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, $2pi$-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 $alpha beta > 1 + 3pi/2$, 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.