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Math Help - two little problems about limsup / liminf.

  1. #1
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    two little problems about limsup / liminf.

    my teacher of Mathematical Analysis gave me these two problems:

    1) let f be a continuous function on (0,1) and such that liminf f(x) < limsup f(x) for x that approaches 0, then for every real value L that belongs to (liminf f(x), limsup f(x)) exists {Xn} convergent sequence to 0 in (0,1) such that lim f(Xn) = L for n that approaches to infinite.

    2) let f(n), g(n) be two sequences such that f(n) >= g(n) for every n natural number, is true or false that liminf f(n) >= limsup g(n) ? Give a proof of that.

    i don't know how to give a proof of the first fact, about the second one i proved that is true only if the two limits exist but i didn't find a counter-exemple to justify the assumption (so it could be wrong the entire proof).

    Do you get any suggestions?
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  2. #2
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    Re: two little problems about limsup / liminf.

    1) What does it mean that f is continuous? What happens to sequences that converge in (0,1) when they are passed through f? So, you want to find a sequence x_{(n,1)} that converges very close to zero such that f(x_{(n,1)}) converges very close to L. Then you want a sequence of sequences x_{(n,m)} that converges closer and closer to 0 as m and n approach infinity such that f(x_{(n,m)}) approaches closer and closer to L as m and n approach infinity. Use the results of liminf and limsup.

    2) I would need to think about it.
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  3. #3
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    Re: two little problems about limsup / liminf.

    Quote Originally Posted by SlipEternal View Post
    1) What does it mean that f is continuous? What happens to sequences that converge in (0,1) when they are passed through f? So, you want to find a sequence x_{(n,1)} that converges very close to zero such that f(x_{(n,1)}) converges very close to L. Then you want a sequence of sequences x_{(n,m)} that converges closer and closer to 0 as m and n approach infinity such that f(x_{(n,m)}) approaches closer and closer to L as m and n approach infinity. Use the results of liminf and limsup.
    No, i want to prove the opposite: that for every L in (liminf f, limsup f) exists a sequence x(n) in (0,1) so that f(x(n)) = L.
    If i do so i only prove that for every sequence of points in (0,1) , through f, i get a value in (liminf f, limsup f), not that i get every value in that bounded set.
    I miss the opposite inclusion, in other words that (liminf f, limsup f) is included in A = {values given by f(x(n)) for all x(n) in (0,1) that goes to 0}.

    P.S. I think i found one for the 2nd question: f(n) = (-1)^n + 1 and g(n) = (-1)^n, liminf f(n) = 0; limsup g(n) = 1
    Last edited by ayrcast; October 10th 2013 at 10:23 AM.
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  4. #4
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    Re: two little problems about limsup / liminf.

    Quote Originally Posted by ayrcast View Post
    No, i want to prove the opposite: that for every L in (liminf f, limsup f) exists a sequence x(n) in (0,1) so that f(x(n)) = L.
    If i do so i only prove that for every sequence of points in (0,1) , through f, i get a value in (liminf f, limsup f), not that i get every value in that bounded set.
    I miss the opposite inclusion, in other words that (liminf f, limsup f) is included in A = {values given by f(x(n)) for all x(n) in (0,1) that goes to 0}.
    This is why you want a sequence of sequences. For any point L \in \left(\liminf f, \limsup f\right), find a sequence of sequences such that \lim_{n\to \infty} x_{(n,m)} = \dfrac{1}{m+1} and  \lim_{n \to \infty}\lim_{m\to \infty} f\left( x_{(n,m)} \right) = L
    Last edited by SlipEternal; October 10th 2013 at 11:15 AM.
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  5. #5
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    Re: two little problems about limsup / liminf.

    Another way to try problem #1 is:

    For each L \in \left(\liminf_{x \to 0}f(x), \limsup_{x \to 0} f(x) \right), let A_L = \{x \in (0,1) \mid f(x)=L \}. Show that for for every \varepsilon>0, the open ball B(0;\varepsilon)\cap A_L \neq \emptyset. Then, let x_n = \sup\left( B\left(0; \dfrac{1}{n} \right) \cap A_L \right). Now, f(x_n) \to L as n \to \infty.
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  6. #6
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    Re: two little problems about limsup / liminf.

    thanks a lot, i succeeded in proving it thanks to your last suggestion.
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