Dear Friends,

I need help with the following proof for uniform continuity.

Claim. Prove or disprove that the function is uniformly continuous on .

Thanks.

bkarpuz

Answer byLaurent

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- February 19th 2010, 12:58 AMbkarpuz[Solved] Uniform Continuity of $\sqrt{|x|}$ on $\mathbb{R}$
Dear Friends,

I need help with the following proof for uniform continuity.

**Claim**. Prove or disprove that the function is uniformly continuous on .

Thanks.

**bkarpuz**

Answer by**Laurent** - February 19th 2010, 01:51 AMmathemagister
The only domain restriction of the function is that whatever is under the square root cannot be negative: This is always true, by definition. Therefore, there are no domain restrictions.This implies that is uniformly continuous on .

Does that help? :)

I'm sure you see how to apply the here? - February 19th 2010, 02:27 AMbkarpuz
- February 19th 2010, 03:09 AMHallsofIvy
No, it doesn't. The fact that a function is

**defined**for all x does NOT imply that it is uniformly continuous for all x.

Quote:

Does that help? :)

I'm sure you see how to apply the here?

**compact**set, it is uniformly continuous on that set. Since any closed and**bounded**set of real numbers is compact, the only problem is what happens as x goes to infinity. How does the you would use for a given depend on x as x goes to infinity? - February 19th 2010, 04:18 AMmathemagister
- February 19th 2010, 04:35 AMbkarpuz
Thanks

**HallsofIvy**,

I was thinking the same with you and I of course know about the case for compact sets.

In a book, it says that "*Prove that...*", and I could not.

But I still think that it is uniformly continuous since the book's author is a good mathematician.

Also as the function will act slower than the identity function, and identity is uniformly continuous.

So that this makes me feel that the claim is true.

All of your comments are welcome, thanks.

**bkarpuz** - February 19th 2010, 05:21 AMLaurent
Hi,

this function is uniformly continuous.

The " way":

If , then . Indeed, if , ; take , . Thus, if and , then . And if with , then , hence by the above argument. This settles all the cases.

The more sophisticated way :

On , the function is continuous and bounded, hence uniformly continuous. On , the function is differentiable and its derivative is less than , hence it is -Lipschitz and thus uniformly continuous. By symmetry, and gluing the parts together, this concludes. - February 19th 2010, 06:21 AMbkarpuz
I could give the proof in the same way as you have shown in the latter way,

but I was looking for a proof in the first way.

Thanks for the anwer**Laurent**.

**Summary**. Roughly talking, suppose that we are talking about a differentiable function,

first find the largest interval such that the derivative function is unbounded (this is related to the function itself being Lipschitz continuous or not).

If this interval is bounded, then the function is uniformly continuous.

If not, showing the uniform continuity may be a little bit complicated. :)