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Math Help - Not Uniform Continuous?

  1. #1
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    Not Uniform Continuous?

    Consider the function f(x)=xsinx

    Now, f'(x)=xcosx+sinx, which is not bounded, so it is not uniformly continuous.

    So pick  \epsilon = 1 , let  \delta > 0, for every x,y \in \mathbb {R} , |x-y| < \delta , we need  |xsinx-ysiny| \geq \epsilon , but how should I go about that? Thanks.
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  2. #2
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    " f is not uniformly continuous" means:
    \exists\varepsilon_0>0, \forall \delta>0, \exists x,y such that |x-y|<\delta and |f(x)-f(y)|>\varepsilon_0.

    I let you get convinced yourself that a proof would consist in finding a sequence a couples (x_n,y_n)_{n\geq 1} such that |x_n-y_n|\to_n 0 and |f(x_n)-f(y_n)|>\varepsilon_0 for some \varepsilon_0 (or even |f(x_n)-f(y_n)|\to_n+\infty, if possible).

    In order to choose those couples, look at (or just imagine) what the graph looks like: the x_n,y_n must be close but the f(x_n),f(y_n) must be way apart from each other.

    x_n=n\pi is a possible start.
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  3. #3
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    Thanks for the hint. So for the choice that x_n = n \pi , we would have f(x_n) = 0 \ \ \ \forall n , but any point right afterward would get larger as n increases.

    Pick y_n = x_n + \frac { \delta}{2} , then  |x_n - y_n | < \delta , but |f(x_n)-f(y_n)| \rightarrow \infty

    Is this right?
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    Pick y_n = x_n + \frac { \delta}{2} , then  |x_n - y_n | < \delta , but |f(x_n)-f(y_n)| \rightarrow \infty

    Is this right?
    No, because for every \delta>0 you must have |x_n-y_n|<\delta for some n\geq 0. This is why I said you must have |x_n-y_n|\to_n 0.

    You can write y_n=x_n+u_n=n\pi+u_n where u_n\to_n 0^+ is to be determined.

    You have |f(y_n)-f(x_n)|=|f(y_n)|=|(n\pi+u_n)\sin(n\pi+u_n)|=|(n\pi  +u_n)(-1)^n\sin(u_n)| \sim n\pi\sin(u_n)\sim n\pi u_n as n\to\infty. So that you must choose u_n\to_n 0 such that \liminf_n n u_n>0. For instance, u_n=\frac{1}{n} (hence \varepsilon_0 must be small enough), or u_n=\frac{1}{\sqrt{n}} (and any \varepsilon_0 is adequate).

    You can do a more explicit proof using the convexity inequality \sin(u_n)\geq\frac{2u_n}{\pi} which holds as soon as 0\leq u_n\leq \frac{\pi}{2}. This yields to the same possibilities for the choice of (u_n)_n.
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  5. #5
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    Thanks for the help. But one thing I'm not clear is, why is it enough to show that <br />
|x_n-y_n|\to_n 0<br />
implies <br />
|f(x_n)-f(y_n)|>\varepsilon_0<br />
?
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  6. #6
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    Explicitating my first post as much as I can:

    You know that " f is not uniformly continuous" means:
    \exists\varepsilon_0>0, \forall \delta>0, \exists x,y such that |x-y|<\delta and |f(x)-f(y)|>\varepsilon_0.
    You obtain this by taking the logical contrary of the definition of "uniformly continuous".

    Suppose we found a sequence of couples (x_n,y_n)_{n\geq 1} and an \varepsilon_0>0 such that |x_n-y_n|\to_n 0 and, for all n\geq 1, |f(x_n)-f(y_n)|>\varepsilon_0.

    Then, for every \delta>0, there is an n such that |x_n-y_n|<\delta (because of the convergence to 0). And for this same n, we have |f(x_n)-f(y_n)|>\varepsilon_0. As a consequence, we have realized the above definition of "not uniformly continuous" with x_n,y_n in place of x,y.
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