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.

Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word. How to write better comments?
1. Rich Lipton says:

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

1. 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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