# Thread: Uniform continuity of square root on [0,1]

1. ## Uniform continuity of square root on [0,1]

Hi,

I'm trying to show that $\displaystyle f\left(x\right)=\sqrt{x}$ is uniformly continuous on $\displaystyle \left[0,1\right]$.
The problem is I can't use the fact it is continuous on the closed interval so it must be uniformly continuous there. I have to show it directly from definition. I would be really grateful for any hints

2. Originally Posted by airavana
Hi,

I'm trying to show that $\displaystyle f\left(x\right)=\sqrt{x}$ is uniformly continuous on $\displaystyle \left[0,1\right]$.
The problem is I can't use the fact it is continuous on the closed interval so it must be uniformly continuous there. I have to show it directly from definition. I would be really grateful for any hints

Assume $\displaystyle x,y>0$ (if $\displaystyle x=0\,\,or\,\,y=0$ the proof is almost immediate...can you see it?).

Now, let $\displaystyle \epsilon > 0$ be given. We must find $\displaystyle \delta > 0$ so that $\displaystyle |x-y|<\delta \Longrightarrow |\sqrt{x}-\sqrt{y}|<\epsilon\,,\,\,\forall x,y\in [0,1]$ .

Lemma: for $\displaystyle x,y>0\,,\,\,|\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$ .

Prove the lemma (2 cases) and then the election of $\displaystyle \delta$ is straightforward.

Tonio

3. Hi,

thank you very much for your kind help. Would you mind to say if my proof is ok?

Originally Posted by tonio
Assume $\displaystyle x,y>0$ (if $\displaystyle x=0\,\,or\,\,y=0$ the proof is almost immediate...can you see it?).
Let $\displaystyle x=0$. Then $\displaystyle y<\epsilon^2 \Longrightarrow \sqrt{y}< \epsilon$

Originally Posted by tonio
Lemma: for $\displaystyle x,y>0\,,\,\,|\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$ .
Prove the lemma (2 cases)
Let $\displaystyle x>y$
$\displaystyle |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$

$\displaystyle \sqrt{x}-\sqrt{y}\leq\sqrt{x-y}$

$\displaystyle \sqrt{x}\leq\sqrt{x-y}+\sqrt{y}$

$\displaystyle x\leq x-y+y+2\sqrt{\left(x-y\right)y}$

$\displaystyle x\leq x+2\sqrt{\left(x-y\right)y}$

In the case of y>x:
$\displaystyle |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$

$\displaystyle \sqrt{y}-\sqrt{x}\leq\sqrt{y-x}$

$\displaystyle \sqrt{y}\leq\sqrt{y-x}+\sqrt{x}$

$\displaystyle y\leq y-x+x+2\sqrt{\left(y-x\right)x}$

$\displaystyle y\leq y+2\sqrt{\left(y-x\right)x}$

Originally Posted by tonio
and then the election of $\displaystyle \delta$ is straightforward.
$\displaystyle \delta=\epsilon^2$ because
$\displaystyle |x-y|<\epsilon^2 \Longrightarrow |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}<\epsilon$

Is it ok?

Thanks

4. Originally Posted by airavana
Hi,

thank you very much for your kind help. Would you mind to say if my proof is ok?

Let $\displaystyle x=0$. Then $\displaystyle y<\epsilon^2 \Longrightarrow \sqrt{y}< \epsilon$

Let $\displaystyle x>y$
$\displaystyle |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$

$\displaystyle \sqrt{x}-\sqrt{y}\leq\sqrt{x-y}$

$\displaystyle \sqrt{x}\leq\sqrt{x-y}+\sqrt{y}$

$\displaystyle x\leq x-y+y+2\sqrt{\left(x-y\right)y}$

$\displaystyle x\leq x+2\sqrt{\left(x-y\right)y}$

In the case of y>x:
$\displaystyle |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$

$\displaystyle \sqrt{y}-\sqrt{x}\leq\sqrt{y-x}$

$\displaystyle \sqrt{y}\leq\sqrt{y-x}+\sqrt{x}$

$\displaystyle y\leq y-x+x+2\sqrt{\left(y-x\right)x}$

$\displaystyle y\leq y+2\sqrt{\left(y-x\right)x}$

$\displaystyle \delta=\epsilon^2$ because
$\displaystyle |x-y|<\epsilon^2 \Longrightarrow |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}<\epsilon$

Is it ok?

Thanks

Yep

Tonio

5. Thank you again!