# Functional equations with nowhere differentiable solutions

As an example, the functional equation $f(x+y)=f(x)f(y)$, by declaring that $f$ is continuous and differentiable, we can arrive at the unique solution $f(x)=a^x$, by first showing that $f'(x)=f(0)f(x)$ and then using basic ODE theory to declare such an $f$ exists.

I'm interested in an example of a functional equation whose only solution is a continuous nowhere differentiable function. While some gymnastics with trigonometry might arrive at a functional equation for something like the Weierstrass functions, uniqueness might be hard to prove. Are there any good "easy" examples of such functional equations? I'd strongly prefer something that uses basic operations and classical functions and not say, derivatives in the sense of formal power series.

OK, let's do the said gymnastics to arrive at $$f(x)=\cos x + a\cdot f(bx)\qquad(a<1,\;b>1)$$ Now we may expand $f(bx)$ on the right to get $f(b^2x)$, then expand that too, and so on. Unless $f$ grows too fast (let's say we require the solution to be bounded), we'll get the series that looks mighty like the definition of the Weierstrass function.

Looks classical enough to me.

Also, it doesn't have to be trigonometry; I guess the triangle wave instead of cosine would do just as well.