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Thread: sin(x^2) not uniformly continuous

  1. #1
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    sin(x^2) not uniformly continuous

    Can someone please explain to me why sin(x^2) is not uniformly continuous.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by summerset353 View Post
    Can someone please explain to me why sin(x^2) is not uniformly continuous.
    On what set?
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    the set being R
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    MHF Contributor chisigma's Avatar
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    The reason is that the function is contionuos and bounded but its derivative is unbounded...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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    that's false.

    consider $\displaystyle f(t)=\sqrt t$ being uniformly continuous on $\displaystyle [0,\infty)$ but its derivative is not bounded.
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  6. #6
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    Krizalid is right. Having bounded derivative is a sufficient condition for being uniformly continuous, but it is not necessary.

    Consider the sequences $\displaystyle x_n=\sqrt{2\pi n}$, $\displaystyle y_n=\sqrt{2\pi n+\pi/2}$. We have that $\displaystyle |y_n-x_n|$
    tends to 0 and $\displaystyle f(y_n)-f(x_n)=1$. From this fact it follows that $\displaystyle f$ is not uniformly continuous.
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  7. #7
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    Quote Originally Posted by summerset353 View Post
    Can someone please explain to me why sin(x^2) is not uniformly continuous.
    Let $\displaystyle f(x) = \sin(x^2)$. Let $\displaystyle x_n = \sqrt{(2n+\tfrac12)\pi}$ and $\displaystyle y_n = \sqrt{(2n+\tfrac32)\pi}$. Notice that $\displaystyle x_n$ and $\displaystyle y_n$ can be arbitrarily close (for n large enough) and yet $\displaystyle |f(x_n) - f(y_n)|=2$.
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