{p}$

Translation of a function on $R^{n}$

The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$

The space $C_{0} (R^{n})$ of continuous functions vanishing at infinity

Convolution of two functions on $R^{n}$

An $L^{1}$ approximate identity on $R^{n}$

Simple functions with finite-measure support are dense in $L^{p} (μ)$ for $1 \leq p < \infty$

Simple functions are dense in $L^{\infty} (μ)$ in the essential-supremum norm

A finite-measure measurable set in $R^{n}$ is approximable in measure by a finite union of boxes

Finite linear combinations of box indicators are dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$

A finite-measure measurable set in $R^{n}$ has a compact core and a bounded open neighbourhood of arbitrarily small excess

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

The $L^{\infty}$ -closure of $C_{c} (R^{n})$ is $C_{0} (R^{n})$ , not all of $L^{\infty} (R^{n})$

Rational box-step functions form a countable dense subset of $L^{p} (R^{n})$ for $1 \leq p < \infty$

If $μ$ is sigma-finite and $A$ is countably generated, then $L^{p} (μ)$ is separable for $1 \leq p < \infty$

$L^{\infty} [0, 1]$ is not separable

Continuous compactly supported functions are translation-continuous in $L^{p}$

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

Convolution on $L^{1} (R^{n})$ is independent of the chosen Borel representatives

If $f, g \in L^{1} (R^{n})$ , then $f * g$ exists almost everywhere, belongs to $L^{1}$ , and $∥ f * g ∥_{1} \leq ∥ f ∥_{1} ∥ g ∥_{1}$

Convolution on $L^{1} (R^{n})$ is bilinear, commutative, and associative

For $1 < p < \infty$ , the $L^{p}$ norm of a nonnegative function is the supremum of its pairings with $L^{q}$ unit vectors

If $1 < p < \infty$ and $q$ is conjugate to $p$ , then $f * g$ is continuous and vanishes at infinity

A unit-mass smooth bump generates an $L^{1}$ approximate identity

Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$

$L^{1}$ approximate identities converge uniformly on compacta for bounded continuous functions

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