Here is Theorem 7.18 from Baby Rudin:
There exists a real continuous function on the real line which is nowhere differentiable.
Here is a proof of the theorem:
Define
$$\tag{34} \varphi(x) = \lvert x \rvert \qquad \qquad (-1 \leq x \leq 1) $$
and extend the definition of $\varphi(x)$ to all real $x$ by requiring that
$$ \tag{35} \varphi(x+2) = \varphi(x). $$
Then, for all $s$ and $t$,
$$\tag{36} \lvert \varphi(s) – \varphi(t) \rvert \leq \lvert s-t \rvert. $$
In particular, $\varphi$ is continuous on $\mathbb{R}^1$. Define
$$ \tag{37} f(x) = \sum_{n=0}^\infty \left( \frac{3}{4} \right)^n \varphi \left( 4^n x \right). $$
Since $0 \leq \varphi \leq 1$, Theorem 7.10 shows that the series (37) converges uniformly on $\mathbb{R}^1$. By Theorem 7.12, $f$ is continuous on $\mathbb{R}^1$.
Now fix a real number $x$ and a positive integer $m$. Put
$$ \tag{38} \delta_m = \pm \frac{1}{2} \cdot 4^{-m} $$
where the sign is so chosen that no integer lies between $4^m x$ and $4^m \left( x + \delta_m \right)$. This can be done, since $4^m \left\lvert \delta_m \right\rvert = \frac{1}{2}$. Define
$$ \tag{39} \gamma_n = \frac{ \varphi \left( 4^n \left( x + \delta_m \right) \right) – \varphi \left( 4^n x \right) }{ \delta_m }. $$
When $n > m$, then $4^n \delta_m$ is an even integer, so that $\gamma_n = 0$. When $0 \leq n \leq m$, (36) implies that $\left\lvert \gamma_n \right\rvert \leq 4^n$.
Since $\left\lvert \gamma_m \right\rvert = 4^m$, we conclude that
$$
\begin{align}
\left\lvert \frac{ f \left( x + \delta_m \right) – f(x) }{ \delta_m } \right\rvert &= \left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\
&\geq 3^m – \sum_{n=0}^{m-1} 3^n \\
&= \frac{1}{2} \left( 3^m + 1 \right).
\end{align}
$$
As $m \to \infty$, $\gamma_m \to 0$. It follows that $f$ is not differentiable at $x$.
I have two questions regarding the proof.
-
At $(38)$ it's stated that the sign is so chosen that no integer lies between $4^m x$ and $4^m \left( x + \delta_m \right)$. Why is it necessary that no integer should be within that range and what would happen if they were?
-
At the later part of the proof we have:
$$
\begin{align}
\left\lvert \frac{ f \left( x + \delta_m \right) – f(x) }{ \delta_m } \right\rvert &= \left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\
&\geq 3^m – \sum_{n=0}^{m-1} 3^n \\
&= \frac{1}{2} \left( 3^m + 1 \right).
\end{align}
$$
I want to know why is the following part true:
$$
\begin{align}
\left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n \right\rvert
&\geq 3^m – \sum_{n=0}^{m-1} 3^n \\
\end{align}
$$
Any help is appreciated!
Best Answer
If you look at a picture of $\varphi$, it is a saw blade with a slope of $1$ in intervals of the form $(2k,2k+1)$, and a slope of $-1$ in intervals of the form $(2k+1,2(k+1))$. The approach to the proof is essentially to take an arbitrary $x$ and show that there exists a sequence $\{\delta_m\}$ such that $\delta_m \to 0$, but $$ \left|\frac{f(x+\delta_m)-f(x)}{\delta_m}\right| \to +\infty. $$ This method sufficiently demonstrates that the difference quotient cannot converge. The reason we guarantee that no integer lies between $4^mx$ and $4^m(x+\delta_m)$ is to guarantee that we are comparing points on the "saw blade" that correspond to the same branch (either one with slope of $1$ or $-1$, but no mixing). There's no reason to not do this because $\delta_m \to 0$ anyways, and it makes the proof easier.
Here's the full work for that step. The first step of reducing the sum from an infinite sum to one up to $m$ is because $\gamma_n=0$ for $n>m$. Later, we use that $\gamma_m=4^m$. \begin{align} \left\lvert \frac{ f \left( x + \delta_m \right) - f(x) }{ \delta_m } \right\rvert &= \left\lvert \sum_{n=0}^\infty \left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\ &= \left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\ &= \left\lvert \left( \frac{3}{4} \right)^m \gamma_m+ \sum_{n=0}^{m-1}\left( \frac{3}{4} \right)^n \gamma_n \right\rvert \\ &\geq \left\lvert \left(\frac{3}{4}\right) ^m \gamma_m \right\rvert-\left\lvert \sum_{n=0}^{m-1} \left( \frac{3}{4} \right)^n \gamma_n\right\rvert \\ &= 3^m -\left\lvert \sum_{n=0}^{m-1} \left( \frac{3}{4} \right)^n \gamma_n\right\rvert \\ &\geq 3^m - \sum_{n=0}^{m-1} \left( \frac{3}{4} \right)^n |\gamma_n| \\ &\stackrel{|\gamma_n|\leq 4^n}{\geq} 3^m - \sum_{n=0}^{m-1} 3^n \\ &= \frac{1}{2} \left( 3^m + 1 \right). \end{align}