convergence Watch

sonic7899
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I am finding the question below quite hard if anyone can help a bit?

Let f_n (x)=2^n I_{(0,2^{-n})}(x) for n >=1. Show that this converges to 0 as n tends to infinity for each x in (0,1) but that the integral of f_n(x) equals 1 for n>=1.


I can explain in words that as n tends to infinity the interval on which the indicator function, I, is non zero will get smaller but I can't show this?
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ForGreatJustice
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Perhaps this is a silly question, but what is I?
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sonic7899
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Sorry its the indicator function
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nuodai
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(Original post by ForGreatJustice)
Perhaps this is a silly question, but what is I?
It's the indicator function, I_A(x) = \begin{cases} 0 & \text{if } x \not \in A \\ 1 & \text{if } x \in A \end{cases}.

OP: integrate f_n(x) between x=-\infty and x=\infty. This is the same as integrating between x=0 and x=2^{-n} since the function is elsewhere (and is constant in that interval); you'll notice that you get an answer which is independent of n, and so the sequence is constant.

As for showing that the function converges to zero, pick any point and show that either the function is always zero there, or that it will eventually be zero there if n is big enough.
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IrrationalNumber
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You won't be able to show uniform convegence (it doesn't converge uniformly!) but you can show pointwise convergence.
You want to use the fact that for every y there is an n such that 2^n>y.
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sonic7899
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(Original post by nuodai)
It's the indicator function, I_A(x) = \begin{cases} 0 & \text{if } x \not \in A \\ 1 & \text{if } x \in A \end{cases}.

OP: integrate f_n(x) between x=-\infty and x=\infty. This is the same as integrating between x=0 and x=2^{-n} since the function is elsewhere (and is constant in that interval); you'll notice that you get an answer which is independent of n, and so the sequence is constant.

As for showing that the function converges to zero, pick any point and show that either the function is always zero there, or that it will eventually be zero there if n is big enough.
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?
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sonic7899
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(Original post by IrrationalNumber)
You won't be able to show uniform convegence (it doesn't converge uniformly!) but you can show pointwise convergence.
You want to use the fact that for every y there is an n such that 2^n>y.
So if I fix x in (0,1), then there will always exist an N such that x>2^-n. Then |fn(x)-0|=0<e for e>0 and n>N. Hence the sequence converges to zero?
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IrrationalNumber
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(Original post by sonic7899)
So if I fix x in (0,1), then there will always exist an N such that x>2^-n. Then |fn(x)-0|=0<e for e>0 and n>N. Hence the sequence converges to zero?
I would write it like this: Fix x in (0,1). Fix epsilon bigger than 0. We know there exists an N in natural numbers such that for n>N  2^n&gt;2^N&gt;\frac{1}{x} . Now for any such n, we have  x&gt;2^{-n} and so x is not in (0,2^-n). Hence for n>N we know that f_n(x)=0 and so |f_n(x)-0|=0<epsilon. So f_n(x) converges pointwise to the 0 function.
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sonic7899
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(Original post by IrrationalNumber)
I would write it like this: Fix x in (0,1). Fix epsilon bigger than 0. We know there exists an N in natural numbers such that for n>N  2^n&gt;2^N&gt;\frac{1}{x} . Now for any such n, we have  x&gt;2^{-n} and so x is not in (0,2^-n). Hence for n>N we know that f_n(x)=0 and so |f_n(x)-0|=0<epsilon. So f_n(x) converges pointwise to the 0 function.
yeah your right, that is a bit more formal. Just to make sure I understand, if I changed the question so that f_n(x) = I_{(n,n+1)}(x) instead. To show fn is bounded for n>=1 with fn(x) tending to 0 as n tend to infinity for each real x.

Would I just say again that fixing that there is an n+2>N this time?
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IrrationalNumber
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(Original post by sonic7899)
Would I just say again that fixing that there is an n+2>N this time?
What? I don't think that's what you meant to write.
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sonic7899
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(Original post by IrrationalNumber)
What? I don't think that's what you meant to write.
Sorry I am trying to say that there is an N>n+1 such that the indicator is 0?
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IrrationalNumber
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(Original post by sonic7899)
Sorry I am trying to say that there is an N>n+1 such that the indicator is 0?
Fix x in reals. At what point do the indicators start being 0?
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sonic7899
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(Original post by IrrationalNumber)
Fix x in reals. At what point do the indicators start being 0?
For numbers outside of n and n+1 interval?
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IrrationalNumber
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(Original post by sonic7899)
For numbers outside of n and n+1 interval?
You mean the interval n and n+1 surrounding x? So how large do we need n to be in order for f_n(x) to be 0?
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sonic7899
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(Original post by IrrationalNumber)
You mean the interval n and n+1 surrounding x? So how large do we need n to be in order for f_n(x) to be 0?
This is what I don't think I understand because would a large n not still allow for the function to be 1?
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IrrationalNumber
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(Original post by sonic7899)
This is what I don't think I understand because would a large n not still allow for the function to be 1?
The function would be 1 in certain places, but it wouldn't be 1 at x. Since that trick works for any x, the function converges POINTWISE to 0. You seem to be thinking about convergence in some sort of uniform sense, and the function does not converge uniformly to 0.
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IrrationalNumber
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Do this: draw a graph of the x axis, then mark the point x=3.5, and then draw the functions f_1, f_2,f_3,f_4,f_5 in different colours, and tell me how large n needs to be before f_n(x) is 0.
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sonic7899
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(Original post by IrrationalNumber)
Do this: draw a graph of the x axis, then mark the point x=3.5, and then draw the functions f_1, f_2,f_3,f_4,f_5 in different colours, and tell me how large n needs to be before f_n(x) is 0.
Right I have drawn the graph like you said, but the answer I have doesn't make sense. For f_n(x) to be 0 does n not have to be bigger than n+1 somehow?
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IrrationalNumber
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(Original post by sonic7899)
Right I have drawn the graph like you said, but the answer I have doesn't make sense. For f_n(x) to be 0 does n not have to be bigger than n+1 somehow?
You don't seem to understand the difference between f_n(x)=0 and f_n = 0. When x=3.5, all we need is f_n(3.5)=0. What number N gives us that if n>N then f_n(3.5)=0?
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sonic7899
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(Original post by IrrationalNumber)
You don't seem to understand the difference between f_n(x)=0 and f_n = 0. When x=3.5, all we need is f_n(3.5)=0. What number N gives us that if n>N then f_n(3.5)=0?
I think I understand now, so you are saying that if N=x and n>3.5 then f_n(3.5)=0?
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