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Math Help - Uniformly continuous function-two variables

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Uniformly continuous function-two variables

    The question:

    Given f(x,y)=\cos(\frac{1}{1-x^2-y^2}), prove:

    a) f(x,y) isn't uniformly continuous functions on D=\{(x,y)|x^2+y^2<1\}

    b) f(x,y) is uniformly continuous functions on D=\{(x,y)| 3 \leqslant x^2+y^2\leqslant 4\}


    My solution(check me please):

    a) First I say,

    x^2+y^2=t,  0 \leqslant t<1

    now,

    f(x,y) becomes to be function with one variable- t:

    f(t)=\cos(\frac{1}{1-t}.

    Now I pick two points t_0,t_1 \in  [0,1):

    t_0=1-\frac{1}{2\pi n},

    t_1=1-\frac{1}{(2n+1)\pi}

    We can see that:

    \lim_{n\to\infty}|t_0-t_1|=\lim_{n\to\infty}|\frac{1}{(2n+1)\pi}-\frac{1}{2\pi n}|=\lim_{n\to\infty}|...|=0

    But:

    |f(t_0)-f(t_1)|=|\cos(t_0)-\cos(t_1)|=|\cos(2\pi n)-\cos((2n+1)\pi)|=2

    Therefor for every 0<\varepsilon<2 \nexists \delta , so that for all |t-t'|<\delta will be held: |f(t)-f(t')|<\varepsilon


    b)

    D=\{(x,y)| 3 \leqslant x^2+y^2\leqslant 4\} is closed and bounded, the function f(x,y) is continuous function in D, hence from Cantor's theorem we can deduce that f(x,y) is uniformly continuous function.
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  2. #2
    Super Member girdav's Avatar
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    It's correct. In a), I think you have to denote the new f by g because it's not the same function (in particular f has two variables whereas g has only one). You can denote s_n instead of t_0 and t_n instead of t_1 to show that those two numbers depend on n.
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