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It is known that nonconstant Hoelder functions $f:R \rightarrow R$ with power $>1$ non exist.

Are there functions $f: R\rightarrow R$ satisfying $$ |f(x)-f(y)| \leq C|x-y|^p +D$$ for all $x,y \in R$, where $C,D>0$, $p>1$?

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Obviously any bounded function will suffice. Are you looking for an unbounded example? –  Alex Becker Aug 1 '12 at 19:55
    
Thanks. Yes, If there are. –  Alex Aug 1 '12 at 19:58
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1 Answer

up vote 3 down vote accepted

Any Lipschitz continuous (Hölder continuous with power $1$) function will do. Suppose $f:\mathbb R\to\mathbb R$ is such that $|f(x)-f(y)|\leq C|x-y|$ for all $x,y\in\mathbb R$. Then for any $p>1$ we have $$|f(x)-f(y)|\leq C(|x-y|^p+1)=C|x-y|^p+C$$ for any $x,y\in\mathbb R$, since if $|x-y|\leq 1$ then clearly $|x-y|< |x-y|^p+1$ while otherwise $|x-y|<|x-y|^p$.

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