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Lipschitz continuity problem

  1. Jun 20, 2006 #1
    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]|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: Jun 21, 2006
  2. jcsd
  3. Jun 21, 2006 #2
    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??
  4. Jun 21, 2006 #3


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    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.
  5. Jun 21, 2006 #4
    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 :)
    Last edited: Jun 21, 2006
  6. Jun 21, 2006 #5
    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]
    Last edited: Jun 21, 2006
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