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Math Help - Uniform Continuity of $\sqrt{|x|}$ on $\mathbb{R}$

  1. #1
    Senior Member bkarpuz's Avatar
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    Talking [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
    Last edited by bkarpuz; February 20th 2010 at 12:10 AM. Reason: Solved
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by bkarpuz View Post
    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?
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by mathemagister View Post
    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!
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  4. #4
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    Quote Originally Posted by mathemagister View Post
    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?
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    Member mathemagister's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    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!
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    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
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    Quote Originally Posted by bkarpuz View Post
    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.
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Laurent View Post
    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.
    Last edited by bkarpuz; February 20th 2010 at 12:08 AM.
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