Lecture 33 - Supplemental Material, Weirstrass Function

We talked about:

L'Hopital's Rule:

\frac{0}{0}

case

Let $f, g$ be continuous on an interval containing $a$ and suppose $f, g$ are differentiable on this interval with the possible exception of the point $a$ . If $lim_{x \to a} f (x) = lim_{x \to a} g (x) = 0$ and $g^{'} (x) \neq 0$ for all $x \neq a$ , then:
$lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L \in R \Rightarrow lim_{x \to a} \frac{f (x)}{g (x)} = L$

Proof

Apply the Generalized Mean Value Theorem via the following method. For argument let $(b, c)$ be our interval.

Consider the case that $a = b$ . Let $ε > 0$ . Supposing $lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L$ , choose $δ > 0$ such that $0 < | x - b | < δ$ implies that $| \frac{f^{'} (x)}{g^{'} (x)} - L | < \frac{ε}{2}$ . Let $x \in (b, b + δ)$ . Let $y \in (b, x)$ . By the Generalized Mean Value Theorem, then $\exists t \in (y, x)$ such that:
$f^{'} (t) (g (x) - g (y)) = g^{'} (t) (f (x) - f (y))$
and because $g (x) \neq g (y)$ and $g^{'} (t) \neq 0$ :
$\frac{f^{'} (t)}{g^{'} (t)} = \frac{f (x) - f (y)}{g (x) - g (y)}$
Now since $b < t < b + δ$ then our implication we started with gives:
$L - \frac{ε}{2} < \frac{f (x) - f (y)}{g (x) - g (y)} < L + \frac{ε}{2}$
for all $y \in (a, x)$ . Let $y \to b^{+}$ then by the Algebraic Limit Theorem for Functional Limits and Limits and Order (Order Limit Theorem) then taking a limit and using $lim_{x \to a} f (x) = 0 = lim_{x \to a} g (x)$ then:
$L - ε < \frac{f (x)}{g (x)} < L + ε$
Doing the cases for $a = c$ and $c \neq a, b$ are done very similarly.

☐

L'Hopital's Rule:

\frac{\infty}{\infty}

case

Suppose $f, g$ are differentiable on $(a, b)$ and that $g^{'} (x) \neq 0$ for all $x \in (a, b)$ . If $lim_{x \to a} g (x) = \infty$ (defined here) (or $- \infty$ ), then:
$lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L \Rightarrow lim_{x \to a} \frac{f (x)}{g (x)} = L$

Notice that we only need the denominator to go to

\infty

in this case. This is different than the

\frac{0}{0}

case.

Proof

Let $ε > 0$ . Because $lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L$ then $\exists δ_{1} > 0$ such that:
$| \frac{f^{'} (x)}{g^{'} (x)} - L | < \frac{ε}{2}$
for all $a < x < a + δ_{1}$ . For convenience of notation, let $t = a + δ_{1}$ and note that $t$ is fixed for the remainder of the argument.

Our functions are not defined at $a$ , but for any $x \in (a, t)$ we can apply the Generalized Mean Value Theorem on the interval $[x, t]$ to get:
$\frac{f (x) - f (t)}{g (x) - g (t)} = \frac{f^{'} (c)}{g^{'} (c)}$
for some $c \in (x, t)$ . Our choice of $t$ then implies that:
$L - \frac{ε}{2} < \frac{f (x) - f (t)}{g (x) - g (t)} < L + \frac{ε}{2}$
for all $x \in (a, t)$ .

In an effort to isolate the fraction $\frac{f (x)}{g (x)}$ , the strategy is to multiply the inequality by $\frac{g (x) - g (t)}{g (x)}$ . We need to be sure, however, that this quantity is positive, which amounts to insisting that $1 \geq \frac{g (t)}{g (x)}$ . Now, because $t$ is fixed and $lim_{x \to a} g (x) = \infty$ then we can choose $δ_{2} > 0$ so that $g (x) \geq g (t)$ for all $a < x < a + δ_{2}$ (because $t$ is fixed, so eventually we pass $g (t)$ as $g (x) \to \infty$ ). Carrying out the desired multiplication results in:
$(L - \frac{ε}{2}) (1 - \frac{g (t)}{g (x)}) < \frac{f (x) - f (t)}{g (x)} < (L + \frac{ε}{2}) (1 - \frac{g (t)}{g (x)})$
which after some manipulation becomes:
$L - \frac{ε}{2} + \frac{- L g (t) + \frac{ε}{2} g (t) + f (t)}{g (x)} < \frac{f (x)}{g (x)} < L + \frac{ε}{2} + \frac{- L g (t) - \frac{ε}{2} g (t) + f (t)}{g (x)}$
Notice the $g(x)

Weirstrass Function

Let's construct the function that is everywhere continuous but nowhere differentiable.

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.

☐

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.
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.
☐

s in the denominator. Again, because $t$ is fixed then $\lim_{ x \to a } g(x) = \infty$. Thus, we can choose a $\delta_{3}$ such that $x \in (a,a + \delta_{3})$ implies that $g(x)$ is large enough to ensure that the terms: $ \frac{-Lg(t) + \frac{\varepsilon}{2}g(t) + f(t)}{g(x)}, \frac{-Lg(t) - \frac{\varepsilon}{2}g(t) + f(t)}{g(x)} $ are less than $\frac{\varepsilon}{2}$ in absolute value (because the numerators are fixed, and $g(x) \to \infty$), so using that in our original inequality and choosing $\delta = \min\{ \delta_{1}, \delta_{2}, \delta_{3} \}$ guarantees that: $ L - \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \lt \frac{f(x)}{g(x)} \lt L + \frac{\varepsilon}{2} + \frac{\varepsilon}{2} $ Thus: $ \left| \frac{f(x)}{g(x)} - L \right| \lt \varepsilon $ for all $a \lt x \lt a + \delta$.

☐

Weirstrass Function

Let's construct the function that is everywhere continuous but nowhere differentiable.

!Wierestrass Function