Limsup and Liminf

Let an (n = 1, 2,…) be a sequence of real numbers, and define the sequence bn as

Подпись: (II.1)bn = sup am

m>n

Подпись: limsup an == lim n—то n—TO Подпись: sup am m>n Подпись: (II.2)

Then bn is a nonincreasing sequence: bn > bn+1 because, if an is greater than the smallest upper bound of an+1, an+2, an+3,…, then an is the maximum of an, an+1, an+2, an+3,..hence, bn = an > bn+1 and, if not, then bn = bn+1 ■ Nonincreasing sequences always have a limit, although the limit may be – to. The limit of bn in (II.1) is called the limsup of an:

image917 Подпись: sup am m>n Подпись: (II.3)

Note that because bn is nonincreasing, the limit of bn is equal to the infimum of bn. Therefore, the limsup of an may also be defined as

Note that the limsup may be +to or – to, for example, in the cases an = n and an = —n, respectively.

Similarly, the liminf of an is defined by

Подпись: (II4)Подпись: (II5)liminfan = lim inf am

n — то n —— to V m>n

or equivalently by

liminfan = sup inf am

n — TO n>1 m>n

Again, it is possible that the liminf is +to or —to.

Note that liminfn—TO an < limsupn—TO an because infm>n am < supm>n am for all indices n > 1, and therefore the inequality must hold for the limits as well.

Theorem II.1:

(a) If liminf n—TO an = limsupn—TO an, then limn—TO an = limsupan, and if

n — TO

liminfn—TO an < limsupn—TO an, then the limit of an does not exist.

(b) Every sequence an contains a subsequence ank such that limt—TO ant = limsupn—TO an, and an also contains a subsequence anm such that

limm — TO anm = liminf n—— TO an.

Proof: The proof of (a) follows straightforwardly from (II.2), (II.4), and the definition of a limit. The construction of the subsequence ant in part (b) can be done recursively as follows. Let b = limsupn—TO an < to. Choose n1 = 1, and suppose that we have already constructed a^. for j = 1t > 1. Then there exists an index nt+1 > nt such that ant+1 > b — 1/(t + 1) because, otherwise, am < b — 1/(t + 1) for all m > nt, which would imply that limsupn—TO an < b — 1/(t + 1). Repeating this construction yields a subsequence ant such that, from large enough t onwards, b — 1/1 < ant < b. If we let t — to, the limsup case of part (b) follows. If limsupn—TO an = to, then, for each index nt we can find an index nt+i > nt such that ant+l > t + 1; hence, limt—TO ant = to. The subsequence in the case limsupn—TO an = —to and in the liminf case can be constructed similarly. Q. E.D.

The concept of a supremum can be generalized to sets. In particular, the countable union U^TO=1 Aj may be interpreted as the supremum of the sequence of sets Aj, that is, the smallest set containing all the sets Aj. Similarly, we may interpret the countable intersection n^TO=1 Aj as the infimum of the sets Aj, that is, the largest set contained in each of the sets Aj. Now let Bn = UTO=n Aj for

n = 1, 2, 3,____ This is a nonincreasing sequence of sets: Bn+1 c Bn; hence,

nn=1 Bn = Bn. The limit ofthis sequence ofsets is the limsup of An for n — to, that is, as in (II.3) we have

def. to / to

limsup An = n I U Aj

n — TO n = 1 j =n

Next, let Cn = nTO=n Aj for n = 1, 2, 3,____ This is a nondecreasing sequence

of sets: Cn c Cn+1; hence, Un=1 Cn = Cn. The limit of this sequence of sets is the liminf of An for n — to, that is, as in (II.5) we have

_ def. TO(TO

liminf An = U n A j

n — TO n=1 у j =n

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