Weierstrass had drawn attention to the fact that there exist functions which are continuous for every value of 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
where is an odd positive integer, and , then the function is continuous but not finitely derivable for any value of .
G.H. Hardy improved this result to allow .
We have and is convergent. Thus, by Wierstrass’s -Test for uniform Convergence the series (1), is uniformly convergent in every interval. Hence is continuous .
Again, we have
Let, now, be any positive integer. Also let denote the sum of the terms and , the remainder after terms, of the series (2), so that
. By Lagrange’s mean value theorem, we have
. We shall now consider .
So far we have taken as an arbitrary but we shall now choose it as follows:
We write , where is the integer nearest to and .
Therefore . We choose, , so that
i.e., which for
for , is an odd integer and is an integer.
Now each term of series in (4) is greater than or equal to 0 and, in particular, the first term is positive,
As , therefore is positive.
Thus we see that when so that , the expression takes arbitrary large values. Hence, does not exist or is at least not finite.
A course of mathematical analysis
S. Chand Co.