Regularity of lipschitz and derivable function

Question

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

Answer 1

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 $x$.

Answer 2

Unless I'm overlooking something, you're simply asking whether a function $f:[0,1] \rightarrow \mathbb R$ with a bounded derivative must be continuously differentiable. This can fail quite spectacularly, as described in this MSE answer and summarized below. Although I don't think I pointed this out in that MSE answer, everything there holds for bounded derivatives, as the constructions cited are based on Volterra functions (see also this MSE answer).

Given any $F_{\sigma}$ first (Baire) category (i.e. meager) subset $E$ of $[0,1],$ there exists such a function whose derivative is discontinuous at each point of $E,$ and such sets can have cardinality continuum in each subinterval of $[0,1].$ Indeed, it is possible that the intersection of $E$ with each subinterval of $[0,1]$ has Hausdorff dimension $1$. Stronger still, given any Hausdorff measure function $h,$ the set $[0,1]-E$ can have Hausdorff $h$-measure zero. (Here, "subinterval" means an interval containing at least $2$ points.)

Moreover, almost all bounded derivatives are discontinuous in the strongest way just stated, where "almost all" is in the Baire category sense for a certain "natural" metric on the set of bounded derivatives. (The size/density you want $E$ to satisfy needs to be fixed in advance, after which a co-meager set of bounded derivatives -- each of which having the desired size/density-discontinuity property -- is found.)