I know that $\displaystyle x^2$is 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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- Apr 10th 2011, 06:18 AMalice8675309uniform continuity of x^2
I know that $\displaystyle x^2$is 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?

- Apr 10th 2011, 08:02 AMPlato
If a function is continuous on any closed interval $\displaystyle [a,b]$ 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*$\displaystyle [a,b]$ there is a__finite subcollection__which also covers $\displaystyle [a,b]$.

Do you have that theorem? - Apr 11th 2011, 04:17 PMDrexel28
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 $\displaystyle \left|x^2-y^2\right|=\left|x-y\right|\left|x+y\right|\leqslant 2|x-y|$ on $\displaystyle [0,1]$ from where uniform continuity follows immediately.

Also, is the first part of your question asking why $\displaystyle x^2$ isn't uniformly continuous on $\displaystyle \mathbb{R}$? 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 $\displaystyle \mathbb{R}$ is sublinear (i.e. $\displaystyle |f(x)|\leqslant Ax+B$ for some constants $\displaystyle A,B$) from where it follows that $\displaystyle x^{1+\delta}$ is not uniformly continuous on $\displaystyle \mathbb{R}$ for every $\displaystyle \delta>0$.