Re: Limit of a sequence

On 27 Jan 2006 03:25:32 -0800, "Mate" <mmatica@xxxxxxxxxxx> wrote:

>Let f : [0,1] --> [0,1] be continuous and strictly decreasing such that
>f(0)=1, f(1)=0.
>Is it true that
>
>lim_(n-->oo) [ int (f(x))^(n+1) dx ] / [ int (f(x))^n dx ] = 1 (?)

Yes.

Lemma If 0 < a < 1 then

int_0^a f^n / int_a^1 f^n -> infinity.

Proof: Choose b with 0 < b < a. Then

int_0^a f^n > b f(b)^n

int_a^1 f^n < (1-a) f(a)^n,

and (f(b)/f(a))^n -> infinity. QED.

Now it's clear that all of your ratios are
less than 1, so we only need to show that the
lim sup of the ratios is >= 1, right.
But for any a as in the lemma, the lemma
shows that

lim sup int_0^1 f^{n+1} / int_0^1 f^n

= lim sup int_0^a f^(n+1) / int_0^a f^n,

and it's clear that

int_0^a f^(n+1) / int_0^a f^n > f(a),

hence its lim sup is >= f(a). So the lim sup
of the original ratio is >= f(a) for all a in (0,1).

>If the limit in (?) exists, then it must be 1 ( = ||f||_oo ),
>but I think that the sequence diverges for some f.
>I could not find such an f. Can you help?
>
>Thank you.


************************

David C. Ullrich
.

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