Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limsup and bounded above

  1. Apr 5, 2012 #1
    Claim (?):
    limsup xk < ∞
    k->∞

    IF AND ONLY IF

    the sequence {xk} is bounded above.



    Does anyone know if this is true or not? (note that the claim is "if and only if")
    If it is true, why?

    Thanks!
     
  2. jcsd
  3. Apr 6, 2012 #2
    I think I'm OK with the direction:
    limsup x_k < ∞ implies the sequence {x_k} is bounded above.
    k->∞
    (becuase if the sequence is not bounded above, then limsup=∞)




    But is the converse direction also true?

    The sequence {x_k} is bounded above

    implies

    limsup x_k < ∞ ???
    k->∞
     
  4. Apr 6, 2012 #3

    chiro

    User Avatar
    Science Advisor

    Hey kingwinner.

    Think about the idea of the lowest upper bound. Suppose there is a value that is above this upper bound. Then this means the lowest upper bound is that new value.

    Now imagine that all values are finite. What does this say about the lowest upper bound? What happens if one value is infinite (positive infinity)?
     
  5. Apr 6, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "lim sup xn" is, by definition, the supremum of the set of all subequential limits. If it were not finite, then, given any M, there would have to exist subsequences of {xn having limit larger than M so M could not an upper bound of {xn}.
     
  6. Apr 6, 2012 #5
    Does this imply that
    the sequence {x_k} is bounded above => limsup x_k < ∞ ?

    Are you using contrapositive?
     
    Last edited: Apr 6, 2012
  7. Apr 6, 2012 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that is proving "if a then b" by proving the contrapositive, "if b is not true then a is not true".
     
  8. Apr 6, 2012 #7
    Is it possible to prove "directly" that

    sequence {x_k} is bounded above => limsup x_k < ∞ ??
     
  9. Apr 12, 2012 #8
    Yes. Using the definition above, let [itex] \{x_{k_n}\}_{n=1}^{\infty} [/itex] be any convergent subsequence. By assumption, there is an [itex] M > 0 [/itex] so that [itex]x_k \le M [/itex]. In particular, [itex] M \ge x_{k_n}[/itex]. Thus [itex] M \ge \lim_{n \to \infty} x_{k_n} [/itex]. This shows that M is an upper bound to any subsequential limit. By the definition HoI gave for the limsup as the supremum of all subsequential limits, this shows that limsup x_k < \infty.
     
  10. Apr 14, 2012 #9
    How about this? limsup means the limit of the supremum of terms with greater and greater index. If the whole set is bounded above, then sup_{k>0} is finite, right? sup(k>0), sup(k>1), sup(k>2),... forms a decreasing sequence right? The limit is either well defined real number or -infinity, isn't it?


    So in some problem, say if we were to show that limsup x_k < infinity, it suffices to prove that sequence {x_k} is bounded above, and vice versa, is that correct?

    Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limsup and bounded above
  1. Compact -> bounded (Replies: 2)

Loading...