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Math Help - Some uniform continuous problems

  1. #1
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    Some uniform continuous problems

    Discussing the uniform continuity of the following:

    1. f(x)=x^2, \ \ \ x \in (-1,1)

    Claim: f is uniformly continuous on (-1,1)

    Proof.

    Given  \epsilon > 0, pick  \delta = \frac { \epsilon }{2} , then for each x,y \in (-1,1) with |x-y|< \delta , we would have |x^2-y^2|=|x-y||x+y|< \delta |x+y| < \delta |1+1| = \epsilon

    Q.E.D.

    2. f(x)=e^{-x} \ \ \ x \in [0,\infty )

    Claim: f is uniformly continuous.

    Proof.

    Given  \epsilon > 0, pick  \delta >0, then for each x,y \in [0, \infty) with |x-y|< \delta , we would have |e^{-x} - e^{-y}| \leq |e^0-e^0|=|1-1|=0 < \epsilon

    Q.E.D.

    3. f(x)=x^{ \frac {1}{3} } \ \ \ x \in [0, \infty )

    Claim: f is not uniformly continuous.

    So far, I know that f'(x)= \frac {1}{3x^{ \frac {2}{3}}} , which is not below above as x approaches to 0.

    But is it possible to prove this by defintion?

    Proof so far.

    Pick  \epsilon =1 , then  \forall \delta > 0 with |x-y|< \delta , I need to show that |x^{ \frac {1}{3}} - y^ { \frac {1}{3} } | < \epsilon But I'm stuck in here, any hints?

    Thanks!!!
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  2. #2
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    Correct on nos. 1 and 2. For 3, split the interval into two subintervals, [0,1] and [1,∞], and ask why the function should be uniformly continuous on both of them.
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  3. #3
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    For the first interval of  [0,1] : f is continuous on [0,1], a compact set, meaning f is also uniformly continuous.

    For the second interval of  [1, \infty ) : The derivative is bounded by 0 below and 1/3 above, therefore it is uniformly continuous.

    Is this right? So if a function is uniformly continuous on two subinterval, then it is also uniformly continuous on the whole thing?
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  4. #4
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    Yes, and yes.
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    I was rewriting my solutions, and I found a problem in my answer to the function of e^{-x}

    Is |e^{-x}-e^{-y}| \leq |e^0-e^0| ? I'm thinking now it is not, since y can be something really big, that would bring the distance out.

    Should I consider the series  e^{-x} = \sum ^ \infty _{k=0} \frac {1}{k!}(-x)^k and prove this series is uniformly convergence?

    Then I use the M test, since  \frac {1}{k!}(-x)^k \leq \frac {2^kx^k}{(k+1)!} , and  \sum ^ \infty _{k=0} \frac {2^kx^k}{(k+1)!} converges by the ratio test, so f is uniformly convergence, is that right? Thanks.
    Last edited by tttcomrader; November 9th 2008 at 09:31 AM.
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  6. #6
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    Quote Originally Posted by tttcomrader View Post
    2. f(x)=e^{-x} \ \ \ x \in [0,\infty )

    Claim: f is uniformly continuous.
    Quote Originally Posted by tttcomrader View Post
    I was rewriting my solutions, and I found a problem in my answer to the function of e^{-x}

    Should I consider the series  e^{-x} = \sum ^ \infty _{k=0} \frac {1}{k!}(-x)^k and prove this series is uniformly convergence?

    Then I use the M test, since  \frac {1}{k!}(-x)^k \leq \frac {2^kx^k}{(k+1)!} , and  \sum ^ \infty _{k=0} \frac {2^kx^k}{(k+1)!} converges by the ratio test, so f is uniformly convergence, is that right? Thanks.
    You seem to be confusing uniform continuity with uniform convergence. If you want to show that the series  e^{-x} = \sum ^ \infty _{k=0} \frac {1}{k!}(-x)^k is uniformly convergent on the interval [0,∞) then you use the M-test. But the question asks you to show that the function e^{-x} is uniformly continuous on the interval [0,∞). The easiest way to do that is to observe that the derivative -e^{-x} is bounded on that interval.
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