Lipschitz continuity problem

  • Thread starter Azael
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  • #1
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Let the function
[tex]f:[0,\infty) \rightarrow \mathbb{R}[/tex] be lipschitz continous with lipschits constant K. Show that over small intervalls [tex] [a,b] \subset [0,\infty) [/tex] 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 [tex]|f(x)-f(y)| \leq k|x-y|[/tex]

[tex]b>a[/tex]
[tex]|f(b)-f(a)| \leq k(b-a) \Leftrightarrow -k(b-a) \leq f(b)-f(a) \leq k(b-a)[/tex]

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

Answers and Replies

  • #2
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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??
 
  • #3
StatusX
Homework Helper
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Replace b by x in your equations, and restrict x to lie between a and b. Then there is just one step left.
 
  • #4
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like this??

continuing from the last step of post one

=>
[tex] -k(x-a) \leq f(x)-f(a) \leq k(x-a) [/tex] with [tex] a \leq x \leq b[/tex]
<=>
[tex] f(a)-k(x-a) \leq f(x) \leq k(x-a)+f(a) [/tex]

thanks alot!
that seems to have solved it :)
 
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  • #5
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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
[tex] f(a)-K(a,x)(x-a) \leq f(x) \leq f(a)+K(a,x)(x-a) [/tex]
 
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