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I want to prove that the weak limit of a disjoint normalized (pairwise disjoint supports, elements of norm $1$) sequence $(f_n)$ in $L^p$ for $p >1$ is zero ?

I started with the measure of the support of each function $f_n$ , but is it true that it goes to zero ?

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what is a disjoint normalized sequence? –  Emanuele Paolini Feb 11 '13 at 5:19
    
@ Emanuele it means that each function in this sequence has norm 1 and each two functions have disjoint supports –  user60184 Feb 11 '13 at 5:24
    
Do you know that every bounded sequence in $L_p$ with $1<p<\infty$ has a weakly convergent subsequence? –  Jonas Meyer Feb 11 '13 at 5:26
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Here is a suggested approach with disclaimer that I've not thought through it well: Try to show that if $(g_n)$ converges weakly to $g$ and $(g_n)$ converges pointwise to $0$, then $g=0$ a.e. Because every bounded sequence in $L_p$ with $1<p<\infty$ has a weakly convergent subsequence, this implies that every subsequence of $(f_n)$ has a subsequence that converges weakly to $0$. In any topological space, given a sequence $(x_n)$, if there exists $x$ such that every subsequence of $(x_n)$ has a subsequence that converges to $x$, then $(x_n)$ converges to $x$. –  Jonas Meyer Feb 11 '13 at 5:38
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@Jonas thank you very much ! –  user60184 Feb 18 '13 at 21:26

2 Answers 2

Let $S_j$ be the support of $f_j$, $S:=\bigcup_jS_j$ and $q$ the conjugate exponent of $p$ (that is $p^{-1}+q^{-1}=1$). Then for each $g\in L^q$ and each $n_0$, we have $$\int f_n\cdot \operatorname{sgn}(f)|g|\chi_{S_{n_0}}\to \int |f|\cdot|g|\chi_{n_0}$$ as $n\to +\infty$. But the LHS is $0$ when $n>n_0$ so for each $n_0$, $f\chi_{S_{n_0}}=0$ hence $f\chi_S=0$.

As $\{f_n\chi_S\}=\{f_n\}$, this sequence converges weakly to both $f$ and $f\chi_S=0$, and a weak limit is unique, so we can conclude.

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One way to prove it is by using the following 2 facts 1- a normalized disjoint sequence in $Lp$ is basic, and 2- every basic sequence in reflexive space $X$ is weakly null !

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