0
$\begingroup$

Let $X, Y$ be Banach spaces and let $T : (X^*, w^*) \rightarrow (Y^*,w^*)$ be a linear map. Suppose that $T$ is sequentially continuous. Must $T$ be weak*-to-weak*-continuous ?

$\endgroup$
3
  • $\begingroup$ in what topology? ​ ​ $\endgroup$
    – user5810
    Commented Jun 2, 2016 at 23:21
  • $\begingroup$ In the case where $Y$ is the scalar field, the magic words to search for are "Mazur's property" $\endgroup$
    – Yemon Choi
    Commented Jun 3, 2016 at 2:48
  • $\begingroup$ Sorry to have to reach you this way, M.G. I had put in an account merger request to the SE Community Team; after some delay they asked that you contact them directly (via 'Contact Us', which you'll see by scrolling to the bottom of this page) so that they could perform the necessary ownership checks. $\endgroup$
    – Todd Trimble
    Commented Jun 29, 2016 at 13:57

1 Answer 1

2
$\begingroup$

The answer is sometimes "no", even when $Y$ is one-dimensional.

Let's say the scalars are $\mathbb R$; the same argument will work for $\mathbb C$.

Let $\omega_1$ be the least uncountable ordinal. Let $K = [0,\omega_1]$ with the order topology, a compact Hausdorff space. This is chosen so that the subset $[0,\omega_1)$ is sequentially closed but not closed. The characteristic function of the singleton $\{\omega_1\}$ is a sequentially continuous but not continuous function from $K$ to $\mathbb R$.

Now Banach spaces $X = C(K)$ and $Y = \mathbb R$ will be used in our example.

Note that every $f \in C(K)$ is constant in some neighborhood of the endpoint $\omega_1$. Compute the dual $X^* = M(K) = l^1(K)$; that is: countably-additive signed measures $\mu$ on $K$ with countable support. Define the map $T : M(K) \to \mathbb R$ mapping the signed measure $\mu$ to the value $\mu(\{\omega_1\})$. So $T$ takes the $\omega_1$-component of the $l_1(K)$ function.

Claim: $T$ is w* sequentially continuous. Let $\mu_n, \mu \in M(K)$ with $\mu_n \to \mu$ for the w* topology. We must show $T(\mu_n) \to T(\mu)$. There exists a countable ordinal $\alpha$ so that $\mu$ and all $\mu_n$ have support in $[0,\alpha] \cup \{\omega_1\}$. [The supremum of countably many countable ordinals is countable.] There is a continuous function $f \in C(K)$ that vanishes on $[0,\alpha]$ and has $f(\omega_1)=1$. By w* convergence we have $\int f\;d\mu_n \to \int f\;d\mu$. Thus we get $T(\mu_n) \to T(\mu)$.

Finally we claim that $T$ is not w* continuous. For each $\alpha \in [0,\omega_1]$, let $e_\alpha \in X^*$ be the point-mass at $\alpha$. That is, $e_\alpha(f) = f(\alpha)$ for all $f \in C(K)$. Then the net $(e_\alpha)_{\alpha < \omega_1}$ converges w* to $e_{\omega_1}$. [Because the elements of $C(K)$ are continuous at the point $\omega_1$.] But $T(e_\alpha)=0$ for all $\alpha < \omega_1$ and $T(e_{\omega_1}) = 1$. So $T$ is not w* continuous.

$\endgroup$
2
  • $\begingroup$ I am confused about something. I seem to recall that every finite-rank operator between TVS's is continuous. If your example is valid, that means my recollection has failed me. In particular it would mean that there are rank-1 operators between dual spaces which fail to admit a pre-adjoint. $\endgroup$
    – Ben W
    Commented Jun 7, 2016 at 9:35
  • $\begingroup$ For any non-reflexive Banach space $X$, there are elements of $X^{**}$ not continuous in the w* topology of $X^*$. $\endgroup$ Commented Jun 7, 2016 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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