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Under ab>1+3π/2, the classical Weierstrass function is continuous everywhere and differentiable nowhere

Statement

Let 0<a<1, let b>1 be an odd integer, and let Wa,b be the classical Weierstrass function (The classical Weierstrass function). If ab>1+3π/2, then Wa,b is continuous at every real point and differentiable at no real point.

Facts & Assumptions

Given: Parameters 0<a<1 and an odd integer b>1 satisfying ab>1+3π/2, and an arbitrary point x0∈R.

[L1]
[L2]

For every m≥1, the low-frequency increment at the probes satisfies ∣Am∣<π(ab)m(xm−x0)/(ab−1) (Low-frequency bound for the Weierstrass difference quotient).

[L3]

For the tail increment at the same probes, ∣Bm∣≥(2/3)(ab)m(xm−x0) (The Weierstrass tail has one sign and dominates at the probe points).

[L5]

Differentiability at x0 requires the finite limit of (Wa,b(x)−Wa,b(x0))/(x−x0) as x→x0 (The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set).

[L6]

The probes satisfy xm>x0 and xm→x0 (Nearest-integer probe points for the Weierstrass function).

[L7]

The number π=2γ is positive because the smallest positive zero of cosine satisfies γ∈(0,2) (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two).

Proof

technique · direct
1.1givenL1L6

Fix x0 and use [L6] for its probe sequence (xm). Continuity at x0 already follows from [L1].

1.2givenL7algebra

The hypothesis ab>1+3π/2 and positivity of π give ab>1 and c:=23−πab−1>0.

2.1step 1.1L2L3algebra

For m≥1, splitting the series increment at frequency m gives Wa,b(xm)−Wa,b(x0)=Am+Bm. The reverse triangle inequality and [L2] to [L3] yield ∣Wa,b(xm)−Wa,b(x0)∣≥(ab)m(23−πab−1)(xm−x0).

3.1step 1.2step 2.1L4algebra

Since xm−x0>0, divide step 2.1 by it. Step 1.2 and [L4] show that the absolute values of the selected difference quotients are at least c(ab)m and tend to +∞.

4.1step 1.1step 3.1L1L5∎

Although xm→x0, the difference quotients along this sequence have no finite limit by step 3.1, so [L5] rules out differentiability at x0. The point x0 was arbitrary, while [L1] gives continuity everywhere. Then Wa,b is continuous at every real point and differentiable at no real point.

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