# Weak-* compactness in L^1

Hey I'm really stuck on what I think is an interesting 'paradox'. Consider the sequence of functions $f_n = 1_{[n,n+1]}$ (indicator functions of the interval $[n,n+1]$.

These are uniformly bounded in the $L^1$ norm. It follows that, considering $L^1 \subset (L^1)^{**}$, that this belongs to a weak-* compact set (by the banach alaoglu theorem). This should mean that there is a weak-* convergent NET. You can see easily there is no weak-* convergent subsequence: consider just $g=1 \in L^{\infty}$ then $\int f_n g = 1$ always.

My question is, what is going on here? Compactness of the weak-* unit ball is still true, but what does it mean in this case? Does it mean that every neighborhood of 0 in the weak-* topology intersects some of my functions f_n?

• It seems paradoxical perhaps because you can't see the limit points. They lie in $(L^\infty)^*\setminus L^1$. If you were considering $L^1\subset C_0(\mathbb{R})^*$, on the other hand, then of course your sequence would converge weak-$*$ to 0. – Jonas Meyer Jul 19 '10 at 1:44
• It may also help to solve the "paradox", comparing the behaviour of your $f_n$ in the $L^p$ spaces for $p > 1.$ These are dual spaces, and the Banach-Alaoglu theorem applies. Moreover, they are dual spaces of separable spaces, so that their closed unit ball is weak* compact and metrizable. Indeed, $f_n$ is weak* convergent to zero for all $p > 1$. – Pietro Majer Jul 19 '10 at 6:35

Of course the cluster points of the sequence (the limits of subnets) are not in $L_1$ but in its second dual $L_1^{**} = L_\infty^*$.
For a simpler example, take $l_1$ and consider the unit vectors $e_n$ in the second dual. Each cluster point of that sequence corresponds to a free ultrafilter on $\mathbb{N}$. And again, no subsequence converges weak* in $l_1^{**}$. Only subnets.
• These weak-* cluster points are interesting objects; they generalize the idea of the limit of a sequence, similar to the notion of a Banach limit. If $f$ is such a cluster point, for any $x \in l^\infty$, $f(x)$ is a cluster point of the sequence $\{x(n)\}$ in $\mathbb{R}$, so there is a subsequence $x(n_k) \to f(x)$. In particular, $\liminf x(n) \le f(x) \le \limsup x(n)$, $f$ is a positive linear functional, and if $\{x(n)\}$ converges, $f(x) = \lim x(n)$. Unlike a Banach limit, though, $f$ cannot be invariant under shifts (this is incompatible with being a subsequential limit). – Nate Eldredge Jul 20 '10 at 17:10