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Homework Statement
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->[tex]\infty[/tex] ([tex]\int[/tex][tex]^{1}_{0}[/tex]fn(x)dx) = [tex]\int[/tex][tex]^{1}_{0}[/tex] (limn->[tex]\infty[/tex] fn(x))dx
Homework 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->[tex]\infty[/tex] ([tex]\int[/tex][tex]^{a}_{x}[/tex]fn(t)dt) = [tex]\int[/tex][tex]^{a}_{x}[/tex] (limn->[tex]\infty[/tex] fn(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]. 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 [tex]\epsilon[/tex] >0 and an index N, so that
|fn(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 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.