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Let $X$ and $Y$ be Banach spaces, and let $S$ be any bounded linear function between the dual spaces of $Y$ and $X$, i.e. $S: Y^* \to X^*$. I need to come up with some $S$ such that it is not the predual of any bounded linear function $T: X \to Y$.

I have seen this post about it. Here it proves that $S$ will only have some $T$ as a predual if and only if it is weak* continuous. I do not understand this concept and in my class we have not discussed it, so I guess it is not neccessarry to find such an example.

Can anybody suggest some functions? Thanks!

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    see also https://math.stackexchange.com/questions/55679/weak-to-weak-continuous-operator-which-is-not-norm-continuous – daw Dec 11 '21 at 19:37

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Take $X=Y=c_0$, $X^*=Y^*=l^1$. Define $$ S(x) = (\sum x_n) e_1, $$ where $e_1$ is the unit vector $(1,0,0,\dots)$. $S$ is clearly bounded. It is not weak-star continuous: $Se_k = e_1$ for all $k$ but $e_k \rightharpoonup^* 0$ in $l^1=(c_0)^*$.

daw
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  • Thank you for your answer. Could you maybe show there is no $T$ such that $T^=S$ without using the notion of weak continuity? –  Dec 11 '21 at 19:40