Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
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
1  
Related: en.wikipedia.org/wiki/Kirszbraun_theorem –  commenter Sep 27 '12 at 9:28
    
@user38773: I think that's an answer? –  joriki Sep 27 '12 at 10:32
add comment

1 Answer

up vote 0 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.