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!!
in order for your initial statement to be true. I didn't check your counterexample in the $p=1$ case, but it looks right.
– 2012-04-23case (you can first prove your $f$ is in $L_p$ using Fatou).
– 2012-04-23follows up to subsequence from the weak compactness of the unit ball of Banach spaces. The pointwise convergence shows that the limit is for the whole sequence.
– 2012-04-23