- 18 Oct '03 01:17 / 1 editI made one up. Let f(x) = 1+x for x in [-2,0] and 1-x if x is in [0,2] and continuing elsewhere where f(x+4) = f(x). Then let:

F(x) = SUM (n=1 to infinity) 2^-n f(2^2^n x)

F(x) is everywhere continuous and nowhere differentiable. In general, just find a function with regular points where the derivative fails to exist and add them up in such a way that at least one term is guaranteed to have an undefined derivative. I think many others work.

I've got a proof; thanks for this, twas fun to work out. - 23 Oct '03 17:04In the stochastic calculus there are functions called wiener processes. Those are almost surely nowhere differentiable. "almost surely" is because of the randomness of the function. The only problem is that they are too complicated to explain here. But they look like the graph of stockprices. The reason they are nowheree differentiable is because you can construct one taking the sum of tent functions (functions like f(x)=x for x in [0,1] f(x)=-x+1 for x in (1,2] and f(x)=0 elsewhere) Then of course you need to add incountable infinite tent functions.

But it works