Let be lipschitz $f$ on $[0,1]$ and everywhere derivable. Is it true that $f\in C^1([0,1])$ ?



Regularity of lipschitz and derivable function
I claim that a function with these properties need not be $C^1$.


We start with the function $f: t \in (-1,1)\setminus \{ 0 \} \mapsto \operatorname{sin}(1/t)$, and we also set $f(0) = 0$. The antiderivative of $f$, namely $F: x \in (-1,1) \mapsto \int\_0^x \operatorname{sin}(1/t) \mathrm{d} t$ is Lipschitz, but not $C^1$ because of the discontinuity of $f$ at the origin.


It remains to check that $F$ is indeed differentiable at $x = 0$. This is just an integral evaluation: specifically we need only show that
$F(x)/x = \frac{1}{x} \int\_0^x \operatorname{sin}(1/t) \mathrm{d}t \to 0$ as $x \to 0$.


For the sake of completeness, this would go as follows. Ignore the factor $1/x$ for now, and do the change of variable $\theta = 1/t$ in the integral. This becomes $-\int\_{1/x}^{\infty} \operatorname{sin}(\theta)/\theta^2 \mathrm{d}\theta = [\operatorname{cos}(\theta)/\theta^2]\_{1/x}^\infty + \int\_{1/x}^\infty \operatorname{cos}(\theta)/\theta^3 \mathrm{d}t$. After taking absolute values, one gets $\lvert F(x) \rvert \leq 2 x^2$, whence the conclusion follows after dividing by $x$.


