Interesting Integral problem

1. May 7, 2010

MatthewD

1. The problem statement, all variables and given/known data

Suppose $$f$$ is continuous and $$F(x)=\int_a^x f(t)dt$$ bounded on $$[a,b)$$. Given $$g>0, g'\geq 0$$ and $$g'$$ locally integrable on $$[a,b)$$ and $$lim_{ x\rightarrow b^-} g(x) =$$ infinity. prove
for p>1
$$\displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = 0}$$

2. Relevant equations

3. The attempt at a solution
If you know $$lim_{ x\rightarrow b^-} g(x) =\infty$$, don't you also know $$lim_{ x\rightarrow b^-} \frac{1}{[g(x)]} = 0$$ and therefore $$lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p}=0$$...so we're done? But the hint says to use parts so I dont know :(
Any help would be greatly appreciated

2. May 8, 2010

lanedance

thats only the case if the integral is bounded, so you should try & show that...

3. May 8, 2010

MatthewD

How would I show an integral is bounded? If I know obviously $$\int_a^x f(t)dt$$ is bounded, how would I know anything about the g function. The information given basically denies that its bounded right? Since the limit is infinity? I'm so lost!

4. May 8, 2010

lanedance

have you tried the suggested hint?

5. May 8, 2010

lanedance

also, bounded would show it, but I think a lesser constraiint may also work, either way I would try simplifying things by using integration by parts

6. May 8, 2010

MatthewD

When I tried parts, i let u= g(t) and dv = f(t)dt so du = g'(t)dt and v = F(t) and I got

$$\displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} (g(x)F(x) -g(a)F(a)- \int^x_a F(t)g'(t)dt)}$$

$$\displaystyle{= lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} (g(x)F(x) - \int^x_a F(t)g'(t)dt)}$$
since F(a) = 0

Can I reduce further...?

7. May 8, 2010

MatthewD

i see it now, using parts helped a lot, thanks

assume further that $$\int^b_a f(t)dt$$ converges and show $$lim_{x \rightarrow b^-} \frac{1}{g(x)} \int_a^x f(t)g(t)dt = 0$$

i used parts again and came down to the fact:

$$lim_{x \rightarrow b^-} \frac{1}{g(x)} \int_a^x f(t)g(t)dt = lim_{x \rightarrow b^-} \frac{1}{g(x)} \int_a^x F(t)g'(t)dt$$

how do i show, for sure, this =0?

8. May 9, 2010

lanedance

how about usng the fact that F is bounded, g'>0 and setting up an inequality which squeezes your expression to zero?

9. May 9, 2010

MatthewD

The entire expression, or just the integral part? Could I do:

$$lim_{x\rightarrow b^-} \frac{1}{g(x)} \leq lim_{x\rightarrow b^-} \frac{1}{g(x)} \int^x_a F(t)g'(t)dt \leq lim_{x\rightarrow b^-} \int^x_a M g'(t)dt$$

where the right hand side becomes

$$=lim_{x\rightarrow b^-} \frac{1}{g(x)}M(g(x)-g(a))$$
$$=lim_{x\rightarrow b^-} M(1-\frac{g(a)}{g(x)})$$

But how does this squeeze to 0? or am i looking at it wrong?

10. May 9, 2010

lanedance

the 2nd half of the equality is looking good, but i think you dropped a factor of p somewhere, maybe you could assume $p = 1+\epsilon, \epsilon>0$

you may need to tighten it up by accounting for the case when M<0, ie by starting from $\exists M > 0 : |F(x)|<M,\forall x \in [a,b)$

11. May 9, 2010

MatthewD

but there's no p in the second part of the problem...?

12. May 9, 2010

MatthewD

This is the second part...

13. May 9, 2010

lanedance

ok sorry, i missed the 2nd part, so you're ok with the first bit now?

so if i integrate by parts I get a little different:
$$\lim_{x \to b^-} \frac{1}{g(x)}\int_a^x dt f(t)g(t)$$
$$= \lim_{x \to b^-}\frac{1}{g(x)}(F(t)g(t)|_a^x - \int_a^x dt F(t)g'(t))$$
$$= \lim_{x \to b^-}(F(x)-\frac{F(a)g(a)}{g(x)} - \frac{1}{g(x)}\int_a^x dt F(t)g'(t))$$

using convergence of F as x->b, then
$$= F(b) - \lim_{x \to b^-}\frac{1}{g(x)}\int_a^x dt F(t)g'(t)$$
so working this line, i guess you need to try & show the limit of the end expression is F(b)...

Last edited: May 9, 2010
14. May 9, 2010

lanedance

i haven't totally worked it out, but it may help to try and show for any a<x0<b
$$\lim_{x \to b^-}\frac{1}{g(x)}\int_a^{x_0} dt F(t)g'(t) = 0$$
noting that as g' is locally integrable, it means g is defined on the interval [a,b), ie. it doesnt explode anywhere

then try splitting the integral into separate portions & considering the continuity of the function f on a small region near to b. (though i could see this might get tricky with double limits....)