LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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 ${f} \cup {f_{n} : n \in N}$ is equicontinuous.

Facts & Assumptions

Given: $a \in K$ and $ε > 0$ .

[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 $C (K, R)$ ).

Proof

technique · direct

1.1

Choose $N$ such that $∣ f_{n} (x) - f (x) ∣ < ε / 3$ for every $n \geq N$ and $x \in K$ .

givenchoose

1.2

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

L1choose

2.1

On that neighbourhood, the triangle inequality gives $∣ f_{n} (x) - f_{n} (a) ∣ < ε$ for $n \geq N$ , 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 · two levels

7 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

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