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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 $x<y$. When $|x-y| \geq T$ then there are $n,m\in N$ such that $x-mT, y-nT\in [0,T]$ and $|g(x)-g(y)|=|f(x-mT)-f(y-nT)| \leq K \cdot T \leq K|x-y|$. When $|x-y|<T$ and, for some integer $n$, is $x,y \in [nT, (n+1)T]$ we have $|g(x)-g(y)|=|f(x-nT)-f(y-nT)|\leq K |x-y|$.

It remains the case when $|x-y| <T$ and, for some integer $n$, is $x\in [nT,(n+1)T]$, $y\in [(n+1)T, (n+2)T]$.

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Then use the triangular inequality $|f(y)-f(x)|\leqslant|f(y)-f((n+1)T)|+|f((n+1)T)-f(x)|$. – Did Jul 21 '12 at 8:56
Thanks! Now I understand. – Alex Jul 21 '12 at 9:01

1 Answer 1

up vote 1 down vote accepted

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 $x\in [nT,(n+1)t]$, $y\in [(n+1)T,(n+2)T]$.

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