Limit Superior and Limit InferiorDate: 03/27/2005 at 17:18:40 From: Joshua Subject: lim sup and lim inf Hello, I'm reading a real analysis text and I'm having trouble understanding completely what lim sup and lim inf are. First of all, is "lim sup" supposed to be looked at as a single "operation"? Second, I don't really understand the notation and how the mechanism works according to the notations. You can look here for what I have in mind: http://en.wikipedia.org/wiki/Limit_inferior Lastly, I just don't understand the concept at all. Can you first give me an intuitive understanding and then a more formal way of explaining it? Thank you very much! Date: 03/30/2005 at 20:20:13 From: Doctor Fenton Subject: Re: lim sup and lim inf Hi Joshua, Thanks for writing to Dr. Math. Yes, "lim sup" is a single process (and the same is true of lim inf); it is related to the usual limit process. If you have a bounded sequence x_1,x_2,x_3,..., then you can form a new sequence y_n by defining y_n = sup{x_k: k >= n}. I assume that you understand what the least upper bound or supremum of a set is: it is the smallest number which is at least as large as every element of the set. (In particular, if the set has a largest element, it is the supremum of the set.) Notice that y_(n+1) = sup {x_(n+1),x_(n+2),x_(n+3),...} is the least upper bound of a smaller set than y_n = sup {x_n,x_(n+1),x_(n+2),...} so y_(n+1) <= y_n , and the sequence y_n is a decreasing sequence which is bounded below (by any element of the sequence). Since the y_n sequence is decreasing and bounded below, it has a limit, call it L, and L = lim sup x_n = lim y_n , n->oo n->oo (or, since y_n is decreasing, the limit of the sequence must also be the infimum or greatest lower bound of the set {y_n}, and L = lim sup x_n = inf y_n . ) n->oo n>0 With this latter characterization of the lim sup as the greatest lower bound of the y_n, any larger number cannot be a lower bound for the y_n, so if E is any positive number, L+E is not a lower bound for the y_n, and there must be some y_N < L+E, so that L <= y_N < L+E . Since y_N is the least upper bound of {x_N,x_(N+1),x_(N+2),..}, then y_N - E cannot be an upper bound for this set, which means that there must be an x_m with m > N such that y_N - E < x_m , so that L - E <= y_N - E < x_m < y_N < L + E . Since E is arbitrary (i.e. can be as small as we wish), this shows that there is a subsequence of the x_n 's (formed from the x_m terms) which converges to L. For each E > 0, there can only be a finite number of x_n 's which are larger than L+E, so no subsequence of x_n can converge to a larger value than L. So, in summary, the lim sup of the x_n is the largest number L such that there is a subsequence of x_n's which converge to L. A similar analysis leads to the conclusion that the lim inf x_n is the smallest number for which there is a subsequence of the x_n which converges to it. If lim x_n = L exists, then lim sup x_n = lim inf x_n = L . n->oo n->oo n->oo If the two are different, then in a sense they measure how much the sequence fails to have a limit: there is one subsequence which converges to lim sup x_n, and another which converges to lim inf x_n . As an example, let x_n = (-1)^n * (n+1)/n , so x_1 = -2, x_2 = 3/2, x_3 = -4/3, x_4 = 5/4 , x_5 = -6/5, x_6 = 7/6 and so on. The odd terms x_1,x_3,x_5,... increase to -1, while the even terms x_2,x_4,x_6,... decrease to +1, so lim sup (-1)^2 (n+1)/n = +1 and lim inf (-1)^n (n+1)/n = -1 . If you have any questions, please write back and I will try to explain further. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
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