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does anyone know the following result? If it holds in this form and any source which presents it? Thanks a lot.

Consider metric space $(X,d_{X})$. Let $f:A \subset (X,d_{X}) \rightarrow \mathbb{R}$ be a Lipschitz continuous function i.e. $\exists K > 0$ such that \begin{align*} |f(x)-f(y)| \leq Kd_{X}(x,y) ~~~~\forall x,y \in A \end{align*} then define functions

$$\bar{f_{1}}(x) := \sup\lbrace f(y) - Kd_{X}(x,y): y\in A \rbrace ~~ \forall x \in X$$ and $$\bar{f_{2}}(x) := \inf\lbrace f(y) + Kd_{X}(x,y): y\in A \rbrace ~~ \forall x \in X$$ it follows that \begin{align*} \bar{f_{1}}, \bar{f_{2}}: X \rightarrow \mathbb{R} \end{align*} are Lipschitz continuous with \begin{align*} |\bar{f_{i}}(x)-\bar{f_{i}}(z)| \leq Kd_{X}(x,z) ~~~\forall x,z \in X \text{ and } \forall i = 1,2 \end{align*} and \begin{align*} \bar{f_{i}}(x) = f(x) ~~~ \forall x \in A \text{ and } \forall i =1,2 \end{align*}

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Take $\epsilon>0$ and choose $y_x\in A$ such that $$\overline{f}_1(x)\ge f(y_x)-Kd_X(x,y_x)\ge \overline{f}_1(x)-\epsilon\tag{1}.$$

It follow from $(1)$ and the definition of $\overline{f}_1$ that $$\overline{f}_1(x)-\overline{f}_1(z)\le f(y_x)-Kd_X(x,y_x)+\epsilon-f(y_x)+Kd_X(z,y_x).\tag{2}$$

We conclude from $(2)$ that $$\overline{f}_1(x)-\overline{f}_1(z)\le Kd_X(x,z)+\epsilon.$$

Once $\epsilon$ is arbitrary, we conclude that $\overline{f}_1(x)-\overline{f}_1(z)\le Kd_X(x,z)$. The other inequality is similar, while the fact that $\overline{f}_1$ is a extension of $f$ is trivial. Analogous arguments works also for $\overline{f}_2$. As an reference, take a look in Kirszbraun theorem and in section 2.10.43. of Federer's book.

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  • $\begingroup$ So then this would be a good result to use for the last part of my proof in the post, where I extend from Lipschitz continuity almost everywhere to everywhere. $\endgroup$ – user100431 Aug 29 '14 at 20:30
  • $\begingroup$ Yes, you can use this result too. But I think that, only with the extension by coninutiy, you will get a Lipschitz function. Have you tried to prove it? $\endgroup$ – Tomás Aug 29 '14 at 22:42
  • $\begingroup$ But this result gives a Lipschitz continuous extension on the domain everywhere, that's exactly what I want. What exactly do you mean by "extension by continuity", do you mean by defining the extension to be $f(a) = \lim_{x \rightarrow a}f(x)$ for all $a \in \partial I$? I haven't tried that extension since I thought this one to be sufficient. What do you think? $\endgroup$ – user100431 Aug 29 '14 at 22:49
  • $\begingroup$ I think that extension by continuity is more fast to state and more easy to prove. $\endgroup$ – Tomás Aug 29 '14 at 23:07
  • $\begingroup$ Could you give a small hint as to how you would approach this proof(just a hint), it doesn't seem that you can simply use an inequality argument with the limits. $\endgroup$ – user100431 Aug 30 '14 at 16:26

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