Let $W(x)$ be a real-vlued function defined on a (possibly infinite) interval $\text{T}\subseteq\mathbb{R}$ containing $0$ that is continuous everywhere differentiable nowhere on $\text{T}$.
Define the sequence of function $f_n:\text{T}\to\mathbb{R}$ as follows:
$$f_0(x)\triangleq{W(x)}$$ and $$f_n(x)\triangleq\int_{0}^{x}f_{n-1}(u)du$$ for $n=1,2,3,4,5,...$ .
Then it follows the fact that $f_n$ is differentiable exactly $n$-times everywhere on $\text{T}$.
Question:
Does there exist such $W(x)$ and $\text{T}$ so that the sequence of function $f_n$ on $\text{T}$ "converges" to a limit function $f_{\infty}$ on $\text{T}$ with $f_{\infty}$ being infinitely many times differentiable (i.e., smooth) on $\text{T}$?
If $T$ is a bounded interval and $W:T\to \mathbb{R}$ a bounded continuous function, then your sequence $\{f_n\}_{n\in \mathbb{N}}$ uniformly converges to the zero-function. Indeed, one check's by induction that $\|W\|_{\infty}=M\Rightarrow |f_n(x)| \leq M\frac{|x|^n}{n!}$. Since $T$ is bounded, $|x|<N$ for a certain $0<N<+\infty$. Hence $\|f_n\|_{\infty}\leq M\frac{N^n}{n!}$ which converges to zero as $n \to \infty$. Lastly, the zero-function is as smooth as you can get.