# Thread: Fourier analysis

1. ## Fourier analysis

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

Thanks.

2. Hello,
Originally Posted by asi123
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 ?

3. Originally Posted by asi123
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:

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

or:

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

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

4. Originally Posted by CaptainBlack
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.

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

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

CB

6. 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.

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

$
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 $L^1$ 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:

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

CB