TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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.

The Takagi series converges uniformly to a continuous nowhere differentiable function

Statement

The Takagi series $T (x) = \sum_{n \geq 0} 2^{- n} ϕ (2^{n} x)$ converges uniformly on $[0, 1]$ . Its sum is continuous and has no finite derivative at any point of $[0, 1]$ , with one-sided derivatives at the endpoints understood.

Facts & Assumptions

Given: The tent function and series are as in The tent function $ϕ (t) = dist (t, Z)$ and the Takagi series $T (x) = \sum_{n \geq 0} 2^{- n} ϕ (2^{n} x)$ .

[L1]

The Weierstrass M-test gives uniform convergence from a summable uniform majorant (The Weierstrass M-test gives absolute pointwise convergence and uniform convergence of a function series).

[L2]

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

[L3]

Differentiability requires convergence of the relevant difference quotients (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).

Proof

technique · direct

1.1

Since $0 \leq ϕ \leq 1 / 2$ , the $n$ th summand is bounded by $2^{- n - 1}$ ; its majorant series converges. Thus $T$ converges uniformly by [L1].

givenL1algebra

1.2

For $x \in [0, 1)$ and each $N$ , let $u_{N} \leq x < v_{N}$ be the adjacent dyadic rationals of order $N$ ; at $x = 1$ use the left adjacent interval. Every summand of index at least $N$ vanishes at both endpoints, while each earlier summand is affine on that interval with slope $ε_{k} \in {- 1, 1}$ .

givenalgebra

2.1

Every summand is continuous, so $T$ is continuous by [L2].

step 1.1L2algebra

2.2

Hence $(T (v_{N}) - T (u_{N})) / (v_{N} - u_{N}) = \sum_{k < N} ε_{k}$ . These secant slopes cannot converge to a finite real number, because consecutive partial sums differ by $ε_{N}$ of absolute value one.

step 1.2algebra

3.1

If a finite derivative existed at an interior point, both endpoint difference quotients and therefore their secant combination would tend to it; at a dyadic point or endpoint the same argument uses the nested one-sided dyadic intervals. This contradicts step 2.2 and [L3].

step 2.2L3algebra∎

Depends on

Used by

A continuous nowhere differentiable singular one simplex Counterexample
Fixed-time assertions do not yield a pathwise nowhere statement Counterexample
FALSE: every continuous function is differentiable almost everywhere False statement
Classical counterparts for the trigonometry-free oscillators Remark

Dependency tree · two levels

24 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 Takagi function: a survey (standard reference, not scraped)