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.