5
$\begingroup$

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!!

  • 0
    Thanks @DavidMitra ! I see what you mean. Your simpler example uses a stretched rectangle while I made the detour into a triangle...2012-04-23

1 Answers 1

2

If 1, the statement is true. This is rather standard, see for instance Lemma 4.8 in Kavian, Introduction à la théorie des points critiques, Springer-Verlag. You example assumes $p=1$, where the statement is actually false. A more compact example is the sequence of functions $f_n(x)=n \mathrm{e}^{-nx}$, $x \in [0,1]$. The reason is, essentially, that $L^1$ is not a reflexive Banach space, while $L^p$ is, for every finite $p >1$.

  • 1
    In a reflexive Banach space, every ball is weakly compact. This is a consequence of the Banach-Alaoglu theorem. It is Corollary 9.4.2 of Larsen, "Functional analysis. An introduction", Marcel Dekker. It turns out that $L^p(0,1)$ is reflexive if 1, but it is not reflexive is $p=1$ or $p=\infty$. A bounded sequence in $L^p$ weakly sub-converges if 1. This i false if $p=1$ or $p=\infty$.2012-04-23