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

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$f_{k} (x) = x^{k + 1}$ 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∎

Depends on

Used by

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

Dependency tree · two levels

37 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

J. Lebl, Basic Analysis I, §6.1 (standard reference, not scraped)
W. Trench, Introduction to Real Analysis (standard reference, not scraped)

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