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
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