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Hölder Continuity

  1. Dec 12, 2005 #1
    Hello guys, I am trying to prove that the function

    [tex]f(u)=-\frac{1}{(1+u)^2}[/tex]
    is Hölder continuous for [itex]-1<u \le 0[/itex] but I am stuck. Here is what I have done:

    If [itex]|u_1-u_0| \le \delta[/itex] then

    [tex]\left|-\frac{1}{(1+u_1)^2}+\frac{1}{(1+u_0)^2}\right| \le \left|\frac{(u_1+u_0)+2}{(1+u_1)^2(1+u_0)^2}\right||u_1-u_0| \le \frac{2|u_1-u_0|}{(1+u_1)^2(1+u_0)^2}[/tex]

    and I dont know how to continue... Any suggestions?
     
    Last edited: Dec 12, 2005
  2. jcsd
  3. Dec 13, 2005 #2

    Galileo

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    If [itex]u_0[/itex] and [itex]u_1[/itex] are between -1 and 0, what can you say about the bounds on [itex]2/[(1+u_0)^2(1+u_1)]^2[/itex]?
     
  4. Dec 13, 2005 #3
    I dont understand what you mean.

    The worst behavior of the bound is when [itex]u_1[/itex] and [itex]u_0[/itex] are close to [itex]-1[/itex], but that should reflect in the bounding constant [itex]H(u_0)[/itex] right?

    Let me put it this way (maybe im saying stupid things but, that woulndt be new :P )

    If [itex]-a <u \le 0 [/itex] and [itex] 0<a<1 [/itex], then

    [tex]|f(u_1)-f(u_0)| \le \frac{2}{(1-a)^4}|u_1-u_0|[/tex]

    so it is Hölder as long as im far from [itex]-1[/itex] right?

    Now, what i have to do is instead of finding a bound depending on [itex]a[/itex] is find a bound depending on [itex]u_0[/itex] right?


    EDIT:

    In fact,

    [tex]|f(u_1)-f(u_0)| \le \frac{2}{(1-a)^3}|u_1-u_0|.[/tex]
     
    Last edited: Dec 13, 2005
  5. Dec 13, 2005 #4
    How about this...

    [tex]|f(u_1)-f(u_0)| \le H(u_0,u_1) |u_1-u_0| [/tex]

    where

    [tex]H(u_0,u_1)=\left\{\begin{array}{cc} \dfrac{2}{(1+u_0)^3} & \hbox{if} \quad u_0<u_1 \\ & \\ \dfrac{2}{(1+u_1)^3} & \hbox{if} \quad u_0>u_1 \end{array}\right.[/tex]

    So [itex]f(u) \in C^{0,1}[/itex] in [itex]-1<u_0,u_1 \le 0[/itex], just not uniformly Hölder...

    If you were my analysis teacher, would you flunk me?
     
    Last edited: Dec 14, 2005
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