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ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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For real continuous functions modulo constants, the quotient norm is half the oscillation

Example

Let K be a nonempty compact metric space, let C(K,R) carry the supremum norm, and let M be the subspace of constant functions. For fC(K,R) write

osc(f):=maxKfminKf.

Then in the quotient C(K,R)/M,

f+M=12osc(f).

Facts & Assumptions

Given: A nonempty compact metric space K, a real-valued continuous function f on K, and the constant-function subspace M.

[L1]

The quotient seminorm is f+M=infcRfc (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L2]

A continuous real-valued function on a nonempty compact metric space attains its maximum and minimum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L3]

The quotient seminorm is a norm when the subspace is closed (The quotient seminorm is a norm exactly when the subspace is closed).

Verification

technique · direct
1.1

By [L2], let m:=minKf and M0:=maxKf, and put c0:=(M0+m)/2. Then for every xK, mf(x)M0, so f(x)c0(M0m)/2. Hence fc0(M0m)/2.

L2algebra
2.1

For any real constant c, both M0c and mc are bounded above by fc. Since the distance between M0 and m is M0m, at least one of those two numbers is at least (M0m)/2. Therefore fc(M0m)/2 for every c.

step 1.1L2algebra
3.1

Steps 1.1 and 2.1 give infcRfc=(M0m)/2, so [L1] yields f+M=osc(f)/2. The constant subspace is closed because a uniform limit of constant functions is constant, so [L3] confirms that this is an honest norm on the quotient.

step 1.1step 2.1L1L3

Depends on

Used by

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Dependency tree · two levels

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Sources