Wierestrass Function

One function (we don't actually cover this) defined the function:

\sum_{n = 0}^{\infty} a^{n} \cos (b^{n} x)

where $a, b \in R$ are constants. Specifically there were more restrictions on $a, b$ not discussed here. This is one function that would have the property of being everywhere continuous but nowhere differentiable, but the $\cos$ here is making such high oscillations that it makes the derivative not exist.

A different function that (only starts to) take on this role is:

ϕ (x) = {\begin{cases} | x | & - 1 \leq x \leq 1 \\ \dots \end{cases}

Where we extend this function to all of $R$ by requiring that $ϕ (x + 2) = ϕ (x)$ .

Now this function is differentiable at any $x \notin N$ while $x \in R$ , but the idea is to try to fill in the gaps. We'll consider the function of the form:

g (x) = \sum_{n = 0}^{\infty} a^{n} ϕ (b^{n} x)

where $a, b$ have the similar restraints ( $a b > 1$ and $a < 1$ ). We'll specifically want to use:

f (x) = \sum_{n = 0}^{\infty} {(\frac{3}{4})}^{n} ϕ (4^{n} x)

We'll show that $f$ is continuous, and then next that $f$ is differentiable nowhere.

Continuity

$f (x)$ above is defined for all of $x \in R$ . Namely we could compare our series defined via $f$ converges since it's less than the geometric series:

\sum_{n = 0}^{\infty} {(\frac{3}{4})}^{n}

which itself is a convergent geometric series.

For continuity at every $x \in R$ , notice that $\forall N \in N$ that:

f (x) = S_{N} (x) + R_{N} (x)

where:

S_{N} (x) = \sum_{n = 0}^{N} {(\frac{3}{4})}^{n} ϕ (4^{n} x)

is clearly continuous, and:

R_{N} (x) = \sum_{n = N + 1}^{\infty} {(\frac{3}{4})}^{n} ϕ (4^{n} x) \leq \sum_{n = N + 1}^{\infty} {(\frac{3}{4})}^{n} = \frac{{(\frac{3}{4})}^{N + 1}}{\frac{1 - 3}{4}} = 3 \cdot {(\frac{3}{4})}^{N}

So thus $0 \leq R_{N} (x) \leq 3 {(\frac{3}{4})}^{N}$ .

So the argument is the following:

Proof

Let $x \in R$ and let $ε > 0$ . Choose some $N \in N$ such that $3 {(\frac{3}{4})}^{N} < \frac{ε}{2}$ . Using the argument above, then:

f (x) = S_{N} (x) + R_{N} (x)

where $S_{N} (x)$ is continuous. Thus then by definition then choose $\exists δ > 0$ such that $| x - y | < δ$ implies $| S_{N} (x) - S_{N} (y) | < \frac{ε}{2}$ . Now let $y$ be such that $| x - y | < δ$ . Then:

\begin{aligned} | f (x) - f (y) | & = | S_{N} (x) + R_{N} (x) - S_{N} (y) - R_{N} (y) | \\ \leq | S_{N} (x) - S_{N} (y) | + | R_{N} (x) - R_{N} (y) | \\ < \frac{ε}{2} + \frac{ε}{2} \\ = ε \end{aligned}

Thus $f$ is uniformly continuous.

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Not Differentiable

Notice that for any $x, y \in R$ then:

| ϕ (x) - ϕ (y) | \leq | x - y |

with equality only when $∄$ an integer between $x, y$ . Let's prove the failure for differentiability at any $x \in R$ .

Proof

Let $x \in R$ . and fix $m \in N$ .

What may seem weird, let $δ_{m} = \pm \frac{1}{2} \cdot \frac{1}{4^{m}}$ , where:

Use $+$ if $∄$ an integer between $4^{m} x$ and $4^{m} x + \frac{1}{2}$
Use $-$ if $∄$ an integer between $4^{m} x - \frac{1}{2}$ and $4^{m} x$ .
Thus using our $δ_{m}$ then $4^{m} (x + δ_{m})$ and $4^{m} x$ have no integers between them.

Now we'll show that $\frac{f (x + δ_{m}) - f (x)}{δ_{m}}$ has no limit as $m \to \infty$ (which makes $δ_{m} \to 0$ ). This is the $h$ -definition of a derivative. Doing this:

\begin{aligned} \frac{f (x + δ_{m}) - f (x)}{δ_{m}} & = \frac{1}{δ_{m}} (\sum_{n = 0}^{\infty} {(\frac{3}{4})}^{n} ϕ (4^{n} (x + δ_{m})) - \sum_{n = 0}^{\infty} {(\frac{3}{4})}^{n} ϕ (4^{n} x)) \\ = \sum_{n = 0}^{\infty} {(\frac{3}{4})}^{n} \frac{ϕ (4^{n} (x + δ_{m})) - ϕ (4^{n} x)}{δ_{m}} \end{aligned}

Now notice that the term in the box is some $γ_{n}$ so making that replacement:

\begin{aligned} = \sum_{n = 0}^{\infty} {(\frac{3}{4})}^{n} γ_{n} \end{aligned}

We have the following cases:

Say $n > m$ . Then into $ϕ$ we get:

4^{n} (x + δ_{m}) = 4^{n} x \pm \frac{1}{2} \cdot 4^{n - m}

now since $n > m$ then the $4^{n - m}$ has at least one factor of 4, and the $4^{n}$ also has a factor of $4$ , so the whole term is an even integer. Since $ϕ$ has period 2, then:

ϕ (4^{n} (x + δ_{m})) - ϕ (4^{n} x) = 0

Thus for $n > m$ then $γ_{n} = 0$ . Thus the infinite series becomes a finite series.
2. Say $n = m$ . Then notice:

4^{n} (x + δ_{m}) = 4^{m} (x + δ_{m}) \Rightarrow 4^{n} x = 4^{m} x

Now there cannot be an integer between these values:

ϕ (4^{m} (x + δ_{m})) - ϕ (4^{m} x) = \pm 4^{m} δ_{m} \Rightarrow γ_{n} = \pm 4^{m}

Say $0 \leq n < m$ . We won't get the exact value but we will approximate $γ_{n}$ . Notice that using what we found at the start of the differentiability section:

\begin{aligned} | ϕ (4^{n} (x + δ_{m})) - ϕ (4^{n} x) | & \leq | 4^{n} (x + δ_{m}) - 4^{n} x | \\ \leq 4^{n} | δ_{m} | \end{aligned}

so then $| γ_{n} | \leq 4^{n}$ .

Now:

\begin{aligned} \frac{f (x + δ_{m}) - f (x)}{δ_{m}} & = \sum_{n = 0}^{\infty} {(\frac{3}{4})}^{n} γ_{n} \\ = \sum_{n = 0}^{m} {(\frac{3}{4})}^{n} γ_{n} & (γ_{n} = 0 \forall n > m) \\ = {(\frac{3}{4})}^{m} γ_{m} + \sum_{n = 0}^{m - 1} {(\frac{3}{4})}^{n} γ_{n} \end{aligned}

Let the left value be $A$ and for the right denote it $B_{n}$ (the $n$ is chosen to say that it is a sum for a sequence, not that it depends on $n$ ). Notice:

| A | = | ({\frac{3}{4}}^{m} γ_{m}) | = | {(\frac{3}{4})}^{m} (\pm 4^{m}) | = 3^{m}

\begin{aligned} | B | & = | \sum_{n = 0}^{m - 1} {(\frac{3}{4})}^{n} γ_{n} | \\ \leq \sum_{n = 0}^{m - 1} {(\frac{3}{4})}^{n} | γ_{n} | \\ \leq \sum_{n = 0}^{m - 1} 3^{n} \\ = \frac{3^{m} - 1}{3 - 1} \\ \leq \frac{1}{2} \cdot 3^{m} \end{aligned}

Therefore:

\begin{aligned} | \frac{f (x + δ_{m}) - f (x)}{δ_{m}} | & = | A + B_{n} | \\ \geq | | A | - | B_{n} | | \\ \geq 3^{m} - \frac{1}{2} \cdot 3^{m} \\ = \frac{1}{2} \cdot 3^{m} \\ \to \infty \end{aligned}

so the limit doesn't exist.
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