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lim fn(x) = f(x) but lim ∫ |f(x)-fn(x)|dx ≠ 0 ? |
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| Jan13-11, 07:44 PM | #1 |
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lim fn(x) = f(x) but lim ∫ |f(x)-fn(x)|dx ≠ 0 ?
1. The problem statement, all variables and given/known data
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 -> ∞) (with integral from - to + infinity)but lim ∫ |f(x)-fn(x)|dx ≠ 0 (as n -> ∞) 3. 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? |
| Jan13-11, 08:06 PM | #2 |
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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.
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| Jan13-11, 08:21 PM | #3 |
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Thanx LeonhardEuler:
I tried coming up with a nother function, what about fn(x) = |1/n| for x ∈ [0, n] and fn(x) will converge to 0 function uniformly fn(x) = 0 elsewhere in [-infinity, infinity] however, while ∫ |fn(x)| dx = 2 for all n in N ??? |
| Jan13-11, 08:23 PM | #4 |
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lim fn(x) = f(x) but lim ∫ |f(x)-fn(x)|dx ≠ 0 ?
Yeah, that works.
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| Jan13-11, 08:24 PM | #5 |
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Except the integral is 1, not 2.
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| Jan13-11, 08:28 PM | #6 |
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oh yeah, my mistake.
Thanks! |
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| functions sequence, integrability |
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