# Sequence of functions; uniform convergence and integrating

1. Feb 7, 2009

### Appa

1. The problem statement, all variables and given/known data
The sequence fn: [-1,1] -> R, fn(x)= nxe-nx2 converges pointwise to f(x)= 0, x in [-1,1]. Can you verify the following:

limn->$$\infty$$ ($$\int$$$$^{1}_{0}$$fn(x)dx) = $$\int$$$$^{1}_{0}$$ (limn->$$\infty$$ fn(x))dx

2. Relevant equations
Theorem: If fn is continuous on the interval D for every n and fn converges uniformly to f on D=[a,b], then

limn->$$\infty$$ ($$\int$$$$^{a}_{x}$$fn(t)dt) = $$\int$$$$^{a}_{x}$$ (limn->$$\infty$$ fn(t))dt = $$\int$$$$^{a}_{x}$$f(t)dt) for every x in D.

3. The attempt at a solution
The main idea is to find out whether the sequence is uniformly convergent on D= [0,1]. fn = nxe-nx2 is continuous for every n because et is always continuous. I tried something like this to verify uniform convergence:
fn(x) is uniformly convergent on D if there is a positive number $$\epsilon$$ >0 and an index N, so that
|fn(x) -f(x)|<$$\epsilon$$ for every n$$\geq$$N and every x in [0,1]
$$\Leftrightarrow$$ |nxe-nx2-0|<$$\epsilon$$
$$\Leftrightarrow$$ nxe-nx2 <$$\epsilon$$

But I don't know how to go forward. How can I show that the greater the index n gets, the closer fn gets to 0? I tried comparison but couldn't find any sequence, the term of which would be greater than fn so that the sequence would still converge.

2. Feb 7, 2009

### Dick

Finding a uniform limit for f_n isn't that hard. You just need to find the extrema of the functions as a function of n. Take the derivative, set it equal to zero etc. But I'm going to guess that what you are really expected to do is actually calculate the integral of f_n and verify that the limit of the integrals is zero.