- #1

Appa

- 15

- 0

## Homework Statement

The sequence f

_{n}: [-1,1] -> R, f

_{n}(x)= nxe

^{-nx2}converges pointwise to f(x)= 0, x in [-1,1]. Can you verify the following:

lim

_{n->[tex]\infty[/tex]}([tex]\int[/tex][tex]^{1}_{0}[/tex]f

_{n}(x)dx) = [tex]\int[/tex][tex]^{1}_{0}[/tex] (lim

_{n->[tex]\infty[/tex]}f

_{n}(x))dx

## Homework Equations

Theorem: If f

_{n}is continuous on the interval D for every n and f

_{n}converges uniformly to f on D=[a,b], then

lim

_{n->[tex]\infty[/tex]}([tex]\int[/tex][tex]^{a}_{x}[/tex]f

_{n}(t)dt) = [tex]\int[/tex][tex]^{a}_{x}[/tex] (lim

_{n->[tex]\infty[/tex]}f

_{n}(t))dt = [tex]\int[/tex][tex]^{a}_{x}[/tex]f(t)dt) for every x in D.

## The Attempt at a Solution

The main idea is to find out whether the sequence is uniformly convergent on D= [0,1]. f

_{n}= nxe

^{-nx2}is continuous for every n because e

^{t}is always continuous. I tried something like this to verify uniform convergence:

f

_{n}(x) is uniformly convergent on D if there is a positive number

_{[tex]\epsilon[/tex]}>0 and an index N, so that

|f

_{n}(x) -f(x)|<

_{[tex]\epsilon[/tex]}for every n[tex]\geq[/tex]N and every x in [0,1]

[tex]\Leftrightarrow[/tex] |nxe

^{-nx2}-0|<

_{[tex]\epsilon[/tex]}

[tex]\Leftrightarrow[/tex] nxe

^{-nx2}<

_{[tex]\epsilon[/tex] }

But I don't know how to go forward. How can I show that the greater the index n gets, the closer f

_{n}gets to 0? I tried comparison but couldn't find any sequence, the term of which would be greater than f

_{n}so that the sequence would still converge.