Let $M([0,1])$ be the set of finite signed measures on $[0,1]$ (with the topology generated by the sets $\left\{ \mu \in M([0,1]) : \left| \int f(x) \mu(dx)- a\right| \leq \delta\right\}$ for all $\delta>0$, $a \in R$ and $f \in C_b([0,1])$ (continuous and bounded). (hence weak-*-topology)

Here it is answered that $M([0,1])$ is not first countable.

But is it also not sequential; that is, does sequential continuity of a function $F : M([0,1]) \rightarrow \mathbb R$ imply continuity of $F$ in general or is that not the case?


It is not sequential. Here is an explicit counterexample.

Notation: $\|\mu\|$ is the total variation norm of $\mu$, and $\bar{B}_r \subset M([0,1])$ is the closed norm ball of radius $r$ centered at $0$. $\delta_x$ is the Dirac measure which places a unit point mass at $x \in [0,1]$.

For each positive integer $n > 0$, let $P_n \subset [0,1]$ be any finite set of size at least $(2n)^{2n}$, and let $$E_n = \{ n(\delta_{x} - \delta_y) : x,y \in P_n, \, x \ne y\} \subset M([0,1]).$$

Notice that $E_n$ is finite, and that $\|\mu\| = 2n$ for every $\mu \in E_n$.

Set $E = \bigcup_{n=1}^{\infty} E_n$. I claim that 0 is a weak-* limit point of $E$. Let $f_1, \dots, f_m \in C([0,1])$ and let $\epsilon > 0$. I will produce $\mu \in E$ with $\left|\int f_k\,d\mu\right| < \epsilon$ for every $k$. Without loss of generality, assume $\|f_k\|_\infty \le 1$ for all $k$ (otherwise divide $\epsilon$ by $\max_k \|f_k\|_\infty$). Define $F : [0,1] \to [-1,1]^m$ by $F = (f_1, \dots, f_m)$. Choose $n > \max(m, 1/\epsilon)$.

We can cover $[-1,1]^m$ with $(2n^2)^{m}$ cubes of side length $1/n^2$. Since $$|P_n| \ge (2n)^{2n} > (2n)^{2m} = (4n^2)^m > (2n^2)^m$$ by the pigeonhole principle there must exist two distinct $x,y \in P_n$ with $F(x), F(y)$ in the same cube. This means that $|f_k(x) - f_k(y)| \le 1/n^2$ for every $k$. So if we take $\mu = n(\delta_x - \delta_y) \in E_n$, we have $\left|\int f_k\,d\mu\right| \le 1/n < \epsilon$ for every $k$. This proves the claim that 0 is a weak-* limit point of $E$.

Now for any $n$, we know that $\bar{B}_n$ is weak-* closed and disjoint from the finite set $E_n$. The weak-* topology is completely regular, so there exists a weak-* continuous function $G_n : M([0,1]) \to [0,1]$ with $G_n = 0$ on $\bar{B}_n$ and $G_n = 1$ on $E_n$. Set $G = \sum_{n=1}^\infty G_n$. Note that on any ball $\bar{B}_n$, only finitely many terms of the sum are nonzero, so the sum makes sense, and the restriction of $G$ to any ball is weak-* continuous. If $\mu_k$ is a weak-* convergent sequence with limit $\mu$, then by the uniform boundedness principle all the $\mu_k$ and $\mu$ lie in some ball $B$. The restriction of $G$ to $B$ is weak-* continuous, so $G(\mu_k) \to G(\mu)$. Thus we have shown $G$ is weak-* sequentially continuous.

On the other hand, $G$ is not weak-* continuous, since $G(0) = 0$ but $G \ge 1$ on $E$.

(The same uniform boundedness argument shows that $E$ is weak-* sequentially closed, but not weak-* closed since it does not contain 0.)

More generally, we can replace $M([0,1])$ by the dual of any separable Banach space. See the paper:

Humphrey, A. James and Simpson, Stephen G. Separable Banach space theory needs strong set existence axioms. Trans. Amer. Math. Soc. 348 (10), 4231-4255, 1996. Open access full text

Theorem 2.5 of that paper shows that for any infinite-dimensional separable Banach space $X$, there is a countable subset $Z \subset X^*$ which is weak-* sequentially closed but weak-* dense. In particular, $Z$ is not weak-* closed, so the weak-* topology on $X^*$ is not sequential in the usual sense of the word.

To show it is not sequential in your sense: As an intermediate step of their construction, they get a sequence which has $0$ as a weak-* limit point but intersects every ball in only finitely many points; we could proceed as we did above to construct a function $G : X^* \to \mathbb{R}$ which is weak-* sequentially continuous but not weak-* continuous. (Their set $Z$ is formed by translating such a sequence by a countable weak-* dense set.)


This query has been answered definitively but it might be of interest to point out that it can be solved in a very general setting using a very old theorem of Banach (often referred to as the Banach-Dieudonné theorem to disinguish it from the many other theorems of the former and to acknowledge the fact that the latter extended it to Fréchet spaces). This can be paraphrased as follows: if $E$ is a separable infinite dimensional Banach space, then the topology of uniform convergence on the compacta of $E$ provides its dual with a locally convex topology which coincides with the bounded weak star topology. The latter is, by definition, the finest locally convex topology which agrees with the weak star topology on the unit ball, even the finest topology which agrees with it on bounded sets. Since this topology is always strictly stronger than the weak star topology (it is complete) but agrees with it on bounded sets, which are metrisable for both topologies), this answers your question even in this very general situation.

  • $\begingroup$ Could you explain a little bit more, why this answers my question? $\endgroup$
    – Ori
    Jun 25 '15 at 14:52
  • $\begingroup$ It answers your question since it shows that the property you quote fails in the dual of ANY separable, infinite dimensional Banach space, not just that of $C[0,1]$. $\endgroup$
    – priel
    Jul 26 '15 at 21:09

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