convergence Watch

IrrationalNumber
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#21
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#21
(Original post by sonic7899)
I think I understand now, so you are saying that if N=x and n>3.5 then f_n(3.5)=0?
Right (but you want N to be natural, so in this case N=4 or N=3 works)! So if instead I asked you to do it for 10.3, what would you say to me? How bout 11? In general, if I gave you an x and asked you to find the N, what would you say?
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sonic7899
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#22
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#22
(Original post by IrrationalNumber)
Right (but you want N to be natural, so in this case N=4 or N=3 works)! So if instead I asked you to do it for 10.3, what would you say to me? How bout 11? In general, if I gave you an x and asked you to find the N, what would you say?
Ok so in general n>N>=x
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IrrationalNumber
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#23
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#23
(Original post by sonic7899)
Ok so in general n>N>=x
So to structure the proof, I'd write this
Fix x in real. Fix epsilon >0. Take any natural N bigger than or equal to x. Then for n>N we have |f_n(x)-0|=|0-0|<epsilon. Thus f_n(x) converges pointwise to the 0 function (notice the convergence is not uniform. See if you can find a sequence of functions for which f_n converges uniformly to the 0 function but the integrals do not converge to 0).
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sonic7899
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#24
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#24
(Original post by IrrationalNumber)
So to structure the proof, I'd write this
Fix x in real. Fix epsilon >0. Take any natural N bigger than or equal to x. Then for n>N we have |f_n(x)-0|=|0-0|<epsilon. Thus f_n(x) converges pointwise to the 0 function (notice the convergence is not uniform. See if you can find a sequence of functions for which f_n converges uniformly to the 0 function but the integrals do not converge to 0).
Can I just check, to show that f_n is bounded is is just the fact that it can't be more than 1?
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IrrationalNumber
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#25
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#25
(Original post by sonic7899)
Can I just check, to show that f_n is bounded is is just the fact that it can't be more than 1?
Yep.
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sonic7899
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#26
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#26
(Original post by IrrationalNumber)
Yep.
Can I just check, are both of these sequences not measurable functions as I am trying to show why they disagree with the bounded convergence theorem?
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IrrationalNumber
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#27
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#27
(Original post by sonic7899)
Can I just check, are both of these sequences not measurable functions as I am trying to show why they disagree with the bounded convergence theorem?
Definitely not. Every function in these sequences is measurable. What made you think otherwise?
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DFranklin
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#28
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#28
(Original post by sonic7899)
Can I just check, to show that f_n is bounded is is just the fact that it can't be more than 1?
Not sure how you can say f_n can't be more than 1? Unless there's some assumption you're missing out, that's clearly not true.
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nuodai
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#29
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#29
(Original post by sonic7899)
Ah ok so would it be enough to say that as n tends to infinity, the interval becomes so small that the indicator will be always zero and so the function converges to zero?
Sort of. It's more that if you pick any point and let n be large enough, then the function evaluated at that point will eventually be zero.
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sonic7899
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#30
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#30
(Original post by DFranklin)
Not sure how you can say f_n can't be more than 1? Unless there's some assumption you're missing out, that's clearly not true.
You are right, I forgot that I was looking at a sequence of functions rather than a single function. So if I am trying to show that f_n is a bounded sequence of functions, am I trying to find an M such that for any n>=1 fn will always be less than it?
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sonic7899
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#31
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#31
(Original post by nuodai)
Sort of. It's more that if you pick any point and let n be large enough, then the function evaluated at that point will eventually be zero.
I have tried to show it using pointwise convergence like IrrationalNumber said, is that ok or should it be done your way?
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DFranklin
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#32
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#32
(Original post by sonic7899)
You are right, I forgot that I was looking at a sequence of functions rather than a single function. So if I am trying to show that f_n is a bounded sequence of functions, am I trying to find an M such that for any n>=1 fn will always be less than it?
It's somewhat dangerous to speak of 'bounded' without defining the metric you're using.

For the function under discussion:

\displaystyle \lim_{n \to \infty} \sup_{0\leq x \leq 1} |f_n(x)| = \infty, so f isn't bounded under the sup norm.

At the same time,

\lim_{n \to \infty} \int |f_n(x)| \,dx = 1 so f is bounded under the L^{1} norm.

More generally, one of the key points with these questions is to clearly distinguish (in your mind, and ideally in your answer!) the difference between what's happening at a particular point, and what's happening over the range of the entire function. It's something you seem to be confusing a lot in your posts here.
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IrrationalNumber
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#33
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#33
(Original post by DFranklin)
Not sure how you can say f_n can't be more than 1? Unless there's some assumption you're missing out, that's clearly not true.
It depends what f_n we're talking about, there are two questions being solved at the same time here. The second one is about indicator over (n,n+1). That's certainly uniformly bounded (the question is talking about the boundedness convergence theorem : http://en.wikipedia.org/wiki/Dominat...rgence_theorem)
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DFranklin
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#34
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#34
Sorry, I missed that. (Well, I'm still not sure which function sonic was talking about in that post, but whatever...)
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sonic7899
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#35
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#35
Can I check, I know that the bounded convergence theorem doesn't apply to the second example as the real numbers are not a bounded interval. However, for the first example does the theorem not apply because the function is not bounded?
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