The Weierstrass Function

Prove via the Weierstrass M-test that the Weierstrass function $W(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x)$ ( $a = \frac12$ , $b = 13$ ) is continuous on all of $\mathbb{R}$ . (Hint: the m-test scene in the series chapter.)

u_n(x) = a^n \cos(b^n \pi x), \qquad |u_n(x)| \le \left(\tfrac12\right)^n =: M_n

\sum_{n=0}^{\infty} M_n = 2 < \infty \;\xrightarrow{\text{M-test}}\; \sum u_n \text{ converges uniformly on } \mathbb{R}

|W(x) - W_N(x)| \le \sum_{n=N+1}^{\infty} M_n = \left(\tfrac12\right)^N \quad \text{for every } x

n = 6

: tail bound

(1/2)^6 = 0.015625

— this curve sits within 0.016 of the true

W

at every

x

(

W_6(0) = 1.984375

).

Show proof

Each term $u_n(x) = a^n \cos(b^n \pi x)$ is continuous. Since $|\cos| \le 1$ , $|u_n(x)| \le a^n = \left(\frac12\right)^n =: M_n$ for every $x$ , and $\sum_{n=0}^{\infty} M_n = \sum_{n=0}^{\infty} \left(\frac12\right)^n = 2 < \infty$ . By the Weierstrass M-test (the m-test scene in the series chapter), $\sum u_n$ converges uniformly on $\mathbb{R}$ .
A uniform limit of continuous functions is continuous: fix $x_0$ ; then $|W(x) - W(x_0)| \le |W(x) - W_N(x)| + |W_N(x) - W_N(x_0)| + |W_N(x_0) - W(x_0)|$ . The first and last terms are small by the tail bound $(1/2)^N$ , the middle one by the continuity of the finite sum $W_N$ . Hence $W$ is continuous at $x_0$ , and $x_0$ was arbitrary, so $W$ is continuous on all of $\mathbb{R}$ .
Check the numbers: at $N = 6$ , $|W(x) - W_6(x)| \le (1/2)^6 = 0.015625$ for every $x$ — the curve on screen sits within 0.016 of the true $W$ . Raising $N$ drives the tail bound $(1/2)^N$ to 0, so the on-screen $W_N$ approaches $W$ uniformly.

State the theorem: if $b$ is odd and $ab > 1 + \frac{3\pi}{2}$ , then the Weierstrass function $W$ is differentiable at no point (our $a = \frac12$ , $b = 13$ satisfy $ab = 6.5 > 5.712$ ). The argument below is an idea-level sketch — it conveys the mechanism faithfully but does not reproduce the classical proof's constant-tracking line by line.

ab = 6.5 > 1 + \frac{3\pi}{2} \approx 5.712, \qquad b = 13 \text{ odd}

\text{at scale } b^{-n}\text{: the } n\text{-th harmonic completes a full spike, so some } x_n \to x_0 \text{ has}

\left|\frac{W(x_n) - W(x_0)}{x_n - x_0}\right| \gtrsim (ab)^n \to \infty \quad \text{(idea sketch, not the full proof)}

n = 8

: tail bound

(1/2)^8 = 0.00390625

— eight smooth terms added, yet the curve is jagged at every scale (

W_8(0) = 1.99609375

,

W_8(1) = -1.99609375

).

Show proof

Theorem (Weierstrass–Hardy): if $b$ is an odd integer and $ab > 1 + \frac{3\pi}{2}$ , then $W(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x)$ is differentiable at no point. We use $a = \frac12$ , $b = 13$ , so $ab = 6.5 > 1 + \frac{3\pi}{2} \approx 5.712$ — the condition holds.
The mechanism — zooming in at scale $b^{-n}$ : the $n$ -th harmonic $\cos(b^n \pi x)$ completes a full period over every interval of length about $b^{-n}$ , so near any point $x_0$ there is a spike point $x_n$ with $|x_n - x_0| \lesssim b^{-n}$ where the $n$ -th cosine swings nearly trough-to-peak (about $2$ ). Across this spike, the $n$ -th term's difference quotient is $a^n \cdot (\approx 2) / (\approx b^{-n}) \approx 2(ab)^n = 2 \cdot 6.5^n \to \infty$ (about $6.37 \times 10^6$ at $n = 8$ ).
The remaining terms cannot beat the spike: the terms with $k < n$ are nearly flat at this scale (differentiating brings out $a^k b^k \pi$ , a geometric sum with ratio $ab$ and a strictly smaller constant than the spike's), and the terms with $k > n$ are uniformly tiny by the tail bound $2a^{n+1}/(1-a)$ . The classical proof chooses the spike points delicately, tracks all three constants, and shows that $ab > 1 + \frac{3\pi}{2}$ is exactly the inequality saying "the spike's constant wins."
Honest label: this is an idea-level sketch. The full proof is the classical Weierstrass–Hardy construction — building the spike points and tracking all three constants to the end — which we do not reproduce line by line here. The takeaway it conveys is this: the difference quotient blows up to infinity along the spike sequence $x_n \to x_0$ , so no finite number can serve as $W'(x_0)$ ; $x_0$ was arbitrary, so $W$ is differentiable nowhere.

Each partial sum $W_N$ is a finite sum of cosines, hence smoothly (infinitely) differentiable on $\mathbb{R}$ . Yet $W_N \to W$ uniformly (problem 1) while $W$ is differentiable nowhere (problem 2). Conclude that a uniform limit of differentiable functions need not be differentiable.

W_N(x) = \sum_{n=0}^{N} a^n \cos(b^n \pi x) \quad \text{smooth on } \mathbb{R}

W_N \xrightarrow{\text{uniformly}} W, \qquad |W - W_N| \le \left(\tfrac12\right)^N \to 0

W \text{ differentiable nowhere (problem 2)}

n = 2

: a sum of three smooth terms (

W_2(0) = 1.75

, tail bound

(1/2)^2 = 0.25

) — a uniform limit of such smooth things can fail to be differentiable.

Show proof

Each $W_N$ is a finite sum, so termwise differentiation is legitimate: $W_N'(x) = -\sum_{n=0}^{N} a^n b^n \pi \sin(b^n \pi x)$ exists at every $x$ . For instance $W_2'(x) = -\pi\sin(\pi x) - \frac{13\pi}{2}\sin(13\pi x) - \frac{169\pi}{4}\sin(169\pi x)$ (since $a^1 b^1 \pi = \frac{13\pi}{2}$ and $a^2 b^2 \pi = \frac{169\pi}{4}$ ). Every curve on screen is a smooth function.
But by problem 1's tail bound $|W - W_N| \le (1/2)^N \to 0$ , we have $W_N \to W$ uniformly on $\mathbb{R}$ . And by problem 2, the limit $W$ is differentiable at no point.
Hence a uniform limit of differentiable functions need not be differentiable — continuity is preserved under uniform limits (problem 1's argument), but differentiability is not. This is why analysis demands stronger convergence (e.g. uniform convergence of the derivatives) before differentiating termwise.

The Weierstrass Function

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