Lim fn(x) = f(x) but lim ∫ |f(x)-fn(x)|dx ≠ 0 ?

  • Thread starter pulin816
  • Start date
  • #1
pulin816
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Homework Statement



Hey, I have another questions,
I need to find an example of a sequence of integrable functions fn:R -> R, n =1, 2, ...
such that

lim fn(x) = f(x) (as n -> ∞)
but lim ∫ |f(x)-fn(x)|dx ≠ 0 (as n -> ∞)​
(with integral from - to + infinity)

The Attempt at a Solution



I've tried
fn = (x + x/n)
and f = x

the first conditions would be satisfied, but on the other hand,
will the limits and the integral be interchangeable? I've read that it is only permitted if the expression inside is bounded. |x/n| can't be bounded since it has an absolute sign wrapped around or would it?

Any suggestions? Thank you !!!

p.s. Would the term 'integrable' here mean a function that is reinmann integrable?
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
860
1
I wouldn't worry about the interchange of limits, at least not right now. The function you suggested would not work because the limit of the integrals is 0 anyway.
 
  • #3
pulin816
6
0
Thanx LeonhardEuler:

I tried coming up with a nother function, what about

fn(x) = |1/n| for x ∈ [0, n] and
fn(x) = 0 elsewhere in [-infinity, infinity]​
fn(x) will converge to 0 function uniformly
however, while ∫ |fn(x)| dx = 2 for all n in N

???
 
  • #4
LeonhardEuler
Gold Member
860
1
Yeah, that works.
 
  • #5
LeonhardEuler
Gold Member
860
1
Except the integral is 1, not 2.
 
  • #6
pulin816
6
0
oh yeah, my mistake.
Thanks!
 

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