Results 1 to 7 of 7

Thread: Fourier analysis

  1. #1
    Member
    Joined
    May 2008
    Posts
    171

    Fourier analysis

    Hey guys, I'm kind a stuck with this one, any ideas?

    Thanks.
    Attached Thumbnails Attached Thumbnails Fourier analysis-scan0004.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by asi123 View Post
    Hey guys, I'm kind a stuck with this one, any ideas?

    Thanks.
    Have you ever heard of $\displaystyle L^p$ spaces ? Because I know that there are specific conditions for the convergence in these spaces (which have the norms you're using)

    Also, is that the complete text ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by asi123 View Post
    Hey guys, I'm kind a stuck with this one, any ideas?

    Thanks.
    To paraphrase:

    Prove that if $\displaystyle f_n \to f$ as $\displaystyle n \to \infty$ in $\displaystyle L^2_{(a,b)}$ norm then $\displaystyle f_n \to f$ as $\displaystyle n \to \infty$ in $\displaystyle L^1_{(a,b)}$ norm.

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

    Introduce a function $\displaystyle h_n(x)=|f_n(x)-f(x)|$ and the constant function $\displaystyle c(x)=1$ on $\displaystyle (a,b)$ .

    Now the Cauch-Scwartz inequality tell us that:

    $\displaystyle
    \langle h_n,c \rangle^2 \le \|h_n\|_2\ \|c\|_2
    $

    or:

    $\displaystyle
    \left( \int_a^b h_n(x)\ dx \right)^2 \le (b-a) \int h(x)^2 \ dx
    $

    So there exists a $\displaystyle k>0$ such that:

    $\displaystyle \left(\int_a^b |f_n(x)-f(x)| \ dx\right)^2 \le k \int_a^b |f_n(x)-f(x)|^2\ dx$

    But the sequence of positive terms on the right is a null sequence and hence the sequence of positive terms on the left is a null sequence, hence:

    $\displaystyle \int_a^b |f_n(x)-f(x)| \ dx \to 0$ , as $\displaystyle n \to \infty$

    Hence convergence in $\displaystyle L^2_{(a,b)}$ norm implies convergence in $\displaystyle L^1_{(a,b)}$ norm

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2008
    Posts
    171
    Quote Originally Posted by CaptainBlack View Post
    To paraphrase:

    $\displaystyle \left(\int_a^b |f_n(x)-f(x)| \ dx\right)^2 \le k \int_a^b |f_n(x)-f(x)|^2\ dx$

    But the sequence of positive terms on the right is a null sequence and hence the sequence of positive terms on the left is a null sequence, hence:

    $\displaystyle \int_a^b |f_n(x)-f(x)| \ dx \to 0$ , as $\displaystyle n \to \infty$

    Hence convergence in $\displaystyle L^2_{(a,b)}$ norm implies convergence in $\displaystyle L^1_{(a,b)}$ norm

    CB
    Can you please tell me what you mean in the sentence I marked. I don't speak fluent English and defiantly not "math" English so it will be much appreciate if you clarify on that.

    Thanks a lot.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by asi123 View Post
    Can you please tell me what you mean in the sentence I marked. I don't speak fluent English and defiantly not "math" English so it will be much appreciate if you clarify on that.

    Thanks a lot.
    A null sequence is a sequence that goes to zero as the index goes to infinity.

    That is the sequence $\displaystyle \{ S_n,\ n=1,2,\ ..\ \}$ is a null sequence if and only if:

    $\displaystyle \lim_{n \to \infty}S_n=0$

    CB
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2008
    Posts
    171
    The next part of the question is, is it vice verse?

    I don't think so, but I cant find an example, any ideas?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by asi123 View Post
    The next part of the question is, is it vice verse?

    I don't think so, but I cant find an example, any ideas?

    Thanks.
    Consider the sequence of functions $\displaystyle \{f_n,\ n=1,2,\ .. \}$ :

    $\displaystyle
    f_n(x)=\begin{cases}
    0&x<-1/n\\
    \sqrt{n}&-1/n \le x<0\\
    -\sqrt{n}&0 \le x < 1/n\\
    0&1/n \le x
    \end{cases}
    $

    If I have set this up right this tends to the zero function in $\displaystyle L^1$ norm as $\displaystyle n$ goes to infinity. However it does not do so in $\displaystyle L^2$ norm.

    You will need to confirm this in detail, as I have just roughted out the argument and there may be some remaining errors hanging about.

    THis can be simplified to:

    $\displaystyle
    f_n(x)=\begin{cases}
    \sqrt{n}&0 \le x <1/n\\
    0& {\rm{otherwise}}
    \end{cases}
    $

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Fourier Analysis
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 6th 2011, 04:49 AM
  2. fourier analysis
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: May 8th 2011, 07:48 PM
  3. fourier analysis
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Aug 18th 2010, 03:21 AM
  4. Some Fourier Analysis
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 16th 2010, 05:11 AM
  5. Fourier analysis
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 28th 2006, 01:20 PM

Search Tags


/mathhelpforum @mathhelpforum