Everywhere Continuous Non-differentiable Function

Weierstrass introduced a shocking idea in 1872. There exist functions that are continuous everywhere but differentiable nowhere. Not just at a few points. At no points whatsoever.

This was a direct attack on mathematical intuition. Before Weierstrass, most mathematicians assumed that continuous functions were “smooth” except at isolated trouble spots. Weierstrass proved them wrong with a concrete counterexample.

The Weierstrass Function

The function is defined as an infinite series:

$$f(x) = \sum_{n=0}^{\infty} b^n \cos(a^n \pi x) \tag{1}$$

Written out explicitly:

$$f(x) = \cos(\pi x) + b\cos(a\pi x) + b^2\cos(a^2\pi x) + b^3\cos(a^3\pi x) + \ldots$$

The parameters must satisfy three conditions:

• \( a \) is an odd positive integer
• \( 0 < b < 1 \)
• \( ab > 1 + \frac{3}{2}\pi \)

Under these conditions, \( f(x) \) is continuous for all \( x \) but has no finite derivative at any point.

A common example uses \( a = 21 \) and \( b = 0.5 \). The resulting function looks like a jagged, self-similar fractal curve.

Why the Function Is Continuous

The continuity proof is straightforward. Each term in the series is bounded:

$$|b^n \cos(a^n \pi x)| \leq b^n$$

Since \( 0 < b < 1 \), the series \( \sum b^n \) is a convergent geometric series.

By the Weierstrass M-Test for uniform convergence, the series in equation (1) converges uniformly on every interval. A uniformly convergent series of continuous functions is itself continuous.

So \( f(x) \) is continuous everywhere. No surprises here.

Why the Function Is Nowhere Differentiable

This is the hard part. We need to show that the difference quotient

$$\frac{f(x+h) – f(x)}{h}$$

has no finite limit as \( h \to 0 \) for any value of \( x \).

The difference quotient expands to:

$$\frac{f(x+h) – f(x)}{h} = \sum_{n=0}^{\infty} b^n \frac{\cos[a^n\pi(x+h)] – \cos(a^n\pi x)}{h} \tag{2}$$

We split this into two parts. Let \( S_m \) be the sum of the first \( m \) terms. Let \( R_m \) be the remainder after \( m \) terms.

Bounding the First m Terms

Using the Mean Value Theorem on each cosine term:

$$\frac{|\cos[a^n\pi(x+h)] – \cos(a^n\pi x)|}{|h|} \leq a^n\pi$$

So the partial sum satisfies:

$$|S_m| \leq \sum_{n=0}^{m-1} b^n a^n \pi = \pi \frac{(ab)^m – 1}{ab – 1} < \frac{\pi(ab)^m}{ab – 1}$$

This gives us an upper bound on \( |S_m| \).

The Remainder Term Blows Up

Here’s the clever part. We choose \( h \) strategically to make \( R_m \) large.

Write \( a^m x = \alpha_m + \xi_m \), where \( \alpha_m \) is the nearest integer to \( a^m x \) and \( -\frac{1}{2} \leq \xi_m < \frac{1}{2} \).

Choose \( h \) so that:

$$h = \frac{1 – \xi_m}{a^m} \tag{3}$$

As \( m \to \infty \), this \( h \to 0 \). But with this specific choice, the remainder term satisfies:

$$|R_m| > \frac{2(ab)^m}{3} \tag{4}$$

The Punchline

Combining our bounds on \( S_m \) and \( R_m \):

$$\left|\frac{f(x+h) – f(x)}{h}\right| = |R_m + S_m| \geq |R_m| – |S_m| > \left(\frac{2}{3} – \frac{\pi}{ab-1}\right)(ab)^m$$

Since \( ab > 1 + \frac{3}{2}\pi \), the coefficient \( \left(\frac{2}{3} – \frac{\pi}{ab-1}\right) \) is positive.

As \( m \to \infty \), the term \( (ab)^m \to \infty \).

So the difference quotient grows without bound. The limit doesn’t exist. The derivative \( f'(x) \) doesn’t exist at any point \( x \). ∎

Key Takeaways

• The Weierstrass function is continuous everywhere but differentiable nowhere. It’s the canonical example of a “pathological” function.
• Uniform convergence guarantees continuity. But the high-frequency oscillations (controlled by \( a^n \)) prevent any tangent line from existing.
• G.H. Hardy later proved that non-differentiability holds for the weaker condition \( ab \geq 1 \), not just \( ab > 1 + \frac{3}{2}\pi \).
• The function is fractal-like. It has the same jagged structure at every scale, which is why no tangent line can ever “catch” it.

Frequently Asked Questions