I know that 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, 07:18 AMalice8675309uniform continuity of x^2
I know that 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, 09:02 AMPlato
If a function is continuous on any closed interval 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*there is a__finite subcollection__which also covers .

Do you have that theorem? - Apr 11th 2011, 05:17 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 on from where uniform continuity follows immediately.

Also, is the first part of your question asking why isn't uniformly continuous on ? 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 is sublinear (i.e. for some constants ) from where it follows that is not uniformly continuous on for every .