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.
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 $h$.
