# Function extending a Lipschitz function

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 $F: X \rightarrow \mathbb R$).

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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? – Nonliapunov Sep 27 '12 at 8:44
From this condition follows that $F$ is finite but on $E$. – R.S Sep 27 '12 at 8:50
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 $f(y)-Md(x,y)\leq f(z)+Md(z,x)$. – Nonliapunov Sep 27 '12 at 9:05
– commenter Sep 27 '12 at 9:28
@user38773: I think that's an answer? – joriki Sep 27 '12 at 10:32

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 $F(x)\leq f(z)+Md(z,x)$.