Suppose is continuous and bounded on . Given and locally integrable on and infinity. prove

for p>1

I think I should use integration by parts... but I am not sure. Any help would be greatly appreciated.

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- May 7th 2010, 11:33 AMAidenMatthewIntegrability
Suppose is continuous and bounded on . Given and locally integrable on and infinity. prove

for p>1

I think I should use integration by parts... but I am not sure. Any help would be greatly appreciated. - May 7th 2010, 11:40 AMAidenMatthew
In addition to this, I want to show that if converges, then we have

- May 7th 2010, 05:20 PMBBrown
- May 8th 2010, 12:41 PMOpalg
I agree, integration by parts is definitely the right way to go about this. In fact, . When you divide by , the term in square brackets will go to 0 as . So you just need to estimate the size of the integral on the right.

You know that F(t) is bounded, say for some constant M. Then . Again, when you divide by , that expression will go to 0 as . - May 8th 2010, 01:45 PMBBrown
- May 8th 2010, 01:47 PMAidenMatthew
Thats what I was thinking, and then use the fact that ... I'm still a little uneasy about how to set it up with this difference though... but I think the arguments are similar... doesnt convergence imply boundedness so the same argument goes?

- May 8th 2010, 09:15 PMcasanova
using the same integration by parts, at the very end of moving things around with what you said above you have

right? you know what is and you know is bounded, so you're done unless i'm missing something - May 9th 2010, 02:04 AMOpalg
This is a bit more subtle than the case where p>1. But you have the additional information that exists as the limit . So F is continuous on the left at b, and given there exists a point such that whenever x and t lie in the interval .

To estimate , split it up as . The first of those two integrals is independent of x, so when we divide by g(x) and let , it will go to 0. Thus we need only look at the second of the two integrals, .

Integrating by parts as previously, we get . Use the fact that to write this as . The first two terms on the right side of that equation are bounded, so when we divide by g(x) and let they will go to 0. Thus we need only estimate the size of the integral term.

For that, using the inequality , we get . When we divide by g(x), that gives a result that is essentially less than .

Putting everything together, that should show that .

I don't think it's as simple as that, because you don't know that is bounded. In fact, as , so that integral might well be unbounded. - May 9th 2010, 06:57 AMAidenMatthew
- May 9th 2010, 07:22 AMOpalg
It's a constant as far as t is concerned, and the integral is with respect to t. After all, x is the upper limit of integration in the integral , so x has to be regarded as a constant for the purposes of that integral.

The reason that F is continuous at b comes from the definition of an improper integral. To say that the integral*converges*means that , which is the same as saying that F is continuous at b.