# Do the notions of weak and weak* convergence coincide for $\ell^1(\mathbb{N})$?

As my friends and I were studying for our real analysis final exam yesterday, we were playing with various examples and found ourselves asking this question:

The space $\ell^1(\mathbb{N})$ is the dual of $c_0(\mathbb{N})$, and the dual of $\ell^1(\mathbb{N})$ is $\ell^\infty(\mathbb{N})$.

Is it possible to have a sequence $\{b_n\}\in\ell^1(\mathbb{N})$ converge to $b\in\ell^1(\mathbb{N})$ weakly*, but not weakly?

We knew that the weak and weak* topologies agree on a reflexive space, and because $\ell^1(\mathbb{N})$ is non-reflexive, we were wondering whether the weak and weak* topologies would agree.

Parsing the definitions, this is just asking whether there is a sequence $\{b_n\}\in\ell^1(\mathbb{N})$ such that, for any $r\in c_0(\mathbb{N})$, we have $$\sum_{k=1}^\infty (b_n)_kr_k\to\sum_{k=1}^\infty b_kr_k\quad \text{ as }n\to\infty,$$ but for some $s\in \ell^\infty(\mathbb{N})$, $$\sum_{k=1}^\infty (b_n)_ks_k\nrightarrow\sum_{k=1}^\infty b_ks_k\quad \text{ as }n\to\infty.$$ WLOG, we can let take $b=0$ (just subtract $b$ from all the $b_n$), so this becomes:

Is there a sequence $\{b_n\}\in\ell^1(\mathbb{N})$ such that, for any $r\in c_0(\mathbb{N})$, we have $$\sum_{k=1}^\infty (b_n)_kr_k\to 0\quad \text{ as }n\to\infty,$$ but for some $s\in \ell^\infty(\mathbb{N})$, $$\sum_{k=1}^\infty (b_n)_ks_k\nrightarrow 0\quad \text{ as }n\to\infty ?$$

We weren't able to come up with any examples, but of course that doesn't mean there aren't any. Also, just to double-check, were we correct in assuming that it would suffice to check whether weak and weak* convergence of sequences agreed in order to determine whether the weak and weak* topologies agreed?

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The unit vectors converge weak* to 0 but not weakly to 0. The weak and weak* topologies agree iff $X$ is reflexive. – David Mitra May 12 '12 at 2:53
Ah, thank you - that's certainly a sequence we should have checked :) Could you provide a reference or proof for the "only if" direction? It intuitively makes sense, in that if $X$ is not reflexive then there would then (seem to) be more demands made of a weakly convergent sequence than a weakly* convergent sequence, but I'm not sure how to show that is actually the case. – Zev Chonoles May 12 '12 at 2:57
It's in Diestel's Sequences and Series in Banach Spaces. It should be in any FA text. By the way, it does not suffice to check sequences only. This is in Diestel as well. – David Mitra May 12 '12 at 3:01
We didn't have an assigned book for the course; I'd been looking at Eidelman, Milman, and Tsolomitis's Functional Analysis: An Introduction but couldn't find the result (it is a slightly lower-level book however). I'll be sure to get Diestel. You're welcome to post your comments as an answer of course :) – Zev Chonoles May 12 '12 at 3:15

The unit vectors in $\ell_1$ converge weak* to 0 (this is easy to see, with $c_0$ as the pre-dual) but not weakly to 0 (look at the action of $(1,1,\ldots)\in\ell_\infty$ on them; this can also be seen since no sequence of convex combinations of the unit vectors in $\ell_1$ can coverge in norm to 0).
One space where this happens is in $\ell_\infty^*$ (c.f., Joseph Diestel, Sequences and Series in Banach Spaces, Theorem 15, page 103). This result is attributed to A. Grothendieck from his 1953 paper: Sur Les applications linéaires faiblement compactes d'espaces du type C(K), Canadian J. Math., 5, 129-173.