Construct a continuous function $f: (0, 1) \rightarrow \mathbb R$ but is not differentiable at any point of $\{\frac{1}{1+n}: n \in\mathbb N\}$

Question

I can construct many function continuous in this interval . I have to choose one function that is not differentiable in $(0,1)$ because all point of $\{\frac{1}{1+n}: n \in\mathbb N\}$ is in the interval $(0,1)$,but i don't know that function choose that is not differentiable,moreover I don't know like prove that one function is not differentiable, with the definition or have to use other theorem?

Answer 1

I am not sure that I understood the question, but I suspect that $\left|\sin\left(\frac\pi x\right)\right|$ will do.

Answer 2

You may apply condensation of singularities. The function $f(x)=\left\{\begin{array}{rcl} 0 & \text{at } x=0\\ x\sin\frac{1}{x} & \text{otherwise}\end{array}\right.$ is everywhere continuous and differentiable everywhere except at the origin. We have $\left|f(x)\right|\leq 1$, hence

$$ g(x)=\sum_{n\geq 1}\frac{1}{2^n}\,f\left(x-\frac{1}{n}\right) $$ is a continuous function on $[0,1]$, but it is not differentiable at any point of the form $\frac{1}{n}$.