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Thread: Continuous functions, Lifchitz's condition...

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

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

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


    Thank you!
    Last edited by Also sprach Zarathustra; Jun 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 $\displaystyle u=f(x,y)$ defined in $\displaystyle D$, continuous by $\displaystyle x$ and holds Lifchitz's condition for $\displaystyle y$, in other words: $\displaystyle |f(x,y_1)-f(x,y_2)|\leq A|y_1-y_2|$ where $\displaystyle (x,y_1),(x,y_2)$ are points in $\displaystyle D$ and $\displaystyle A$ is constant. Prove $\displaystyle f(x,y)$ continuous in $\displaystyle D$.

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


    Thank you!
    Note that for any [tex](x_0,y_0)\in D[tex] you have that $\displaystyle \left\|f(x,y)-f(x_0,y_0)\|\leqslant \|f(x,y_0)-f(x,y)\|+\|f(x_0,y_0)-f(x,y_0)\|$ 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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    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 $\displaystyle \left\|f(x,y)-f(x_0,y_0)\|\leqslant \|f(x,y_0)-f(x,y)\|+\|f(x_0,y_0)-f(x,y_0)\|$ 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 $\displaystyle ||f(x_0,y_0)-f(x,y_0) || $ is less than epsilon/(2A) by the Lipschitz condition?( $\displaystyle |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 $\displaystyle ||f(x_0,y_0)-f(x,y_0) || $ is less than epsilon/(2A) by the Lipschitz condition?( $\displaystyle |f(x_0,y_0)-f(x,y_0) | $\leq A|y_0-y_0|=0 )

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