Re: lim inf and lim sup

David C. Ullrich wrote:
> On 20 Nov 2005 06:47:45 -0800, "Ryan Reich" <ryan.reich@xxxxxxxxx>
> wrote:
> >You can do it directly, you just have to avoid one particular catch.
> >We have by definition
> >
> >limsup_n a_n = inf_N S_N, where S_N = sup { a_n | n >= N }
> >liminf_n a_n = sup_N I_N, where I_N = inf { a_n | n >= N }
> >
> >Pick any N. Then by definition I_N <= a_n for all n >= N; then for any
> >M > N surely this holds for n >= M as well, so I_N <= S_M for all such
> >M, and hence
> >
> >I_N <= inf_{M > N} S_M.
> >
> >Now observe that if P < N and M > N, { a_n | n >= P } contains { a_n |
> >n >= M } and hence has at least as high a supremum; i.e. S_P >= S_M.
> >Thus
> >
> >limsup a_n = inf_{M > N} S_M, for any N
> >
> >(i.e. the first N terms do not add anything to the "eventual
> >supremum"). Thus we have shown
> >
> >I_N <= limsup a_n.
> >
> >Then surely
> >
> >liminf a_n = sup_N I_N <= limsup a_n
> >
> >by definition of supremum. The "catch" was that it is very tempting to
> >say that clearly I_N <= S_N, so taking limits, QED. But of course you
> >are not taking limits, but suprema and infima, and on top of that the
> >indices on each side of the inequality are varying independently, so
> >that this argument completely dodges the point of the question.
>
> There is no catch - the proof is entirely trivial.
>
> In fact it's clear that I_N <= S_N, right? It's also
> clear that the sequence S(N) is monotone, which implies
> that lim sup a_n = lim_N S_N (actually lim_N S_N is
> the usual definition, don't know where you got yours.)
> And lim inf a_n = lim_N I_N, so there you are.

I agree the proof is easy, but in fact yours is the same as mine. We
both started essentially in the same place, though I gave a slightly
broader statement. We then both invoked monotonicity and derived an
equivalent expression for the limsup or liminf, and from there the
answer was trivial. All I did differently was use a slightly different
definition and explain the work in greater detail, with more
whitespace.

And the catch is still a catch if you are using my definition, unless
you invoke monotonicity again. That is, clearly, the key point here.
Here's an alternate proof that tries to proceed along the lines of the
catch: since I_N <= S_N for all N, and since the I_N are increasing and
the S_N decreasing, in fact the set { I_N } lies entirely below the set
{ S_N }. Then clearly the sup of the former is below the inf of the
latter. Of course, that is nearly word-for-word what you wrote and on
top of that, the first sentence essentially establishes what I
established in the first part of my original proof. I guess my point
is that there is only one proof of this statement, though you can say
it slightly differently in many ways.

The definition I gave is the definition I learned (I think it's in
Spivak's "Calculus", though that's not where I learned it. I suspect
it's due to Darboux, since in defining the integral by Darboux sums one
uses supremum and infimum extensively). It's equivalent to yours
since, as we both noted, S_N is a decreasing sequence and I_N an
increasing sequence, so the limits are the infimum and supremum,
respectively. My definition seems preferable (this is of course
subjective) since one of the benefits of using limsup and liminf is
that they always exist (at least, if you count infinities as reasonable
values) and, if they are equal, then the limit itself exists. Defining
limsup in terms of limit has the appearance of circularity from this
perspective (actually, this is a rather attractive way to define the
limit in the first place, as it completely elides the epsilons). In
fact, in some ways it is actually circular, in that one proves that a
monotone (say, increasing) sequence has a limit specifically by
exhibiting the supremum of that sequence as the limit.

--
Ryan Reich
ryan.reich@xxxxxxxxx

.

Relevant Pages

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