# Theorem 7.18 in Baby Rudin: Existence of an everywhere continuous but nowhere differentiable real function on the real line

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.

• Have check the so call Weierstrass function? – Guy Fsone Sep 4 '17 at 12:04
• @GuyFsone no, I have yet to. Can you please suggest a place where I can read for these? – Saaqib Mahmood Sep 4 '17 at 12:06
• Concerning (36): Did you draw the graph of $\varphi$? – Jochen Sep 4 '17 at 12:10
• – Guy Fsone Sep 4 '17 at 12:13
• 36 is just the reverse triangle inequality. – Guacho Perez Jan 3 at 5:11

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, $$\left\lvert \gamma_m \right\rvert = 4^m$$ .