# Fourier analysis

• Dec 6th 2008, 01:20 PM
asi123
Fourier analysis
Hey guys, I'm kind a stuck with this one, any ideas?

Thanks.
• Dec 6th 2008, 02:31 PM
Moo
Hello,
Quote:

Originally Posted by asi123
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 ?
• Dec 7th 2008, 12:01 AM
CaptainBlack
Quote:

Originally Posted by asi123
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
• Dec 7th 2008, 05:02 AM
asi123
Quote:

Originally Posted by CaptainBlack
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.
• Dec 7th 2008, 05:24 AM
CaptainBlack
Quote:

Originally Posted by asi123
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
• Dec 7th 2008, 05:55 AM
asi123
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.
• Dec 7th 2008, 06:41 AM
CaptainBlack
Quote:

Originally Posted by asi123
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