Can you give two elementary examples that show:
- a bounded sequence in $L^1$ with no weakly convergent subsequence;
- a bounded sequence in $L^\infty$ with no weakly convergent subsequence??
For the $L^1$ case, I was thinking about an approximation of the Dirac delta, say rectangles of area 1 that get higher and smaller around the origin, but I cannot see if I can make it work. I'm totally lost about the $L^\infty$ case.
Best Answer
For the case of $L^1$ your example does work. Let $f_n= n\chi_{(0,1/n)}.$ Suppose, to reach a contradiction, that $f_{n_k}$ converges weakly in $L^1.$ Passing to a further subsequence, which I'll still denote by $f_{n_k},$ we can assume $n_{k+1}/n_k \to \infty.$ Define $g\in L^\infty$ by setting $g=(-1)^k$ on $(1/n_k, 1/n_{k+1}).$ Then $\int_0^1 f_{n_k}g \to -1$ as $k\to \infty$ through odd integers, while $\int_0^1 f_{n_k}g \to 1$ through even integers. Thus $\int_0^1 f_{n_k}g$ fails to have a limit as $k\to \infty,$ contradiction.
Added later: Why do we need $n_k/n_{k+1} \to \infty?$ Hopefully this will help:
$$\int_0^1 f_{n_k}g = n_k\int_0^{1/n_k} g = n_k\sum_{j=k}^{\infty}(-1)^j(1/n_j- 1/n_{j+1}) = (-1)^k(1- n_k/n_{k+1}) + r_k.$$
Verify that $|r_k| \le n_k/n_{k+1}.$ Thus the above integrals have the behavior described.