How Does Lipschitz Continuity Constrain a Function's Graph Between Two Lines?

[Azael]

Let the function
f:[0,\infty) \rightarrow \mathbb{R} be lipschitz continuous 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 don't know how to continue

 

[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 don't know if that is proof enough??

 

[StatusX]

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 :)

 

[Azael]

If I want to show the same thing for a local lipschitz continuous 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)