A Continuous, Nowhere Differentiable Function: Part 1

When studying calculus, we learn that every differentiable function is continuous, but a continuous function need not be differentiable at every point. A standard example is $f(x) = |x|$, which is continuous everywhere, and differentiable everywhere except at $x=0$.

In this series of articles, we will construct a function that is continuous everywhere, but differentiable nowhere.

It is possible to construct such a function by summing a sequence of scaled and shifted copies of the absolute value function: $g(x) = \sum_{n=0}^{\infty}a_n |x – b_n|$. By carefully choosing the sequences $(a_n)$ and $(b_n)$, one can make $g$ continuous and nowhere differentiable.

We will use a different approach which will allow us to have some fun with Fourier series. Let us start by writing down two general expressions for the Fourier series of an even, $2\pi$-periodic, real-valued function with zero mean:

$$h(x) = \sum_{n=1}^{\infty} a_n \cos(nx) = \text{Re}\left(\sum_{n=1}^{\infty} a_n e^{inx} \right)$$

where $(a_n)$ is a sequence of real numbers. The terms of this series are continuous and differentiable (infinitely many times) at every point, yet as we will see, it is still possible to choose $(a_n)$ such that $h$ is continuous but nowhere differentiable.

Let’s examine what properties we will require from $(a_n)$ in order to achieve this. First, a sufficient condition to ensure continuity of $h$ is to require $\sum_{n = 1}^{\infty} |a_n| < \infty$. Indeed, if this condition holds, then the inequality $|a_n \cos(nx)| \leq |a_n|$ implies that the series $\sum_{n=1}^{\infty} a_n \cos(nx)$ is absolutely convergent at every point. Furthermore, by the Weierstrass M-test, the series is also uniformly convergent, so $h$ is continuous.

Now, what about nowhere differentiability? If we define the power series

$$p(z) = \sum_{n=1}^{\infty} a_n z^n$$

then $h(x) = \text{Re}(p(e^{ix}))$. The radius of convergence of this power series is

$$R = \frac{1}{\limsup |a_n|^{1/n}}$$

If $R < 1$, then the power series will diverge on the unit circle $|z| = 1$, so the Fourier series $h(x)$ will also diverge at every point. If $R > 1$, then the unit circle $|z| = 1$ is contained in the interior of the disk of convergence, so $p$ is analytic on the unit circle, and consequently, $h$ is not only continuous but differentiable infinitely many times at every point. In other words, if $R > 1$, then $h$ will be a $C^\infty$ function, also known as a smooth function. So, if we want $h$ to be continuous but nowhere differentiable (or, more generally, to be anything but a smooth function), we need to choose $(a_n)$ such that $R = 1$. This is a necessary, but not sufficient condition.

Summarizing, a sufficient condition for $h$ to be continuous is $\sum_{n=1}^{\infty} |a_n| < \infty$, and a necessary condition for $h$ to be nowhere differentiable is $\limsup |a_n|^{1/n} = 1$. We will restrict our attention to functions satisfying both of these criteria. One such function is the Weierstrass function (more precisely, family of functions):

$$w(x) = \sum_{k=0}^{\infty} \alpha^k \cos(\beta^k x)$$

where $0 < \alpha < 1$, and $\beta \geq 2$ is an integer. Comparing this with the Fourier series expression

$$\sum_{n=1}^{\infty}a_n \cos(nx),$$

we see that the series defining $w$ is indeed a Fourier series, with the property that the only nonzero coefficients are the ones associated with the frequencies $n=\beta, \beta^2, \beta^3, \ldots$ Thus there are increasingly large gaps between nonzero coefficients as $n$ increases. Specifically,

$$a_n = \begin{cases}\alpha^k & \text{if }n = \beta^k \\ 0 & \text{otherwise}\end{cases}$$

Such a series is called a lacunary Fourier series. (“Lacuna” is simply a fancy word for “gap.”) Since

$$\sum_{n=1}^{\infty}|a_n| = \sum_{k=1}^{\infty}\alpha^k =\frac{\alpha}{1-\alpha} < \infty,$$

the convergence is uniform, so $w$ is continuous for any fixed $\alpha \in (0,1)$. Also,

$$|a_n|^{1/n} = \begin{cases} |\alpha|^{k / \beta^k} & \text{if }n = \beta^k \\ 0 & \text{otherwise}\end{cases}$$

Since $\alpha > 0$ and $k / \beta^k \to 0$ as $k \to \infty$, it follows that $\limsup |a_n|^{1/n} = 1$ and therefore the radius of convergence is $R = 1$. This means that the necessary condition for $w$ to be nowhere differentiable is satisfied.

Can we find a sufficient condition for $w$ to be nowhere differentiable? A general principle with respect to the Fourier series is that smoother functions have coefficients that decay more rapidly as $n \to \infty$. A nowhere differentiable function will be very un-smooth, so we want the Fourier coefficients $a_n$ to decay slowly (we want more contribution from the high-frequency components), while maintaining the constraint $\sum_{k=1}^{\infty}|\alpha^k| < \infty$ to preserve continuity.

The form of the function $w(x) = \alpha^k \cos(\beta^k x)$ gives us two knobs to play with: $\alpha$ and $\beta$. As long as $0<\alpha<1$, we maintain uniform convergence, hence continuity. For a given choice of $\alpha$, we can slow down the decay by increasing $\beta$. In 1872, Weierstrass proved that as long as $\beta$ is large enough to satisfy $\alpha \beta > 1 + 3\pi/2$, then $w$ will be nowhere differentiable. This condition was later improved by Hardy to $\alpha \beta \geq 1$.

You can see a plot of what this function looks like for a suitable choice of $\alpha, \beta$ at the Wikipedia page for Weierstrass function.

The remaining parts of this article will present a proof of Weierstrass’s result, with just enough background material to make it self-contained without having to read 100+ pages to get to the nice proof in Stein and Shakarchi’s Fourier Analysis, which we will follow in spirit.