# Lipschitz continuity problem

Azael
Let the function
$$f:[0,\infty) \rightarrow \mathbb{R}$$ be lipschitz continous with lipschits constant K. Show that over small intervalls $$[a,b] \subset [0,\infty)$$ the graph has to lie betwen two straight lines with the slopes k and -k.

This is how I have started:

Definition of lipschits continuity $$|f(x)-f(y)| \leq k|x-y|$$

$$b>a$$
$$|f(b)-f(a)| \leq k(b-a) \Leftrightarrow -k(b-a) \leq f(b)-f(a) \leq k(b-a)$$

But after this I am a bit stumped. I dont know how to continue:grumpy:

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## Answers and Replies

Azael
Is it enough to just divide all sides with (b-a) to show that the slope of the grap is always less than K and more than -K?? To me it feels like it but I dont know if that is proof enough??

Homework Helper
Replace b by x in your equations, and restrict x to lie between a and b. Then there is just one step left.

Azael
like this??

continuing from the last step of post one

=>
$$-k(x-a) \leq f(x)-f(a) \leq k(x-a)$$ with $$a \leq x \leq b$$
<=>
$$f(a)-k(x-a) \leq f(x) \leq k(x-a)+f(a)$$

thanks alot!
that seems to have solved it :)

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Azael
If I want to show the same thing for a local lipschitz continous function(that it lies betwen two straight lines with slope K(a,b) and -K(a,b) where K can be different for different intervalls) how do I proced. It is exactly the same as above but with K replaced with K(a,b) in all places?

so that I instead get as last line
$$f(a)-K(a,x)(x-a) \leq f(x) \leq f(a)+K(a,x)(x-a)$$

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