# 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.