Suppose that $f$ is a continuous, nonconstant function on $[0,1]$. Fix some $0<a<1$. Is it possible to establish the following inequality
$$ |f(x+h)-f(x)| \leq C \left[ |h|^a + |2f(x)-f(x+h)-f(x-h)| \right] ~~~~~\forall x,x-h,x+h\in I, $$
where $C$ only depends on $f$ and $a$?


$|h|^a$ in the RHS of the inequality is added to restrict the scenario when $f$ is differentiable.



Can the second-order difference control the first-order difference for nowhere differentiable functions?
No. E.g., take any $b\in(0,1)$ such that $1-b<a$. Take any strictly decreasing sequence $(x\_n)$ in $[0,1]$ converging to $0$.


Then there clearly exists a continuous function $f$ on $[0,1]$ such that $f(0)=0$ and for all natural $k$ and all $x\in[x\_{2k},x\_{2k-1}]$ we have
$$f(x)=2h\_{2k}^{-b}(x-x\_{2k}),$$
where $h\_n:=x\_{n-1}-x\_n>0$; for instance, for each natural $k$, you can define $f$ on $[x\_{2k+1},x\_{2k}]$ by linear interpolation: $f(x):=2h\_{2k+1}^{-b}(x\_{2k}-x)$ for $x\in[x\_{2k+1},x\_{2k}]$. Note that $f(x)\to0$ for $x\in[x\_{2k},x\_{2k-1}]$ and $k\to\infty$.


For each natural $k$, let now $x=x\_{2k}$ and $h=h\_{2k}/2$. Then $|f(x+h)-f(x)|=h\_{2k}^{1-b}$ and $|h|^a + |2f(x)-f(x+h)-f(x-h)|=h^a=h\_{2k}^a/2^a=o(|f(x+h)-f(x)|)$ as $k\to\infty$. So, there is no real $C$ such that $|f(x+h)-f(x)|\le C[|h|^a + |2f(x)-f(x+h)-f(x-h)|]$ for all such $x$ and $h$.


