ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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 $f \in C (K, R)$ write

$osc (f) : = \max_{K} f - \min_{K} f .$

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

$∥ f + M ∥ = \frac{1}{2} osc (f) .$

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 $∥ f + M ∥ = \inf_{c \in R} ∥ f - c ∥_{\infty}$ (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

1.1L2algebra

By [L2], let $m : = \min_{K} f$ and $M_{0} : = \max_{K} f$ , and put $c_{0} : = (M_{0} + m) / 2$ . Then for every $x \in K$ , $m \leq f (x) \leq M_{0}$ , so $∣ f (x) - c_{0} ∣ \leq (M_{0} - m) / 2$ . Hence $∥ f - c_{0} ∥_{\infty} \leq (M_{0} - m) / 2$ .

2.1step 1.1L2algebra

For any real constant $c$ , both $∣ M_{0} - c ∣$ and $∣ m - c ∣$ are bounded above by $∥ f - c ∥_{\infty}$ . Since the distance between $M_{0}$ and $m$ is $M_{0} - m$ , at least one of those two numbers is at least $(M_{0} - m) / 2$ . Therefore $∥ f - c ∥_{\infty} \geq (M_{0} - m) / 2$ for every $c$ .

3.1step 1.1step 2.1L1L3∎

Steps 1.1 and 2.1 give $\inf_{c \in R} ∥ f - c ∥_{\infty} = (M_{0} - m) / 2$ , 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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

28 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

Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)
Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)