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Math Help - Uniform continuity of square root on [0,1]

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

    Hi,

    I'm trying to show that f\left(x\right)=\sqrt{x} is uniformly continuous on \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
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  2. #2
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    Quote Originally Posted by airavana View Post
    Hi,

    I'm trying to show that f\left(x\right)=\sqrt{x} is uniformly continuous on \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 x,y>0 (if x=0\,\,or\,\,y=0 the proof is almost immediate...can you see it?).

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


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

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

    Tonio
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  3. #3
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    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 x,y>0 (if x=0\,\,or\,\,y=0 the proof is almost immediate...can you see it?).
    Let x=0. Then y<\epsilon^2 \Longrightarrow \sqrt{y}< \epsilon

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

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

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

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

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

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

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

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

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

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

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

    Is it ok?

    Thanks
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  4. #4
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    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 x=0. Then y<\epsilon^2 \Longrightarrow \sqrt{y}< \epsilon


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

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

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

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

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

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

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

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

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

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


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

    Is it ok?

    Thanks

    Yep

    Tonio
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  5. #5
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    Thank you again!
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