In my homework for real-analysis I was asked to prove the following statement:
On $[0,1]$, for 1\leq{}p<\infty, If $f_{n}\rightarrow{}f$ a.e. and $||f_{n}||_{p}\leq{}M \space\space\forall\space n$, show that $f_{n}\rightarrow{}f$ weakly in $L^{p}$.
I thought about it for some time, but I could not come up with a proof. Then suddenly it seemed that I found a counter-example for this statement. Can anybody help me judge whether my counter-example is valid?
$ f_{n} = \begin{cases} n^{2}x, & x\in [0,\frac{1}{n}] \\ 2n-n^{2}x, & x\in[\frac{1}{n},\frac{2}{n}] \\ 0, & x\in[\frac{2}{n},1] \end{cases} $
$ g=1 \text{ on } [0,1] $
$ f=0 \text{ on } [0,1] $
Then it seems to me that $f_{n}\rightarrow{}f$ a.e. on $[0,1]$, $f\in{}L^{1},g\in{L^{\infty}}, ||f||_{1}=1\leq2$. But $\int{f_{n}g=1\nrightarrow0=\int{fg}}$. So $f_n$ does not converge weakly to $f$ in $L^{1}$.
Is my counter-example valid? If not, how can I prove the statement?
Thank you!!