I am working on an exercise that introduces the Hölder condition and uses it to proove some results.
Let $f : [0,1] \mapsto \mathbb{R}$ be a continuous function that satisfies $f(0) = f(1) = 0$ and for a fixed $s\in [0,1[$ and $x_0 \in [0,1]$, $$\forall x\in[0,1],|f(x)-f(x_0)| \le C|x-x_0|^s $$ let $I = \{(j,k) \in \mathbb{N}^2 | j \in \mathbb{N}, 0 \le k \lt 2^j \}$ and $$\forall (j,k) \in I, c_{j,k}(f) = f((k+1/2)2^{-j})-\cfrac{f(k2^{-j})+f((k+1)2^{-j})}{2}$$ proove there exists $c_1\gt0$ for which $$ \forall (j,k)\in I, |c_{j,k}(f)|\le c_1(2^{-j}+|k2^{-j}-x_0|)^s$$ is true
Here is what I tried to do: First I tried to introduce $f(x_0)$ in the formula of $c_{j,k}$ in order to use the Hölder condition by adding and substracting it. So $$c_{j,k}(f) = f((k+1/2)2^{-j})-f(x_0)-\frac{1}{2}[f(k2^{-j})-f(x_0)+f((k+1)2^{-j})-f(x_0)]$$ then $$\begin{align*}|c_{j,k}(f)| &\le |f((k+1/2)2^{-j})-f(x_0)|+\frac{1}{2}[|f(k2^{-j})-f(x_0)|+|f((k+1)2^{-j})-f(x_0)|] \\ &\le |(k+1/2)2^{-j} - x_0|^s+\frac{1}{2}\left[|k2^{-j} - x_0|^s+|(k+1)2^{-j} - x_0|^s\right]\end{align*}\label{1}\tag{1}$$
We had prooven previously that for $a,b\gt 0$, $a^s +b^s \le 2^{1-s}(a+b)^s$ So i tried to use that but in vain since I didn't find a way to combine the absolute value operators to get rid of $(k+1/2)2^{-j}$ and $(k+1)2^{-j}$. I think I've lost a lot of precision by using the triangular inequality in \eqref{1} but I don't see another way to use the Hölder condition. I've also tried to visualize $c_{j,k}(f)$ geometrically and it seems to me it is the distance between the value taken by the function $f$ at the middle of the segment $[k2^{-j},(k+1)2^{-j}]$ and what it would have been if it was affine.
NB: I know this isn't really the Hölder condition since it isn't for all $x,y \in [0,1]$ But let's just call it that for the sake of simplicity. I also don't have a lot of experience with hölder spaces, the exercise only explains the Hölder condition and gives nothing more than that