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Let $X$ be a metric space with a metric $d$, let $E\subset X$. We have a function $f:E \rightarrow \mathbb R$ satisfying for some $M>0$: $$ |f(x)-f(y)|\leq M d(x,y) \quad \text{for } x,y \in E. $$

I wish to show that a function $$ F(x)=\sup_{y \in E} [f(y)-M d(x,y)] $$ for $x \in X$, is finite (that is $F: X \rightarrow \mathbb R$).

  • 0
    From the Lipschitz condition, you have $f(y)-f(x)\leq M d(x,y)$, so that $f(y)-Md(x,y)\leq f(x)$. Doesn't this solve it?2012-09-27
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    From this condition follows that $F$ is finite but on $E$.2012-09-27
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    If $x\notin E$, let $z\in E$ be arbitrary, then $f(y)-Md(x,y)=f(y)-Md(z,y)+M(d(z,y)-d(x,y))$, but from the triangle inequality $d(z,y)-d(x,y)\leq d(z,x)$ so hat $f(y)-Md(x,y)\leq f(z)+Md(z,x)$.2012-09-27
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    Related: http://en.wikipedia.org/wiki/Kirszbraun_theorem2012-09-27
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    @user38773: I think that's an answer?2012-09-27

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If $x\in E$, then $f(y)-f(x)\leq Md(x,y)$ so that $f(y)-Md(x,y)\leq f(x)$ and hence $F(x)\leq f(x)$.

If $x\notin E$, let $z\in E$ be arbitrary, then $f(y)-Md(x,y)=f(y)-Md(z,y)+M(d(z,y)-d(x,y))$, but from the triangle inequality $d(z,y)-d(x,y)\leq d(z,x)$ so that $f(y)-Md(x,y)\leq f(z)+Md(z,x)$ and hence $F(x)\leq f(z)+Md(z,x)$.