Everywhere Continuous Non-differentiable Function

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


$ \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.


$ 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. 

First time here? Try these resources...

  1. Best VPN Services
  2. Best WordPress Hostings
  3. Best WordPress Themes
  4. Best WordPress Plugins
  5. Best Gutenberg Block Plugins
  6. Best Email Marketing Plugins
  7. Best WordPress Caching Plugins
  8. Best WooCommerce Plugins
  9. Email Marketing Guide for Beginners
  10. Best Small Business Apps
  11. Best Business Name Generators
  12. Top Plagiarism Checkers
  13. Free Web Hosting Services
  14. Best Online Businesses to Start
  15. Best Online Course Platforms
  16. Best Online Casinos in India
  17. Best Affiliate Marketing Programs
  18. More Resources...

Get useful blogging, marketing and learning resources, delivered to your mailbox. Plus, get regularly updated with extra tools & guides to help you learn, grow and earn better.

Get 17+ exclusive e-books & templates for free, to begin with. 🎁

%d bloggers like this: