Weierstrass had drawn attention to the fact that there exist functions which are continuous for every value of $ x$ but do not possess a derivative for any value. We now consider the celebrated function given by Weierstrass to show this fact. It will be shown that if

$ f(x)= \displaystyle{\sum_{n=0}^{\infty} } b^n \cos (a^n \pi x) \ \ldots (1) \\ = \cos \pi x +b \cos a \pi x + b^2 \cos a^2 \pi x+ \ldots $ where $ a$ is an odd positive integer, $ 0 < b <1$ and $ ab > 1+\frac{3}{2} \pi$ , then the function $ f$ is continuous $ \forall x$ but not finitely derivable for any value of $ x$ .

G.H. Hardy improved this result to allow $ ab \ge 1$ .

We have $ |b^n \cos (a^n \pi x)| \le b^n$ and $ \sum b^n$ is convergent. Thus, by Wierstrass’s $ M$ -Test for uniform Convergence the series (1), is uniformly convergent in every interval. Hence $ f$ is continuous $ \forall x$ .

Again, we have $ \dfrac{f(x+h)-f(x)}{h} = \displaystyle{\sum_{n=0}^{\infty}} b^n \dfrac{\cos [a^n \pi (x+h)]-\cos a^n \pi x}{h} \ \ \ldots (2)$

Let, now, $ m$ be any positive integer. Also let $ S_m$ denote the sum of the $ m$ terms and $ R_m$ , the remainder after $ m$ terms, of the series (2), so that

$ \displaystyle{\sum_{n=0}^{\infty}} b^n \dfrac{\cos [a^n \pi (x+h)]-\cos a^n \pi x}{h} = S_m+R_m $ . By Lagrange’s mean value theorem, we have

$ \dfrac{|\cos {[a^n \pi (x+h)]} -\cos {a^n \pi x|}}{|h|}=|a^n \pi h \sin {a^n \pi(x+\theta h)}| \le a^n \pi |h|$ ,

$ |S_m| \le \displaystyle{\sum_{n=0}^{m-1}} b^n a^n \pi = \pi \dfrac {a^m b^m -1}{ab-1} < \pi \dfrac {a^m b^m}{ab-1}$ . We shall now consider $ R_m$ .

So far we have taken $ h$ as an arbitrary but we shall now choose it as follows:

We write $ a^m x=\alpha_m+\xi_m$ , where $ \alpha_m$ is the integer nearest to $ a^m x$ and $ -1/2 \le \xi_m < 1/2$ .

Therefore $ a^m(x+h) = \alpha_m+\xi_m+ha^m$ . We choose, $ h$ , so that $ \xi_m+ha^m=1$

i.e., $ h=\dfrac{1-\xi_m}{a^m}$ which $ \to 0 \ \text{as} \ m \to \infty$ for $ 0< h \le \dfrac{3}{2a^m} \ \ldots (3)$

Now, $ a^n \pi (x+h) = a^{n-m} a^m (x+h.) \\ \ =a^{n-m} \pi [(\alpha_m +\xi_m)+(1-\xi_m)] \\ \ =a^{n-m} \pi(\alpha_m+1)$

Thus $ \cos[a^n \pi (x+h)] =cos [a^{n-m} (\alpha_m-1) \pi] =(-1)^{\alpha_{m+1}}$ .

$ \cos (a^n \pi x) = \cos [a^{n-m} (a^m \pi x)] \\ \ =\cos [a^{n-m} (\alpha_m+\xi_m) \pi] \\ \ =\cos a^{n-m} \alpha_m \pi \cos a^{n-m} \xi_m \pi – \sin a^{n-m} \alpha_m \pi \sin a^{n-m} \xi_m \pi \\ \ = (-1)^{\alpha_m} \cos a^{n-m} \xi_m \pi$ for $ a$ , is an odd integer and $ \alpha_m$ is an integer.

Therefore, $ R_m =\dfrac{(-1)^{\alpha_m}+1}{h} \displaystyle{\sum_{n=m}^{\infty}} b^n [2+\cos (a^{n-m} \xi_m \pi] \ \ldots (4)$

Now each term of series in (4) is greater than or equal to 0 and, in particular, the first term is positive, $ |R_m| > \dfrac{b^m}{|h|} > \dfrac{2a^m b^m}{3} \ \ldots (3)$

Thus $ \left| {\dfrac{f(x+h) -f(x)}{h}} \right| = |R_m +S_m| \\ \ \ge |R_m|-|S_m| > \left({\frac{2}{3} -\dfrac{\pi}{ab-1}} \right) a^mb^m$

As $ ab > 1+\frac{3}{2}\pi$ , therefore $ \left({\frac{3}{2} -\dfrac{\pi}{ab-1}} \right) $ is positive.

Thus we see that when $ m \to \infty$ so that $ h \to 0$ , the expression $ \dfrac{f(x+h)-f(x)}{h}$ takes arbitrary large values. Hence, $ f’(x)$ does not exist or is at least not finite.

### Reference

A course of mathematical analysis

SHANTI NARAYAN

PK MITTAL

S. Chand Co.