Here is Theorem 7.18 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
There exists a real continuous function on the real line which is nowhere differentiable.
And, here is Rudin's proof ( steps wherein I've been unable to figure out on my own and hence would appreciate the help of the Math SE community):
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 $t$,
$$\tag{36} \lvert \varphi(s) – \varphi(t) \rvert \leq \lvert s-t \rvert. $$
[ How to obtain the inequality in (36)? ]
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 $\left\lvert \gamma_m \right\rvert = 4^m$ [ How? ], 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 $x$.
Here is Theorem 7.10 in Baby Rudin, 3rd edition:
Suppose $\left\{ f_n \right\}$ is a sequence of functions defined on $E$, and suppose $$ \left\lvert f_n (x) \right\rvert \leq M_n \qquad \qquad (x \in E, \ n = 1, 2, 3, \ldots \ ). $$
Then $ \sum f_n $ converges uniformly on $E$ if $ \sum M_n$ converges.Note that the converse is not asserted ( and is, in fact, not true).
And, here is Theorem 7.12:
If $\left\{ f_n \right\}$ is a sequence of continuous functions on $E$, and if $f_n \to f$ uniformly on $E$, then $f$ is continuous on $E$.
The rest of the proof I understand, I think.
However, I would appreciate if someone could give the crux of the procedure involved in the construction of this particular example and also give a blueprint for constructing this class of functions.
Best Answer
Note that the the sign is so chosen that no integer lies between $4^m x$ and $4^m \left( x + \delta_m \right)$. Because of this, and $4^n \delta_m = \pm \frac{1}{2}$, we have $\left\lvert \varphi \left( 4^n \left( x + \delta_m \right) \right) - \varphi \left( 4^n x \right) \right\rvert = \frac{1}{2} $. Thus, $\left\lvert \gamma_m \right\rvert = 4^m$ .