# Weak convergence on L^p [closed]

Let Let $X=[0,1]$ with the Lebesgue measure, find a sequence $\{f_n\}$ of measurable functions $f_n:X \rightarrow{ \mathbb{R} }$ such that:

1. $f_n(x)\rightarrow{0}$ almost everywhere $x∈[0,1]$

2. $f_n$ converge to $0$ in measure.

3. $f_n$ not converge weakly to $0$ in $L^p([0,1])$ for any $p$, $1≤p< \infty$

## closed as off-topic by Davide Giraudo, user91500, JonMark Perry, R_D, Daniel W. FarlowJul 30 '16 at 13:18

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• What is a necessary condition for a sequence to be weakly convergent? – Daniel Fischer Nov 5 '14 at 1:16

Let $$f_n(x)= \begin{cases} n^2&\text{for 0\leq x\leq\frac1n},\\ 0&\text{for \frac1n< x\leq1}. \end{cases}$$ The sequence is unbounded, hence not weakly convergent, but convergent in the sense of 1 and 2.
• Of note: One can verify that the sequence does not converge weakly to $0$ directly. Integrate against $\chi_{[0,1]}\in L_P^*$. – David Mitra Nov 5 '14 at 15:38