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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 $x,y \in [0,2c]$, $x<y$. It is obvious that if $x,y \in [0,c]$ or $x,y \in [c,2c]$ then $|g(x)-g(y)|\leq M |x-y|^p$. For $x\in [0,c]$, $y\in [c,2c]$ we have

$$ |g(x)-g(y)| \leq |g(x)-g(c)|+|g(c)-g(y)| =|f(x)-f(c)|+|f(0)-f(y-c)|\leq M [(x-c)^p+(y-c)^p]\leq M 2^{1-p} (c-x+y-c)^p= M 2^{1-p} |y-x|^p. $$

(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

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1 Answer 1

up vote 1 down vote accepted

Suppose $f(x) = x^p$ for $0 \le x \le \epsilon$ and $f(x) = -(c-x)^p$ for $c-\epsilon \le x \le c$, which (at least if $c$ is large enough and you define $f$ suitably on $[\epsilon, c-\epsilon]$) has constant $M=1$. Then taking $x=c+\epsilon$, $y=c-\epsilon$, $$ |g(x) - g(y)| = 2 \epsilon^p = 2^{1-p} |x-y|^p$$ and you can't get rid of the $2^{1-p}$.

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