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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 believe the RHS should read $(g(|b-a|))^c$.

My thoughts: for $h > 0$ I can write the above as $$\frac{f(x+h)-f(x)}{h}\le \frac{[g(h)]^c-g(0)^c}{h}$$

and taking $\lim$ as $h\to 0+$ seems to give $f'_+(x)\le ((g(x))^c)'|_{x=0}=cg(0)^{c-1}g'(0)=0$. (Since $g$ and $t\mapsto t^c$ are differentiable at the origin, we know that the right derivative equals the derivative). But there are a couple of concerns:

The major concern is what to do with $h < 0$? If the same were true for $h < 0$, it would follow that $f$ is constant. But the inequality is reversed if $h<0$...

user557
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  • Small note: what you believe and what is written is the same. $g(x)^c = (g(x))^c = ((g(x)))^c$ (the parentheses are superfluous). There is no confusion here with $g(x^c)$ under the standard precedence rules. – Clement C. Jul 11 '18 at 23:53
  • Looks to me like you should say $g\ge 0$, otherwise raising to the $c$ power could be problematic – zhw. Jul 12 '18 at 01:13

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The answer is based on your idea after some modification.

We have: $$ f(a)-f(b)\leq(g(|a-b|))^c \,\land \, f(b)-f(a)\leq(g(|a-b|))^c \implies |f(a)-f(b)|\leq (g(|a-b|))^c $$ From which for $h\neq 0$, letting $h\to 0$: $$\left |\frac{f(x+h)-f(x)}{h}\right | \leq\frac{(g(|h|))^c-(g(0))^c}{|h|} \to (g^c)'(0+) = 0 $$ So that by the squezee theorem $f'(x)$ exists and we have $$ f'(x)=0 \implies f(x)=c $$ For some real $c$. Which means the function $f$ is infinitely many times differentiable.

Jakobian
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  • Why is the limit of the RHS of your inequality the derivative of $g(x)^c$ at $x=0$? I thought the derivative of $g(x)^c$ at $x=0$ is $$\lim_{h\to 0} \frac{g(h)^c-g(0)^c}{h},$$ and that's not the limit of the expression on the RHS (you have absolute value as well). – user557 Jul 12 '18 at 01:03
  • Yes, but if we go with $h$ to $0$, it's the same for us, as if we evaluated the limit from the right side. In other words, it's equal to $\lim_{h\to 0^+} \frac{g(h)^c-g(0)^c}{h}$. To see this you can consider when $h\to 0^+$ and when $h\to 0^-$ – Jakobian Jul 12 '18 at 01:06
  • It would be better to not plop down the symbol $f'(x)$ until you prove it exists. Just show the RHS $\to 0,$ hence so does the LHS. Also, you are using $c$ with two different meanings. – zhw. Jul 12 '18 at 01:15
  • @zhw. you're right. Thanks – Jakobian Jul 12 '18 at 01:16
  • @user5579085 You know $(g^c)'(0)= 0.$ Hence the derivative equals $0$ from the right. – zhw. Jul 12 '18 at 01:19
  • @zhw. Do you mean I can take the limit as $h\to 0-$ in the inequality in my question? (If so, I don't understand why, since $h>0$ is assumed.) Or which argument do you mean? – user557 Jul 12 '18 at 02:24
  • @user5579085 No, from the right is $h\to 0^+.$ – zhw. Jul 12 '18 at 02:29
  • @zhw. Which derivative equals $0$ from the right? The derivative of $g^c$ at the origin? (Yes, it is.) Or that of $f(x)$ at an arbitrary $x$? (I can only see that $f_{+}'(x)\le 0$.) I'm not sure what you meant by that comment. – user557 Jul 12 '18 at 02:40
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    @user5579085 I don't understand where you have difficulty. Do you see that $$\left |\frac{f(x+h)-f(x)}{h}\right | \leq\frac{(g(|h|))^c-(g(0))^c}{|h|} \to (g^c)'(0+) = 0?$$ This proves $f'(x)=0!!!$ – zhw. Jul 12 '18 at 03:02
  • @zhw. Oh, now I see, your original comment was a response to my concern that there is $|h|$ (and not just $h$) in the formula from the answer. I got your point, thanks a lot! – user557 Jul 12 '18 at 03:41