Re: Double limits at the infinity.

I'm not sure if this is related to what you asked, but there's a
condition that implies the existence of the double limit. Suppose that
for every x there exists Ly(x) = lim (y -> oo) (f(x,y) and that there
exists L = lim (x -> oo) Ly(x). In addition, suppose the "function of
functions" f(x,y) converges uniformly to Ly, in the sense that, for
every eps>0, there exists a real K, depending only on eps, such that,
if y > K, then |f(x,y) - Ly(x)| < eps for every real x. This implies
the double limit exists and equals L, that is, for every eps >0, there
exists A such that x > A and y> A implies |f(x,y - L| < eps.

Im not quite sure, but I think this also implies the existence of the
function Lx(y) = lim ( x-> oo) f(x,y) and that lim (y -> oo) Lx(y) = L.

Amanda

David C. Ullrich escreveu:
> On Thu, 21 Jul 2005 13:46:25 +0200, Jesús Cid-Sueiro
> <jcid@xxxxxxxxxxx> wrote:
>
> >Given function f(x,y) I would need to know under which conditions over
> >f, the limit of f as x,y go to +infinity does not depend on the way to
> >go, for instance
> >
> >lim_{y-->inf} lim_{x-->inf) f(x,y) =
> >lim_{x-->inf) lim_{y-->inf} f(x,y) =
> >lim_{x-->inf} f(x,x) =
> >
> >
> >I guess that the above relations are true if the limit exists, f is
> >continuous and df(x,y)/dx and df/dy always go to zero as x and y go to
> >+inf (in any way).
> >
> >Is this correct?
>
> I'm not sure what you mean by "the limit exists".
>
> Of course if you mean "if lim_{(x,y) -> (inf,inf)} f(x,y)
> exists then the answer is yes, all the above are equal.
> But that's probably not what you mean, since that's
> pretty obvious, and also has nothing to do with the
> conditions on the derivatives.
>
> If you mean
>
> ">I guess that the above relations are true if
> lim_{y-->inf} lim_{x-->inf) f(x,y) exists, f is continuous
> and df(x,y)/dx and df/dy always go to zero as x and y go to
> +inf (in any way)"
>
> then the answer is no.
>
> >Jesús.
>
>
> ************************
>
> David C. Ullrich

.

Relevant Pages

  • Re: Inconsistency of the usual axioms of set theory
    ... exists a Dedekind-infinite set" implies the existence of N? ... one can prove that "there exists an infinite ... particular axiom of infinity ...
    (sci.math)
  • Re: Exception to the rule? (Tarski´s T-scheme)
    ... >>can substitute any proposition you want in for p, ... have to get the existence of a sentence from elsewhere. ... implies that, if there is one sentence, there are two things. ... It's not the T-schema which gives you that. ...
    (sci.logic)
  • Re: Inconsistency of the usual axioms of set theory
    ... exists a Dedekind-infinite set" implies the existence of N? ... Therefore tommy1729 was not babbling. ... ZF-Infinity that if an infinite set exists -- and we ...
    (sci.math)
  • Re: Q: One-way functions
    ... Consequences of One-Way Permutations", ... that the existence of one-way permutations implies the existence or the ... to this question (what are the implications of UP = coUP verses not?) ...
    (sci.crypt)
  • semifinite measures
    ... there's a subset F of E with c< u< inf. ... Since u is semifinite, it follows E/F contains a subset G ... If, by way of contradiction, we assume there's a c>0 such that u<= ... this implies s is not in S. But this ...
    (sci.math)