LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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A uniformly convergent sequence of continuous functions, together with its limit, is equicontinuous

Statement

If $f_n\to f$ uniformly on a compact metric space and every $f_n$ is continuous, then $\{f\}\cup\{f_n:n\in\mathbb N\}$ is equicontinuous.

Facts & Assumptions

Given: $a\in K$ and $\varepsilon>0$ .

[L1]

A uniform limit of continuous real-valued functions is continuous (The uniform limit of continuous real-valued functions on a metric space is continuous).

[L2]

Equicontinuity is the common-radius condition for a family of functions (Equicontinuity, pointwise boundedness, and uniform boundedness for families in $C(K,\mathbb R)$ ).

Proof

technique · direct

1.1

Choose $N$ such that $|f_n(x)-f(x)|<\varepsilon/3$ for every $n\ge N$ and $x\in K$ .

givenchoose

1.2

By [L1], choose a neighbourhood of $a$ on which $|f(x)-f(a)|<\varepsilon/3$ ; by continuity of the finitely many $f_n$ with $n<N$ , shrink it so the same inequality with $\varepsilon$ holds for all of them.

L1choose

2.1

On that neighbourhood, the triangle inequality gives $|f_n(x)-f_n(a)|<\varepsilon$ for $n\ge N$ , while step 1.2 covers $f$ and $n<N$ .

step 1.1step 1.2algebra

3.1

Thus the displayed family is equicontinuous at arbitrary $a$ , hence equicontinuous.

step 2.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 23 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

The Ascoli--Arzelà Theorem (MIT) (standard reference, not scraped)