# 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)$

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.

You can either start a new conversation or continue an existing one.Please don't use this comment form just to build backlinks. If your comment is not good enough and if in some ways you are trying to just build links — your comment will be deleted. Use this form to build a better and cleaner commenting ecosystem. Students are welcome to ask for help, freebies and more. Your email will not be published or used for any purposes.I believe that Hardy improved the result to allow ab > 1. I may be wrong…

Yes Sir! Hardy showed that the function of the above construction (Cosine Function) is non-derivable with the assumptions $ 0 \lt a \lt 1$ and $ ab \ge 1$.

Corrected!

I believe that Hardy improved the result to allow ab > 1. I may be wrong…

Yes Sir! Hardy showed that the function of the above construction (Cosine Function) is non-derivable with the assumptions $ 0 < a <1$ and $ ab ge 1$.

Corrected!