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Sep 20, 2018 at 10:37 vote accept Giulia S-A.
Sep 20, 2018 at 6:50 comment added Jochen Wengenroth They are equivalent for convergence of sequences because of compactology ($\mathcal E'$ is a Montel space, i.e., the closed bounded sets are compact, and weakly convergent sequences are bounded).
Sep 19, 2018 at 18:42 comment added Abdelmalek Abdesselam It seems your argument proves something stronger because you are using the weak-$\ast$ topology on $\mathcal{E}'$ instead of the strong topology as in Jochen's answer. Although, I don't remember if the two topologies are equivalent from the point of view of convergence of sequence because of more Bairology.
Sep 19, 2018 at 15:45 history answered Mizar CC BY-SA 4.0