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Math Help - Help find counterexample: multiple limits

  1. #1
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    Help find counterexample: multiple limits

    A true/false question from a quiz a while back. It's still bugging me because I'm told it's a false statement, but after three weeks I can't think of a counterexample. Maybe I just haven't quite made sense of it. A suitable counterexample or even a clarifying explanation of what this is saying would be of great help. Thanks in advance--here's the statement:

    If  \int _0 ^{\infty} f_k(x) dx = \lim _{A\to \infty} \int _0 ^A f_k(x) dx exists for each  k \in \mathbb{N} and  f_k \to 0 uniformly on  [0,\infty ) , then  \lim _{k\to \infty } \left(\lim _{A\to \infty } \int _0 ^A f_k(x) dx \right) = 0 .
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  2. #2
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    I am very unsure so maybe somebody else can answer too, but I think it is about order of taking limits. I thought true at first too but I remember sometimes the order of limits can give different result. If the order of limits did not mater then you can change to take first the limit as k goes to infinity of the finite integral from 0 to A of f_k. This will be 0 for any A because f_k ---> 0 uniformly and it is ok to switch order of limit and integral when there is uniform convergence of sequence of functions. Then taking other limit as A goes to infinity will still be 0. You must find example that order of taking limits is different. I am sorry I can not make one I have been away from math many years that is why my name is memory loss LOL
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  3. #3
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    Fading pulse example (updated)

    Try this one.
    Define <br />
f_k(x) = \left\lbrace<br />
\begin{array}{cl}<br />
\frac{1}{k} & x \in [0,k] \\<br />
0 & \text{otherwise}<br />
\end{array}<br />
\right.<br />
then  f_k(x) \rightarrow 0 uniformly on  [0,\infty) but for each k,  \int_{0}^{\infty}{f_k(x)} = \frac{1}{k} \times k = 1 so  \lim_{k \rightarrow \infty}{\int_{0}^{\infty}{f_k(x)}} = 1 .
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  4. #4
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    Quote Originally Posted by huram2215 View Post
    Try this one.
    Define <br />
f_k(x) = \left\lbrace<br />
\begin{array}{cl}<br />
\frac{1}{k} & x \in [0,k] \\<br />
0 & \text{otherwise}<br />
\end{array}<br />
\right.<br />
then  f_k(x) \rightarrow 0 uniformly on  [0,\infty) but for each k,  \int_{0}^{\infty}{f_k(x)} = \frac{1}{k} \times k = 1 so  \lim_{k \rightarrow \infty}{\int_{0}^{\infty}{f_k(x)}} = 1 .
    Ooooh, that was slick. Much obliged!

    So memory loss was right--it's about the order of the limiting processes mattering.
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