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Math Help - Fourier analysis

  1. #1
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    Fourier analysis

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

    Thanks.
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  2. #2
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    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 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 ?
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  3. #3
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    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 f_n \to f as n \to \infty in L^2_{(a,b)} norm then f_n \to f as n \to \infty in L^1_{(a,b)} norm.

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

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

    Now the Cauch-Scwartz inequality tell us that:

     <br />
\langle h_n,c \rangle^2 \le \|h_n\|_2\ \|c\|_2<br />

    or:

     <br />
\left( \int_a^b h_n(x)\ dx \right)^2 \le (b-a) \int h(x)^2 \ dx<br />

    So there exists a k>0 such that:

     \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:

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

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

    CB
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    Quote Originally Posted by CaptainBlack View Post
    To paraphrase:

     \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:

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

    Hence convergence in L^2_{(a,b)} norm implies convergence in 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.
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    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 \{ S_n,\ n=1,2,\ ..\ \} is a null sequence if and only if:

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

    CB
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    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.
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  7. #7
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    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 \{f_n,\ n=1,2,\ .. \} :

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

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

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

    CB
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