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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.

user557
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2 Answers2

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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.)

Clement C.
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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 $r >c$.

Delta-u
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  • $c>1$, not $c=1$: for $c=3$ for instance your choice still leads to $c/2>1$, and thus $f$ being infinitely differentiable. (Actually, constant). And the condition with absolute values is not "stronger", it is equivalent. (Apply the original condition to both $a,b$ and $b,a$ to see it.) – Clement C. Jul 12 '18 at 16:46
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    Good point for the equivalence, for the $c$ it is indeed true that you need to adapt the constant $1/2$ to $c$ but any $|x|^r$ with $r<1/c$ ( in the same spirit as your answer :-)) – Delta-u Jul 12 '18 at 16:49