An example of $g$ such that $f(b)-f(a) \le g(|b-a|)^c$ does not imply the differentiability of $f$ 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: 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.)
A: With the square root for $g$ you are looking for functions such that:
$$f(b)-f(a) \leq |b-a|^\frac{c}{2}$$
as noted in the comments the condition is equivalent to:
$$|f(b)-f(a)| \leq |b-a|^\frac{c}{2}$$
you obtain for $c <2$ the Holdër functions which are not necessarily differentiable, see for example here.
If $c >2 $ you can replace $g(x)=\sqrt{x}$ by $g(x)=\sqrt[r]{x}$ for any $r >c$.
