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Math Help - Is \sqrt{1-x^2} uniformly continuous?

  1. #1
    Super Member redsoxfan325's Avatar
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    Is \sqrt{1-x^2} uniformly continuous?

    Is f(x)=\sqrt{1-x^2} uniformly continuous?

    Here is my proposed solution.

    Break [-1,1] into three partitions: [-1,-4/5]\cup(-4/5,4/5)\cup[4/5,1]

    On [-1,-4/5], f(x) is defined on a compact set, so \forall~\epsilon>0, \exists~\delta_{-1}>0 such that |x-y|<\delta_{-1} \implies |f(x)-f(y)|<\epsilon.

    On (-4/5,4/5), f(x) has a bounded derivative: |f'(x)| \leq \frac{4}{3}. Let \delta_0 = \frac{3}{4}\epsilon. Thus, when |x-y|<\delta_0, by the MVT, |f(x)-f(y)|=|x-y|f'(t) \leq M|x-y| < \frac{4}{3}\cdot\frac{3}{4}\epsilon = \epsilon.

    On [4/5,1], f(x) is defined on a compact set, so \forall~\epsilon>0, \exists~\delta_1>0 such that |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon.

    Let \delta = \min\{\delta_{-1},\delta_0,\delta_1\} and it follows that \forall~x\in[-1,1], \forall~y\in[-1,1], |f(x)-f(y)|<\epsilon when |x-y|<\delta, so f(x) is uniformly continuous.

    Is that correct?
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  2. #2
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    Quote Originally Posted by redsoxfan325 View Post
    Is f(x)=\sqrt{1-x^2} uniformly continuous?

    Here is my proposed solution.

    Break [-1,1] into three partitions: [-1,-4/5]\cup(-4/5,4/5)\cup[4/5,1]

    On [-1,-4/5], f(x) is defined on a compact set, so \forall~\epsilon>0, \exists~\delta_{-1}>0 such that |x-y|<\delta_{-1} \implies |f(x)-f(y)|<\epsilon.

    On (-4/5,4/5), f(x) has a bounded derivative: |f'(x)| \leq \frac{4}{3}. Let \delta_0 = \frac{3}{4}\epsilon. Thus, when |x-y|<\delta_0, by the MVT, |f(x)-f(y)|=|x-y|f'(t) \leq M|x-y| < \frac{4}{3}\cdot\frac{3}{4}\epsilon = \epsilon.

    On [4/5,1], f(x) is defined on a compact set, so \forall~\epsilon>0, \exists~\delta_1>0 such that |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon.

    Let \delta = \min\{\delta_{-1},\delta_0,\delta_1\} and it follows that \forall~x\in[-1,1], \forall~y\in[-1,1], |f(x)-f(y)|<\epsilon when |x-y|<\delta, so f(x) is uniformly continuous.

    Is that correct?
    why did you divide the interval? f(x)=\sqrt{1-x^2} is continuous over [-1,1], which is a compact set. thus f is uniformly continuous over [-1,1].
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  3. #3
    Super Member redsoxfan325's Avatar
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    That's a good point. Man, do I feel like an idiot...
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