Proving a Limit Involving Integrals and a Locally Integrable Function

[MatthewD]

Homework Statement

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
$$\displaystyle{lim_{ x\rightarrow b^-} \frac{1}{[g(x)]^p} \int^x_a f(t)g(t)dt = 0}$$

Homework Equations

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 don't know :(
Any help would be greatly appreciated

 

[lanedance]

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

 

[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!

 

[lanedance]

have you tried the suggested hint?

 

[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

 

[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...?

 

[MatthewD]

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

i'm suppose to follow it up with this addition:

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?

 

[lanedance]

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

 

[MatthewD]

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

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?

 

[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 $\exists M > 0 : |F(x)|<M,\forall x \in [a,b)$

 

[MatthewD]

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

 

[MatthewD]

i'm suppose to follow it up with this addition:

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

This is the second part...

 

[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)...

 

[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 doesn't 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...)