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Let $ f: M \to \mathbf{R}$ a $k$-Lipschitz function, i,e $ |f(x)-f(y)| \le k \cdot d(x,y) $ for every $x,y \in M$. Prove that $ \forall x\in M$ :

$ f\left( x \right) = \inf\limits_{y \in M} \{{f\left( y \right) + k \cdot d\left( {x,y} \right)}\} = \sup\limits_{y \in M} \{ {f\left( y \right) - k \cdot d\left( {x,y} \right)} \}. $ I have no idea how to prove this :/

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    yes you are right!2012-03-28

1 Answers 1

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Hint: You have to suppose Start defining the functions H, G

H(x)=\inf...$ and $G(x)=\sup...$ then first note that $H(x)\leq f(x)$ (just apply $ y=x$), then for the other inequality, consider $y_n$ such that $f(y_n)+kd(x,y_n)

$f(x)=[f(x)-f(y_n)]-kd(x,y_n)+[f(y_n)+kd(x,y_n)]<|f(x)-f(y_n)|-kd(x,y_n)+H(x)+\frac{1}{n}<0+H(x)+\frac{1}{n}

Since $n is arbitrary we get

f(x)\leq H(x)$ ergo $f(x)=H(x)$.

The other equality is left as an exercise!