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Let $f: [0,T] \rightarrow \mathbb{R}$, where $T>0$, be a Lipschitz with constant $K$ and $f(0)=f(T)$. Let us define $g(x)=f(x)$ for $x \in [0,T]$ and $g(x+T)=g(x)$ for $x \in \mathbb{R}$.

Does $g$ satisfies $$|g(x)-g(y)| \leq K |x-y|$$ for $x,y \in \mathbb{R}$?

It is clear that we may assume that $x

It remains the case when $|x-y|

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    Then use the triangular inequality $|f(y)-f(x)|\leqslant|f(y)-f((n+1)T)|+|f((n+1)T)-f(x)|$.2012-07-21

1 Answers 1

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Let $x\in [0,T]$, $y\in [T,2T]$. Then $$ |g(y)-g(x)|\le |g(y)-g(T)|+|g(T)-g(x)|=|f(y-T)-f(0)|+|f(T)-f(x)|\le K(y-T)+K(T-x)=K(y-x) $$ Of course the same argument works if $x\in [nT,(n+1)t]$, $y\in [(n+1)T,(n+2)T]$.