If a function $f: [0, 1] \to \mathbb R$ is differentiable at a point $p$, then there exists some unique linear function $L_p$, and some unique function $r_p$ such that $f(x) = f(p) + L_p (x - p) + r_p (x - p)$, where $r_p$ is $o(|x-p|)$ as $x$ approaches $p$.
Given $e > 0$, and a function $f$ differentiable at $p$, define $t(e, f, p)$ to be
$$\sup \{d:\ \min(p, 1-p) > d > 0;\ r_p(x-p) < e|x-p| \text{ whenever } |x-p| < d\} $$
Now for $0 < p < 1$, define the quantity $N(f, p)$ to be
$$\limsup_{e \to 0+} \frac{1}{2et} \int |r_p (y)/(y - p)| \ dy$$
where $t$ is short for $t(e, f, p)$, and the integral is taken over all $y \in [p - t, p + t]\setminus p$ with respect to Lebesgue measure.
We call $N(f, p)$ the roughness of $f$ at $p$. Note that $N(f, p)$ is between $0$ and $1$ inclusive.
Are there any differentiable functions $f: [0, 1] \to R$ such that $N(f, p) = 1$ for every $ p \in (0, 1)$? What about continuously differentiable functions?
