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Math Help - limits and uniformly continuous

  1. #1
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    limits and uniformly continuous

    I need to show that if f is uniformly continuous on (a,b], then the right hand limit of f(x) exists.

    This is what i have so far:
    If f is uniformly continuous on (a,b], given epsilon >0, there is a delta > 0 s.t. |x-y| < delta -> |f(x)-f(y)| < epsilon for all x,y in (a,b]
    Let x be in (a, a + delta).
    |f(x) - f(y)|
    <= |f(x) - L|+|L-f(y)|
    < |f(x) - L|
    < epsilon.

    Thus, the right hand limit of f exist

    Does this make sense?
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  2. #2
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    Quote Originally Posted by inthequestofproofs View Post
    I need to show that if f is uniformly continuous on (a,b], then the right hand limit of f(x) exists.

    This is what i have so far:
    If f is uniformly continuous on (a,b], given epsilon >0, there is a delta > 0 s.t. |x-y| < delta -> |f(x)-f(y)| < epsilon for all x,y in (a,b]
    Let x be in (a, a + delta).
    |f(x) - f(y)|
    <= |f(x) - L|+|L-f(y)|
    < |f(x) - L|
    < epsilon.

    Thus, the right hand limit of f exist

    Does this make sense?
    Not really. First, we don't know what your L is; second, |f(x)-L|+|L-f(y)|<|f(x)-L| is more than dubious.

    Since you have to prove the existence of a limit and your function is not monotone, you should use Cauchy condition. This should be pretty straightforward (no computation at all).
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  3. #3
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    sorry for so many questions, but the cauchy condition is for sequences. I don't know how to apply it for a function. Could you explain?
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  4. #4
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    Quote Originally Posted by inthequestofproofs View Post
    sorry for so many questions, but the cauchy condition is for sequences. I don't know how to apply it for a function. Could you explain?
    You have the same criterion as for sequences.

    Fact (Cauchy property for functions) : f(t) has a limit when t\to a^+ if, and only if, for every \epsilon>0, there is t_0>a such that, for any t,t'\in(a,t_0), |f(t)-f(t')|<\epsilon.

    Proof:
    a) Assume f(t)\to_{t\to a^+}\ell. Let \epsilon>0. There is t_0>a such that, if a<t<t_0, |f(t)-\ell|<\epsilon/2. Then, for all a<t,t'<t_0, |f(t)-f(t')|\leq |f(t)-\ell|+|\ell-f(t')|\leq \epsilon.

    b) Assume Cauchy condition : for every \epsilon>0, there is t_0>a such that, for any a<t,t'<t_0, |f(t)-f(t')|<\epsilon. Choose a particular sequence (u_n)_{n\in\mathbb{N}} such that u_n\to_n a^+. Then the assumption shows that (f(u_n))_{n\in\mathbb{N}} is a Cauchy sequence (I let you check that). Therefore, it has a limit \ell\in\mathbb{R} : f(u_n)\to_n \ell.
    Let us show that f(t)\to_{t\to a^+} \ell. Let \epsilon>0. Let t_0 be given by the assumption. There is n_0 such that n\geq n_0 implies a<u_n<t_0 (because u_n\to_n a^+) and |f(u_n)-\ell|<\epsilon (because f(u_n)\to_n\ell). For every a<t<t_0, we have |f(t)-\ell|\leq |f(t)-f(u_n)|+|f(u_n)-\ell|\leq 2\epsilon (first term because of Cauchy condition since a<u_n,t<t_0). This concludes.

    --
    In your problem, the uniform continuity immediately gives the Cauchy condition.
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