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Suppose $f,\omega:{\bf R}\to{\bf R}$ are functions with $\omega(0)=0$. Suppose for some $\alpha>1$, we have $$ f(b)\leq f(a)+\omega(|b-a|)^\alpha\quad\hbox{for all } a,b\in{\bf R}\tag{*} $$ If $\omega$ is differentiable at $x=0$, show that $f\in C^\infty({\bf R})$.

The original problem is given as follows:

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I think $\omega(|b-a|)^\alpha$ should be understood as $[\omega(|b-a|)]^\alpha$.

The condition (*) can be written as $$ \frac{|f(x+h)-f(x)|}{h}\leq \frac{\omega(|h|)^\alpha-\omega(0)^\alpha}{h}. $$ This seems to imply the differentiability of $f$. But how would one expect that $f$ could be smooth?

  • It seems that if one can show (similarly as in https://math.stackexchange.com/q/1223377/9464) that $f'=0$, then we are done. –  Jul 07 '17 at 16:26

1 Answers1

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Enlightened by this question:

$f$ defined $|f(x) - f(y)| \leq |x - y|^{1+ \alpha}$ Prove that $f$ is a constant.

I have an answer.

Since $\alpha>1$, the map $x\mapsto x^\alpha$ is differentiable at $x=0$. The inequality $$ \frac{|f(x+h)-f(x)|}{h}\leq \frac{\omega(|h|)^\alpha-\omega(0)^\alpha}{h}. $$ implies that for any $x\in{\bf R}$, $$ |f'(x)|\leq \omega(0)^{\alpha-1}\cdot\omega'(0)=0 $$ which implies that $f$ is a constant.