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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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The uniform closure of a unital real function algebra is closed under absolute value, maximum, and minimum

Statement

If AC(K,R)A\subseteq C(K,\mathbb R) is a unital real function algebra and A\overline A is its uniform closure, then uAu\in\overline A implies uA|u|\in\overline A; consequently uvu\vee v and uvu\wedge v lie in A\overline A whenever u,vu,v do.

Facts & Assumptions

Given: A unital real function algebra AA and u,vAu,v\in\overline A.

[L1]

The algebra operations and constants are those in A unital point-separating real subalgebra of C(K,R)C(K,\mathbb R).

[L2]

Polynomials uniformly approximate t|t| on every bounded closed interval (Polynomials are uniformly dense in C([a,b],R)C([a,b],\mathbb R) for every closed interval).

Proof

technique · direct
1.1

Choose anAa_n\in A converging uniformly to uu. Their ranges, together with that of uu, lie in one bounded interval.

givenchoose
1.2

By [L2], choose polynomials pjp_j with pj(0)=0p_j(0)=0 converging uniformly to t|t| on that interval. Then pj(an)Ap_j(a_n)\in A by [L1].

L1L2choose
2.1

A diagonal choice of j,nj,n makes pj(an)p_j(a_n) uniformly converge to u|u|, so uA|u|\in\overline A.

step 1.1step 1.2algebra
3.1

The identities uv=(u+v+uv)/2u\vee v=(u+v+|u-v|)/2 and uv=(u+vuv)/2u\wedge v=(u+v-|u-v|)/2 and [L1] give the remaining closure.

step 2.1L1algebra

Depends on

Used by

Dependency tree · next 3 levels

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