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## Everywhere Continuous Non-derivable 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.

### Reference

A course of mathematical analysis
SHANTI NARAYAN
PK MITTAL
S. Chand Co.