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Math Help - Find limit of sequence of functions

  1. #1
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    Find limit of sequence of functions

    Let f_n\in C[0,1] be defined by
    f_n =\begin{cases}n^{2/3} & \mbox{if }0\le x\le \frac{1}{n}\\2n^{2/3}-n^{5/3}x & \mbox{if }\frac{1}{n} < x< \frac{2}{n}\\0 & \mbox{if }\frac{2}{n} \le x\le 1\\\end{cases}

    Let \parallel f \parallel _1 = \int_{0}^{1}\mid f(x)\mid\, dx and \parallel f \parallel _2 = \left (\int_{0}^{1}\mid f(x)\mid ^2\, dx \right )^\frac{1}{2}

    Show that f_n \rightarrow 0 with respect to \parallel \cdot \parallel _1, but f_n is not convergent with respect to \parallel \cdot \parallel _2







    ---------------------


    Now, I did the following, but it must be wrong since it doesn't depend on the choice of norm, so I get that f_n \rightarrow 0 in both cases.


    Choose x \in [0,\tfrac{1}{n}] for some small n.
    \exists \, n_0 > n st x \notin [0,\tfrac{1}{n} ] but rather x \in (\tfrac{1}{n} ,\tfrac{2}{n} )
    \exists \, \tilde{n}_0 > n_0 st x \notin (\tfrac{1}{n},\tfrac{2}{n} ) but rather x \in [\tfrac{2}{n} ,1 )
    \Rightarrow \, \exists \, \hat{n}_0 st f_n = 0 \, \forall \, n \ge \hat{n}_0

    So, clearly, we have that \forall \, \epsilon >0 \, \exists \, \hat{n}_0 \in \mathbb{N} st \parallel 0 - f_n \parallel _1 < \epsilon \, \forall \, n \ge \hat{n}_0







    So what have I missed that stops the above from working, and what is a better way of doing it?
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  2. #2
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    [mistaken]
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  3. #3
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    Quote Originally Posted by wglmb View Post
    Let f_n\in C[0,1] be defined by
    f_n =\begin{cases}n^{2/3} & \mbox{if }0\le x\le \frac{1}{n}\\2n^{2/3}-n^{5/3}x & \mbox{if }\frac{1}{n} < x< \frac{2}{n}\\0 & \mbox{if }\frac{2}{n} \le x\le 1\\\end{cases}

    Let \parallel f \parallel _1 = \int_{0}^{1}\mid f(x)\mid\, dx and \parallel f \parallel _2 = \left (\int_{0}^{1}\mid f(x)\mid ^2\, dx \right )^\frac{1}{2}

    Show that f_n \rightarrow 0 with respect to \parallel \cdot \parallel _1, but f_n is not convergent with respect to \parallel \cdot \parallel _2
    Why not just compute \|f_n\|_1 and \|f_n\|_2? You can notice that f_n(x)\geq 0 hence

    \int |f_n(x)|^p dx =( n^{2/3})^p  \frac{1}{n}+\int_{1/n}^{2/n} (n^{2/3}-2n^{5/3}x)^p dx

    for p=1,2. For p=1, if you get 0, you're done. For p=2, you should not get 0; if you get an infinite limit, there is no limit in L^2. If there is a finite limit... it may need some more work to find it or prove there is none. (I think here you get an infinite limit, just because of the first term (n^{2/3})^p  \frac{1}{n}, so you're done)

    In fact, since the space is [0,1], if f_n\to f in L^2, then f_n\to f in L^1 hence the limit should be 0 and it suffices to check if \lim_n \|f_n\|_2=0 to know if there is convergence in L^2.

    Now, I did the following, but it must be wrong since it doesn't depend on the choice of norm, so I get that f_n \rightarrow 0 in both cases.


    Choose x \in [0,\tfrac{1}{n}] for some small n.
    \exists \, n_0 > n st x \notin [0,\tfrac{1}{n} ] but rather x \in (\tfrac{1}{n} ,\tfrac{2}{n} )
    \exists \, \tilde{n}_0 > n_0 st x \notin (\tfrac{1}{n},\tfrac{2}{n} ) but rather x \in [\tfrac{2}{n} ,1 )
    \Rightarrow \, \exists \, \hat{n}_0 st f_n = 0 \, \forall \, n \ge \hat{n}_0

    So, clearly, we have that \forall \, \epsilon >0 \, \exists \, \hat{n}_0 \in \mathbb{N} st \parallel 0 - f_n \parallel _1 < \epsilon \, \forall \, n \ge \hat{n}_0
    What you proved here is that f_n converges to 0 almost-everywhere (in fact, everywhere). As you probably know, you may have f_n(x)\to f(x) for almost all (or all) x, and \int f_n \not\rightarrow \int f. This is however true under a few conditions: cf. the bounded convergence theorem. In this case it is not straightforward to apply it to prove convergence in L^1, and anyway it wouldn't suffice for the L^2 case.
    Last edited by Laurent; November 5th 2009 at 03:26 AM. Reason: addendum (2nd part)
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  4. #4
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    Thanks very much, that was great help!

    wglmb
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