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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Jonas Meyer Jul 19 '10 at 1:44
  • $\begingroup$ 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 $p > 1$. $\endgroup$ – 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.

  • $\begingroup$ 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). $\endgroup$ – Nate Eldredge Jul 20 '10 at 17:10

Some people like nets (Moore–Smith convergence) because they superficially resemble sequences. I think your example shows that this resemblance is superficial indeed, and that nets are more complicated beasts than they seem to be. On the other hand, the theory of filters (due to Cartan) seem more complicated at first, but are more natural to work with once you get used to them. And they have the extra advantage of not seeming to be what they are not, i.e., souped-up sequences.

(If this seems subjective and argumentative, so be it.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.