Recall the problem from my previous question:
Let $f,g: \mathbb R\to \mathbb R$ be functions such that $g(x)\ge 0,\ g(0)=0$ and $g$ is differentiable at the origin. Suppose for some $c > 1$, $$f(b)-f(a) \le g(|b-a|)^c$$ for all $a,b\in \mathbb R$. Prove that $f$ is infinitely differentiable on $\mathbb R$.
I've been trying to construct a counterexample when the statement fails if the assumption on the differentiability of $g$ is omitted. That is, an example of $f,g$ such that $g$ is only continuous and $f$ is not differentiable at all points of the real line.
Since $g$ is assumed to be non-negative, the first candicate for $g$ is $g(t)=\sqrt t$ - it's continuous at the origin but not differentiable there. But there are so many candidates for $f$ that I don't know how I can choose one so that the inequality is satisfied.
A silly (but valid) counterexample would be $$ g(x) = \begin{cases} 0 & \text{ if } x=0\\ 1 & \text{ otherwise.} \end{cases} $$ It's not even continuous, but satisfies the assumption; and with this it is easy to find examples of $f$ not even continuous themselves.
A more interesting one would be to "reverse-engineer" the dependence on $c\geq 1$ to "beat" the use of it in the proof. Namely, set $$ g(x) = \lvert x\rvert^{1/{c}}, \qquad x\in\mathbb{R} $$ Then $g$ is continous, and differentiable on $\mathbb{R}\setminus\{0\}$. However, the assumption $$ f(b) - f(a) \leq \lvert b-a\rvert, \qquad a,b\in\mathbb{R} $$ leads to continuity (Lipschitz continuity, even) of $f$, but not to its differentiability.
(You can of course do more fine-tuned things: for instance, with the exponent $\alpha/c$ you can restrict $f$ to be $\alpha$-Hölder continuous.)