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Find region for which F(x,y) = (x+y)^2 is Lipschitz in y

  1. Apr 6, 2014 #1
    As the title says, I need to find such a region.

    Taking any x, and any y1 and y2 I used the expression |F(x,y1) - F(x,y2)| and plugged in the function respectively for y1 and y2.

    Now I have to find values for x and y such that the following condition (Lipschitz condition) is satisfied:

    | 2x + (y1 + y2) | 0 (indeed after having simplified out the previous expression w.r.t the Lipschitz condition)

    My initial idea was to find x for which y = 0 and then the same thing for y1 and y2. This method though is not enough since for x = 0 the region of y1 and y2 for which the condition is satisfied, will have to depend on y1 and y2 directly. It will be better to attain a region such that it will not depend on the variables (obviously). I hope I am not incorrect here.

    I cannot seem to find a way to get this region and I would very much appreciate any insight given.

    Thank you.
     
    Last edited: Apr 6, 2014
  2. jcsd
  3. Apr 6, 2014 #2

    pasmith

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    For each [itex]x \in \mathbb{R}[/itex] and each [itex]K > 0[/itex] there exists an interval [tex]L_K(x) = \left[-\tfrac12K - x,\tfrac12K - x\right][/tex] such that if [itex]y_1 \in L_K(x)[/itex] and [itex]y_2 \in L_K(x)[/itex] then [itex]|F(x,y_1) - F(x,y_2)| \leq K|y_1 - y_2|[/itex], so [itex]f_x : y \mapsto F(x,y)[/itex] is lipschitz with respect to [itex]y[/itex] with lipschitz constant [itex]K[/itex].

    Note that you must fix [itex]x[/itex] and then determine the interval; if you want a region of the [itex](x,y)[/itex]-plane in which [itex]F[/itex] is lipschitz with respect to [itex]y[/itex] with constant [itex]K[/itex] then the condition is
    [tex]
    |F(x_1,y_1) - F(x_2,y_2)|
    = |x_1^2 - x_2^2 + 2x_1y_1 - 2x_2y_2 + y_1^2 - y_2^2| \leq K|y_1 - y_2|.
    [/tex]
     
  4. Apr 6, 2014 #3

    micromass

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    Please post such questions in the homework forum in the future :smile: I'll move it to there now!
     
  5. Apr 7, 2014 #4
    Shouldn't the interval not contain variables though, to have a definite region? I know you can choose any x and fix it, but it seems indefinite to me.

    x should remain fixed instead of having x1 and x2, no? I am unsure how to attain values for both x and y such that the following inequality is satisfied:

    | 2x + (y1 + y2) | <= K
     
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