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Math Help - Continuous functions, Lifchitz's condition...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Continuous functions, Lifchitz's condition...

    Let u=f(x,y) defined in D, continuous by x and holds Lifchitz's condition for y, in other words: |f(x,y_1)-f(x,y_2)|\leq  A|y_1-y_2| where (x,y_1),(x,y_2) are points in D and A is constant. Prove f(x,y) continuous in D.

    I know that f(x,y) is continuous by y (from Lifchitz's condition) and it's given that f(x,y) is continuous by x, but now what?


    Thank you!
    Last edited by Also sprach Zarathustra; June 14th 2011 at 04:33 PM. Reason: now=know
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    MHF Contributor Drexel28's Avatar
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    Re: Continuous functions, Lifchitz's condition...

    Quote Originally Posted by Also sprach Zarathustra View Post
    Let u=f(x,y) defined in D, continuous by x and holds Lifchitz's condition for y, in other words: |f(x,y_1)-f(x,y_2)|\leq  A|y_1-y_2| where (x,y_1),(x,y_2) are points in D and A is constant. Prove f(x,y) continuous in D.

    I know that f(x,y) is continuous by y (from Lifchitz's condition) and it's given that f(x,y) is continuous by x, but now what?


    Thank you!
    Note that for any [tex](x_0,y_0)\in D[tex] you have that \left\|f(x,y)-f(x_0,y_0)\|\leqslant \|f(x,y_0)-f(x,y)\|+\|f(x_0,y_0)-f(x,y_0)\| now if you make a neighborhood small enough to make the first less than epsilon over two by continuity and the second epsilon/(2A) by the Lipschitz condition you're done.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Continuous functions, Lifchitz's condition...

    Great!!! Thank you!
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Continuous functions, Lifchitz's condition...

    Quote Originally Posted by Drexel28 View Post
    Note that for any [tex](x_0,y_0)\in D[tex] you have that \left\|f(x,y)-f(x_0,y_0)\|\leqslant \|f(x,y_0)-f(x,y)\|+\|f(x_0,y_0)-f(x,y_0)\| now if you make a neighborhood small enough to make the first less than epsilon over two by continuity and the second epsilon/(2A) by the Lipschitz condition you're done.
    I don't understanding something...

    How is that second term  ||f(x_0,y_0)-f(x,y_0) || is less than epsilon/(2A) by the Lipschitz condition?(  |f(x_0,y_0)-f(x,y_0) | \leq A|y_0-y_0|=0 )

    Hmmm...
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    MHF Contributor Drexel28's Avatar
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    Re: Continuous functions, Lifchitz's condition...

    Quote Originally Posted by Also sprach Zarathustra View Post
    I don't understanding something...

    How is that second term  ||f(x_0,y_0)-f(x,y_0) || is less than epsilon/(2A) by the Lipschitz condition?(  |f(x_0,y_0)-f(x,y_0) | \leq A|y_0-y_0|=0 )

    Hmmm...
    Let (x_0,y_0)\in D be arbitrary. By continuity in the x variable there exists some \gamma>0 such that if |x-x_0|<\gamma we have that \displaystyle f(x,y)-f(x_0,y)|<\frac{\varepsilon}{2}. Let \displaystyle \delta=\min\left\{\gamma,\frac{\varepsilon}{2A}\ri  ght\}. Let then (x,y)\in B_\delta(x_0,y_0). We know that |f(x_0,y_0)-f(x,y)|\leqslant |f(x,y_0)-f(x_0,y_0)|+|f(x,y_0)-f(x,y)|. Now, since |x-x_0|\leqslant \|(x,y)-(x_0,y_0)\|<\gamma we have that this first term is less than \displaystyle \frac{\varepsilon}{2} and this second term is, by Lipschitz, less than \displaystyle A|y-y_0|\leqslant A\|(x,y)-(x_0,y_0)\|<A\frac{\varepsilon}{2A}=\frac{ \varepsilon}{2}.
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