Let $f:[0,c] \rightarrow \mathbb R$ be Hölder continuous with constant $M>0$ and power $p \in (0,1)$ and satisfies $f(0)=f(c)$. Let $g:[0,2c] \rightarrow \mathbb R$ be given by:
$ g(x)=f(x) \ \ \textrm{for} \ \ x \in [0,c] $ and $ g(x)=f(x-c) \ \ \textrm{for} \ \ x \in [c,2c]. $
Then $g$ is Hölder continuous with the same power but I don't know if constant $M$ is the same.
I try in this way. Let $x,y \in [0,2c]$, $x
$ |g(x)-g(y)| \leq |g(x)-g(c)|+|g(c)-g(y)| =|f(x)-f(c)|+|f(0)-f(y-c)|\leq M [(x-c)^p+(y-c)^p]\leq M 2^{1-p} (c-x+y-c)^p= M 2^{1-p} |y-x|^p. $
(I have used inequality: $2^{p-1}(u^p+v^p)\leq (u+v)^p$ for $u,v \geq 0$, $p\in (0,1)$, which follows from concavity of $[0,\infty) \ni t\mapsto t^p$.)
Does constant $M 2^{1-p}$ can be improved to $M$ ?
Thanks