Function extending a Lipschitz function

Question

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

Answer 1

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