Results 1 to 4 of 4

Thread: Find limit of sequence of functions

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    16

    Find limit of sequence of functions

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

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

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







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


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


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

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







    So what have I missed that stops the above from working, and what is a better way of doing it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2006
    From
    Florida
    Posts
    228
    [mistaken]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by wglmb View Post
    Let $\displaystyle f_n\in C[0,1]$ be defined by
    $\displaystyle f_n =\begin{cases}n^{2/3} & \mbox{if }0\le x\le \frac{1}{n}\\2n^{2/3}-n^{5/3}x & \mbox{if }\frac{1}{n} < x< \frac{2}{n}\\0 & \mbox{if }\frac{2}{n} \le x\le 1\\\end{cases}$

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

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

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

    for $\displaystyle p=1,2$. For $\displaystyle p=1$, if you get 0, you're done. For $\displaystyle p=2$, you should not get 0; if you get an infinite limit, there is no limit in $\displaystyle L^2$. 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 $\displaystyle (n^{2/3})^p \frac{1}{n}$, so you're done)

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

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


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

    So, clearly, we have that $\displaystyle \forall \, \epsilon >0 \, \exists \, \hat{n}_0 \in \mathbb{N}$ st $\displaystyle \parallel 0 - f_n \parallel _1 < \epsilon \, \forall \, n \ge \hat{n}_0$
    What you proved here is that $\displaystyle f_n$ converges to $\displaystyle 0$ almost-everywhere (in fact, everywhere). As you probably know, you may have $\displaystyle f_n(x)\to f(x)$ for almost all (or all) $\displaystyle x$, and $\displaystyle \int f_n \not\rightarrow \int f$. 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 $\displaystyle L^1$, and anyway it wouldn't suffice for the $\displaystyle L^2$ case.
    Last edited by Laurent; Nov 5th 2009 at 02:26 AM. Reason: addendum (2nd part)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2009
    Posts
    16
    Thanks very much, that was great help!

    wglmb
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Find the limit of sequence
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Nov 12th 2011, 11:56 AM
  2. Prove limit of sequence of functions is continuous
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 27th 2009, 12:44 PM
  3. Replies: 1
    Last Post: Jan 16th 2009, 01:00 PM
  4. Replies: 2
    Last Post: Jan 13th 2009, 12:17 AM
  5. find the limit of the sequence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 1st 2007, 02:27 PM

Search Tags


/mathhelpforum @mathhelpforum