1. ## L2 space

Let $\displaystyle f_n,f,g \in L^2(X,F,\mu)$ and $\displaystyle |f_n|\leq g$ for all $\displaystyle n$.
Show that $\displaystyle f_n\rightarrow f$ in $\displaystyle L^2$ $\displaystyle iff$ $\displaystyle f_n \rightarrow f$ in measure and $\displaystyle \int f_n^2d\mu\rightarrow \int f^2d\mu$.
i could show that if $\displaystyle f_n\rightarrow f$ in $\displaystyle L^2$ , $\displaystyle \int f_n^2d\mu\rightarrow \int f^2d\mu$.
but im stuck on showing if $\displaystyle f_n\rightarrow f$ in $\displaystyle L^2$, $\displaystyle f_n \rightarrow f$ in measure. please help me on this. thank you.

2. Originally Posted by GTO
Let $\displaystyle f_n,f,g \in L^2(X,F,\mu)$ and $\displaystyle |f_n|\leq g$ for all $\displaystyle n$.
Show that $\displaystyle f_n\rightarrow f$ in $\displaystyle L^2$ $\displaystyle iff$ $\displaystyle f_n \rightarrow f$ in measure and $\displaystyle \int f_n^2d\mu\rightarrow \int f^2d\mu$.
i could show that if $\displaystyle f_n\rightarrow f$ in $\displaystyle L^2$ , $\displaystyle \int f_n^2d\mu\rightarrow \int f^2d\mu$.
but im stuck on showing if $\displaystyle f_n\rightarrow f$ in $\displaystyle L^2$, $\displaystyle f_n \rightarrow f$ in measure. please help me on this. thank you.
Since $\displaystyle f_n \rightarrow f$ in $\displaystyle L^2$, we have $\displaystyle \int_X |f-f_n|^2 d\mu \rightarrow 0$. Let $\displaystyle E=\{x:|f-f_n|>\epsilon\}$ and $\displaystyle D= \{x:|f-f_n|<\epsilon\}$. Then $\displaystyle \int_X |f-f_n|^2 d\mu=\int_E |f-f_n|^2 d\mu +\int_D |f-f_2|^2 d\mu \rightarrow 0$. So $\displaystyle \int_E |f-f_n|^2 d\mu\rightarrow 0$.

i got this far but i dont know how i can show $\displaystyle |f-f_n|<\infty$. i am thinking that the given assumption $\displaystyle |f_n|<g$ can be used here but i dont know how i can use it. please help me .

3. Originally Posted by GTO
Since $\displaystyle f_n \rightarrow f$ in $\displaystyle L^2$, we have $\displaystyle \int_X |f-f_n|^2 d\mu \rightarrow 0$. Let $\displaystyle E=\{x:|f-f_n|>\epsilon\}$ and $\displaystyle D= \{x:|f-f_n|<\epsilon\}$. Then $\displaystyle \int_X |f-f_n|^2 d\mu=\int_E |f-f_n|^2 d\mu +\int_D |f-f_2|^2 d\mu \rightarrow 0$. So $\displaystyle \int_E |f-f_n|^2 d\mu\rightarrow 0$.

i got this far but i dont know how i can show $\displaystyle |f-f_n|<\infty$. i am thinking that the given assumption $\displaystyle |f_n|<g$ can be used here but i dont know how i can use it. please help me .
Can i assume that $\displaystyle f_n$ is bounded since $\displaystyle \int |f_n|^2 <\infty$?

4. Originally Posted by GTO
Since $\displaystyle f_n \rightarrow f$ in $\displaystyle L^2$, we have $\displaystyle \int_X |f-f_n|^2 d\mu \rightarrow 0$. Let $\displaystyle E=\{x:|f-f_n|>\epsilon\}$ and $\displaystyle D= \{x:|f-f_n|<\epsilon\}$. Then $\displaystyle \int_X |f-f_n|^2 d\mu=\int_E |f-f_n|^2 d\mu +\int_D |f-f_2|^2 d\mu \rightarrow 0$. So $\displaystyle \int_E |f-f_n|^2 d\mu\rightarrow 0$.

i got this far but i dont know how i can show $\displaystyle |f-f_n|<\infty$. i am thinking that the given assumption $\displaystyle |f_n|<g$ can be used here but i dont know how i can use it. please help me .
i am not really sure but since $\displaystyle \int |f-f_n|^2 < \epsilon$ for all $\displaystyle n>N$, $\displaystyle |f-f_n|<\infty$. otherwise this integral will not go to 0. can someone tell me if it is correct?