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Let $X$ and $Y$ be Banach spaces, with dual spaces $X'$ and $Y'$, and assume that $T:X' \rightarrow Y'$ is a sequentially continuous operator for the strong topologies of $X'$ and $Y'$.

Is it true that if $u_N \overset{\ast}{\rightharpoonup} u$ in $X'$, then $T(u_N) \overset{\ast}{\rightharpoonup} T(u)$ in $Y'$?

If $X$ and $Y$ are reflexive Banach spaces, it seems to me that it is. Indeed in that case, $X'' = X$ and $Y'' = Y$, so weak star convergence is equivalent to weak convergence, and so for $v \in Y$ $$(T u_N, v)_{Y',Y} = (u_N,T^*v)_{X',X} \rightarrow (u,T^*v)_{X',X} = (Tu,v)_{Y',Y}.$$

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    $\begingroup$ Norm continuity does not imply weak$^{\ast}$-continuity, see here for example. In your question body you ask about weak$^{\ast}$-sequential continuity, which may be different from weak$^{\ast}$ continuity (in general topological spaces, there can be sequentially continuous maps that are discontinuous), I don't know off the top of my head whether that can happen here. If the spaces are reflexive, then weak$^{\ast}$ is the same as weak, and a continuous linear map is always weakly continuous. $\endgroup$ – Daniel Fischer Dec 7 '17 at 16:11
  • $\begingroup$ Thank you very much, the link to the other questions is exactly what I was looking for, and I probably couldn't have formulated the search terms to come across it. $\endgroup$ – Roberto Rastapopoulos Dec 7 '17 at 16:18
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    $\begingroup$ It is the case if $T$ is the adjoint of an operator from $Y$ to $X$. $\endgroup$ – gerw Dec 7 '17 at 16:29
  • $\begingroup$ Hi! Just to let you know that I have posted a follow-up question here, and I would be very glad to hear any feedback you might have. $\endgroup$ – Roberto Rastapopoulos Dec 8 '17 at 15:16
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Not true in general. Take $X = \ell^1$ and $Y = \mathbb{R}$. Let $T$ be any continuous "Banach limit" functional on $X' = \ell^\infty$, such that $Tx = \lim_{n \to \infty} x(n)$ for any $x$ such that the limit exists. (Such functionals $T$ exist by Hahn-Banach.) Then let $x_N = e_1 + \dots + e_N \in \ell^\infty$. It is clear that $x_N$ converges weak-* to the constant sequence 1 (dominated convergence), but $Tx_N = 0$ for all $N$ while $T1=1$.

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