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✓ 15 results · all verified · 0 also independently AI-judged

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Sequences and Series of Functions; Uniform Convergence

1 · Prerequisites

2 · Summary

Real convergence and the Cauchy criterion supply the pointwise and completeness arguments, while metric continuity and compactness control functions on metric domains. The supremum metric for bounded functions, Heine-Borel compactness, the Riemann criterion, the fundamental theorem of calculus, scalar series tests, and Abel summation provide the estimates used for completeness, integration, differentiation, and function series.

Pointwise, uniform, and uniformly Cauchy convergence are defined by their quantifier order, followed by the uniform Cauchy criterion and its function-series form. Algebraic permanence, continuity of uniform limits, completeness of $C (K, R)$ , interchange with integration, and the derivative-limit theorem develop the principal consequences. The M-test and uniform Dirichlet and Abel tests give convergence criteria for function series, while the closed-interval Dini theorem converts monotone pointwise convergence of continuous functions into uniform convergence.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions

Definition

Let $X$ be a set and, for each $k \in N$ , let $f_{k} : X \to R$ be a real-valued function (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ ). Let $f : X \to R$ .

The sequence $(f_{k})$ converges pointwise to $f$ on $X$ when, for every $x \in X$ , the real sequence $(f_{k} (x))$ converges to $f (x)$ (Limits and Cauchy sequences of reals). Thus the index after which $∣ f_{k} (x) - f (x) ∣ < ε$ may depend on both $ε$ and $x$ .

The sequence $(f_{k})$ converges uniformly to $f$ on $X$ when

$(\forall ε > 0) (\exists N \in N) (\forall k \geq N) (\forall x \in X) ∣ f_{k} (x) - f (x) ∣ < ε,$

where $ε$ ranges over the positive reals. Here one index $N$ serves every point of $X$ .

The sequence $(f_{k})$ is uniformly Cauchy on $X$ when

$(\forall ε > 0) (\exists N \in N) (\forall m, n \geq N) (\forall x \in X) ∣ f_{m} (x) - f_{n} (x) ∣ < ε .$

For each of the three notions above, restricting the error to positive rationals gives an equivalent condition. The real-error condition immediately implies the rational-error condition. Conversely, given a real $η > 0$ , choose $n \geq 1$ with $1 / n < η$ by For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ; the condition for the positive rational $1 / n$ implies the condition for $η$ . The real-error form is used because it makes the uniform quantifiers transparent.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Uniform convergence of real-valued functions implies pointwise convergence

Statement

Let $X$ be a set. If a sequence of functions $f_{k} : X \to R$ converges uniformly to $f : X \to R$ , then it converges pointwise to $f$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Facts & Assumptions

Given: A set $X$ , functions $f_{k}, f : X \to R$ , and uniform convergence $f_{k} \to f$ on $X$ .

[A1]

Uniform convergence means that for every real $ε > 0$ there is $N \in N$ such that $∣ f_{k} (x) - f (x) ∣ < ε$ for every $k \geq N$ and every $x \in X$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Proof

technique · direct

1.1

Fix $x \in X$ and a real $ε > 0$ . By [A1] choose $N \in N$ such that $∣ f_{k} (y) - f (y) ∣ < ε$ for every $k \geq N$ and every $y \in X$ .

A1choose

2.1

In particular, $∣ f_{k} (x) - f (x) ∣ < ε$ for every $k \geq N$ .

step 1.1

3.1

Since $x$ and $ε$ were arbitrary, $f_{k} (x) \to f (x)$ for every $x \in X$ , which is pointwise convergence.

step 2.1A1∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy

Statement

Let $X$ be a set and let $f_{k} : X \to R$ for every $k \in N$ . Then $(f_{k})$ converges uniformly on $X$ to some $f : X \to R$ if and only if $(f_{k})$ is uniformly Cauchy on $X$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Facts & Assumptions

Given: A set $X$ and a sequence of functions $f_{k} : X \to R$ .

[A1]

Uniform convergence to $f$ means that for every real $ε > 0$ there is $N$ such that $∣ f_{k} (x) - f (x) ∣ < ε$ for every $k \geq N$ and every $x \in X$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A2]

Uniform Cauchyness means that for every real $ε > 0$ there is $N$ such that $∣ f_{m} (x) - f_{n} (x) ∣ < ε$ for every $m, n \geq N$ and every $x \in X$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A3]

Pointwise convergence as defined through real sequences can equivalently be tested with every positive real error (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[L1]

For reals $u, v$ , $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

[L2]

Every Cauchy sequence of reals converges to a real (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).

Proof

technique · direct

1.1

Suppose first that $f_{k} \to f$ uniformly on $X$ , and let $ε > 0$ be real. By [A1] choose $N$ with $∣ f_{j} (x) - f (x) ∣ < ε / 2$ for every $j \geq N$ and $x \in X$ . Thus, for $m, n \geq N$ and $x \in X$ , $∣ f_{m} (x) - f_{n} (x) ∣ \leq ∣ f_{m} (x) - f (x) ∣ + ∣ f (x) - f_{n} (x) ∣ < ε$ . Since $ε$ was arbitrary, $(f_{k})$ is uniformly Cauchy.

A1A2L1choosealgebra

1.2

Conversely, suppose that $(f_{k})$ is uniformly Cauchy on $X$ . For each $x \in X$ , [A2] makes $(f_{k} (x))$ a Cauchy real sequence; by [L2] it has a real limit $f (x)$ . These values define a function $f : X \to R$ .

A2L2construct

1.3

Under this converse assumption, let $ε > 0$ be real and choose $N$ such that $∣ f_{m} (x) - f_{n} (x) ∣ < ε / 2$ for every $m, n \geq N$ and every $x \in X$ .

A2choose

2.1

Fix $k \geq N$ and $x \in X$ . Pointwise convergence at $x$ gives a threshold $M$ such that $∣ f_{n} (x) - f (x) ∣ < ε / 2$ for $n \geq M$ . Choose $n \geq N, M$ . Then $∣ f_{k} (x) - f (x) ∣ \leq ∣ f_{k} (x) - f_{n} (x) ∣ + ∣ f_{n} (x) - f (x) ∣ < ε$ .

step 1.2step 1.3A3L1choose

3.1

The index $N$ in step 1.3 is independent of $k$ and $x$ , so step 2.1 proves $f_{k} \to f$ uniformly. Together with step 1.1 this proves both directions.

step 1.1step 2.1A1∎

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

A series of real-valued functions and its pointwise and uniform convergence through its partial sums

Definition

Let $X$ be a set and let $f_{k} : X \to R$ for $k \in N$ . The series of real-valued functions $\sum f_{k}$ is studied through its partial-sum functions

$S_{n} (x) : = \sum_{k < n} f_{k} (x) (n \in N, x \in X),$

where the sum on the right is the finite sum of Series, partial sums, convergence and the sum, divergence, and the tail series. Thus $S_{0}$ is the zero function and $S_{n + 1} = S_{n} + f_{n}$ under the pointwise operations of The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ .

The series $\sum f_{k}$ converges pointwise to $S : X \to R$ when $S_{n} \to S$ pointwise, and it converges uniformly to $S$ when $S_{n} \to S$ uniformly (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

The series is absolutely convergent at $x \in X$ when the scalar series $\sum ∣ f_{k} (x) ∣$ converges. It is absolutely pointwise convergent when this holds for every $x \in X$ .

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

A series of real-valued functions converges uniformly if and only if its tails are uniformly small

Statement

Let $X$ be a set and let $f_{k} : X \to R$ . The function series $\sum f_{k}$ converges uniformly on $X$ if and only if, for every real $ε > 0$ , there is $N \in N$ such that

$∣ \sum_{k = m + 1}^{n} f_{k} (x) ∣ < ε$

for every $n > m \geq N$ and every $x \in X$ .

Facts & Assumptions

Given: A set $X$ , functions $f_{k} : X \to R$ , and partial-sum functions $S_{r} (x) = \sum_{k < r} f_{k} (x)$ .

[L1]

The series $\sum f_{k}$ converges uniformly exactly when its partial-sum sequence $(S_{r})$ converges uniformly (A series of real-valued functions and its pointwise and uniform convergence through its partial sums).

[L2]

A sequence of real-valued functions converges uniformly exactly when it is uniformly Cauchy (A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy).

[L3]

For $n > m$ , $\sum_{k = m + 1}^{n} f_{k} (x) = S_{n + 1} (x) - S_{m + 1} (x)$ , and $∣ - u ∣ = ∣ u ∣$ for every real $u$ (A series of real-valued functions and its pointwise and uniform convergence through its partial sums, Basic properties of the absolute value).

Proof

technique · direct

1.1

Suppose first that $\sum f_{k}$ converges uniformly, and let $ε > 0$ be real.

1.2

Conversely, suppose the displayed tail condition holds, and fix a real $ε > 0$ and a corresponding index $N$ .

given

2.1

By [L1] and [L2], choose $K$ such that $∣ S_{p} (x) - S_{q} (x) ∣ < ε$ for every $p, q \geq K$ and every $x \in X$ .

step 1.1L1L2choose

2.2

Put $K : = N + 1$ . If $p, q \geq K$ and $x \in X$ , then the difference is $0$ when $p = q$ ; if $p > q$ , set $m = q - 1 \geq N$ and $n = p - 1 > m$ , so the tail condition and [L3] give $∣ S_{p} (x) - S_{q} (x) ∣ < ε$ ; the case $q > p$ follows by symmetry of absolute value.

step 1.2L3algebra

3.1

For $n > m \geq K$ and $x \in X$ , the indices $n + 1, m + 1$ are at least $K$ , so [L3] and step 2.1 give $∣ \sum_{k = m + 1}^{n} f_{k} (x) ∣ < ε$ .

step 2.1L3

3.2

Thus $(S_{r})$ is uniformly Cauchy, hence converges uniformly by [L2], and therefore $\sum f_{k}$ converges uniformly by [L1].

step 2.2L1L2

4.1

Steps 3.1 and 3.2 prove the two implications, hence the equivalence.

step 3.1step 3.2∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Uniform limits respect sums and scalar multiples

Statement

Let $X$ be a set. Suppose $f_{k} \to f$ and $g_{k} \to g$ uniformly on $X$ , where all functions are real valued. Then, for all $α, β \in R$ ,

$α f_{k} + β g_{k} ⟶ α f + β g$

uniformly on $X$ . In particular, uniform convergence is preserved by sums, differences, and scalar multiples.

Facts & Assumptions

Given: A set $X$ , uniformly convergent sequences $f_{k} \to f$ and $g_{k} \to g$ , and reals $α, β$ .

[A1]

Uniform convergence gives, for every real $η > 0$ , one index after which $∣ f_{k} (x) - f (x) ∣ < η$ at every $x$ , and likewise for $g_{k} \to g$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[L1]

For reals $u, v$ , $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ , while $∣ c u ∣ = ∣ c ∣ ∣ u ∣$ and absolute values are nonnegative (The triangle inequality, Basic properties of the absolute value).

Proof

technique · direct

1.1

Let $ε > 0$ be real and put $η : = ε / (2 (∣ α ∣ + ∣ β ∣ + 1)) > 0$ .

constructL1

1.2

By uniform convergence choose $N_{f}, N_{g}$ such that $∣ f_{k} (x) - f (x) ∣ < η$ for $k \geq N_{f}$ and all $x$ , and $∣ g_{k} (x) - g (x) ∣ < η$ for $k \geq N_{g}$ and all $x$ .

A1choose

2.1

Choose an index $N$ at least as large as $N_{f}$ and $N_{g}$ . For every $k \geq N$ and $x \in X$ , one has $∣ α (f_{k} (x) - f (x)) + β (g_{k} (x) - g (x)) ∣ \leq ∣ α ∣ η + ∣ β ∣ η < ε$ .

step 1.1step 1.2L1choosealgebra

3.1

The expression in step 2.1 is $∣ (α f_{k} + β g_{k}) (x) - (α f + β g) (x) ∣$ , and the index $N$ serves every $x$ , so the asserted convergence is uniform.

step 2.1A1∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Products converge uniformly when both factors converge uniformly and one limiting factor and one approximating family are uniformly bounded

Statement

Let $X$ be a set, and suppose $f_{k} \to f$ and $g_{k} \to g$ uniformly on $X$ . Assume there are reals $B, C \geq 0$ such that

$∣ f (x) ∣ \leq B and ∣ g_{k} (x) ∣ \leq C$

for every $x \in X$ and every $k \in N$ . Then $f_{k} g_{k} \to f g$ uniformly on $X$ .

The same conclusion holds after interchanging the two factors: it is enough that one limit function and the approximating sequence of the other factor have uniform bounds.

Facts & Assumptions

Given: Uniform convergence $f_{k} \to f$ and $g_{k} \to g$ on $X$ , with bounds $∣ f (x) ∣ \leq B$ and $∣ g_{k} (x) ∣ \leq C$ for all $x, k$ .

[A1]

Uniform convergence gives one index serving all points for any prescribed positive real error (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A2]

A subset of $R$ is bounded when it has real lower and upper bounds; the displayed absolute-value inequalities are the corresponding uniform bounds on the ranges (Lower bound, bounded below, bounded set).

[L1]

For reals $u, v, c$ , $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ and $∣ c u ∣ = ∣ c ∣ ∣ u ∣$ (The triangle inequality, Basic properties of the absolute value).

Proof

technique · direct

1.1

Let $ε > 0$ be real and put $η : = ε / (B + C + 1) > 0$ .

construct

1.2

Choose $N$ such that, for every $k \geq N$ and every $x \in X$ , both $∣ f_{k} (x) - f (x) ∣ < η$ and $∣ g_{k} (x) - g (x) ∣ < η$ .

A1choose

2.1

For $k \geq N$ and $x \in X$ , add and subtract $f (x) g_{k} (x)$ to obtain $∣ f_{k} (x) g_{k} (x) - f (x) g (x) ∣ \leq ∣ g_{k} (x) ∣ ∣ f_{k} (x) - f (x) ∣ + ∣ f (x) ∣ ∣ g_{k} (x) - g (x) ∣ < (B + C) η < ε$ .

step 1.1step 1.2A2L1algebra

3.1

Since $N$ is independent of $x$ , step 2.1 proves $f_{k} g_{k} \to f g$ uniformly. Interchanging the names of the factors gives the symmetric clause.

step 2.1A1∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The uniform limit of continuous real-valued functions on a metric space is continuous

Statement

Let $(X, d)$ be a metric space and let $f_{k} : X \to R$ be continuous for every $k \in N$ , where $R$ has its usual metric $d_{R} (s, t) = ∣ s - t ∣$ . If $f_{k} \to f$ uniformly on $X$ , then $f : X \to R$ is continuous.

Facts & Assumptions

Given: A metric space $(X, d)$ , continuous functions $f_{k} : X \to R$ , and uniform convergence $f_{k} \to f$ .

[A1]

Uniform convergence gives, for every real $η > 0$ , one index $N$ such that $∣ f_{k} (x) - f (x) ∣ < η$ for every $k \geq N$ and every $x \in X$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A2]

Continuity of $f_{N}$ at $a$ means that for every real $η > 0$ there is $δ > 0$ such that $d (x, a) < δ$ implies $∣ f_{N} (x) - f_{N} (a) ∣ < η$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded).

[L1]

For reals $u, v$ , $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

Proof

technique · direct

1.1

Fix $a \in X$ and a real $ε > 0$ . By uniform convergence choose $N$ such that $∣ f_{N} (x) - f (x) ∣ < ε / 3$ for every $x \in X$ .

A1choose

1.2

By continuity of $f_{N}$ at $a$ , choose $δ > 0$ such that $d (x, a) < δ$ implies $∣ f_{N} (x) - f_{N} (a) ∣ < ε / 3$ .

A2choose

2.1

If $d (x, a) < δ$ , then $∣ f (x) - f (a) ∣ \leq ∣ f (x) - f_{N} (x) ∣ + ∣ f_{N} (x) - f_{N} (a) ∣ + ∣ f_{N} (a) - f (a) ∣ < ε$ .

step 1.1step 1.2L1algebra

3.1

Thus $f$ is continuous at the arbitrary point $a$ , and hence continuous on $X$ .

step 2.1A2∎

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The uniform limit of uniformly continuous real-valued functions is uniformly continuous

Statement

Let $(X, d)$ be a metric space. If each $f_{k} : X \to R$ is uniformly continuous and $f_{k} \to f$ uniformly on $X$ , then $f$ is uniformly continuous.

Facts & Assumptions

Given: A metric space $(X, d)$ , uniformly continuous functions $f_{k} : X \to R$ , and uniform convergence $f_{k} \to f$ .

[A1]

Uniform convergence gives one index serving every point for any prescribed positive real error (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[A2]

Uniform continuity of $f_{N}$ means that for every real $η > 0$ there is $δ > 0$ such that $d (x, y) < δ$ implies $∣ f_{N} (x) - f_{N} (y) ∣ < η$ for all $x, y \in X$ (Uniform continuity of a map of metric spaces: one $δ$ serving every point, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded).

[L1]

For reals $u, v$ , $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

Proof

technique · direct

1.1

Let $ε > 0$ be real. Choose $N$ such that $∣ f_{N} (x) - f (x) ∣ < ε / 3$ for every $x \in X$ .

A1choose

1.2

By uniform continuity of $f_{N}$ , choose $δ > 0$ such that $d (x, y) < δ$ implies $∣ f_{N} (x) - f_{N} (y) ∣ < ε / 3$ for every $x, y \in X$ .

A2choose

2.1

If $d (x, y) < δ$ , then $∣ f (x) - f (y) ∣ \leq ∣ f (x) - f_{N} (x) ∣ + ∣ f_{N} (x) - f_{N} (y) ∣ + ∣ f_{N} (y) - f (y) ∣ < ε$ .

step 1.1step 1.2L1algebra

3.1

The same $δ$ serves every pair $x, y$ , so $f$ is uniformly continuous.

step 2.1A2∎

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The space $C (K, R)$ of continuous real-valued functions on a nonempty compact metric space

Definition

Let $(K, d)$ be a nonempty compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space). Define

$C (K, R) : = {f \in R^{K} : f : (K, d) \to (R, d_{R}) is continuous},$

where $R^{K}$ is the function space of The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ and $d_{R} (s, t) = ∣ s - t ∣$ is the usual metric (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).

This definition introduces the set of continuous functions only. Boundedness and the supremum metric are assertions to be proved, not clauses of the definition.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

$C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$

Statement

Let $(K, d)$ be a nonempty compact metric space. Every member of $C (K, R)$ is bounded, so the supremum metric

$d_{\infty} (f, g) : = \sup_{x \in K} ∣ f (x) - g (x) ∣$

is defined on $C (K, R)$ . With this metric, $C (K, R)$ is complete.

Facts & Assumptions

Given: A nonempty compact metric space $(K, d)$ and the set $C (K, R)$ of continuous real-valued functions on it.

[L1]

Every continuous real-valued function on a nonempty compact metric space has a bounded range (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L2]

If $S$ is nonempty, the formula $d_{\infty} (f, g) = \sup_{x \in S} ∣ f (x) - g (x) ∣$ defines a metric on the set of bounded functions $S \to R$ (The supremum metric $d_{\infty} (f, g) = \sup_{x} ∣ f (x) - g (x) ∣$ is a metric on the bounded real-valued functions on a nonempty set).

[L3]

A sequence is Cauchy in a metric $d$ when, for every positive error, all pairwise distances sufficiently far out are below that error; it converges to $p$ when its distances to $p$ tend to zero (Cauchy sequence in a metric space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[L4]

A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy (A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy).

[L5]

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

[L6]

A metric space is complete when every Cauchy sequence in it converges to one of its points (Complete metric space: every Cauchy sequence converges in the space).

Proof

technique · direct

1.1

By [L1], every $f \in C (K, R)$ is bounded. Thus $C (K, R)$ is a subset of the bounded functions on $K$ , and the restriction of the metric in [L2] is a metric on $C (K, R)$ .

L1L2

1.2

Let $(f_{j})$ be a Cauchy sequence in this supremum metric.

givenL3

2.1

Given a real $ε > 0$ , choose $J$ such that $d_{\infty} (f_{m}, f_{n}) < ε$ for all $m, n \geq J$ . Then $∣ f_{m} (x) - f_{n} (x) ∣ \leq d_{\infty} (f_{m}, f_{n}) < ε$ for all such $m, n$ and every $x \in K$ , so $(f_{j})$ is uniformly Cauchy.

step 1.2L2L3

3.1

By [L4] there is a function $f : K \to R$ such that $f_{j} \to f$ uniformly on $K$ .

step 2.1L4

4.1

The function $f$ is continuous by [L5], hence belongs to $C (K, R)$ and is bounded by [L1].

step 3.1L1L5

5.1

Let $ε > 0$ . Uniform convergence gives $J$ such that $∣ f_{j} (x) - f (x) ∣ < ε / 2$ for every $j \geq J$ and $x \in K$ ; hence $d_{\infty} (f_{j}, f) \leq ε / 2 < ε$ , so $f_{j} \to f$ in the supremum metric.

step 3.1step 4.1L2L3

6.1

Every Cauchy sequence in $C (K, R)$ therefore converges in the supremum metric to a member of $C (K, R)$ , so the metric space is complete.

step 1.1step 5.1L6∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error

Statement

Let $u, v \in R$ , and let $f$ and $g$ be integrable between $u$ and $v$ . If $η \geq 0$ and

$∣ f (x) - g (x) ∣ \leq η$

throughout the closed interval with endpoints $u$ and $v$ , then

$∣ \int_{u}^{v} f - \int_{u}^{v} g ∣ \leq η ∣ v - u ∣ .$

Facts & Assumptions

Given: Reals $u, v$ , functions $f, g$ integrable between them, and a real $η \geq 0$ with $∣ f - g ∣ \leq η$ on the interval between them.

[L1]

Linear combinations of integrable functions are integrable and their integrals are the corresponding linear combinations, including for oriented limits (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L2]

If $a < b$ , an integrable function $h$ satisfying $m \leq h (x) \leq M$ on $[a, b]$ has $m (b - a) \leq \int_{a}^{b} h \leq M (b - a)$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L3]

For every real $z$ , $- ∣ z ∣ \leq z \leq ∣ z ∣$ and $∣ - z ∣ = ∣ z ∣$ ; for $c > 0$ , $∣ z ∣ < c$ exactly when $- c < z < c$ (Basic properties of the absolute value).

Proof

technique · direct

1.1

If $u = v$ , both oriented integrals are $0$ and the asserted inequality holds.

L1algebra

1.2

Suppose $u < v$ and put $h : = f - g$ . Then $h$ is integrable and $\int_{u}^{v} h = \int_{u}^{v} f - \int_{u}^{v} g$ .

2.1

The hypothesis gives $∣ h (x) ∣ \leq η$ , while [L3] gives $- ∣ h (x) ∣ \leq h (x) \leq ∣ h (x) ∣$ ; hence $- η \leq h (x) \leq η$ on $[u, v]$ , and [L2] gives $- η (v - u) \leq \int_{u}^{v} h \leq η (v - u)$ .

step 1.2L2L3

3.1

Hence $∣ \int_{u}^{v} f - \int_{u}^{v} g ∣ = ∣ \int_{u}^{v} h ∣ \leq η (v - u)$ when $u < v$ .

step 1.2step 2.1L3

4.1

If $u > v$ , apply step 3.1 to the ordered pair $(v, u)$ and use antisymmetry of oriented integrals; the same bound results because $∣ u - v ∣ = ∣ v - u ∣$ .

step 3.1L1L3

5.1

The alternatives $u = v$ , $u < v$ , and $u > v$ are exhaustive, and steps 1.1, 3.1, and 4.1 give the claimed inequality.

step 1.1step 3.1step 4.1∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

A uniform limit of Riemann-integrable functions is Riemann integrable, and its integral is the limit of their integrals

Statement

Let $a < b$ be reals. Suppose every $f_{k} : [a, b] \to R$ is Riemann integrable and $f_{k} \to f$ uniformly on $[a, b]$ . Then $f$ is Riemann integrable and

$\int_{a}^{b} f_{k} ⟶ \int_{a}^{b} f .$

Facts & Assumptions

Given: Reals $a < b$ , integrable functions $f_{k} : [a, b] \to R$ , and uniform convergence $f_{k} \to f$ .

[A1]

Uniform convergence means that for every real $η > 0$ one index makes $∣ f_{k} (x) - f (x) ∣ < η$ for every later $k$ and every $x \in [a, b]$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[L1]

An integrable function on $[a, b]$ is bounded; conversely, a bounded function $h$ there is Riemann integrable exactly when, for every real $ε > 0$ , some partition $P$ satisfies $U (h, P) - L (h, P) < ε$ (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ ).

[L2]

Darboux upper and lower sums are finite sums of the subinterval suprema and infima times the subinterval lengths; finite sums preserve inequalities and split and telescope in the usual way (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , Laws of finite sums and finite products).

[L3]

If two integrable functions differ by at most $η$ uniformly, then their integrals differ by at most $η (b - a)$ (Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

Proof

technique · direct

1.1

Let $ε > 0$ be real, put $η : = ε / (4 (b - a)) > 0$ , and choose an index $j$ such that $∣ f_{j} (x) - f (x) ∣ < η$ for every $x \in [a, b]$ .

A1choose

1.2

By integrability of $f_{j}$ and [L1], choose a partition $P$ with $U (f_{j}, P) - L (f_{j}, P) < ε / 2$ .

L1choose

2.1

The integrable function $f_{j}$ is bounded, say $∣ f_{j} (x) ∣ \leq M$ on $[a, b]$ ; then $∣ f (x) ∣ \leq M + η$ , so $f$ is bounded.

step 1.1L1algebra

3.1

On each subinterval of $P$ , step 1.1 gives $\sup f \leq \sup f_{j} + η$ and $\inf f \geq \inf f_{j} - η$ ; these suprema and infima exist by step 2.1. Multiplying by the nonnegative subinterval lengths and summing gives $U (f, P) \leq U (f_{j}, P) + η (b - a)$ and $L (f, P) \geq L (f_{j}, P) - η (b - a)$ .

step 1.1step 1.2step 2.1L2algebra

4.1

Therefore $U (f, P) - L (f, P) \leq U (f_{j}, P) - L (f_{j}, P) + 2 η (b - a) < ε$ , so [L1] makes $f$ integrable.

step 3.1L1algebra

5.1

Now let $ε > 0$ be real and choose $N$ such that $∣ f_{k} (x) - f (x) ∣ < ε / (b - a + 1)$ for every $k \geq N$ and every $x \in [a, b]$ .

step 4.1A1choose

6.1

For $k \geq N$ , both functions are integrable, so [L3] gives $∣ \int_{a}^{b} f_{k} - \int_{a}^{b} f ∣ \leq ε (b - a) / (b - a + 1) < ε$ .

step 4.1step 5.1L3algebra

7.1

Step 6.1 proves $\int_{a}^{b} f_{k} \to \int_{a}^{b} f$ , while step 4.1 proves integrability of $f$ .

step 4.1step 6.1∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

If continuously differentiable functions converge at one point and their derivatives converge uniformly on a closed interval, then the functions converge uniformly to a differentiable function whose derivative is the derivative limit

Statement

Let $a < b$ be reals and let $f_{k} : [a, b] \to R$ be continuously differentiable: each $f_{k}$ is differentiable on $[a, b]$ and each derivative $f_{k}^{'}$ is continuous there (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). Suppose there is $c \in [a, b]$ such that the real sequence $(f_{k} (c))$ converges to $A$ , and suppose $f_{k}^{'} \to g$ uniformly on $[a, b]$ . Then there is a differentiable function $f : [a, b] \to R$ such that

$f_{k} ⟶ f uniformly, f^{'} = g .$

Facts & Assumptions

Given: Reals $a < b$ , a point $c \in [a, b]$ , continuously differentiable functions $f_{k} : [a, b] \to R$ , convergence $f_{k} (c) \to A$ , and uniform convergence $f_{k}^{'} \to g$ .

[L1]

A uniform limit of continuous real-valued functions is continuous, and a continuous function on $[a, b]$ is Riemann integrable (The uniform limit of continuous real-valued functions on a metric space is continuous, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[L2]

Uniform convergence of integrable functions preserves integrability and the limit of the integrals (A uniform limit of Riemann-integrable functions is Riemann integrable, and its integral is the limit of their integrals).

[L3]

If $s < t$ , $G$ is differentiable on $[s, t]$ , and $G^{'}$ is integrable there, then $\int_{s}^{t} G^{'} = G (t) - G (s)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ). Restriction to a closed subinterval preserves differentiability and the derivative at its limit points, and integrability on $[a, b]$ passes to every nondegenerate closed subinterval (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, A function integrable on $[a, b]$ is integrable on every closed subinterval). Finally $\int_{t}^{s} h = - \int_{s}^{t} h$ and $\int_{s}^{s} h = 0$ whenever the displayed integrals are defined (The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L4]

If $g$ is continuous, its integral function $H (x) = \int_{a}^{x} g$ is differentiable with $H^{'} = g$ ; oriented additivity gives $\int_{c}^{x} g = H (x) - H (c)$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

[L5]

Sums and scalar multiples of differentiable functions are differentiable, with the corresponding derivative rules (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L6]

A uniform bound $∣ p - q ∣ \leq η$ on an interval gives $∣ \int_{u}^{v} p - \int_{u}^{v} q ∣ \leq η ∣ v - u ∣$ (Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error).

[L7]

On a subset of $R$ with its usual subspace metric, real-native continuity is equivalent to metric-space continuity (Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace).

Proof

technique · constructive

1.1

By [L7], each real-continuous derivative $f_{k}^{'}$ is metric-continuous. The uniform-limit clause of [L1] makes $g$ metric-continuous, and [L7] makes $g$ real-continuous. The integrability clause of [L1] therefore makes every $f_{k}^{'}$ and $g$ Riemann integrable; [L2] also gives the integrability of the uniform limit.

L1L2L7

1.2

Let $ε > 0$ . Choose $N_{0}$ such that $∣ f_{k} (c) - A ∣ < ε / 2$ for $k \geq N_{0}$ , and choose $N_{1}$ such that $∣ f_{k}^{'} (x) - g (x) ∣ < ε / (2 (b - a + 1))$ for $k \geq N_{1}$ and all $x \in [a, b]$ .

givenchoose

2.1

Fix $k$ and $x \in [a, b]$ . If $c < x$ , restrict $f_{k}$ to $[c, x]$ ; its derivative is $f_{k}^{'}$ there and that derivative is integrable there by steps 1.1 and [L3], so the first clause of [L3] gives $f_{k} (x) = f_{k} (c) + \int_{c}^{x} f_{k}^{'}$ . If $x < c$ , apply that clause on $[x, c]$ and then use orientation; if $x = c$ , use $\int_{c}^{c} f_{k}^{'} = 0$ . Thus in every case $f_{k} (x) = f_{k} (c) + \int_{c}^{x} f_{k}^{'}$ .

step 1.1L3

2.2

Define $H (x) : = \int_{a}^{x} g$ and construct $f : [a, b] \to R$ by $f (x) : = A + H (x) - H (c) = A + \int_{c}^{x} g$ .

step 1.1L4construct

3.1

By [L4] and [L5], $H$ is differentiable with $H^{'} = g$ , and therefore the constructed function $f = A + H - H (c)$ is differentiable with $f^{'} = g$ .

step 2.2L4L5

3.2

Choose $N$ at least as large as $N_{0}, N_{1}$ . For $k \geq N$ and $x \in [a, b]$ , steps 2.2 and 2.1 with [L6] give $∣ f_{k} (x) - f (x) ∣ \leq ∣ f_{k} (c) - A ∣ + ∣ \int_{c}^{x} f_{k}^{'} - \int_{c}^{x} g ∣ < ε / 2 + ε ∣ x - c ∣ / (2 (b - a + 1)) < ε$ .

step 2.2step 2.1step 1.2L6choosealgebra

4.1

The index $N$ in step 3.2 serves every $x \in [a, b]$ , so $f_{k} \to f$ uniformly; step 3.1 gives $f^{'} = g$ . Thus the constructed $f$ has both asserted properties.

step 3.2step 3.1discharge-construct∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

The Weierstrass M-test gives absolute pointwise convergence and uniform convergence of a function series

Statement

Let $X$ be a set and let $f_{k} : X \to R$ . Suppose there is a sequence of nonnegative reals $(M_{k})$ such that

$∣ f_{k} (x) ∣ \leq M_{k} (k \in N, x \in X)$

and the scalar series $\sum M_{k}$ converges. Then $\sum f_{k} (x)$ converges absolutely for every $x \in X$ , and the function series $\sum f_{k}$ converges uniformly on $X$ .

Facts & Assumptions

Given: Functions $f_{k} : X \to R$ and nonnegative reals $M_{k}$ with $∣ f_{k} (x) ∣ \leq M_{k}$ for all $k, x$ , such that $\sum M_{k}$ converges.

[L1]

If $0 \leq a_{k} \leq b_{k}$ eventually and $\sum b_{k}$ converges, then $\sum a_{k}$ converges (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

[L2]

A convergent real series has uniformly small scalar tails: for every real $ε > 0$ there is $N$ such that $∣ \sum_{k = m + 1}^{n} a_{k} ∣ < ε$ whenever $n > m \geq N$ (A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$ ).

[L3]

Repeated triangle inequalities for finite sums give $∣ \sum_{k = p}^{q} u_{k} ∣ \leq \sum_{k = p}^{q} ∣ u_{k} ∣$ , and finite sums preserve termwise inequalities (The triangle inequality, Basic properties of the absolute value, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

A function series converges uniformly exactly when its tails are uniformly small (A series of real-valued functions converges uniformly if and only if its tails are uniformly small, A series of real-valued functions and its pointwise and uniform convergence through its partial sums).

Proof

technique · direct

1.1

Fix $x \in X$ . Since $0 \leq ∣ f_{k} (x) ∣ \leq M_{k}$ for every $k$ , [L1] shows that $\sum ∣ f_{k} (x) ∣$ converges.

givenL1

1.2

Let $ε > 0$ . By [L2] choose $N$ such that $\sum_{k = m + 1}^{n} M_{k} < ε$ whenever $n > m \geq N$ , the absolute value being unnecessary because the terms are nonnegative.

L2choose

2.1

For $n > m \geq N$ and $x \in X$ , one has $∣ \sum_{k = m + 1}^{n} f_{k} (x) ∣ \leq \sum_{k = m + 1}^{n} ∣ f_{k} (x) ∣ \leq \sum_{k = m + 1}^{n} M_{k} < ε$ .

step 1.2L3

3.1

Step 1.1 gives absolute pointwise convergence, while step 2.1 and [L4] give uniform convergence of $\sum f_{k}$ .

step 1.1step 2.1L4∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Uniform Dirichlet test: uniformly bounded partial sums times a uniformly decreasing null family give a uniformly convergent function series

Statement

Let $X$ be a set and let $u_{k}, v_{k} : X \to R$ . Put $A_{n} (x) : = \sum_{k < n} u_{k} (x)$ . Suppose:

there is $M \geq 0$ such that $∣ A_{n} (x) ∣ \leq M$ for every $n$ and $x$ ;
$v_{k} (x) \geq 0$ and $v_{k + 1} (x) \leq v_{k} (x)$ for every $k$ and $x$ ;
$v_{k} \to 0$ uniformly on $X$ .

Then the function series $\sum u_{k} v_{k}$ converges uniformly on $X$ .

Facts & Assumptions

Given: Functions $u_{k}, v_{k} : X \to R$ satisfying the three hypotheses in the Statement, with partial sums $A_{n} (x) = \sum_{k < n} u_{k} (x)$ .

[L1]

Abel summation by parts expresses a finite sum $\sum_{j < r} a_{j} b_{j}$ as $A_{r} b_{r - 1} - \sum_{j < r - 1} A_{j + 1} (b_{j + 1} - b_{j})$ , where $A_{r} = \sum_{j < r} a_{j}$ (Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$ ).

[L2]

Finite sums split and telescope, preserve inequalities, and obey the triangle inequality after repeated use of $∣ s + t ∣ \leq ∣ s ∣ + ∣ t ∣$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality, Basic properties of the absolute value).

[L3]

A function series converges uniformly exactly when its tails are uniformly small (A series of real-valued functions converges uniformly if and only if its tails are uniformly small).

Proof

technique · direct

1.1

Let $ε > 0$ . Uniform convergence $v_{k} \to 0$ gives $N$ such that $0 \leq v_{k} (x) < ε / (2 M + 1)$ for every $k \geq N$ and every $x \in X$ .

givenchoose

1.2

Fix $n > m \geq N$ and $x \in X$ , put $p : = m + 1$ , $q : = n$ , and for $p \leq r \leq q$ put $B_{r} (x) : = \sum_{k = p}^{r} u_{k} (x) = A_{r + 1} (x) - A_{p} (x)$ .

L2construct

2.1

For every $p \leq r \leq q$ , the bound on the $A_{j}$ gives $∣ B_{r} (x) ∣ \leq ∣ A_{r + 1} (x) ∣ + ∣ A_{p} (x) ∣ \leq 2 M$ .

step 1.2L2

2.2

Applying [L1] to the shifted finite list from $p$ through $q$ gives $\sum_{k = p}^{q} u_{k} (x) v_{k} (x) = B_{q} (x) v_{q} (x) + \sum_{k = p}^{q - 1} B_{k} (x) (v_{k} (x) - v_{k + 1} (x))$ .

step 1.2L1L2

3.1

Since $v_{k} (x) - v_{k + 1} (x) \geq 0$ , steps 2.1 and 2.2 with telescoping give $∣ \sum_{k = p}^{q} u_{k} (x) v_{k} (x) ∣ \leq 2 M v_{q} (x) + 2 M \sum_{k = p}^{q - 1} (v_{k} (x) - v_{k + 1} (x)) = 2 M v_{p} (x) < ε$ .

step 1.1step 2.1step 2.2L2algebra

4.1

The estimate in step 3.1 holds for every $n > m \geq N$ and $x \in X$ , so [L3] proves uniform convergence of $\sum u_{k} v_{k}$ .

step 3.1L3∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Uniform Abel test: a uniformly convergent function series times a uniformly bounded pointwise monotone family gives a uniformly convergent product series

Statement

Let $X$ be a set and let $u_{k}, v_{k} : X \to R$ . Suppose $\sum u_{k}$ converges uniformly on $X$ , there is $M \geq 0$ with $∣ v_{k} (x) ∣ \leq M$ for every $k, x$ , and for each fixed $x$ the real sequence $(v_{k} (x))$ is monotone. Its direction may depend on $x$ . Then $\sum u_{k} v_{k}$ converges uniformly on $X$ .

Facts & Assumptions

Given: Functions $u_{k}, v_{k} : X \to R$ satisfying the hypotheses in the Statement.

[L1]

Uniform convergence of $\sum u_{k}$ is equivalent to uniformly small tails (A series of real-valued functions converges uniformly if and only if its tails are uniformly small).

[L2]

For real sequences $(a_{j}), (b_{j})$ and $A_{n} : = \sum_{j < n} a_{j}$ , Abel summation by parts says that, for every $n \geq 1$ , $\sum_{j < n} a_{j} b_{j} = A_{n} b_{n - 1} - \sum_{j < n - 1} A_{j + 1} (b_{j + 1} - b_{j})$ (Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$ ).

[L3]

Finite sums split and telescope, and repeated triangle inequalities bound the absolute value of a finite sum by the sum of the absolute values (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality, Basic properties of the absolute value).

Proof

technique · direct

1.1

Let $ε > 0$ and put $η : = ε / (3 M + 1) > 0$ . By [L1] choose $N$ such that $∣ \sum_{k = m + 1}^{n} u_{k} (x) ∣ < η$ for every $n > m \geq N$ and every $x \in X$ .

L1choose

1.2

For every $x \in X$ and naturals $p \leq q$ , monotonicity makes all successive differences $v_{k} (x) - v_{k + 1} (x)$ have one sign, so $\sum_{k = p}^{q - 1} ∣ v_{k} (x) - v_{k + 1} (x) ∣ = ∣ v_{p} (x) - v_{q} (x) ∣ \leq 2 M$ .

givenL3algebra

2.1

Fix $n > m \geq N$ and $x \in X$ , put $p : = m + 1$ , $q : = n$ , and define $B_{r} (x) : = \sum_{k = p}^{r} u_{k} (x)$ for $p \leq r \leq q$ . Then $∣ B_{r} (x) ∣ < η$ for every such $r$ .

step 1.1construct

3.1

For $0 \leq j \leq q - p$ put $a_{j} : = u_{p + j} (x)$ and $b_{j} : = v_{p + j} (x)$ . Their partial sums satisfy $A_{j + 1} = B_{p + j} (x)$ , so [L2] with $n = q - p + 1$ gives $\sum_{k = p}^{q} u_{k} (x) v_{k} (x) = B_{q} (x) v_{q} (x) + \sum_{k = p}^{q - 1} B_{k} (x) (v_{k} (x) - v_{k + 1} (x))$ .

step 2.1L2L3

4.1

By steps 2.1, 3.1, and 1.2, $∣ \sum_{k = p}^{q} u_{k} (x) v_{k} (x) ∣ \leq η ∣ v_{q} (x) ∣ + η \sum_{k = p}^{q - 1} ∣ v_{k} (x) - v_{k + 1} (x) ∣ \leq 3 M η < ε$ .

step 2.1step 3.1step 1.2L3algebra

5.1

The estimate is uniform in $x$ and holds for every $n > m \geq N$ , so [L1] proves uniform convergence of $\sum u_{k} v_{k}$ .

step 4.1L1∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform

Statement

Let $a < b$ be reals. Suppose $f_{k} : [a, b] \to R$ and $f : [a, b] \to R$ are continuous, $f_{k} \to f$ pointwise, and the sequence is pointwise monotone in one fixed direction:

$f_{k} (x) \leq f_{k + 1} (x) for all k, x,$

$f_{k + 1} (x) \leq f_{k} (x) for all k, x .$

Then $f_{k} \to f$ uniformly on $[a, b]$ .

Facts & Assumptions

Given: Reals $a < b$ , continuous functions $f_{k}, f : [a, b] \to R$ , pointwise convergence $f_{k} \to f$ , and one of the two pointwise monotonicity conditions in the Statement.

[A1]

Uniform convergence means that for every real $ε > 0$ there is $N$ such that $∣ f_{k} (x) - f (x) ∣ < ε$ for every $k \geq N$ and every $x \in [a, b]$ (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[L1]

Sums and scalar multiples of continuous real functions are continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L2]

If $h : A \to R$ is continuous, the inverse image of an open subset of $R$ is relatively open in $A$ : it is $O \cap A$ for some open $O \subseteq R$ ( $f : A \to R$ is continuous on $A$ if and only if the preimage of every open subset of $R$ is the intersection with $A$ of an open subset of $R$ , and dually for closed sets).

[L3]

Every open cover of the closed bounded interval $[a, b]$ has a finite subcover (Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact).

[L4]

Every finite list of natural numbers has a greatest member: apply the finite-real maximum theorem to their canonical images, which preserve the natural-number order; finite choices can be made without any choice axiom (Every nonempty finite set of reals has a maximum and a minimum, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, $\leq$ is a linear order on $N$ , Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Proof

technique · direct

1.1

Fix a real $ε > 0$ . For each $n \in N$ put $h_{n} : = f_{n} - f$ and $U_{n} : = {x \in [a, b] : ∣ h_{n} (x) ∣ < ε}$ . The function $h_{n}$ is continuous by [L1], so [L2] makes $U_{n}$ relatively open in $[a, b]$ .

L1L2construct

1.2

In the nondecreasing case, pointwise convergence forces $f_{n} (x) \leq f (x)$ for every $n, x$ , and the errors $f (x) - f_{n} (x)$ decrease with $n$ ; in the nonincreasing case it forces $f (x) \leq f_{n} (x)$ and the errors $f_{n} (x) - f (x)$ decrease. Thus $U_{n} \subseteq U_{n + 1}$ in either case, and pointwise convergence gives $⋃_{n} U_{n} = [a, b]$ .

givenalgebra

2.1

Let $O$ be the family of all open sets $O \subseteq R$ whose trace $O \cap [a, b]$ equals $U_{n}$ for some $n$ . By step 1.1 each $U_{n}$ has such an open witness, and by step 1.2 the family $O$ covers $[a, b]$ .

step 1.1step 1.2L2construct

3.1

By [L3], choose finitely many $O_{0}, \dots, O_{r} \in O$ covering $[a, b]$ . By finite choice, choose $n_{0}, \dots, n_{r}$ with $O_{i} \cap [a, b] = U_{n_{i}}$ , and let $N : = \max {n_{0}, \dots, n_{r}}$ .

step 2.1L3L4choose

4.1

Since the $U_{n}$ are increasing, every $U_{n_{i}}$ is contained in $U_{N}$ ; the traces of the $O_{i}$ cover $[a, b]$ , so $U_{N} = [a, b]$ , and then $U_{k} = [a, b]$ for every $k \geq N$ .

step 1.2step 3.1

5.1

Therefore $∣ f_{k} (x) - f (x) ∣ < ε$ for every $k \geq N$ and every $x \in [a, b]$ . Since $ε$ was arbitrary, the convergence is uniform.

step 4.1A1∎

RemarkRemark: AI-adaptedProof: Not applicableverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

Agreement of the quantified real-valued definition with the later uniform-metric and uniform-topology formulations

For bounded real-valued functions on a nonempty set, the quantified condition of Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions agrees with convergence in the supremum metric of The supremum metric $d_{\infty} (f, g) = \sup_{x} ∣ f (x) - g (x) ∣$ is a metric on the bounded real-valued functions on a nonempty set. Indeed, uniform error below $ε$ gives supremum distance at most $ε$ , while uniform error below $ε / 2$ gives supremum distance strictly below $ε$ ; the converse follows because every pointwise error is at most the supremum distance.

The general function-space formulation is Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C (X, Y)$ ↗, with its convergence dictionary Convergence in the uniform metric is exactly uniform convergence: one $N$ serving every point ↗. The metric-target uniform limit theorem A uniform limit of continuous functions is continuous, so $C (X, Y)$ is closed in $Y^{X}$ under the uniform metric ↗ and the compact-metric version of Dini's theorem Dini's theorem: on a compact metric space a nondecreasing sequence of continuous real functions converging pointwise to a continuous limit converges uniformly ↗ extend the real-valued results proved here. These links are included only for orientation.

5 · Examples, counterexamples and false statements

None yet.

Sequences and Series of Functions; Uniform Convergence

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions

Definition

Uniform convergence of real-valued functions implies pointwise convergence

Statement

Facts & Assumptions

Proof

A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy

Statement

Facts & Assumptions

Proof

A series of real-valued functions and its pointwise and uniform convergence through its partial sums

Definition

A series of real-valued functions converges uniformly if and only if its tails are uniformly small

Statement

Facts & Assumptions

Proof

Uniform limits respect sums and scalar multiples

Statement

Facts & Assumptions

Proof

Products converge uniformly when both factors converge uniformly and one limiting factor and one approximating family are uniformly bounded

Statement

Facts & Assumptions

Proof

The uniform limit of continuous real-valued functions on a metric space is continuous

Statement

Facts & Assumptions

Proof

The uniform limit of uniformly continuous real-valued functions is uniformly continuous

Statement

Facts & Assumptions

Proof

The space C(K,R)C(K,R) of continuous real-valued functions on a nonempty compact metric space

Definition

C(K,R)C(K,R) is complete in the supremum metric for every nonempty compact metric space KK

Statement

Facts & Assumptions

Proof

Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error

Statement

Facts & Assumptions

Proof

A uniform limit of Riemann-integrable functions is Riemann integrable, and its integral is the limit of their integrals

Statement

Facts & Assumptions

Proof

If continuously differentiable functions converge at one point and their derivatives converge uniformly on a closed interval, then the functions converge uniformly to a differentiable function whose derivative is the derivative limit

Statement

Facts & Assumptions

Proof

The Weierstrass M-test gives absolute pointwise convergence and uniform convergence of a function series

Statement

Facts & Assumptions

Proof

Uniform Dirichlet test: uniformly bounded partial sums times a uniformly decreasing null family give a uniformly convergent function series

Statement

Facts & Assumptions

Proof

Uniform Abel test: a uniformly convergent function series times a uniformly bounded pointwise monotone family gives a uniformly convergent product series

Statement

Facts & Assumptions

Proof

Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform

Statement

Facts & Assumptions

Proof

Agreement of the quantified real-valued definition with the later uniform-metric and uniform-topology formulations

5 · Examples, counterexamples and false statements

Sources

The space $C (K, R)$ of continuous real-valued functions on a nonempty compact metric space

$C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$