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Let $l^{\infty}$ (respectively, $l^{1}$) be the space of bounded (respectively, absolutely summable) real sequences. I need to find out if $l^{\infty}$ equipped with the Mackey topology $\tau(l^{\infty},l^{1})$, i.e. the finest locally convex topology that leads to the topological dual $l^{1}$, is strongly/hereditarily Lindelöf.

This is a curious case because $l^{\infty}$ equipped with weak* topology is strongly Lindelöf (as a countable union of second countable balls), while it is not Lindelöf with respect to the norm-topology. The Mackey topology is finer than the former and coarser than the latter.

Thanks in advance!

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This is true and follows from the fact that in this case the Mackey topology agrees with the weak $\ast$ topology on balls.

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  • $\begingroup$ Thanks so much! At first, it sounded too good to be true. I don't know the proof in your mind, but I was able to prove your claim by using the fact that the Mackey topology agrees with the so called "strict topology" (Conway, TAMS, 1967). This appears to be an exceptional case though. On bounded subsets of $L^{\infty}([0,1],\lambda)$, where $\lambda$ is the Lebesgue measure, Mackey appears to be distinct from weak*. $\endgroup$ – OzE Nov 20 '17 at 19:17
  • $\begingroup$ Yes. The special ingredient in this case is that result of Schur that weak and norm compactness coincide for subsets of $\ell^1$. In the general case of the dual of a separable Banach space you do have a corresponding result, not for the Mackey topology but for the so-called bw$\ast$ topology, i.e., for the strict topology generated by the norm and the weak$\ast$ topology.This is a complete locally convex topology with the same convergent sequences as the weak$\ast$ topology. $\endgroup$ – chaqueyapu Nov 21 '17 at 20:48

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