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Prove directly that sqrt defined on [0,1] is uniformly continuous

  1. Dec 31, 2011 #1
    This is not a homework problem, but it may as well be, so I thought I'd post it here.

    1. The problem statement, all variables and given/known data

    The function [itex]f:[0,1] \to \mathbb{R} [/itex] given by [itex]f(x)=\sqrt{x}[/itex] is continuous on a compact domain, so it is uniformly continuous. Prove that [itex]f[/itex] is uniformly continuous directly (with a [itex]\delta-\epsilon[/itex] proof).

    2. Relevant equations

    Show [itex]\forall \epsilon >0\quad \exists \delta >0 [/itex] s.t. [itex]\forall x,y \in [0,1]\quad |x-y|< \delta \Rightarrow |f(x)-f(y)|< \epsilon [/itex]

    3. The attempt at a solution

    I believe we have to use that [itex]|x-y|=|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}|[/itex]. [itex]\quad |x-y|<\delta \Rightarrow |\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}|< \delta [/itex], but from here I'm kind of lost. It seems [itex]|\sqrt{x}+\sqrt{y}|[/itex] may be arbitrarily small, which gets in our way when we try to bound [itex]|\sqrt{x}-\sqrt{y}|[/itex].

    Intuitively the problem makes perfect sense, just trying to work out the algebra.
    Thanks
     
  2. jcsd
  3. Jan 1, 2012 #2
    abs(x+y) <= abs(x) + abs(y) by triangle inequality, now use the fact that your domain is bounded.
     
  4. Jan 1, 2012 #3
    Sorry, I'm still not seeing it. I know that [itex]|\sqrt{x}+\sqrt{y}|[/itex] is bounded. To me it seems like the problem is that it may be arbitrarily small. Remember I'm trying to bound [itex]|\sqrt{x}-\sqrt{y}|[/itex] by epsilon.
     
  5. Jan 1, 2012 #4
    My mistake.

    Use this inequality:
    abs(sqrt(x)-sqrt(y)) <= sqrt(abs(x-y))
     
  6. Jan 1, 2012 #5

    Hurkyl

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  7. Jan 1, 2012 #6
    You shouldn't do it this way. Remember, you specify epsilon and try to find some conditions to put on delta to guarantee continuity.

    Start with |f(x)-f(y)| is less than epsilon. Multiplying by |f(x)+f(y)| is the right thing to do, but the fact that |f(x)+f(y)| is arbitrarily small not a defect in any respect. The thing that is important is that |f(x)+f(y)| is very easily bounded by a number (Why? Because your domain is also bounded!). So you should get that when you multiply |f(x)-f(y)||f(x)+f(y)| this thing is bounded by an expression in epsilon that you define to be delta. Incidentally, |f(x)-f(y)||f(x)+f(y)| is precisely |x-y| in this case.

    You might think it is a problem that it is arbitrarily small because you think of epsilon as being constant, and therefore some tinkering with |f(x)+f(y)| increases or decreases |f(x)-f(y)|. By choosing x,y you specify closeness of the two values of the function. That is epsilon. If you change x,y after that, you are actually changing epsilon.

    Also, if you don't like epsilons and deltas you can do this by a topological argument. Show that the pre-image f^(-1)[(a,b)] for any (a,b) an open interval of R is an open set of your domain. Be careful though, you then need to consider the subset that f maps to in R.

    ^^^ I don't encourage you to learn topology without knowing about metric spaces though. Just saying.
     
    Last edited: Jan 1, 2012
  8. Jan 1, 2012 #7
     
  9. Jan 1, 2012 #8
    What's the largest that |f(x)+f(y)| can possibly be?

    Think about this for a second. If your expression |f(x)+f(y)| was bounded below, that would mean that |f(x)-f(y)||f(x)+f(y)| would always be LARGER than [itex]B\epsilon[\itex]. Once you have chosen epsilon, you are guaranteed that the smallest possible case for |f(x)-f(y)||f(x)+f(y)| is when |f(x)+f(y)| has value B. Incidentally, inf|f(x)+f(y)| = 0. So what you end up showing there is epsilon is greater than zero. Of course it is.

    Think bigger, not smaller.
     
    Last edited: Jan 1, 2012
  10. Jan 1, 2012 #9
    But this is what I want to show! So if we assume this, our steps from now on must be a chain of equivalencies, not just implications.
    Aren't you taking the implication the wrong direction? We certainly have that [itex]|\sqrt{x}-\sqrt{y}|< \varepsilon \,\Leftrightarrow \,|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}|<\, \varepsilon \,|\sqrt{x}+\sqrt{y}|[/itex], but this is an expression in [itex]x[/itex] and [itex]y[/itex], and epsilon is not allowed to depend on [itex]x[/itex] or [itex]y[/itex].

    The reason I brought up [itex]|\sqrt{x}+\sqrt{y}|[/itex] being arbitrarily small is that if we had [itex]|\sqrt{x}+\sqrt{y}|[/itex] bounded below by [itex]B[/itex], then we'd have [itex]|\sqrt{x}+\sqrt{y}||\sqrt{x}-\sqrt{y}|< \delta \Rightarrow |\sqrt{x}-\sqrt{y}| < \frac{\delta}{|\sqrt{x}+\sqrt{y}|} < \frac{\delta}{B}[/itex] and set [itex]\delta=\varepsilon B[/itex], and we'd be done.
    Right, there is a simple argument by compactness and continuity, but the problem is to do it the hard way. :)

    Thanks for your help, sorry if I'm not understanding what you're saying.
     
  11. Jan 1, 2012 #10
    2, but I don't see how bounding it above helps in this case. Thanks
     
  12. Jan 1, 2012 #11
    Whoops, meant to mention that setting [itex]\delta=2\varepsilon[/itex] does not work in this case, since the equality goes the wrong way. So the leftward implication doesn't hold.
     
  13. Jan 1, 2012 #12
    Whoops, the "inequality" goes the wrong way
     
  14. Jan 1, 2012 #13
    Thanks, I'm working on trying to prove this inequality. Do you have a reference?
     
  15. Jan 1, 2012 #14
    You get |f(x)-f(y)||f(x)+f(y)| is always less than or equal to 2 epsilon. That is the condition you put on delta. You're trying to show that a delta exists that gives you continuity, not that there is an immutable value of delta that is special.
     
  16. Jan 1, 2012 #15
    I'm trying to show uniform continuity.

    [itex]|\sqrt{x}-\sqrt{y}||\sqrt{x}+\sqrt{y}|<2\varepsilon \not\Rightarrow |\sqrt{x}-\sqrt{y}|<\varepsilon[/itex]
     
  17. Jan 1, 2012 #16
    NVM. I suck. Google agrees. :(
     
  18. Jan 1, 2012 #17
    No prob, still trying to figure it out.
     
  19. Jan 2, 2012 #18
    Well i just googled the question to see if there were any epsilon delta proofs, people had suggested using that inequality.

    As for proving the inequality, we only need to consider the case for x, y >= 0. I'm not sure if it works for negative x,y.

    But to prove, just consider each case separately:
    x=0, y=0
    x>y=0
    y>x=0
    x>y>0
    y>x>0
     
  20. Jan 2, 2012 #19

    SammyS

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    You don't have to do all of these cases. Without Loss Of Generality, assume x ≥ y .

    And of course the case x=y works for x=0, y=0, and is trivial.
     
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