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.

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Gaurav Tiwari
Gaurav Tiwari is a professional graphic & web designer from New Delhi, India. gauravtiwari.org is his personal space where he writes on blogging, digital marketing, content writing, learning and business growth. Gaurav has contributed in developing more than 325 brands worldwide and while you are reading this, he's busy building a couple more.

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2 Comments

  1. 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$.

      Hardy G. H., Weierstrass’s nondifferentiable
      function, Transactions of the American Mathematical Society -17 – 1916,

      Corrected!

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