# A Continuous, Nowhere Differentiable Function: Part 2

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This is Part 2 of a series of articles in which the goal is to exhibit a continuous function which is nowhere differentiable, and to explore some interesting properties of a special type of Fourier series.

In Part 1, we defined the Weierstrass function,

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

We showed that the series is absolutely and uniformly convergent, so $w$ is a continuous function. We also showed that the radius of convergence is $1$, which is a necessary (but not sufficient) condition for the function to be non-smooth. In this post, we will continue working toward proving that in fact, $w$ is nowhere differentiable, provided that the condition $\alpha \beta > 1$ is satisfied.

To do this, we will identify a condition which must be satisfied by the Fourier series of a continuous function which is differentiable at a point $x_0$, and we will show that the Weierstrass function does not satisfy this condition, hence it is nowhere differentiable.

The general form for the Fourier series of a $2\pi$-periodic function $f$ is

$$f(x) = \sum_{n = -\infty}^{\infty}c_n e^{inx}$$

where convergence is understood to mean that the sequence of symmetric partial sums

$$f_N(x) = \sum_{n = -N}^{N} c_n e^{inx}$$

satisfies

$$\lim_{N \to \infty}f_n(x) = f(x)$$

for every $x$. In the special case where $f$ is real and even, the partial sums can be rewritten as

$$f_N(x) = c_0 + \sum_{n=1}^{N} c_n(e^{inx} + e^{-inx}) = c_0 + \sum_{n=1}^{N} (2c_n) \cos(nx) = \sum_{n = 0}^{N} a_n \cos(nx)$$

where

$$a_n = \begin{cases} c_n & \text{if }n = 0 \\ 2c_n & \text{otherwise} \\ \end{cases}$$

In general, it’s hard to find useful conditions which must be satisfied by the sequence $f_N$ when $f$ is an arbitrary continuous function; indeed, one can construct a continuous functions whose Fourier series diverges at some or even infinitely many points. (That’s a subject for another blog post!)

However, by a technique called Cesàro summation, we can obtain a more well-behaved sequence of functions which always converges to $f$ in a controlled way which can be quantified.  To do this, let’s define

$$\overline{f}_N(x) = \frac{1}{N}\sum_{k=0}^{N-1}f_k(x) = \frac{1}{N}\sum_{k=0}^{N-1} \sum_{n=-k}^{k}c_n e^{inx}$$

In other words, we form a sequence $(\overline{f}_n)$ where the $N$’th term is the average of the first $N$ partial sums of the Fourier series of $f$. Examining the rightmost expression, we see that $\overline{f}_N$ sums a total of $N$ copies of $c_0 e^{i0x} = c_0$, and $N-1$ copies of $c_1 e^{i1x}$ and $c_{-1}e^{-i1x}$, and in general, $N – |n|$ copies of $c_n e^{inx}$. Therefore, an equivalent way to express $\overline{f}_N(x)$ is

$$\overline{f}_N(x) = \frac{1}{N}\sum_{n=-N}^{N}(N-|n|)c_n e^{inx} = \sum_{n=-N}^{N}\left(1 – \frac{|n|}{N}\right)c_n e^{inx}$$

This shows that $\overline{f}_N$ is formed from the terms of the Fourier series by first weighting the Fourier coefficients $(c_n)$ by the triangular sequence

$$t_n = \begin{cases}1 – \frac{|n|}{N} & \text{if }|n| < N \\ 0 & \text{otherwise} \\ \end{cases}$$

This reduction of the influence of the higher-frequency terms is what allows the better convergence properties. In particular, we will prove the following.

Theorem 1. If $f$ is a continuous function which is differentiable at $x_0$, then the sequence of Cesàro sums $\overline{f}_N(x)$ of its Fourier series satisfies

$$\limsup_{N \to \infty} \frac{|\overline{f}’_N(x)|}{\log N} < \infty$$

where $\overline{f}’_N$ is the derivative of $\overline{f}_N$.

We will then show that the Weierstrass function $w$ does not satisfy this condition at any point $x_0$, and therefore is nowhere differentiable.

The key to proving Theorem 1 is the following characterization of the Cesàro sum:

Lemma 2. If $f$ is a continuous, $2\pi$-periodic function, then

$$\overline{f}_N(x) = \int_{-\pi}^{\pi} K_N(u)f(x-u) du = \int_{-\pi}^{\pi} K_N(x-u)f(u) du$$

where

$$K_N(x) = \frac{1}{N}\left(\frac{\sin\frac{Nx}{2}}{\sin\frac{x}{2}}\right)^2$$

is called the $N$’th Fejér kernel. The equivalent expressions $\int_{-\pi}^{\pi} K_N(u)f(x-u) du = \int_{-\pi}^{\pi} K_N(x-u)f(u) du$ are called the convolution of $K_N$ and $f$, which is often written as $K_N \ast f = f \ast K_N$.

Proof of Lemma 2: The Fourier coefficient $c_n$ is

$$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) e^{-inx} dx$$

Therefore,

\begin{aligned} \overline{f}_N(x) &= \frac{1}{N}\sum_{k=0}^{N-1} \sum_{n=-k}^{k}c_n e^{inx} \\ &= \frac{1}{2\pi N}\sum_{k=0}^{N-1} \sum_{n=-k}^{k}\left(\int_{-\pi}^{\pi}f(u)e^{-inu}du\right) e^{inx}\\ &= \frac{1}{2\pi N}\sum_{k=0}^{N-1} \sum_{n=-k}^{k} \left( \int_{-\pi}^{\pi} f(u) e^{in(x-u)}du\right) \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(u) \left(\frac{1}{N}\sum_{k=0}^{N-1} \sum_{n=-k}^{k} e^{in(x-u)}\right) du \\ \end{aligned}

So it suffices to show that

$$K_N(x) = \frac{1}{N}\sum_{k=0}^{N-1} \sum_{n=-k}^{k} e^{inx}$$

Now, $\sum_{n=-k}^{k} e^{inx} =1 + 2\sum_{n=1}^{k} \cos(nx)$. Multiply both sides by $\sin(x/2)$ and apply the trig identity $\sin(a)\cos(b) = \frac{1}{2}[\sin(a+b) + \sin(a-b)]$ on the right hand side. The sum telescopes, and the right hand side reduces to $\sin((k+1/2)x)$. This yields the identity

$$\sin(x/2)\sum_{n=-k}^{k} e^{inx} = \sin((k+1/2)x)$$

Summing both sides from $k=0$ to $N-1$, we have

$$\sin(x/2)\sum_{k=0}^{N-1} \sum_{n=-k}^{k} e^{inx} = \sum_{k=0}^{N-1}\sin((k+1/2)x)$$

Again multiply both sides by $\sin(x/2)$, and apply the trig identity $\sin(a)\sin(b) = \frac{1}{2}[\cos(a-b) – \cos(a+b)]$ on the right hand side. The sum telescopes again, leaving us with

$$\sin^2(x/2) \sum_{k=0}^{N-1} \sum_{n=-k}^{k} e^{inx} = \frac{1}{2}(1 – \cos(Nx)) = \sin^2(Nx/2)$$

Dividing both sides by $\sin^2(x/2)$ gives us the desired result. $\square$

We will apply Lemma 2 to complete the proof of Theorem 1 in the next post.

2 replies
1. Greg Bernhardt says:

Nice jbunniii! Looking forward to part 3!

2. zwierz says:

note also that the set of differentiable at least at one point functions has the first Baire category in $C[a,b]$. This fact is much simpler to prove than to find nowhere differentiable function explicitly. So that in some sense the set of differentiable functions is very small in $C[a,b]$