Homework Help: Integral Convergence Theorem (limit of integrals)

1. Apr 13, 2010

kingwinner

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

2. Relevant equations
3. The attempt at a solution
I think it is related to the integral convergence theorem.

But how can we deal with the function g(x)? Is this problem supposed to be a direct use of the theorem or do we have to start the proof from scratch? How should we begin the proof?

Any help is appreciated!

2. Apr 13, 2010

Gib Z

You can use the Integral convergence theorem right after you show t_n(x) = f_n(x)g(x) converges uniformly to f(x)g(x).

3. Apr 13, 2010

kingwinner

OK, so we need to show that sup|f_n(x)g(x)-f(x)g(x)|->0.

sup|f_n(x)g(x)-f(x)g(x)|≤sup|f_n(x)-f(x)|+sup|g(x)|

How can we deal with sup|g(x)|?

4. Apr 13, 2010

Gib Z

Sorry, how did you get the second line? Shouldn't the Right hand side be the product of those two terms, not the sum.

5. Apr 13, 2010

kingwinner

oops...you're right.
|f_n(x)g(x)-f(x)g(x)|=|f_n(x)-f(x)||g(x)|

How can we show sup|f_n(x)g(x)-f(x)g(x)|->0 then?

Last edited: Apr 14, 2010
6. Apr 13, 2010

Gib Z

Look at both the terms on the right hand side. The first you know something about f_n and f to help you, and the extreme value theorem says that since g is an element of C[0,1], it attains a maximum value in that interval as well.

7. Apr 13, 2010

kingwinner

OK, that makes sense. But how can we handle the supremum? Will our bounds be preserved after taking the supremum? Why or why not?

8. Apr 13, 2010

Gib Z

The extreme value theorem states g(x) attains both a maximum and a minimum in [0,1]. Do you understand why sup |g(x)| = max{ |min g(x)| , |max g(x)| } ?

Which ever one of them it is, let it be a constant A.

Since f_n converges uniformly to f, you know that for a chosen positive chosen, let it be $\epsilon / A$, then $|f_n (x) - f(x) | < \epsilon / A$ for sufficiently large values of n.

Put this together, theres not much left to finish the question.

9. Apr 13, 2010

kingwinner

But is it necessarily true that
sup{|f_n(x)f(x)||g(x)|} = sup{|f_n(x)f(x)|} sup{|g(x)|} ??

10. Apr 13, 2010

Gib Z

No, but you have learned that for two sets A and B, and their product set AB

sup AB is less or equal to sup A sup B. You can use that to finish this question.