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In my homework for real-analysis I was asked to prove the following statement:

On $[0,1]$, for $1\leq{}p&lt\infty$, 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!!

Vokram
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    You need $1<p<\infty$ in order for your initial statement to be true. I didn't check your counterexample in the $p=1$ case, but it looks right. – David Mitra Apr 23 '12 at 14:42
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    A simpler counterexample for $p=1$, by the way, is $f_n=n\chi_{[0,1/n]}$, $f=0$, $g=\chi_{[0,1]}$. – David Mitra Apr 23 '12 at 14:45
  • Just by curiosity, look at the the Brezis-Lieb theorem, wich states: ${f_n}\subset L^p(\Omega)$ open and bounded, with $i\leqslant p\leqslant \infty$ and $\Omega\subset\mathbb{R}^n$, such that:

    1. $|f|_p\leqslant M$

    2. $f_n\rightarrow f$

    Then $$\lim_n(|f_n|_p-|f_n-f|_p)=|f|_p$$

    If you were able to show $|f_n|_p\rightarrow |f|_p$, you in fact can prove the strong convergence. I coudn't figure out why it doesn't happen here.

     – matgaio Apr 23 '12 at 15:00
  • See here for a proof for the $1<p<\infty$ case (you can first prove your $f$ is in $L_p$ using Fatou). – David Mitra Apr 23 '12 at 15:03
  • Well, weak and strong convergence are rather different. Weak convergence for $1<p<\infty$ follows 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. – Siminore Apr 23 '12 at 15:04
  • Related question: http://math.stackexchange.com/questions/134908/compactness-in-l1/134962#134962 – Nate Eldredge Apr 23 '12 at 16:11
  • Thanks @DavidMitra ! I see what you mean. Your simpler example uses a stretched rectangle while I made the detour into a triangle... – Vokram Apr 23 '12 at 17:09

1 Answers1

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If $1&ltp&lt+\infty$, 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 $p >1$.

Siminore
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  • Thank you so much @Siminore ! Can you elaborate a bit more on your point "The reason is, essentially, that L1 is not a reflexive Banach space, while Lp is, for every finite p>1"? I am also taking functional analysis and I'm particularly interested in it... Would very much appreciate that! – Vokram Apr 23 '12 at 17:14
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    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 $1<p<\infty$, but it is not reflexive is $p=1$ or $p=\infty$. A bounded sequence in $L^p$ weakly sub-converges if $1<p<\infty$. This i false if $p=1$ or $p=\infty$. – Siminore Apr 23 '12 at 17:26