Results 1 to 5 of 5

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

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    3

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by airavana View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2010
    Posts
    3
    Hi,

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

    Quote Originally Posted by tonio View Post
    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 $\displaystyle y<\epsilon^2 \Longrightarrow \sqrt{y}< \epsilon$

    Quote Originally Posted by tonio View Post
    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:
    $\displaystyle |\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}$

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

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

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

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

    Quote Originally Posted by tonio View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by airavana View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2010
    Posts
    3
    Thank you again!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Square root inside square root equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Oct 10th 2011, 04:17 PM
  2. uniform differentiable => uniform continuity
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Nov 30th 2009, 03:19 PM
  3. Replies: 12
    Last Post: Nov 22nd 2008, 12:41 PM
  4. Simplifying Square Root w/in square root
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Sep 7th 2006, 08:01 AM
  5. Replies: 2
    Last Post: Apr 29th 2006, 01:13 AM

Search Tags


/mathhelpforum @mathhelpforum