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Given $f: Y\subset\mathbb{R}\to \mathbb{R}$ a $k$-lipschitz function, (i.e $|f(x)-f(y)|\leq k|x-y|$ for any $x,y\in Y$) I need to prove the existence of a $k$-lipschitz function $g:\mathbb{R}\to \mathbb{R}$ such that $g|_Y=f$.

My answer when $f$ is bounded is considering $g(x)=\inf_{y\in Y}\{f(y)+k|x-y|\}.$

Is it correct?. How do you find $g$ when $f$ is not bounded?.

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    @LeonidKovalev My definition of $g$ does not always work because $f$ not necessarly has a lower bound.2012-05-15

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An alternate explicit construction:

First, you can continuously extend to the closure $\bar{Y}$ using the Lipschitz condition.

Then, since $\bar{Y}$ is closed, for every $x\in \mathbb{R}\setminus \bar{Y}$ one can find $x_- = \max \bar{Y}\cap \{ y < x\}$ and $x_+ = \min \bar{Y}\cap \{y > x\}$. Then just linearly interpolate: $ g(x) = f(x_-) + \frac{f(x_+) - f(x_-)}{x_+ - x_-} (x - x_-) $


But let me explain Leonid Kovalev's comment. Notice that fixing some arbitrary $x' \in Y$, we have that for any $x\in\mathbb{R}$ now chosen to be fixed

$ f(y) - f(x') + k|x-y| \geq f(y) - f(x') + k|x' - y| - k|x-x'| $

from triangle inequality. But using the $k$ lipschitz property you have that

$ f(y) - f(x') + k|x' - y| \geq 0 $

so the expression

$ f(y) - f(x') + k|x-y| \geq -k|x-x'| $

where the right hand side is independent of $y$. Or, in other words

$ f(y) + k|x-y| \geq f(x') - k|x-x'| $

so the expression you want to take the infimum of (in $y\in Y$) is bounded from below by some constant, and hence the infimum exists.

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    Ty for clarification.2012-05-16