# Thread: Uniform Continuity of $\sqrt{|x|}$ on $\mathbb{R}$

1. ## [Solved] Uniform Continuity of $\sqrt{|x|}$ on $\mathbb{R}$

Dear Friends,

I need help with the following $\varepsilon-\delta$ proof for uniform continuity.

Claim. Prove or disprove that the function $\sqrt{|x|}$ is uniformly continuous on $\mathbb{R}$.

Thanks.

bkarpuz

Answer by Laurent

2. Originally Posted by bkarpuz
Dear Friends,

I need help with the following $\varepsilon-\delta$ proof for uniform continuity.

Claim. Prove or disprove that the function $\sqrt{|x|}$ is uniformly continuous on $\mathbb{R}$.

Thanks.

bkarpuz
The only domain restriction of the function is that whatever is under the square root cannot be negative: $|x| \geq 0$ This is always true, by definition. Therefore, there are no domain restrictions.This implies that $\sqrt{|x|}$ is uniformly continuous on $\mathbb{R}$.

Does that help?

I'm sure you see how to apply the $\varepsilon-\delta$ here?

3. Originally Posted by mathemagister
I'm sure you see how to apply the $\varepsilon-\delta$ here?
Actually i am not sure this is why I asked the question.
Note that I am not asking just continuity, I am asking about the uniform continuity!

4. Originally Posted by mathemagister
The only domain restriction of the function is that whatever is under the square root cannot be negative: $|x| \geq 0$ This is always true, by definition. Therefore, there are no domain restrictions.This implies that $\sqrt{|x|}$ is uniformly continuous on [tex]\mathbb{R}[tex].
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.

Does that help?

I'm sure you see how to apply the $\varepsilon-\delta$ here?
However, bkarpuz, it is true that if a function is continuous on a 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 $\delta$ you would use for a given $\epsilon$ depend on x as x goes to infinity?

5. Originally Posted by HallsofIvy
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.

However, bkarpuz, it is true that if a function is continuous on a 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 $\delta$ you would use for a given $\epsilon$ depend on x as x goes to infinity?
Oh, sorry; I guess I was rushing through the problem to quickly. Thanks for pointing that out HallsofIvy!

6. Originally Posted by HallsofIvy
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.

However, bkarpuz, it is true that if a function is continuous on a 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 $\delta$ you would use for a given $\epsilon$ depend on x as x goes to infinity?
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 $x\to\infty$ 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

7. Originally Posted by bkarpuz
Claim. Prove or disprove that the function $\sqrt{|x|}$ is uniformly continuous on $\mathbb{R}$.
Hi,
this function is uniformly continuous.

The " $\epsilon-\delta$ way":

If $0\leq x, y$, then $|\sqrt{y}-\sqrt{x}|\leq \sqrt{|y-x|}$. Indeed, if $0\leq b\leq a$, $(a-b)^2=a^2-2ba+b^2\leq a^2-2b^2+b^2=a^2-b^2$; take $a=\max(\sqrt{x},\sqrt{y})$, $b=\min(\sqrt{x},\sqrt{y})$. Thus, if $0\leq x,y$ and $|x-y|\leq \epsilon^2$, then $|\sqrt{y}-\sqrt{x}|\leq \epsilon$. And if $x\leq 0\leq y$ with $|x-y|\leq \epsilon^2$, then $|y-(-x)|=y+x\leq y\leq \epsilon^2$, hence $|\sqrt{|y|}-\sqrt{|x|}|=|\sqrt{y}-\sqrt{-x}|\leq\epsilon$ by the above argument. This settles all the cases.

The more sophisticated way :

On $[-1,1]$, the function is continuous and bounded, hence uniformly continuous. On $[1,+\infty)$, the function is differentiable and its derivative is less than $\frac{1}{2}$, hence it is $\frac{1}{2}$-Lipschitz and thus uniformly continuous. By symmetry, and gluing the parts together, this concludes.

8. Originally Posted by Laurent
Hi,
this function is uniformly continuous.

The " $\epsilon-\delta$ way":

If $0\leq x, y$, then $|\sqrt{y}-\sqrt{x}|\leq \sqrt{|y-x|}$. Indeed, if $0\leq b\leq a$, $(a-b)^2=a^2-2ba+b^2\leq a^2-2b^2+b^2=a^2-b^2$; take $a=\max(\sqrt{x},\sqrt{y})$, $b=\min(\sqrt{x},\sqrt{y})$. Thus, if $0\leq x,y$ and $|x-y|\leq \epsilon^2$, then $|\sqrt{y}-\sqrt{x}|\leq \epsilon$. And if $x\leq 0\leq y$ with $|x-y|\leq \epsilon^2$, then $|y-(-x)|=y+x\leq y\leq \epsilon^2$, hence $|\sqrt{|y|}-\sqrt{|x|}|=|\sqrt{y}-\sqrt{-x}|\leq\epsilon$ by the above argument. This settles all the cases.

The more sophisticated way :

On $[-1,1]$, the function is continuous and bounded, hence uniformly continuous. On $[1,+\infty)$, the function is differentiable and its derivative is less than $\frac{1}{2}$, hence it is $\frac{1}{2}$-Lipschitz and thus uniformly continuous. By symmetry, and gluing the parts together, this concludes.
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.

### 1/sqrt(x) uniform continuity

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