An inequality about the second-order difference

Question

Fix a continuously differentiable but nowhere twice differentiable function $f$ on $\mathbb{R}$ supported on $[0,1]$. Is it true that for all $x\in[0,1]$ and all $\delta$ sufficiently small \begin{align*} & \sup_{0<h\leq \delta}|f'(x+2h)-2f'(x+h)+f'(x)|\leq \\ & \quad C \delta^{-1}\sup_{0<h\leq \delta}|f(x+2h)-2f(x+h)+f(x)|~~? \end{align*} Here $C$ is allowed to depend on $f$, and does not depend on $\delta$ and $x$.

Answer 1

$\newcommand\de\delta\newcommand\lhs{\text{lhs}}\newcommand\rhs{\text{rhs}}$No. E.g., let $$f(x):=x^3\sin\frac1x$$ for $x\in(0,1/2]$, with $f(0):=0$. Then $f$ can be obviously extended to a continuously differentiable function $f$ on $\mathbb R$ supported on $[0,1]$.

Moreover,
$$\rhs(\de):=\de^{-1}\sup_{0<h\le\de}|f(2h)-2f(h)+f(0)| \\ =\de^{-1}\sup_{0<h\le\de}\Big|-2 h^3 \Big(\sin\frac1h-4 \sin \frac{1}{2h}\Big)\Big|=O(\de^2)=o(\de), $$ while for $\de=\dfrac1{2(2n+1)\pi}$ and natural $n\to\infty$ $$\lhs(\de):=\sup_{0<h\le\de}|f'(2h)-2f'(h)+f'(0)| \\ \ge |f'(2\de)-2f'(\de)+f'(0)| \\ =4 \de \sin ^2\frac{1}{4 \de }\, \Big|6 \de \sin\frac{1}{2 \de }-2 \cos \frac{1}{2 \de }-1\Big| \\ \sim4 \de \sin ^2\frac{1}{4 \de }=4\de. $$ So, $\lhs(\de)$ is not $O(\rhs(\de))$.