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

Thanks.

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- Dec 6th 2008, 01:20 PMasi123Fourier analysis
Hey guys, I'm kind a stuck with this one, any ideas?

Thanks. - Dec 6th 2008, 02:31 PMMoo
- Dec 7th 2008, 12:01 AMCaptainBlack
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.

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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 AMasi123
- Dec 7th 2008, 05:24 AMCaptainBlack
- Dec 7th 2008, 05:55 AMasi123
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 AMCaptainBlack
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