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

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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∎

Depends on

Used by

FALSE: C_c(ℝⁿ) is dense in L^∞(ℝⁿ) False statement

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Walter Rudin, Real and Complex Analysis, 3rd ed. (standard reference, not scraped)

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