How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

Every result on this page is machine-checked by a proof checker and read in full by a delegated reviewing agent on the owner's instruction; the judge is an additional, independent cross-model AI review of the proofs. The 8 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Sequences and Series of Functions; Uniform Convergence: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

$f_{k} (x) = x^{k + 1}$ converges pointwise but not uniformly on $[0, 1]$

Statement refuted

Refuted claim: pointwise convergence of real-valued functions on a closed bounded interval implies uniform convergence.

For $k \in N$ define $f_{k} : [0, 1] \to R$ by

$f_{k} (x) : = x^{k + 1} .$

Then $(f_{k})$ converges pointwise to the endpoint indicator

$χ (x) : = {\begin{cases} 0, & 0 \leq x < 1, \\ 1, & x = 1, \end{cases}$

but the convergence is not uniform.

Facts & Assumptions

Given: The functions $f_{k} (x) = x^{k + 1}$ and the endpoint indicator $χ$ on $[0, 1]$ .

[L1]

If $∣ r ∣ < 1$ , then $r^{k} \to 0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ , Integer powers $a^{m}$ ).

[L2]

Bernoulli's inequality says $(1 + t)^{n} \geq 1 + ι (n) t$ for $t \geq - 1$ and $n \in N$ (Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L3]

The canonical naturals satisfy $ι (k + 1) > 0$ , and positive reciprocals reverse nonstrict inequalities (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L4]

Uniform convergence requires one index after which the error is below every prescribed positive real at every point of the domain (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Counterexample

technique · direct

1.1

If $0 \leq x < 1$ , then $∣ x ∣ < 1$ , so [L1] gives $x^{k + 1} \to 0 = χ (x)$ ; at $x = 1$ , one has $f_{k} (1) = 1 = χ (1)$ for every $k$ . Thus $f_{k} \to χ$ pointwise.

1.2

For each $k$ , put $a_{k} : = ι (k + 1) > 0$ and $y_{k} : = 1 - 1 / (2 a_{k})$ . Then $0 \leq y_{k} < 1$ , so $χ (y_{k}) = 0$ .

L3algebra

2.1

Apply [L2] with $n = k + 1$ and $t = - 1 / (2 a_{k})$ : $f_{k} (y_{k}) = {(1 - 1 / (2 a_{k}))}^{k + 1} \geq 1 - a_{k} / (2 a_{k}) = 1 / 2$ .

step 1.2L2L3algebra

3.1

Hence $∣ f_{k} (y_{k}) - χ (y_{k}) ∣ \geq 1 / 2$ for every $k$ , so no index makes the error smaller than $1 / 2$ at every point; the convergence is not uniform.

step 1.2step 2.1L4

4.1

The functions therefore satisfy the refuted claim's hypothesis and violate its conclusion.

step 1.1step 3.1∎

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

Shrinking rectangles converge pointwise to zero while every integral equals one

Statement refuted

Refuted claim: if Riemann-integrable functions on $[0, 1]$ converge pointwise to $0$ , then their integrals converge to $0$ .

For $k \in N$ put $a_{k} : = ι (k + 1)$ , the positive canonical natural in $R$ , and define

$r_{k} (x) : = {\begin{cases} a_{k}, & 0 < x \leq 1 / a_{k}, \\ 0, & x = 0 or 1 / a_{k} < x \leq 1. \end{cases}$

Then $r_{k} \to 0$ pointwise while $\int_{0}^{1} r_{k} = 1$ for every $k$ .

Facts & Assumptions

Given: The functions $r_{k}$ in the Statement, with $a_{k} = ι (k + 1) > 0$ .

[L1]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ ; canonical naturals increase and their positive reciprocals decrease (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

A bounded function on a closed interval with only finitely many possible discontinuities is Riemann integrable (A bounded function on $[a, b]$ that is continuous except at finitely many points is Riemann integrable, 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 $\int_{a}^{b} f$ ).

[L3]

Changing an integrable function at finitely many points preserves its integrability and integral (Changing an integrable function at finitely many points changes neither its integrability nor its integral).

[L4]

A constant $c$ has integral $c (v - u)$ on $[u, v]$ , and integrals add over adjacent subintervals, including the oriented convention at coincident endpoints (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ , 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 $a, b, c$ , The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L5]

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

Counterexample

technique · direct

1.1

Each $r_{k}$ is bounded and is continuous except possibly at $0$ and $1 / a_{k}$ , so it is integrable by [L2].

1.2

Let $q_{k}$ equal $a_{k}$ on $[0, 1 / a_{k}]$ and $0$ on $(1 / a_{k}, 1]$ . The functions $q_{k}$ and $r_{k}$ differ only at $0$ , so they have the same integral by [L3].

L3construct

1.3

At $x = 0$ one has $r_{k} (0) = 0$ for all $k$ . If $x > 0$ , choose $N$ with $1 / ι (N) < x$ ; for $k \geq N$ , monotonicity of the canonical naturals gives $1 / a_{k} < x$ , hence $r_{k} (x) = 0$ . Thus $r_{k} \to 0$ pointwise.

L1L5choose

1.4

To see explicitly that the convergence is not uniform, take $ε : = 1 / 2$ . For every proposed $N \in N$ , choose $k : = N$ and $x_{N} : = 1 / a_{N}$ ; then $∣ r_{N} (x_{N}) - 0 ∣ = a_{N} \geq 1 > ε$ . Thus the uniform quantifier condition in [L5] fails.

givenL1L5

2.1

By [L3] and [L4], endpoint values do not affect either piece, and splitting at $1 / a_{k}$ when it lies in the interior, with the coincident-endpoint convention otherwise, gives $\int_{0}^{1} q_{k} = a_{k} (1 / a_{k}) + 0 = 1$ .

step 1.2L3L4algebra

3.1

Steps 1.2 and 2.1 give $\int_{0}^{1} r_{k} = 1$ for every $k$ , whereas the integral of the zero function is $0$ .

step 1.2step 2.1L3L4

4.1

The sequence therefore converges pointwise to $0$ but its integrals do not converge to the integral of the limit, refuting the claim.

step 1.3step 3.1∎

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

Dini's theorem fails on $[0, \infty)$ : $x / (ι (k + 1) + x)$ decreases pointwise to zero but not uniformly

Statement refuted

Refuted claim: the compact-domain hypothesis in Dini's theorem can be dropped.

On $[0, \infty)$ define

$f_{k} (x) : = \frac{x}{ι (k + 1) + x} .$

The functions $f_{k}$ and their pointwise limit $0$ are continuous, and $f_{k + 1} (x) \leq f_{k} (x)$ for every $x \geq 0$ , but $f_{k} \to 0$ is not uniform.

Facts & Assumptions

Given: The functions $f_{k}$ in the Statement, with $a_{k} : = ι (k + 1) > 0$ .

[L1]

Constants and the identity are continuous; sums and quotients with nonvanishing denominator preserve continuity (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]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ , and the positive canonical naturals increase while their reciprocals decrease (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L3]

A subset of $R$ is compact exactly when it is closed and bounded; $[0, \infty)$ is unbounded (A subset of $R$ is compact if and only if it is closed and bounded, Lower bound, bounded below, bounded set).

[L4]

Dini's theorem on a closed interval concludes uniform convergence from continuity, pointwise monotonicity, and a continuous pointwise limit (Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform).

Counterexample

technique · direct

1.1

For every $k$ , the denominator $a_{k} + x$ is positive on $[0, \infty)$ , so $f_{k}$ is continuous by [L1]; the zero function is continuous as well.

givenL1

1.2

Since $a_{k + 1} > a_{k} > 0$ , one has $a_{k + 1} + x > a_{k} + x > 0$ , hence $f_{k + 1} (x) \leq f_{k} (x)$ for every $x \geq 0$ .

L2algebra

1.3

Fix $x \geq 0$ . If $x = 0$ then $f_{k} (x) = 0$ ; if $x > 0$ , then $0 \leq f_{k} (x) \leq x / a_{k}$ , and [L2] gives $x / a_{k} \to 0$ . Thus $f_{k} (x) \to 0$ for every $x$ .

L2algebra

1.4

At $x_{k} : = a_{k}$ one has $f_{k} (x_{k}) = a_{k} / (a_{k} + a_{k}) = 1 / 2$ , so the convergence is not uniform.

givenalgebra

1.5

The domain $[0, \infty)$ is not compact by [L3].

2.1

Hence all the listed Dini hypotheses except compactness hold, while the uniform conclusion fails; compactness cannot be dropped.

step 1.1step 1.2step 1.3step 1.4step 1.5L4∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

Dini's theorem fails for discontinuous approximants: shrinking interval indicators decrease pointwise to zero but not uniformly

Statement refuted

Refuted claim: continuity of the approximating functions in Dini's theorem can be dropped.

For $k \in N$ define $h_{k} : [0, 1] \to R$ to be the indicator of

$(0, 1 / ι (k + 1)) .$

Thus $h_{k}$ has value $0$ at both endpoints of that open interval. The sequence decreases pointwise to the continuous zero function but does not converge uniformly.

Facts & Assumptions

Given: The indicator functions $h_{k}$ in the Statement, with $a_{k} : = ι (k + 1) > 0$ .

[L1]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ , and positive canonical naturals increase while their reciprocals decrease (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

Continuity at $c$ requires that every positive output error admit a positive input radius on which all function values remain close to the value at $c$ (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).

[L3]

Dini's theorem on a closed interval assumes that every approximating function and the pointwise limit are continuous (Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform).

Counterexample

technique · direct

1.1

The intervals $(0, 1 / a_{k + 1})$ are contained in $(0, 1 / a_{k})$ , so $h_{k + 1} (x) \leq h_{k} (x)$ for every $x \in [0, 1]$ .

1.2

At $x = 0$ , every $h_{k} (x)$ is $0$ . If $x > 0$ , choose $N$ with $1 / ι (N) < x$ ; then $h_{k} (x) = 0$ for all $k \geq N$ . Thus $h_{k} \to 0$ pointwise.

L1choose

1.3

Each $h_{k}$ is discontinuous at $0$ : for any $δ > 0$ , the point $y : = \min {δ / 2, 1 / (2 a_{k})}$ satisfies $0 < y < δ$ , lies in $(0, 1 / a_{k})$ , and has $∣ h_{k} (y) - h_{k} (0) ∣ = 1$ .

L2algebra

1.4

For each $k$ , the point $x_{k} : = 1 / (2 a_{k})$ lies in $(0, 1 / a_{k})$ and satisfies $h_{k} (x_{k}) = 1$ , so the convergence to $0$ is not uniform.

givenL1

2.1

The compact domain, monotone pointwise convergence, and continuous limit remain, but the approximants are discontinuous and uniform convergence fails; their continuity is indispensable in [L3].

step 1.1step 1.2step 1.3step 1.4L3∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

Dini's theorem fails for a discontinuous limit: powers on $[0, 1]$ decrease pointwise to a discontinuous endpoint indicator but not uniformly

Statement refuted

Refuted claim: continuity of the pointwise limit in Dini's theorem can be dropped.

On $[0, 1]$ let $f_{k} (x) = x^{k + 1}$ . These continuous functions decrease pointwise to the discontinuous endpoint indicator

$χ (x) = {\begin{cases} 0, & 0 \leq x < 1, \\ 1, & x = 1, \end{cases}$

and the convergence is not uniform.

Facts & Assumptions

Given: The functions $f_{k} (x) = x^{k + 1}$ and the endpoint indicator $χ$ on $[0, 1]$ .

[L1]

The powers $x^{k + 1}$ converge pointwise to $χ$ on $[0, 1]$ and do not converge uniformly there ( $f_{k} (x) = x^{k + 1}$ converges pointwise but not uniformly on $[0, 1]$ ).

[L2]

Every polynomial function, hence every natural power, is 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, Integer powers $a^{m}$ , 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).

[L3]

Dini's theorem on a closed interval requires the approximating functions and their pointwise limit to be continuous (Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform).

[L4]

Continuity at $c$ requires that every positive output error admit a positive input radius on which all nearby function values remain close to the value at $c$ (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).

Counterexample

technique · direct

1.1

Each $f_{k}$ is continuous by [L2].

1.2

For $x \in [0, 1]$ , $f_{k + 1} (x) = x^{k + 2} = x x^{k + 1} \leq x^{k + 1} = f_{k} (x)$ , so the sequence is pointwise nonincreasing.

givenalgebra

1.3

The pointwise convergence to $χ$ and the failure of uniform convergence are [L1].

1.4

The function $χ$ is discontinuous at $1$ : for every $δ > 0$ , the point $y : = 1 - \min {δ / 2, 1 / 2}$ lies in $[0, 1)$ with $∣ y - 1 ∣ < δ$ and $∣ χ (y) - χ (1) ∣ = 1$ .

L4algebra

2.1

Thus compactness, continuity of all approximants, and monotonicity hold, but the limit is discontinuous and the uniform conclusion fails; continuity of the limit in [L3] is indispensable.

step 1.1step 1.2step 1.3step 1.4L3∎

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

Continuous triangular spikes on $[0, 1]$ converge pointwise to zero but not uniformly when monotonicity is absent

Statement refuted

Refuted claim: the monotonicity hypothesis in Dini's theorem can be dropped.

For $k \in N$ put $a_{k} : = ι (k + 1)$ and define the triangular spike

$h_{k} (x) : = \max {0, 1 - ∣ 2 a_{k} x - 1 ∣} (0 \leq x \leq 1) .$

Each $h_{k}$ is continuous and $h_{k} \to 0$ pointwise, but the convergence is not uniform.

Facts & Assumptions

Given: The functions $h_{k}$ in the Statement, with $a_{k} = ι (k + 1) > 0$ .

[L1]

Constants, the identity, sums, products, absolute values, and pointwise maxima 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, Maximum and minimum of a set, 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]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ ; canonical naturals increase and positive reciprocals decrease (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L3]

Absolute value is nonnegative and has the usual multiplicative law (Basic properties of the absolute value).

[L4]

Dini's theorem on a closed interval requires one pointwise monotonicity direction for the whole sequence (Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform).

Counterexample

technique · direct

1.1

Every $h_{k}$ is continuous by [L1], and the zero function is continuous.

1.2

If $x = 0$ , then $h_{k} (x) = 0$ . If $x > 0$ , choose $N$ with $1 / ι (N) < x$ ; for all sufficiently large $k$ , $a_{k} x \geq 1$ , so $∣ 2 a_{k} x - 1 ∣ \geq 1$ and $h_{k} (x) = 0$ . Thus $h_{k} \to 0$ pointwise.

L2L3choosealgebra

1.3

At $x_{k} : = 1 / (2 a_{k})$ one has $h_{k} (x_{k}) = 1$ , so the convergence is not uniform.

givenalgebra

1.4

At $x = 1 / 4$ , the values at $k = 0, 1, 2$ are respectively $1 / 2, 1, 1 / 2$ , so the sequence is neither pointwise nondecreasing nor pointwise nonincreasing.

givenalgebra

2.1

All Dini hypotheses except monotonicity hold while uniform convergence fails, so monotonicity cannot be dropped.

step 1.1step 1.2step 1.3step 1.4L4∎

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

The double sequence $(m + 1) / (m + n + 2)$ has unequal iterated limits

Statement refuted

Refuted claim: whenever both iterated limits of a double real sequence exist, they are equal.

For $m, n \in N$ define

$s_{m, n} : = \frac{ι (m + 1)}{ι (m + n + 2)} = \frac{ι (m + 1)}{ι (m + 1) + ι (n + 1)} .$

Then

$\lim_{m} \lim_{n} s_{m, n} = 0, \lim_{n} \lim_{m} s_{m, n} = 1.$

The shifts make the expression defined at the first index $(m, n) = (0, 0)$ .

Facts & Assumptions

Given: The double sequence $s_{m, n}$ in the Statement.

[L1]

The canonical-natural map satisfies $ι (0) = 0$ and $ι (r + 1) = ι (r) + 1$ ; positive canonical naturals increase, and their reciprocals decrease (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L3]

A real sequence converges when its terms are eventually within every positive error of the proposed limit (Limits and Cauchy sequences of reals).

Counterexample

technique · direct

1.1

The denominator $ι (m + 1) + ι (n + 1)$ is positive for all $m, n$ , so every $s_{m, n}$ is defined and lies between $0$ and $1$ .

L1algebra

1.2

Fix $m$ and put $a : = ι (m + 1) > 0$ . Then $0 \leq s_{m, n} \leq a / ι (n + 1)$ ; given $ε > 0$ , [L2] makes the latter smaller than $ε$ for all sufficiently large $n$ . Hence $\lim_{n} s_{m, n} = 0$ .

L1L2L3choosealgebra

1.3

Fix $n$ and put $b : = ι (n + 1) > 0$ . Since $∣ 1 - s_{m, n} ∣ = b / (ι (m + 1) + b) \leq b / ι (m + 1)$ , [L2] makes this smaller than any prescribed $ε > 0$ for all sufficiently large $m$ . Hence $\lim_{m} s_{m, n} = 1$ .

L1L2L3choosealgebra

2.1

By step 1.2 the first inner-limit sequence is constantly $0$ , so its limit in $m$ is $0$ ; by step 1.3 the other inner-limit sequence is constantly $1$ , so its limit in $n$ is $1$ .

step 1.2step 1.3L3

3.1

Thus both iterated limits exist and are unequal, refuting the claim.

step 2.1∎

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

$x / (1 + (k + 1)^{2} x^{2})$ converges uniformly to zero on $R$ while every derivative at zero equals one

Statement refuted

Refuted claim: if differentiable functions converge uniformly, their derivatives must converge to the derivative of the limit.

For $k \in N$ put $a_{k} : = ι (k + 1)$ and define

$f_{k} (x) : = \frac{x}{1 + a_{k}^{2} x^{2}} (x \in R) .$

Then $f_{k} \to 0$ uniformly on $R$ , but $f_{k}^{'} (0) = 1$ for every $k$ , whereas the derivative of the zero function is $0$ .

Facts & Assumptions

Given: The functions $f_{k}$ in the Statement, with $a_{k} = ι (k + 1) > 0$ .

[L1]

Every square in an ordered field is nonnegative, with a nonzero square positive; absolute value is multiplicative (Squares of nonzero elements are positive, Basic properties of the absolute value).

[L2]

Natural powers are differentiable by the power rule, and sums, products, and quotients with nonzero denominator obey the usual derivative rules (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, 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, Integer powers $a^{m}$ ).

[L3]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L4]

Uniform convergence requires one index controlling the error at every point (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Counterexample

technique · direct

1.1

The denominator $1 + a_{k}^{2} x^{2}$ is positive for every $x$ , so $f_{k}$ is differentiable on $R$ by [L2].

L1L2

1.2

From $(a_{k} ∣ x ∣ - 1)^{2} \geq 0$ one obtains $2 a_{k} ∣ x ∣ \leq 1 + a_{k}^{2} x^{2}$ , hence $∣ f_{k} (x) ∣ \leq 1 / (2 a_{k})$ for every $x \in R$ .

L1algebra

1.3

At $x = 0$ , the numerator $x$ has derivative $1$ , the denominator $1 + a_{k}^{2} x^{2}$ has value $1$ and derivative $0$ , so the quotient rule gives $f_{k}^{'} (0) = 1$ .

L2algebra

1.4

The zero function has derivative $0$ by the constant case of the power rule.

2.1

Given $ε > 0$ , [L3] gives $N$ such that $1 / (2 a_{k}) < ε$ for every $k \geq N$ ; step 1.2 then gives $∣ f_{k} (x) ∣ < ε$ for every $x \in R$ .

step 1.2L3choose

3.1

Step 2.1 proves $f_{k} \to 0$ uniformly, while steps 1.3 and 1.4 show that the derivatives at $0$ do not converge to the derivative of the limit.

step 2.1step 1.3step 1.4L4

4.1

The uniformly convergent differentiable sequence therefore refutes the claim.

step 3.1∎

Sequences and Series of Functions; Uniform Convergence: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

fk(x)=xk+1fk​(x)=xk+1 converges pointwise but not uniformly on [0,1][0,1]

Statement refuted

Facts & Assumptions

Counterexample

Shrinking rectangles converge pointwise to zero while every integral equals one

Statement refuted

Facts & Assumptions

Counterexample

Dini's theorem fails on [0,∞)[0,∞): x/(ι(k+1)+x)x/(ι(k+1)+x) decreases pointwise to zero but not uniformly

Statement refuted

Facts & Assumptions

Counterexample

Dini's theorem fails for discontinuous approximants: shrinking interval indicators decrease pointwise to zero but not uniformly

Statement refuted

Facts & Assumptions

Counterexample

Dini's theorem fails for a discontinuous limit: powers on [0,1][0,1] decrease pointwise to a discontinuous endpoint indicator but not uniformly

Statement refuted

Facts & Assumptions

Counterexample

Continuous triangular spikes on [0,1][0,1] converge pointwise to zero but not uniformly when monotonicity is absent

Statement refuted

Facts & Assumptions

Counterexample

The double sequence (m+1)/(m+n+2)(m+1)/(m+n+2) has unequal iterated limits

Statement refuted

Facts & Assumptions

Counterexample

x/(1+(k+1)2x2)x/(1+(k+1)2x2) converges uniformly to zero on RR while every derivative at zero equals one

Statement refuted

Facts & Assumptions

Counterexample

Sources

$f_{k} (x) = x^{k + 1}$ converges pointwise but not uniformly on $[0, 1]$

Dini's theorem fails on $[0, \infty)$ : $x / (ι (k + 1) + x)$ decreases pointwise to zero but not uniformly

Dini's theorem fails for a discontinuous limit: powers on $[0, 1]$ decrease pointwise to a discontinuous endpoint indicator but not uniformly

Continuous triangular spikes on $[0, 1]$ converge pointwise to zero but not uniformly when monotonicity is absent

The double sequence $(m + 1) / (m + n + 2)$ has unequal iterated limits

$x / (1 + (k + 1)^{2} x^{2})$ converges uniformly to zero on $R$ while every derivative at zero equals one