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Math Help - uniform continuity of x^2

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    uniform continuity of x^2

    I know that x^2is NOT uniformly continuous, but isn't it u.c. on [0,1]. How would I prove this directly using the definiton? Unless its not uniformly continuous on [0,1] then how would I prove that using the sequential criterion?
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    Quote Originally Posted by alice8675309 View Post
    I know that x^2is NOT uniformly continuous, but isn't it u.c. on [0,1]. How would I prove this directly using the definiton? Unless its not uniformly continuous on [0,1] then how would I prove that using the sequential criterion?
    If a function is continuous on any closed interval [a,b] then it is uniformly continuous there. The proof this is very simple provided one has the necessary tools, theorems, to use.

    You need to know that from any collection of open intervals that cover [a,b] there is a finite subcollection which also covers [a,b].
    Do you have that theorem?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alice8675309 View Post
    I know that x^2is NOT uniformly continuous, but isn't it u.c. on [0,1]. How would I prove this directly using the definiton? Unless its not uniformly continuous on [0,1] then how would I prove that using the sequential criterion?
    An alternative to the Heine-Cantor theorem as Plato suggested you can also use the fact that every function with bounded derivative is Lipschitz and thus trivially uniformly continuous. As per definition you could note that \left|x^2-y^2\right|=\left|x-y\right|\left|x+y\right|\leqslant 2|x-y| on [0,1] from where uniform continuity follows immediately.


    Also, is the first part of your question asking why x^2 isn't uniformly continuous on \mathbb{R}? I'll leave you to think about it, but for your benefit I'd like to remark that it's (fairly) easy to prove that every uniformly continuous function on \mathbb{R} is sublinear (i.e. |f(x)|\leqslant Ax+B for some constants A,B) from where it follows that x^{1+\delta} is not uniformly continuous on \mathbb{R} for every \delta>0.
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